From 0d4b70aada9bbc982f00c27afb1337e2b314eb43 Mon Sep 17 00:00:00 2001
From: hardythe1
Date: Fri, 25 Jul 2014 12:37:07 +0530
Subject: adding books

---
 .../Chapter_01-checkpoint.ipynb                    |  329 ++++
 .../Chapter_01-checkpoint_1.ipynb                  |  329 ++++
 .../Chapter_01-checkpoint_2.ipynb                  |  329 ++++
 .../Chapter_01-checkpoint_3.ipynb                  |  330 ++++
 .../Chapter_02-checkpoint.ipynb                    |  806 ++++++++
 .../Chapter_02-checkpoint_1.ipynb                  |  806 ++++++++
 .../Chapter_02-checkpoint_2.ipynb                  |  806 ++++++++
 .../Chapter_02-checkpoint_3.ipynb                  |  805 ++++++++
 .../Chapter_07.ipynb                               |   23 +
 .../Chapter_07_1.ipynb                             |   23 +
 .../Chapter_07_2.ipynb                             |   23 +
 .../Chapter_07_3.ipynb                             |   23 +
 .../Chapter_11.ipynb                               |   23 +
 .../Chapter_11_1.ipynb                             |   23 +
 .../Chapter_11_2.ipynb                             |   23 +
 .../Chapter_11_3.ipynb                             |   23 +
 .../Chapter_37.ipynb                               |   23 +
 .../Chapter_37_1.ipynb                             |   23 +
 .../Chapter_37_2.ipynb                             |   23 +
 .../Chapter_37_3.ipynb                             |   23 +
 .../README.txt                                     |   10 +
 .../chapter_03-checkpoint.ipynb                    |  525 ++++++
 .../chapter_03-checkpoint_1.ipynb                  |  525 ++++++
 .../chapter_03-checkpoint_2.ipynb                  |  525 ++++++
 .../chapter_03-checkpoint_3.ipynb                  |  524 ++++++
 .../chapter_04-checkpoint.ipynb                    |  185 ++
 .../chapter_04-checkpoint_1.ipynb                  |  185 ++
 .../chapter_04-checkpoint_2.ipynb                  |  185 ++
 .../chapter_04-checkpoint_3.ipynb                  |  186 ++
 .../chapter_05-checkpoint.ipynb                    |  763 ++++++++
 .../chapter_05-checkpoint_1.ipynb                  |  763 ++++++++
 .../chapter_05-checkpoint_2.ipynb                  |  763 ++++++++
 .../chapter_05-checkpoint_3.ipynb                  |  755 ++++++++
 .../chapter_06-checkpoint.ipynb                    |  873 +++++++++
 .../chapter_06-checkpoint_1.ipynb                  |  873 +++++++++
 .../chapter_06-checkpoint_2.ipynb                  |  873 +++++++++
 .../chapter_06-checkpoint_3.ipynb                  |  863 +++++++++
 .../chapter_08-checkpoint.ipynb                    |  273 +++
 .../chapter_08-checkpoint_1.ipynb                  |  273 +++
 .../chapter_08-checkpoint_2.ipynb                  |  273 +++
 .../chapter_08-checkpoint_3.ipynb                  |  271 +++
 .../chapter_09-checkpoint.ipynb                    |  759 ++++++++
 .../chapter_09-checkpoint_1.ipynb                  |  759 ++++++++
 .../chapter_09-checkpoint_2.ipynb                  |  759 ++++++++
 .../chapter_09-checkpoint_3.ipynb                  |  755 ++++++++
 .../chapter_10-checkpoint.ipynb                    | 1066 +++++++++++
 .../chapter_10-checkpoint_1.ipynb                  | 1066 +++++++++++
 .../chapter_10-checkpoint_2.ipynb                  | 1066 +++++++++++
 .../chapter_10-checkpoint_3.ipynb                  | 1059 +++++++++++
 .../chapter_12-checkpoint.ipynb                    |   85 +
 .../chapter_12-checkpoint_1.ipynb                  |   85 +
 .../chapter_12-checkpoint_2.ipynb                  |   85 +
 .../chapter_12-checkpoint_3.ipynb                  |   85 +
 .../chapter_13-checkpoint.ipynb                    | 1203 ++++++++++++
 .../chapter_13-checkpoint_1.ipynb                  | 1203 ++++++++++++
 .../chapter_13-checkpoint_2.ipynb                  | 1203 ++++++++++++
 .../chapter_13-checkpoint_3.ipynb                  | 1200 ++++++++++++
 .../chapter_14-checkpoint.ipynb                    |  898 +++++++++
 .../chapter_14-checkpoint_1.ipynb                  |  898 +++++++++
 .../chapter_14-checkpoint_2.ipynb                  |  898 +++++++++
 .../chapter_14-checkpoint_3.ipynb                  |  886 +++++++++
 .../chapter_15-checkpoint.ipynb                    | 1677 +++++++++++++++++
 .../chapter_15-checkpoint_1.ipynb                  | 1677 +++++++++++++++++
 .../chapter_15-checkpoint_2.ipynb                  | 1677 +++++++++++++++++
 .../chapter_15-checkpoint_3.ipynb                  | 1642 ++++++++++++++++
 .../chapter_16-checkpoint.ipynb                    |  913 +++++++++
 .../chapter_16-checkpoint_1.ipynb                  |  913 +++++++++
 .../chapter_16-checkpoint_2.ipynb                  |  913 +++++++++
 .../chapter_16-checkpoint_3.ipynb                  |  885 +++++++++
 .../chapter_17-checkpoint.ipynb                    |  895 +++++++++
 .../chapter_17-checkpoint_1.ipynb                  |  895 +++++++++
 .../chapter_17-checkpoint_2.ipynb                  |  895 +++++++++
 .../chapter_17-checkpoint_3.ipynb                  |  865 +++++++++
 .../chapter_18-checkpoint.ipynb                    |  535 ++++++
 .../chapter_18-checkpoint_1.ipynb                  |  535 ++++++
 .../chapter_18-checkpoint_2.ipynb                  |  535 ++++++
 .../chapter_18-checkpoint_3.ipynb                  |  533 ++++++
 .../chapter_19-checkpoint.ipynb                    |  974 ++++++++++
 .../chapter_19-checkpoint_1.ipynb                  |  974 ++++++++++
 .../chapter_19-checkpoint_2.ipynb                  |  974 ++++++++++
 .../chapter_19-checkpoint_3.ipynb                  |  974 ++++++++++
 .../chapter_20-checkpoint.ipynb                    | 1627 ++++++++++++++++
 .../chapter_20-checkpoint_1.ipynb                  | 1627 ++++++++++++++++
 .../chapter_20-checkpoint_2.ipynb                  | 1627 ++++++++++++++++
 .../chapter_20-checkpoint_3.ipynb                  | 1617 ++++++++++++++++
 .../chapter_21-checkpoint.ipynb                    | 1641 ++++++++++++++++
 .../chapter_21-checkpoint_1.ipynb                  | 1641 ++++++++++++++++
 .../chapter_21-checkpoint_2.ipynb                  | 1641 ++++++++++++++++
 .../chapter_21-checkpoint_3.ipynb                  | 1631 ++++++++++++++++
 .../chapter_22-checkpoint.ipynb                    |  791 ++++++++
 .../chapter_22-checkpoint_1.ipynb                  |  791 ++++++++
 .../chapter_22-checkpoint_2.ipynb                  |  791 ++++++++
 .../chapter_22-checkpoint_3.ipynb                  |  782 ++++++++
 .../chapter_23-checkpoint.ipynb                    |  434 +++++
 .../chapter_23-checkpoint_1.ipynb                  |  434 +++++
 .../chapter_23-checkpoint_2.ipynb                  |  434 +++++
 .../chapter_23-checkpoint_3.ipynb                  |  433 +++++
 .../chapter_24-checkpoint.ipynb                    |  843 +++++++++
 .../chapter_24-checkpoint_1.ipynb                  |  843 +++++++++
 .../chapter_24-checkpoint_2.ipynb                  |  843 +++++++++
 .../chapter_24-checkpoint_3.ipynb                  |  837 +++++++++
 .../chapter_25-checkpoint.ipynb                    |  647 +++++++
 .../chapter_25-checkpoint_1.ipynb                  |  647 +++++++
 .../chapter_25-checkpoint_2.ipynb                  |  647 +++++++
 .../chapter_25-checkpoint_3.ipynb                  |  643 +++++++
 .../chapter_26-checkpoint.ipynb                    |  496 +++++
 .../chapter_26-checkpoint_1.ipynb                  |  496 +++++
 .../chapter_26-checkpoint_2.ipynb                  |  496 +++++
 .../chapter_26-checkpoint_3.ipynb                  |  495 +++++
 .../chapter_27-checkpoint.ipynb                    |  205 ++
 .../chapter_27-checkpoint_1.ipynb                  |  205 ++
 .../chapter_27-checkpoint_2.ipynb                  |  205 ++
 .../chapter_27-checkpoint_3.ipynb                  |  204 ++
 .../chapter_28-checkpoint.ipynb                    |  736 ++++++++
 .../chapter_28-checkpoint_1.ipynb                  |  736 ++++++++
 .../chapter_28-checkpoint_2.ipynb                  |  736 ++++++++
 .../chapter_28-checkpoint_3.ipynb                  |  729 ++++++++
 .../chapter_29-checkpoint.ipynb                    |  449 +++++
 .../chapter_29-checkpoint_1.ipynb                  |  449 +++++
 .../chapter_29-checkpoint_2.ipynb                  |  449 +++++
 .../chapter_29-checkpoint_3.ipynb                  |  445 +++++
 .../chapter_30-checkpoint.ipynb                    |  303 +++
 .../chapter_30-checkpoint_1.ipynb                  |  303 +++
 .../chapter_30-checkpoint_2.ipynb                  |  303 +++
 .../chapter_30-checkpoint_3.ipynb                  |  303 +++
 .../chapter_31-checkpoint.ipynb                    |  665 +++++++
 .../chapter_31-checkpoint_1.ipynb                  |  665 +++++++
 .../chapter_31-checkpoint_2.ipynb                  |  665 +++++++
 .../chapter_31-checkpoint_3.ipynb                  |  660 +++++++
 .../chapter_32-checkpoint.ipynb                    |  436 +++++
 .../chapter_32-checkpoint_1.ipynb                  |  436 +++++
 .../chapter_32-checkpoint_2.ipynb                  |  436 +++++
 .../chapter_32-checkpoint_3.ipynb                  |  432 +++++
 .../chapter_33-checkpoint.ipynb                    | 1103 +++++++++++
 .../chapter_33-checkpoint_1.ipynb                  | 1103 +++++++++++
 .../chapter_33-checkpoint_2.ipynb                  | 1103 +++++++++++
 .../chapter_33-checkpoint_3.ipynb                  | 1103 +++++++++++
 .../chapter_34-checkpoint.ipynb                    |  449 +++++
 .../chapter_34-checkpoint_1.ipynb                  |  449 +++++
 .../chapter_34-checkpoint_2.ipynb                  |  449 +++++
 .../chapter_34-checkpoint_3.ipynb                  |  447 +++++
 .../chapter_35-checkpoint.ipynb                    |  637 +++++++
 .../chapter_35-checkpoint_1.ipynb                  |  637 +++++++
 .../chapter_35-checkpoint_2.ipynb                  |  637 +++++++
 .../chapter_35-checkpoint_3.ipynb                  |  627 +++++++
 .../chapter_36-checkpoint.ipynb                    | 1017 ++++++++++
 .../chapter_36-checkpoint_1.ipynb                  | 1017 ++++++++++
 .../chapter_36-checkpoint_2.ipynb                  | 1017 ++++++++++
 .../chapter_36-checkpoint_3.ipynb                  | 1011 ++++++++++
 .../chapter_38-checkpoint.ipynb                    |  631 +++++++
 .../chapter_38-checkpoint_1.ipynb                  |  631 +++++++
 .../chapter_38-checkpoint_2.ipynb                  |  631 +++++++
 .../chapter_38-checkpoint_3.ipynb                  |  631 +++++++
 .../chapter_39-checkpoint.ipynb                    |  200 ++
 .../chapter_39-checkpoint_1.ipynb                  |  200 ++
 .../chapter_39-checkpoint_2.ipynb                  |  200 ++
 .../chapter_39-checkpoint_3.ipynb                  |  200 ++
 .../chapter_40-checkpoint.ipynb                    | 1328 +++++++++++++
 .../chapter_40-checkpoint_1.ipynb                  | 1328 +++++++++++++
 .../chapter_40-checkpoint_2.ipynb                  | 1328 +++++++++++++
 .../chapter_40-checkpoint_3.ipynb                  | 1319 +++++++++++++
 .../chapter_41-checkpoint.ipynb                    | 1005 ++++++++++
 .../chapter_41-checkpoint_1.ipynb                  | 1005 ++++++++++
 .../chapter_41-checkpoint_2.ipynb                  | 1005 ++++++++++
 .../chapter_41-checkpoint_3.ipynb                  | 1000 ++++++++++
 .../chapter_42-checkpoint.ipynb                    |  959 ++++++++++
 .../chapter_42-checkpoint_1.ipynb                  |  959 ++++++++++
 .../chapter_42-checkpoint_2.ipynb                  |  959 ++++++++++
 .../chapter_42-checkpoint_3.ipynb                  |  954 ++++++++++
 .../chapter_43-checkpoint.ipynb                    | 1037 +++++++++++
 .../chapter_43-checkpoint_1.ipynb                  | 1037 +++++++++++
 .../chapter_43-checkpoint_2.ipynb                  | 1037 +++++++++++
 .../chapter_43-checkpoint_3.ipynb                  | 1029 +++++++++++
 .../chapter_44-checkpoint.ipynb                    | 1093 +++++++++++
 .../chapter_44-checkpoint_1.ipynb                  | 1093 +++++++++++
 .../chapter_44-checkpoint_2.ipynb                  | 1093 +++++++++++
 .../chapter_44-checkpoint_3.ipynb                  | 1074 +++++++++++
 .../chapter_45-checkpoint.ipynb                    |  435 +++++
 .../chapter_45-checkpoint_1.ipynb                  |  435 +++++
 .../chapter_45-checkpoint_2.ipynb                  |  435 +++++
 .../chapter_45-checkpoint_3.ipynb                  |  428 +++++
 ...iation_of_Voltage_across_resistor_with_time.png |  Bin 0 -> 41180 bytes
 ...tion_of_Voltage_across_resistor_with_time_1.png |  Bin 0 -> 41180 bytes
 .../screenshots/find_Resistor_and_Capacitor.png    |  Bin 0 -> 20738 bytes
 .../screenshots/find_charge.png                    |  Bin 0 -> 15358 bytes
 .../screenshots/find_force.png                     |  Bin 0 -> 8272 bytes
 .../screenshots/find_power_and_work_done.png       |  Bin 0 -> 11537 bytes
 Stoichiometry_And_Process_Calculations/README.txt  |   10 +
 Stoichiometry_And_Process_Calculations/ch10.ipynb  | 1480 +++++++++++++++
 Stoichiometry_And_Process_Calculations/ch11.ipynb  | 1955 ++++++++++++++++++++
 Stoichiometry_And_Process_Calculations/ch12.ipynb  | 1025 ++++++++++
 Stoichiometry_And_Process_Calculations/ch2.ipynb   |  436 +++++
 Stoichiometry_And_Process_Calculations/ch3.ipynb   |  927 ++++++++++
 Stoichiometry_And_Process_Calculations/ch4.ipynb   |  796 ++++++++
 Stoichiometry_And_Process_Calculations/ch5.ipynb   |  497 +++++
 Stoichiometry_And_Process_Calculations/ch6.ipynb   |  354 ++++
 Stoichiometry_And_Process_Calculations/ch7.ipynb   |  945 ++++++++++
 Stoichiometry_And_Process_Calculations/ch8.ipynb   |  927 ++++++++++
 Stoichiometry_And_Process_Calculations/ch9.ipynb   | 1190 ++++++++++++
 .../screenshots/construction_of_the__cox_chart.png |  Bin 0 -> 13706 bytes
 .../screenshots/mole_fraction.png                  |  Bin 0 -> 19072 bytes
 .../screenshots/temperature_benzene.png            |  Bin 0 -> 12932 bytes
 202 files changed, 138258 insertions(+)
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_1.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_2.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_3.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_1.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_2.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_3.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_07.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_07_1.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_07_2.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_07_3.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_11.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_11_1.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_11_2.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_11_3.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_37.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_37_1.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_37_2.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/Chapter_37_3.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/README.txt
 create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_1.ipynb
 create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_2.ipynb
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diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint.ipynb
new file mode 100755
index 00000000..b33db331
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint.ipynb
@@ -0,0 +1,329 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 1: Units Associated with Basic Electrical Quantities</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 4</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the quantity of electricity transferred.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 5;   # in Ampere\n",
+      "t = 120; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q = I*t\n",
+      "\n",
+      "#results\n",
+      "print \"Charge, Q = \", Q,\"coulomb(C)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charge, Q =  600 coulomb(C)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 5</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the force needed.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 5; # in Kg\n",
+      "a = 2; # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "F = M*a\n",
+      "\n",
+      "#results\n",
+      "print \"Force:\", F,\"Newton(N)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force: 10 Newton(N)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 5</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the force acting vertically downwards\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 0.2; # in Kg\n",
+      "g = 9.81; # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "F = M*g\n",
+      "\n",
+      "#results\n",
+      "print \"Force:\", F,\"Newton(N)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force: 1.962 Newton(N)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 6</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate Work Done and Average Power?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "F = 200; # in Newton\n",
+      "d = 20; # in m\n",
+      "t = 25; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "W = F*d\n",
+      "P = W/t\n",
+      "\n",
+      "#results\n",
+      "print \"Power:\", P,\"watt(W)\"\n",
+      "print \"Work Done:\", W,\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power: 160.0 watt(W)\n",
+        "Work Done: 4000 Nm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 6</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is (a) the work done and (b) the power developed?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 1000; # in Kg\n",
+      "h = 10; # in m\n",
+      "t = 20; # in sec\n",
+      "g = 9.81 # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "W = M*g*h\n",
+      "P = W/t\n",
+      "\n",
+      "#results\n",
+      "print \"Work Done:\", W,\"Joule(J)\"\n",
+      "print \"Power:\", P,\"watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work Done: 98100.0 Joule(J)\n",
+        "Power: 4905.0 watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 7</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the conductance of a conductor of resistances\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 10; # in ohm\n",
+      "R2 = 5000; # in ohm\n",
+      "R3 = 0.1; # in ohm\n",
+      "#calculation:\n",
+      "G1 = 1/R1\n",
+      "G2 = 1/R2\n",
+      "G3 = 1/R3\n",
+      "\n",
+      "#results\n",
+      "print \"conductance(G1):\", G1,\"seimen(S)\"\n",
+      "print \"conductance(G2):\", G2,\"seimen(S)\"\n",
+      "print \"conductance(G3):\", G3,\"seimen(S)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductance(G1): 0.1 seimen(S)\n",
+        "conductance(G2): 0.0002 seimen(S)\n",
+        "conductance(G3): 10.0 seimen(S)"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 7</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#How much energy is provided in this time?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 5; # in Volts\n",
+      "I = 3; # in Ampere\n",
+      "t = 600; # in sec\n",
+      "#calculation:\n",
+      "P = V*I\n",
+      "E = P*t\n",
+      "\n",
+      "#results\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy(E): 9000 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 8</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the power rating of the heater and the current taken from the supply.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "E = 18E5; # in Joule\n",
+      "V = 250; # in Volts\n",
+      "t = 1800; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "P = E/t\n",
+      "I = P/V\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 1000.0 Watt(W)\n",
+        "Current(I): 4.0 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_1.ipynb
new file mode 100755
index 00000000..b33db331
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_1.ipynb
@@ -0,0 +1,329 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 1: Units Associated with Basic Electrical Quantities</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 4</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the quantity of electricity transferred.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 5;   # in Ampere\n",
+      "t = 120; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q = I*t\n",
+      "\n",
+      "#results\n",
+      "print \"Charge, Q = \", Q,\"coulomb(C)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charge, Q =  600 coulomb(C)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 5</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the force needed.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 5; # in Kg\n",
+      "a = 2; # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "F = M*a\n",
+      "\n",
+      "#results\n",
+      "print \"Force:\", F,\"Newton(N)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force: 10 Newton(N)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 5</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the force acting vertically downwards\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 0.2; # in Kg\n",
+      "g = 9.81; # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "F = M*g\n",
+      "\n",
+      "#results\n",
+      "print \"Force:\", F,\"Newton(N)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force: 1.962 Newton(N)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 6</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate Work Done and Average Power?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "F = 200; # in Newton\n",
+      "d = 20; # in m\n",
+      "t = 25; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "W = F*d\n",
+      "P = W/t\n",
+      "\n",
+      "#results\n",
+      "print \"Power:\", P,\"watt(W)\"\n",
+      "print \"Work Done:\", W,\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power: 160.0 watt(W)\n",
+        "Work Done: 4000 Nm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 6</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is (a) the work done and (b) the power developed?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 1000; # in Kg\n",
+      "h = 10; # in m\n",
+      "t = 20; # in sec\n",
+      "g = 9.81 # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "W = M*g*h\n",
+      "P = W/t\n",
+      "\n",
+      "#results\n",
+      "print \"Work Done:\", W,\"Joule(J)\"\n",
+      "print \"Power:\", P,\"watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work Done: 98100.0 Joule(J)\n",
+        "Power: 4905.0 watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 7</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the conductance of a conductor of resistances\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 10; # in ohm\n",
+      "R2 = 5000; # in ohm\n",
+      "R3 = 0.1; # in ohm\n",
+      "#calculation:\n",
+      "G1 = 1/R1\n",
+      "G2 = 1/R2\n",
+      "G3 = 1/R3\n",
+      "\n",
+      "#results\n",
+      "print \"conductance(G1):\", G1,\"seimen(S)\"\n",
+      "print \"conductance(G2):\", G2,\"seimen(S)\"\n",
+      "print \"conductance(G3):\", G3,\"seimen(S)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductance(G1): 0.1 seimen(S)\n",
+        "conductance(G2): 0.0002 seimen(S)\n",
+        "conductance(G3): 10.0 seimen(S)"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 7</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#How much energy is provided in this time?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 5; # in Volts\n",
+      "I = 3; # in Ampere\n",
+      "t = 600; # in sec\n",
+      "#calculation:\n",
+      "P = V*I\n",
+      "E = P*t\n",
+      "\n",
+      "#results\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy(E): 9000 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 8</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the power rating of the heater and the current taken from the supply.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "E = 18E5; # in Joule\n",
+      "V = 250; # in Volts\n",
+      "t = 1800; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "P = E/t\n",
+      "I = P/V\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 1000.0 Watt(W)\n",
+        "Current(I): 4.0 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_2.ipynb
new file mode 100755
index 00000000..b33db331
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_2.ipynb
@@ -0,0 +1,329 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 1: Units Associated with Basic Electrical Quantities</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 4</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the quantity of electricity transferred.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 5;   # in Ampere\n",
+      "t = 120; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q = I*t\n",
+      "\n",
+      "#results\n",
+      "print \"Charge, Q = \", Q,\"coulomb(C)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charge, Q =  600 coulomb(C)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 5</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the force needed.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 5; # in Kg\n",
+      "a = 2; # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "F = M*a\n",
+      "\n",
+      "#results\n",
+      "print \"Force:\", F,\"Newton(N)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force: 10 Newton(N)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 5</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the force acting vertically downwards\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 0.2; # in Kg\n",
+      "g = 9.81; # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "F = M*g\n",
+      "\n",
+      "#results\n",
+      "print \"Force:\", F,\"Newton(N)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force: 1.962 Newton(N)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 6</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate Work Done and Average Power?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "F = 200; # in Newton\n",
+      "d = 20; # in m\n",
+      "t = 25; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "W = F*d\n",
+      "P = W/t\n",
+      "\n",
+      "#results\n",
+      "print \"Power:\", P,\"watt(W)\"\n",
+      "print \"Work Done:\", W,\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power: 160.0 watt(W)\n",
+        "Work Done: 4000 Nm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 6</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is (a) the work done and (b) the power developed?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 1000; # in Kg\n",
+      "h = 10; # in m\n",
+      "t = 20; # in sec\n",
+      "g = 9.81 # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "W = M*g*h\n",
+      "P = W/t\n",
+      "\n",
+      "#results\n",
+      "print \"Work Done:\", W,\"Joule(J)\"\n",
+      "print \"Power:\", P,\"watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work Done: 98100.0 Joule(J)\n",
+        "Power: 4905.0 watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 7</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the conductance of a conductor of resistances\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 10; # in ohm\n",
+      "R2 = 5000; # in ohm\n",
+      "R3 = 0.1; # in ohm\n",
+      "#calculation:\n",
+      "G1 = 1/R1\n",
+      "G2 = 1/R2\n",
+      "G3 = 1/R3\n",
+      "\n",
+      "#results\n",
+      "print \"conductance(G1):\", G1,\"seimen(S)\"\n",
+      "print \"conductance(G2):\", G2,\"seimen(S)\"\n",
+      "print \"conductance(G3):\", G3,\"seimen(S)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductance(G1): 0.1 seimen(S)\n",
+        "conductance(G2): 0.0002 seimen(S)\n",
+        "conductance(G3): 10.0 seimen(S)"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 7</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#How much energy is provided in this time?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 5; # in Volts\n",
+      "I = 3; # in Ampere\n",
+      "t = 600; # in sec\n",
+      "#calculation:\n",
+      "P = V*I\n",
+      "E = P*t\n",
+      "\n",
+      "#results\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy(E): 9000 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 8</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the power rating of the heater and the current taken from the supply.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "E = 18E5; # in Joule\n",
+      "V = 250; # in Volts\n",
+      "t = 1800; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "P = E/t\n",
+      "I = P/V\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 1000.0 Watt(W)\n",
+        "Current(I): 4.0 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_3.ipynb
new file mode 100755
index 00000000..52fbd3f7
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_01-checkpoint_3.ipynb
@@ -0,0 +1,330 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:132479673f217c6ad4667b69d4be561d4eafa6bd7e059501c289648667d366ed"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 1: Units Associated with Basic Electrical Quantities</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 4</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 5;   # in Ampere\n",
+      "t = 120; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q = I*t\n",
+      "\n",
+      "#results\n",
+      "print \"Charge, Q = \", Q,\"coulomb(C)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charge, Q =  600 coulomb(C)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 5</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 5; # in Kg\n",
+      "a = 2; # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "F = M*a\n",
+      "\n",
+      "#results\n",
+      "print \"Force:\", F,\"Newton(N)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force: 10 Newton(N)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 5</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 0.2; # in Kg\n",
+      "g = 9.81; # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "F = M*g\n",
+      "\n",
+      "#results\n",
+      "print \"Force:\", F,\"Newton(N)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force: 1.962 Newton(N)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 6</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "F = 200; # in Newton\n",
+      "d = 20; # in m\n",
+      "t = 25; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "W = F*d\n",
+      "P = W/t\n",
+      "\n",
+      "#results\n",
+      "print \"Power:\", P,\"watt(W)\"\n",
+      "print \"Work Done:\", W,\"Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power: 160.0 watt(W)\n",
+        "Work Done: 4000 Nm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 6</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "M = 1000; # in Kg\n",
+      "h = 10; # in m\n",
+      "t = 20; # in sec\n",
+      "g = 9.81 # in m/s2\n",
+      "\n",
+      "#calculation:\n",
+      "W = M*g*h\n",
+      "P = W/t\n",
+      "\n",
+      "#results\n",
+      "print \"Work Done:\", W,\"Joule(J)\"\n",
+      "print \"Power:\", P,\"watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work Done: 98100.0 Joule(J)\n",
+        "Power: 4905.0 watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 7</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 10; # in ohm\n",
+      "R2 = 5000; # in ohm\n",
+      "R3 = 0.1; # in ohm\n",
+      "#calculation:\n",
+      "G1 = 1/R1\n",
+      "G2 = 1/R2\n",
+      "G3 = 1/R3\n",
+      "\n",
+      "#results\n",
+      "print \"conductance(G1):\", G1,\"seimen(S)\"\n",
+      "print \"conductance(G2):\", G2,\"seimen(S)\"\n",
+      "print \"conductance(G3):\", G3,\"seimen(S)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductance(G1): 0.1 seimen(S)\n",
+        "conductance(G2): 0.0002 seimen(S)\n",
+        "conductance(G3): 10.0 seimen(S)"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 7</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 5; # in Volts\n",
+      "I = 3; # in Ampere\n",
+      "t = 600; # in sec\n",
+      "#calculation:\n",
+      "P = V*I\n",
+      "E = P*t\n",
+      "\n",
+      "#results\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy(E): 9000 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 8</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "E = 18E5; # in Joule\n",
+      "V = 250; # in Volts\n",
+      "t = 1800; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "P = E/t\n",
+      "I = P/V\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 1000.0 Watt(W)\n",
+        "Current(I): 4.0 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint.ipynb
new file mode 100755
index 00000000..3ab84705
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint.ipynb
@@ -0,0 +1,806 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 2: An Introduction to Electric Circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 12</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What current must flow if 0.24 coulombs is to be transferred in 15 ms?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Q = 0.24; # in Coulomb\n",
+      "t = 0.015; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "I = Q/t\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 16.0 Ampere(A)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 12</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the quantity of electricity transferred.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 10; # in Ampere\n",
+      "t = 240; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q = I*t\n",
+      "\n",
+      "#resuts\n",
+      "print \"Charge(Q):\", Q,\"Coulomb(C)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charge(Q): 2400 Coulomb(C)\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 14</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.8; # in Ampere\n",
+      "V = 20; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 25.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the p.d.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.010; # in Ampere\n",
+      "R = 2000; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V = I*R\n",
+      "\n",
+      "#results\n",
+      "print \"p.d.(V):\", V,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "p.d.(V): 20.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is the resistance of the coil?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.050; # in Ampere\n",
+      "V = 12; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 240.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what will be the new value of the current flowing?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.005; # in Ampere\n",
+      "V1 = 100; # in Volts\n",
+      "V2 = 25; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V1/I1\n",
+      "I2 = V2/R\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\"\n",
+      "print \"Current(I):\", I2,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 20000.0 Ohms\n",
+        "Current(I): 0.00125 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is the resistance of a coil\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.050; # in Ampere\n",
+      "I2 = 200E-6; # in Ampere\n",
+      "V = 120; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = V/I1\n",
+      "R2 = V/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R1):\", R1,\"Ohms\"\n",
+      "print \"Resistance(R2):\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R1): 2400.0 Ohms\n",
+        "Resistance(R2): 600000.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 16</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "P = 100; # in Watt\n",
+      "V = 250; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "I = P/V\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.4 Ampere(A)\n",
+        "Resistance(R): 625.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.004; # in ampere\n",
+      "R = 5000; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "P = I*I*R\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 0.08 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What current will flow when it is connected to a 240 V supply?\n",
+      "#Find also the power rating of the kettle.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "R = 30; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = V*I\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 8.0 Ampere(A)\n",
+        "Power(P): 1920.0 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the p.d. across the winding, and (b) the power dissipated by the coil.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 5; # in ampere\n",
+      "R = 100; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V = I*R\n",
+      "P = I*R*I\n",
+      "\n",
+      "#results\n",
+      "print \"p.d(V):\", V,\"Volts(V)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "p.d(V): 500 Volts(V)\n",
+        "Power(P): 2500 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the resistance of each resistor.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.020; # in ampere\n",
+      "V1 = 20; # in Volts\n",
+      "I2 = 0.005; # in ampere\n",
+      "V2 = 16; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = V1/I1\n",
+      "R2 = V2/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R1):\", R1,\"Ohms\"\n",
+      "print \"Resistance(R2):\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R1): 1000.0 Ohms\n",
+        "Resistance(R2): 3200.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the current taken by the lamp and its power rating\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "R = 960; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = I*V\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.25 Ampere(A)\n",
+        "Power(P): 60.0 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the load, the power consumed and the energy dissipated in 2 minutes\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 12; # in Volts\n",
+      "R = 40; # in ohms\n",
+      "t = 120; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = I*V\n",
+      "E = P*t\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.3 Ampere(A)\n",
+        "Power(P): 3.6 Watt(W)\n",
+        "Energy(E): 432.0 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#How much energy is provided in this time?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 15; # in Volts\n",
+      "I = 2; # in ampere\n",
+      "t = 360; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "E = V*I*t\n",
+      "\n",
+      "#results\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy(E): 10800 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Estimate the cost per week of electricity\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "I = 13; # in ampere\n",
+      "t = 30; # in hours\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "P = V*I\n",
+      "E = P*t/1000 # in kWh\n",
+      "C = E*p\n",
+      "\n",
+      "#results\n",
+      "print \"Cost per week:\", C,\" Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cost per week: 655.2  Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the power rating of the heater and the current taken from the supply.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 250; # in Volts\n",
+      "E = 3.6E6; # energy in J\n",
+      "t = 2400; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "P = E/t\n",
+      "I = P/V\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 1500.0 Watt(W)\n",
+        "Current(I): 6.0 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power dissipated\n",
+      "#determine the energy used and the cost\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 20; # in ohms\n",
+      "I = 10; # in ampere\n",
+      "t = 6; # in hours\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "P = I*I*R\n",
+      "E = P*t/1000 # in kWh\n",
+      "C = E*p\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Cost per week:\", C,\"Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 2000 Watt(W)\n",
+        "Cost per week: 84.0 Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the weekly cost of electricity to the business.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "P1 = 3; # in kW\n",
+      "P2 = 150; # in Watt\n",
+      "n1 = 2; # no. of P1 Equips\n",
+      "n2 = 6; # no. of P2 Equips\n",
+      "t1 = 20; # in hours each per week\n",
+      "t2 = 30; # in hours each per week\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "E1 = P1*t1*n1 # in kWh by two P1 eqips\n",
+      "E2 = P2*t2*n2/1000 # in kWh by six P2 eqips\n",
+      "Et = E1 + E2\n",
+      "C = Et * 7\n",
+      "\n",
+      "#results\n",
+      "print \"Cost per week:\", C,\"Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cost per week: 1029.0 Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 20</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#state which is most appropriate for the following appliances which are both connected to a 240 V supply \n",
+      "#(a) Electric toaster having a power rating of 1 kW (b) Electric fire having a power rating of 3 kW\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 5; # in Amp\n",
+      "I2 = 10; # in Amp\n",
+      "I3 = 13; # in Amp\n",
+      "P1 = 1000; # in Watts\n",
+      "P2 = 3000; # in Watts\n",
+      "V = 240; #in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "It = P1/V\n",
+      "If = P2/V\n",
+      "\n",
+      "#results\n",
+      "print \"For the toaster,\", I1,\"A fuse is most appropriate\"\n",
+      "print \"For the Fire,\", I3,\"A fuse is most appropriate\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For the toaster, 5 A fuse is most appropriate\n",
+        "For the Fire, 13 A fuse is most appropriate"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_1.ipynb
new file mode 100755
index 00000000..3ab84705
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_1.ipynb
@@ -0,0 +1,806 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 2: An Introduction to Electric Circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 12</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What current must flow if 0.24 coulombs is to be transferred in 15 ms?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Q = 0.24; # in Coulomb\n",
+      "t = 0.015; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "I = Q/t\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 16.0 Ampere(A)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 12</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the quantity of electricity transferred.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 10; # in Ampere\n",
+      "t = 240; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q = I*t\n",
+      "\n",
+      "#resuts\n",
+      "print \"Charge(Q):\", Q,\"Coulomb(C)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charge(Q): 2400 Coulomb(C)\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 14</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.8; # in Ampere\n",
+      "V = 20; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 25.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the p.d.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.010; # in Ampere\n",
+      "R = 2000; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V = I*R\n",
+      "\n",
+      "#results\n",
+      "print \"p.d.(V):\", V,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "p.d.(V): 20.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is the resistance of the coil?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.050; # in Ampere\n",
+      "V = 12; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 240.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what will be the new value of the current flowing?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.005; # in Ampere\n",
+      "V1 = 100; # in Volts\n",
+      "V2 = 25; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V1/I1\n",
+      "I2 = V2/R\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\"\n",
+      "print \"Current(I):\", I2,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 20000.0 Ohms\n",
+        "Current(I): 0.00125 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is the resistance of a coil\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.050; # in Ampere\n",
+      "I2 = 200E-6; # in Ampere\n",
+      "V = 120; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = V/I1\n",
+      "R2 = V/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R1):\", R1,\"Ohms\"\n",
+      "print \"Resistance(R2):\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R1): 2400.0 Ohms\n",
+        "Resistance(R2): 600000.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 16</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "P = 100; # in Watt\n",
+      "V = 250; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "I = P/V\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.4 Ampere(A)\n",
+        "Resistance(R): 625.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.004; # in ampere\n",
+      "R = 5000; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "P = I*I*R\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 0.08 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What current will flow when it is connected to a 240 V supply?\n",
+      "#Find also the power rating of the kettle.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "R = 30; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = V*I\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 8.0 Ampere(A)\n",
+        "Power(P): 1920.0 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the p.d. across the winding, and (b) the power dissipated by the coil.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 5; # in ampere\n",
+      "R = 100; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V = I*R\n",
+      "P = I*R*I\n",
+      "\n",
+      "#results\n",
+      "print \"p.d(V):\", V,\"Volts(V)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "p.d(V): 500 Volts(V)\n",
+        "Power(P): 2500 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the resistance of each resistor.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.020; # in ampere\n",
+      "V1 = 20; # in Volts\n",
+      "I2 = 0.005; # in ampere\n",
+      "V2 = 16; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = V1/I1\n",
+      "R2 = V2/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R1):\", R1,\"Ohms\"\n",
+      "print \"Resistance(R2):\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R1): 1000.0 Ohms\n",
+        "Resistance(R2): 3200.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the current taken by the lamp and its power rating\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "R = 960; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = I*V\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.25 Ampere(A)\n",
+        "Power(P): 60.0 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the load, the power consumed and the energy dissipated in 2 minutes\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 12; # in Volts\n",
+      "R = 40; # in ohms\n",
+      "t = 120; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = I*V\n",
+      "E = P*t\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.3 Ampere(A)\n",
+        "Power(P): 3.6 Watt(W)\n",
+        "Energy(E): 432.0 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#How much energy is provided in this time?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 15; # in Volts\n",
+      "I = 2; # in ampere\n",
+      "t = 360; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "E = V*I*t\n",
+      "\n",
+      "#results\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy(E): 10800 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Estimate the cost per week of electricity\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "I = 13; # in ampere\n",
+      "t = 30; # in hours\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "P = V*I\n",
+      "E = P*t/1000 # in kWh\n",
+      "C = E*p\n",
+      "\n",
+      "#results\n",
+      "print \"Cost per week:\", C,\" Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cost per week: 655.2  Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the power rating of the heater and the current taken from the supply.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 250; # in Volts\n",
+      "E = 3.6E6; # energy in J\n",
+      "t = 2400; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "P = E/t\n",
+      "I = P/V\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 1500.0 Watt(W)\n",
+        "Current(I): 6.0 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power dissipated\n",
+      "#determine the energy used and the cost\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 20; # in ohms\n",
+      "I = 10; # in ampere\n",
+      "t = 6; # in hours\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "P = I*I*R\n",
+      "E = P*t/1000 # in kWh\n",
+      "C = E*p\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Cost per week:\", C,\"Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 2000 Watt(W)\n",
+        "Cost per week: 84.0 Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the weekly cost of electricity to the business.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "P1 = 3; # in kW\n",
+      "P2 = 150; # in Watt\n",
+      "n1 = 2; # no. of P1 Equips\n",
+      "n2 = 6; # no. of P2 Equips\n",
+      "t1 = 20; # in hours each per week\n",
+      "t2 = 30; # in hours each per week\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "E1 = P1*t1*n1 # in kWh by two P1 eqips\n",
+      "E2 = P2*t2*n2/1000 # in kWh by six P2 eqips\n",
+      "Et = E1 + E2\n",
+      "C = Et * 7\n",
+      "\n",
+      "#results\n",
+      "print \"Cost per week:\", C,\"Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cost per week: 1029.0 Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 20</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#state which is most appropriate for the following appliances which are both connected to a 240 V supply \n",
+      "#(a) Electric toaster having a power rating of 1 kW (b) Electric fire having a power rating of 3 kW\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 5; # in Amp\n",
+      "I2 = 10; # in Amp\n",
+      "I3 = 13; # in Amp\n",
+      "P1 = 1000; # in Watts\n",
+      "P2 = 3000; # in Watts\n",
+      "V = 240; #in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "It = P1/V\n",
+      "If = P2/V\n",
+      "\n",
+      "#results\n",
+      "print \"For the toaster,\", I1,\"A fuse is most appropriate\"\n",
+      "print \"For the Fire,\", I3,\"A fuse is most appropriate\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For the toaster, 5 A fuse is most appropriate\n",
+        "For the Fire, 13 A fuse is most appropriate"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_2.ipynb
new file mode 100755
index 00000000..3ab84705
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_2.ipynb
@@ -0,0 +1,806 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 2: An Introduction to Electric Circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 12</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What current must flow if 0.24 coulombs is to be transferred in 15 ms?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Q = 0.24; # in Coulomb\n",
+      "t = 0.015; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "I = Q/t\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 16.0 Ampere(A)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 12</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the quantity of electricity transferred.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 10; # in Ampere\n",
+      "t = 240; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q = I*t\n",
+      "\n",
+      "#resuts\n",
+      "print \"Charge(Q):\", Q,\"Coulomb(C)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charge(Q): 2400 Coulomb(C)\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 14</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.8; # in Ampere\n",
+      "V = 20; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 25.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the p.d.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.010; # in Ampere\n",
+      "R = 2000; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V = I*R\n",
+      "\n",
+      "#results\n",
+      "print \"p.d.(V):\", V,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "p.d.(V): 20.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is the resistance of the coil?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.050; # in Ampere\n",
+      "V = 12; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 240.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what will be the new value of the current flowing?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.005; # in Ampere\n",
+      "V1 = 100; # in Volts\n",
+      "V2 = 25; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V1/I1\n",
+      "I2 = V2/R\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\"\n",
+      "print \"Current(I):\", I2,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 20000.0 Ohms\n",
+        "Current(I): 0.00125 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is the resistance of a coil\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.050; # in Ampere\n",
+      "I2 = 200E-6; # in Ampere\n",
+      "V = 120; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = V/I1\n",
+      "R2 = V/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R1):\", R1,\"Ohms\"\n",
+      "print \"Resistance(R2):\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R1): 2400.0 Ohms\n",
+        "Resistance(R2): 600000.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 16</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "P = 100; # in Watt\n",
+      "V = 250; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "I = P/V\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.4 Ampere(A)\n",
+        "Resistance(R): 625.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.004; # in ampere\n",
+      "R = 5000; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "P = I*I*R\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 0.08 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What current will flow when it is connected to a 240 V supply?\n",
+      "#Find also the power rating of the kettle.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "R = 30; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = V*I\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 8.0 Ampere(A)\n",
+        "Power(P): 1920.0 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the p.d. across the winding, and (b) the power dissipated by the coil.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 5; # in ampere\n",
+      "R = 100; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V = I*R\n",
+      "P = I*R*I\n",
+      "\n",
+      "#results\n",
+      "print \"p.d(V):\", V,\"Volts(V)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "p.d(V): 500 Volts(V)\n",
+        "Power(P): 2500 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the resistance of each resistor.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.020; # in ampere\n",
+      "V1 = 20; # in Volts\n",
+      "I2 = 0.005; # in ampere\n",
+      "V2 = 16; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = V1/I1\n",
+      "R2 = V2/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R1):\", R1,\"Ohms\"\n",
+      "print \"Resistance(R2):\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R1): 1000.0 Ohms\n",
+        "Resistance(R2): 3200.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the current taken by the lamp and its power rating\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "R = 960; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = I*V\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.25 Ampere(A)\n",
+        "Power(P): 60.0 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the load, the power consumed and the energy dissipated in 2 minutes\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 12; # in Volts\n",
+      "R = 40; # in ohms\n",
+      "t = 120; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = I*V\n",
+      "E = P*t\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.3 Ampere(A)\n",
+        "Power(P): 3.6 Watt(W)\n",
+        "Energy(E): 432.0 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#How much energy is provided in this time?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 15; # in Volts\n",
+      "I = 2; # in ampere\n",
+      "t = 360; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "E = V*I*t\n",
+      "\n",
+      "#results\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy(E): 10800 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Estimate the cost per week of electricity\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "I = 13; # in ampere\n",
+      "t = 30; # in hours\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "P = V*I\n",
+      "E = P*t/1000 # in kWh\n",
+      "C = E*p\n",
+      "\n",
+      "#results\n",
+      "print \"Cost per week:\", C,\" Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cost per week: 655.2  Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the power rating of the heater and the current taken from the supply.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 250; # in Volts\n",
+      "E = 3.6E6; # energy in J\n",
+      "t = 2400; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "P = E/t\n",
+      "I = P/V\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 1500.0 Watt(W)\n",
+        "Current(I): 6.0 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power dissipated\n",
+      "#determine the energy used and the cost\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 20; # in ohms\n",
+      "I = 10; # in ampere\n",
+      "t = 6; # in hours\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "P = I*I*R\n",
+      "E = P*t/1000 # in kWh\n",
+      "C = E*p\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Cost per week:\", C,\"Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 2000 Watt(W)\n",
+        "Cost per week: 84.0 Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the weekly cost of electricity to the business.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "P1 = 3; # in kW\n",
+      "P2 = 150; # in Watt\n",
+      "n1 = 2; # no. of P1 Equips\n",
+      "n2 = 6; # no. of P2 Equips\n",
+      "t1 = 20; # in hours each per week\n",
+      "t2 = 30; # in hours each per week\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "E1 = P1*t1*n1 # in kWh by two P1 eqips\n",
+      "E2 = P2*t2*n2/1000 # in kWh by six P2 eqips\n",
+      "Et = E1 + E2\n",
+      "C = Et * 7\n",
+      "\n",
+      "#results\n",
+      "print \"Cost per week:\", C,\"Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cost per week: 1029.0 Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 20</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#state which is most appropriate for the following appliances which are both connected to a 240 V supply \n",
+      "#(a) Electric toaster having a power rating of 1 kW (b) Electric fire having a power rating of 3 kW\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 5; # in Amp\n",
+      "I2 = 10; # in Amp\n",
+      "I3 = 13; # in Amp\n",
+      "P1 = 1000; # in Watts\n",
+      "P2 = 3000; # in Watts\n",
+      "V = 240; #in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "It = P1/V\n",
+      "If = P2/V\n",
+      "\n",
+      "#results\n",
+      "print \"For the toaster,\", I1,\"A fuse is most appropriate\"\n",
+      "print \"For the Fire,\", I3,\"A fuse is most appropriate\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For the toaster, 5 A fuse is most appropriate\n",
+        "For the Fire, 13 A fuse is most appropriate"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_3.ipynb
new file mode 100755
index 00000000..8bc22a1d
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_02-checkpoint_3.ipynb
@@ -0,0 +1,805 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7f992abe2c8d4f4a69090179e106282aadd94b3277ec90a7828dcefac26e044f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 2: An Introduction to Electric Circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 12</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Q = 0.24; # in Coulomb\n",
+      "t = 0.015; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "I = Q/t\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 16.0 Ampere(A)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 12</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 10; # in Ampere\n",
+      "t = 240; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q = I*t\n",
+      "\n",
+      "#resuts\n",
+      "print \"Charge(Q):\", Q,\"Coulomb(C)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Charge(Q): 2400 Coulomb(C)\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 14</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.8; # in Ampere\n",
+      "V = 20; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 25.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.010; # in Ampere\n",
+      "R = 2000; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V = I*R\n",
+      "\n",
+      "#results\n",
+      "print \"p.d.(V):\", V,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "p.d.(V): 20.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.050; # in Ampere\n",
+      "V = 12; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 240.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.005; # in Ampere\n",
+      "V1 = 100; # in Volts\n",
+      "V2 = 25; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R = V1/I1\n",
+      "I2 = V2/R\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R):\", R,\"Ohms\"\n",
+      "print \"Current(I):\", I2,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R): 20000.0 Ohms\n",
+        "Current(I): 0.00125 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 15/h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.050; # in Ampere\n",
+      "I2 = 200E-6; # in Ampere\n",
+      "V = 120; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = V/I1\n",
+      "R2 = V/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R1):\", R1,\"Ohms\"\n",
+      "print \"Resistance(R2):\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R1): 2400.0 Ohms\n",
+        "Resistance(R2): 600000.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 16</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "P = 100; # in Watt\n",
+      "V = 250; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "I = P/V\n",
+      "R = V/I\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Resistance(R):\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.4 Ampere(A)\n",
+        "Resistance(R): 625.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 0.004; # in ampere\n",
+      "R = 5000; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "P = I*I*R\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 0.08 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Find also the power rating of the kettle.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "R = 30; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = V*I\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 8.0 Ampere(A)\n",
+        "Power(P): 1920.0 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I = 5; # in ampere\n",
+      "R = 100; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V = I*R\n",
+      "P = I*R*I\n",
+      "\n",
+      "#results\n",
+      "print \"p.d(V):\", V,\"Volts(V)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "p.d(V): 500 Volts(V)\n",
+        "Power(P): 2500 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 17</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 0.020; # in ampere\n",
+      "V1 = 20; # in Volts\n",
+      "I2 = 0.005; # in ampere\n",
+      "V2 = 16; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = V1/I1\n",
+      "R2 = V2/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance(R1):\", R1,\"Ohms\"\n",
+      "print \"Resistance(R2):\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance(R1): 1000.0 Ohms\n",
+        "Resistance(R2): 3200.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "R = 960; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = I*V\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.25 Ampere(A)\n",
+        "Power(P): 60.0 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 12; # in Volts\n",
+      "R = 40; # in ohms\n",
+      "t = 120; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "I = V/R\n",
+      "P = I*V\n",
+      "E = P*t\n",
+      "\n",
+      "#results\n",
+      "print \"Current(I):\", I,\"Ampere(A)\"\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Current(I): 0.3 Ampere(A)\n",
+        "Power(P): 3.6 Watt(W)\n",
+        "Energy(E): 432.0 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 15; # in Volts\n",
+      "I = 2; # in ampere\n",
+      "t = 360; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "E = V*I*t\n",
+      "\n",
+      "#results\n",
+      "print \"Energy(E):\", E,\"Joule(J)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy(E): 10800 Joule(J)"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 18</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 240; # in Volts\n",
+      "I = 13; # in ampere\n",
+      "t = 30; # in hours\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "P = V*I\n",
+      "E = P*t/1000 # in kWh\n",
+      "C = E*p\n",
+      "\n",
+      "#results\n",
+      "print \"Cost per week:\", C,\" Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cost per week: 655.2  Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V = 250; # in Volts\n",
+      "E = 3.6E6; # energy in J\n",
+      "t = 2400; # in sec\n",
+      "\n",
+      "#calculation:\n",
+      "P = E/t\n",
+      "I = P/V\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Current(I):\", I,\"Ampere(A)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 1500.0 Watt(W)\n",
+        "Current(I): 6.0 Ampere(A)"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 20; # in ohms\n",
+      "I = 10; # in ampere\n",
+      "t = 6; # in hours\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "P = I*I*R\n",
+      "E = P*t/1000 # in kWh\n",
+      "C = E*p\n",
+      "\n",
+      "#results\n",
+      "print \"Power(P):\", P,\"Watt(W)\"\n",
+      "print \"Cost per week:\", C,\"Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power(P): 2000 Watt(W)\n",
+        "Cost per week: 84.0 Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 19</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "P1 = 3; # in kW\n",
+      "P2 = 150; # in Watt\n",
+      "n1 = 2; # no. of P1 Equips\n",
+      "n2 = 6; # no. of P2 Equips\n",
+      "t1 = 20; # in hours each per week\n",
+      "t2 = 30; # in hours each per week\n",
+      "p = 7; # in paise per kWh\n",
+      "\n",
+      "#calculation:\n",
+      "E1 = P1*t1*n1 # in kWh by two P1 eqips\n",
+      "E2 = P2*t2*n2/1000 # in kWh by six P2 eqips\n",
+      "Et = E1 + E2\n",
+      "C = Et * 7\n",
+      "\n",
+      "#results\n",
+      "print \"Cost per week:\", C,\"Paise(p)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cost per week: 1029.0 Paise(p)"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 20</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "I1 = 5; # in Amp\n",
+      "I2 = 10; # in Amp\n",
+      "I3 = 13; # in Amp\n",
+      "P1 = 1000; # in Watts\n",
+      "P2 = 3000; # in Watts\n",
+      "V = 240; #in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "It = P1/V\n",
+      "If = P2/V\n",
+      "\n",
+      "#results\n",
+      "print \"For the toaster,\", I1,\"A fuse is most appropriate\"\n",
+      "print \"For the Fire,\", I3,\"A fuse is most appropriate\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For the toaster, 5 A fuse is most appropriate\n",
+        "For the Fire, 13 A fuse is most appropriate"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_07.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_07.ipynb
new file mode 100755
index 00000000..e3a59d72
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_07.ipynb
@@ -0,0 +1,23 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:745577b5410a081be371acb7057a8995eea1df4ada3643669e4149982d7e2177"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Magnetic circuits"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_07_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_07_1.ipynb
new file mode 100755
index 00000000..e3a59d72
--- /dev/null
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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:745577b5410a081be371acb7057a8995eea1df4ada3643669e4149982d7e2177"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Magnetic circuits"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_07_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_07_2.ipynb
new file mode 100755
index 00000000..e3a59d72
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_07_2.ipynb
@@ -0,0 +1,23 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:745577b5410a081be371acb7057a8995eea1df4ada3643669e4149982d7e2177"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Magnetic circuits"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_07_3.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_07_3.ipynb
new file mode 100755
index 00000000..e3a59d72
--- /dev/null
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+ "metadata": {
+  "name": "",
+  "signature": "sha256:745577b5410a081be371acb7057a8995eea1df4ada3643669e4149982d7e2177"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Magnetic circuits"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_11.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_11.ipynb
new file mode 100755
index 00000000..3454218e
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_11.ipynb
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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:01039969f14274b6a7f44cb14ba52c884b9505fea9216d7f805101b2dc10a0c1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Semiconductor diodes"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_11_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_11_1.ipynb
new file mode 100755
index 00000000..3454218e
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_11_1.ipynb
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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:01039969f14274b6a7f44cb14ba52c884b9505fea9216d7f805101b2dc10a0c1"
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+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Semiconductor diodes"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_11_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_11_2.ipynb
new file mode 100755
index 00000000..3454218e
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_11_2.ipynb
@@ -0,0 +1,23 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:01039969f14274b6a7f44cb14ba52c884b9505fea9216d7f805101b2dc10a0c1"
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+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Semiconductor diodes"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_11_3.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_11_3.ipynb
new file mode 100755
index 00000000..3454218e
--- /dev/null
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+ "metadata": {
+  "name": "",
+  "signature": "sha256:01039969f14274b6a7f44cb14ba52c884b9505fea9216d7f805101b2dc10a0c1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Semiconductor diodes"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_37.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_37.ipynb
new file mode 100755
index 00000000..de8c7108
--- /dev/null
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@@ -0,0 +1,23 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b6595de1d338f736a24eaed6908c5a0f3d88a34ed3ba7d5f05822513e391d985"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "A numerical method of harmonic analysis"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_37_1.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_37_1.ipynb
new file mode 100755
index 00000000..de8c7108
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_37_1.ipynb
@@ -0,0 +1,23 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b6595de1d338f736a24eaed6908c5a0f3d88a34ed3ba7d5f05822513e391d985"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "A numerical method of harmonic analysis"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_37_2.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_37_2.ipynb
new file mode 100755
index 00000000..de8c7108
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_37_2.ipynb
@@ -0,0 +1,23 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b6595de1d338f736a24eaed6908c5a0f3d88a34ed3ba7d5f05822513e391d985"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "A numerical method of harmonic analysis"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/Chapter_37_3.ipynb b/Electrical_Circuit_Theory_And_Technology/Chapter_37_3.ipynb
new file mode 100755
index 00000000..de8c7108
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/Chapter_37_3.ipynb
@@ -0,0 +1,23 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b6595de1d338f736a24eaed6908c5a0f3d88a34ed3ba7d5f05822513e391d985"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "A numerical method of harmonic analysis"
+     ]
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/README.txt b/Electrical_Circuit_Theory_And_Technology/README.txt
new file mode 100755
index 00000000..e94e265f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Pankaj Goyal
+Course: btech
+College/Institute/Organization: Indian Institute of Technology, Bombay
+Department/Designation: Aerospace
+Book Title: Electrical Circuit Theory And Technology
+Author: J. O. Bird
+Publisher: Newnes, Burlington, Ma
+Year of publication: 2003
+Isbn: 0-7506-5784-7
+Edition: 2
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint.ipynb
new file mode 100755
index 00000000..f79d1396
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint.ipynb
@@ -0,0 +1,525 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 3: Resistance variation</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 24</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resistance of an 8 m length of the same wire,\n",
+      "#and (b) the length of the same wire when the resistance is 420\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 600; # in ohms\n",
+      "L = 5; # in meter\n",
+      "L1 = 8; # in meter\n",
+      "R2 = 420; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = R*L1/L\n",
+      "L2 = R2*L/R\n",
+      "\n",
+      "#results\n",
+      "print \"a)Resistance\", R1,\"Ohms\"\n",
+      "print \"b)Length:\", L2,\"meters(m)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Resistance 960.0 Ohms\n",
+        "b)Length: 3.5 meters(m)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 24</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the resistance of a wire of the same length and material if the cross-sectional area is 5 mm2,\n",
+      "#(b) the cross-sectional area of a wire of the same length and material of resistance 750 \n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 300; # in ohms\n",
+      "A = 2; # in mm2\n",
+      "A1 = 5; # in mm2\n",
+      "R2 = 750; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = R*A/A1\n",
+      "A2 = R*A/R2\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Resistance\", R1,\"Ohms\"\n",
+      "print \"(b)C.S.A:\", A2,\"mm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Resistance 120.0 Ohms\n",
+        "(b)C.S.A: 0.8 mm^2"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the resistance of the wire.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.16; # in ohms\n",
+      "A = 3; # in mm2\n",
+      "L = 8; # in m\n",
+      "A1 = 1; # in mm2\n",
+      "\n",
+      "#calculation:\n",
+      "L1 = L*3\n",
+      "R1 = R*A*L1/(A1*L)\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R1,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 1.44 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the resistance.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "A = 100E-6; # in m2\n",
+      "L = 2000; # in m\n",
+      "p = 0.03E-6; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "R = p*L/A\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 0.6 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the cross-sectional area\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.25; # in ohms\n",
+      "L = 40; # in m\n",
+      "p = 0.02E-6; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "A = p*L*1E6/R\n",
+      "\n",
+      "#results\n",
+      "print \"C.S.A \", A,\"mm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C.S.A  3.2 mm^2"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistivity of the wire.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 150; # in ohms\n",
+      "L = 1500; # in m\n",
+      "A = 0.17E-6; # in m2\n",
+      "\n",
+      "#calculation:\n",
+      "p = R*A*1E6/L\n",
+      "\n",
+      "#results\n",
+      "print \"resistivity\", p,\"uOhm.m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistivity 0.017 uOhm.m"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 26</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing the variables:\n",
+      "d = 0.012; # in m\n",
+      "L = 1200; # in m\n",
+      "p = 1.7E-8; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "A = math.pi*d*d/4\n",
+      "R = p*L/A\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", round(R,3),\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 0.18 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R0 = 100; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 70; # in \u00b0C\n",
+      "a0 = 0.0043; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R70 = R0*(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R70,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 130.1 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 27; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 35; # in \u00b0C\n",
+      "a0 = 0.0038; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R0 = R1/(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", round(R0,2),\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 23.83 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R0 = 1000; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 80; # in \u00b0C\n",
+      "a0 = -0.0005; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R80 = R0*(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", R80,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 960.0 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 28</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the resistance of the coil\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R20 = 10; # in ohms\n",
+      "T0 = 20; # in \u00b0C\n",
+      "T1 = 100; # in \u00b0C\n",
+      "a20 = 0.004; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R100 = R20*(1 + (a20)*(T1 - T0))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", R100,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 13.2 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 28</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the temperature to which the coil has risen\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R18 = 200; # in ohms\n",
+      "R1 = 240; # in ohms\n",
+      "T0 = 18; # in \u00b0C\n",
+      "a18 = 0.0039; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "T1 = (((R1/R18)-1)/a18) + T0\n",
+      "\n",
+      "#results\n",
+      "print \"Temperature\", round(T1,2),\"degC\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature 69.28 degC"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 29</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance of the wire\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R20 = 200; # in ohms\n",
+      "T0 = 20; # in \u00b0C\n",
+      "T1 = 90; # in \u00b0C\n",
+      "a0 = 0.004; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R90 = R20*(1 + (a0*T1))/(1 + (a0*T0))\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", round(R90,0),\"ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 252.0 ohms"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_1.ipynb
new file mode 100755
index 00000000..f79d1396
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_1.ipynb
@@ -0,0 +1,525 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 3: Resistance variation</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 24</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resistance of an 8 m length of the same wire,\n",
+      "#and (b) the length of the same wire when the resistance is 420\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 600; # in ohms\n",
+      "L = 5; # in meter\n",
+      "L1 = 8; # in meter\n",
+      "R2 = 420; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = R*L1/L\n",
+      "L2 = R2*L/R\n",
+      "\n",
+      "#results\n",
+      "print \"a)Resistance\", R1,\"Ohms\"\n",
+      "print \"b)Length:\", L2,\"meters(m)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Resistance 960.0 Ohms\n",
+        "b)Length: 3.5 meters(m)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 24</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the resistance of a wire of the same length and material if the cross-sectional area is 5 mm2,\n",
+      "#(b) the cross-sectional area of a wire of the same length and material of resistance 750 \n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 300; # in ohms\n",
+      "A = 2; # in mm2\n",
+      "A1 = 5; # in mm2\n",
+      "R2 = 750; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = R*A/A1\n",
+      "A2 = R*A/R2\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Resistance\", R1,\"Ohms\"\n",
+      "print \"(b)C.S.A:\", A2,\"mm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Resistance 120.0 Ohms\n",
+        "(b)C.S.A: 0.8 mm^2"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the resistance of the wire.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.16; # in ohms\n",
+      "A = 3; # in mm2\n",
+      "L = 8; # in m\n",
+      "A1 = 1; # in mm2\n",
+      "\n",
+      "#calculation:\n",
+      "L1 = L*3\n",
+      "R1 = R*A*L1/(A1*L)\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R1,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 1.44 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the resistance.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "A = 100E-6; # in m2\n",
+      "L = 2000; # in m\n",
+      "p = 0.03E-6; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "R = p*L/A\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 0.6 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the cross-sectional area\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.25; # in ohms\n",
+      "L = 40; # in m\n",
+      "p = 0.02E-6; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "A = p*L*1E6/R\n",
+      "\n",
+      "#results\n",
+      "print \"C.S.A \", A,\"mm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C.S.A  3.2 mm^2"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistivity of the wire.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 150; # in ohms\n",
+      "L = 1500; # in m\n",
+      "A = 0.17E-6; # in m2\n",
+      "\n",
+      "#calculation:\n",
+      "p = R*A*1E6/L\n",
+      "\n",
+      "#results\n",
+      "print \"resistivity\", p,\"uOhm.m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistivity 0.017 uOhm.m"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 26</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing the variables:\n",
+      "d = 0.012; # in m\n",
+      "L = 1200; # in m\n",
+      "p = 1.7E-8; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "A = math.pi*d*d/4\n",
+      "R = p*L/A\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", round(R,3),\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 0.18 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R0 = 100; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 70; # in \u00b0C\n",
+      "a0 = 0.0043; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R70 = R0*(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R70,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 130.1 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 27; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 35; # in \u00b0C\n",
+      "a0 = 0.0038; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R0 = R1/(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", round(R0,2),\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 23.83 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R0 = 1000; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 80; # in \u00b0C\n",
+      "a0 = -0.0005; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R80 = R0*(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", R80,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 960.0 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 28</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the resistance of the coil\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R20 = 10; # in ohms\n",
+      "T0 = 20; # in \u00b0C\n",
+      "T1 = 100; # in \u00b0C\n",
+      "a20 = 0.004; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R100 = R20*(1 + (a20)*(T1 - T0))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", R100,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 13.2 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 28</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the temperature to which the coil has risen\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R18 = 200; # in ohms\n",
+      "R1 = 240; # in ohms\n",
+      "T0 = 18; # in \u00b0C\n",
+      "a18 = 0.0039; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "T1 = (((R1/R18)-1)/a18) + T0\n",
+      "\n",
+      "#results\n",
+      "print \"Temperature\", round(T1,2),\"degC\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature 69.28 degC"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 29</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance of the wire\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R20 = 200; # in ohms\n",
+      "T0 = 20; # in \u00b0C\n",
+      "T1 = 90; # in \u00b0C\n",
+      "a0 = 0.004; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R90 = R20*(1 + (a0*T1))/(1 + (a0*T0))\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", round(R90,0),\"ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 252.0 ohms"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_2.ipynb
new file mode 100755
index 00000000..f79d1396
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_2.ipynb
@@ -0,0 +1,525 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 3: Resistance variation</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 24</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resistance of an 8 m length of the same wire,\n",
+      "#and (b) the length of the same wire when the resistance is 420\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 600; # in ohms\n",
+      "L = 5; # in meter\n",
+      "L1 = 8; # in meter\n",
+      "R2 = 420; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = R*L1/L\n",
+      "L2 = R2*L/R\n",
+      "\n",
+      "#results\n",
+      "print \"a)Resistance\", R1,\"Ohms\"\n",
+      "print \"b)Length:\", L2,\"meters(m)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Resistance 960.0 Ohms\n",
+        "b)Length: 3.5 meters(m)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 24</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the resistance of a wire of the same length and material if the cross-sectional area is 5 mm2,\n",
+      "#(b) the cross-sectional area of a wire of the same length and material of resistance 750 \n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 300; # in ohms\n",
+      "A = 2; # in mm2\n",
+      "A1 = 5; # in mm2\n",
+      "R2 = 750; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = R*A/A1\n",
+      "A2 = R*A/R2\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Resistance\", R1,\"Ohms\"\n",
+      "print \"(b)C.S.A:\", A2,\"mm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Resistance 120.0 Ohms\n",
+        "(b)C.S.A: 0.8 mm^2"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the resistance of the wire.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.16; # in ohms\n",
+      "A = 3; # in mm2\n",
+      "L = 8; # in m\n",
+      "A1 = 1; # in mm2\n",
+      "\n",
+      "#calculation:\n",
+      "L1 = L*3\n",
+      "R1 = R*A*L1/(A1*L)\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R1,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 1.44 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the resistance.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "A = 100E-6; # in m2\n",
+      "L = 2000; # in m\n",
+      "p = 0.03E-6; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "R = p*L/A\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 0.6 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the cross-sectional area\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.25; # in ohms\n",
+      "L = 40; # in m\n",
+      "p = 0.02E-6; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "A = p*L*1E6/R\n",
+      "\n",
+      "#results\n",
+      "print \"C.S.A \", A,\"mm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C.S.A  3.2 mm^2"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistivity of the wire.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 150; # in ohms\n",
+      "L = 1500; # in m\n",
+      "A = 0.17E-6; # in m2\n",
+      "\n",
+      "#calculation:\n",
+      "p = R*A*1E6/L\n",
+      "\n",
+      "#results\n",
+      "print \"resistivity\", p,\"uOhm.m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistivity 0.017 uOhm.m"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 26</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing the variables:\n",
+      "d = 0.012; # in m\n",
+      "L = 1200; # in m\n",
+      "p = 1.7E-8; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "A = math.pi*d*d/4\n",
+      "R = p*L/A\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", round(R,3),\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 0.18 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R0 = 100; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 70; # in \u00b0C\n",
+      "a0 = 0.0043; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R70 = R0*(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R70,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 130.1 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 27; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 35; # in \u00b0C\n",
+      "a0 = 0.0038; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R0 = R1/(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", round(R0,2),\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 23.83 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its resistance\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R0 = 1000; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 80; # in \u00b0C\n",
+      "a0 = -0.0005; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R80 = R0*(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", R80,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 960.0 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 28</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the resistance of the coil\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R20 = 10; # in ohms\n",
+      "T0 = 20; # in \u00b0C\n",
+      "T1 = 100; # in \u00b0C\n",
+      "a20 = 0.004; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R100 = R20*(1 + (a20)*(T1 - T0))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", R100,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 13.2 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 28</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the temperature to which the coil has risen\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R18 = 200; # in ohms\n",
+      "R1 = 240; # in ohms\n",
+      "T0 = 18; # in \u00b0C\n",
+      "a18 = 0.0039; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "T1 = (((R1/R18)-1)/a18) + T0\n",
+      "\n",
+      "#results\n",
+      "print \"Temperature\", round(T1,2),\"degC\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature 69.28 degC"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 29</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance of the wire\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R20 = 200; # in ohms\n",
+      "T0 = 20; # in \u00b0C\n",
+      "T1 = 90; # in \u00b0C\n",
+      "a0 = 0.004; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R90 = R20*(1 + (a0*T1))/(1 + (a0*T0))\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", round(R90,0),\"ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 252.0 ohms"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_3.ipynb
new file mode 100755
index 00000000..d9b6cd03
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_03-checkpoint_3.ipynb
@@ -0,0 +1,524 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c029dc52a72cffc31b077ad9cf4a482b5fa3b43cb728ab8a80727f5c08bb19ce"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 3: Resistance variation</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 24</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 600; # in ohms\n",
+      "L = 5; # in meter\n",
+      "L1 = 8; # in meter\n",
+      "R2 = 420; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = R*L1/L\n",
+      "L2 = R2*L/R\n",
+      "\n",
+      "#results\n",
+      "print \"a)Resistance\", R1,\"Ohms\"\n",
+      "print \"b)Length:\", L2,\"meters(m)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Resistance 960.0 Ohms\n",
+        "b)Length: 3.5 meters(m)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 24</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 300; # in ohms\n",
+      "A = 2; # in mm2\n",
+      "A1 = 5; # in mm2\n",
+      "R2 = 750; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1 = R*A/A1\n",
+      "A2 = R*A/R2\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Resistance\", R1,\"Ohms\"\n",
+      "print \"(b)C.S.A:\", A2,\"mm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Resistance 120.0 Ohms\n",
+        "(b)C.S.A: 0.8 mm^2"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.16; # in ohms\n",
+      "A = 3; # in mm2\n",
+      "L = 8; # in m\n",
+      "A1 = 1; # in mm2\n",
+      "\n",
+      "#calculation:\n",
+      "L1 = L*3\n",
+      "R1 = R*A*L1/(A1*L)\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R1,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 1.44 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "A = 100E-6; # in m2\n",
+      "L = 2000; # in m\n",
+      "p = 0.03E-6; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "R = p*L/A\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 0.6 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.25; # in ohms\n",
+      "L = 40; # in m\n",
+      "p = 0.02E-6; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "A = p*L*1E6/R\n",
+      "\n",
+      "#results\n",
+      "print \"C.S.A \", A,\"mm^2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C.S.A  3.2 mm^2"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 25</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 150; # in ohms\n",
+      "L = 1500; # in m\n",
+      "A = 0.17E-6; # in m2\n",
+      "\n",
+      "#calculation:\n",
+      "p = R*A*1E6/L\n",
+      "\n",
+      "#results\n",
+      "print \"resistivity\", p,\"uOhm.m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistivity 0.017 uOhm.m"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 26</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing the variables:\n",
+      "d = 0.012; # in m\n",
+      "L = 1200; # in m\n",
+      "p = 1.7E-8; # in ohm m\n",
+      "\n",
+      "#calculation:\n",
+      "A = math.pi*d*d/4\n",
+      "R = p*L/A\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", round(R,3),\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 0.18 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R0 = 100; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 70; # in \u00b0C\n",
+      "a0 = 0.0043; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R70 = R0*(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", R70,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 130.1 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 27; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 35; # in \u00b0C\n",
+      "a0 = 0.0038; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R0 = R1/(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", round(R0,2),\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 23.83 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 27</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R0 = 1000; # in ohms\n",
+      "T0 = 0; # in \u00b0C\n",
+      "T1 = 80; # in \u00b0C\n",
+      "a0 = -0.0005; # in per\u00b0C\n",
+      "\n",
+      "#calculation:\n",
+      "R80 = R0*(1 + (a0*T1))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", R80,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 960.0 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 28</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R20 = 10; # in ohms\n",
+      "T0 = 20; # in \u00b0C\n",
+      "T1 = 100; # in \u00b0C\n",
+      "a20 = 0.004; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R100 = R20*(1 + (a20)*(T1 - T0))\n",
+      "\n",
+      "#results\n",
+      "print \"resistance\", R100,\"Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resistance 13.2 Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 28</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R18 = 200; # in ohms\n",
+      "R1 = 240; # in ohms\n",
+      "T0 = 18; # in \u00b0C\n",
+      "a18 = 0.0039; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "T1 = (((R1/R18)-1)/a18) + T0\n",
+      "\n",
+      "#results\n",
+      "print \"Temperature\", round(T1,2),\"degC\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature 69.28 degC"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 29</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R20 = 200; # in ohms\n",
+      "T0 = 20; # in \u00b0C\n",
+      "T1 = 90; # in \u00b0C\n",
+      "a0 = 0.004; # in per\u00b0C\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R90 = R20*(1 + (a0*T1))/(1 + (a0*T0))\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", round(R90,0),\"ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 252.0 ohms"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint.ipynb
new file mode 100755
index 00000000..0c38cd30
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint.ipynb
@@ -0,0 +1,185 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 4: Chemical effects of\n",
+      "electricity</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 35</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. and the internal resistance of the batteries formed by (a)Series arrangement and (b)Parallel Arrangement.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.2; # in ohms\n",
+      "n = 8; # no. of cells\n",
+      "e = 2.2; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "es = n*e\n",
+      "ep = e\n",
+      "Rs = n*R\n",
+      "Rp = R/n\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Resistance\", Rs,\"ohms; e.m.f\", es,\"Volts(V)\"\n",
+      "print \"(b)Resistance\", Rp,\"ohms; e.m.f\", ep,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Resistance 1.6 ohms; e.m.f 17.6 Volts(V)\n",
+        "(b)Resistance 0.025 ohms; e.m.f 2.2 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 35</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate its terminal p.d. if it delivers (a) 5 A, (b) 50 A\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "r = 0.02; # in ohms\n",
+      "e = 2; # in volts\n",
+      "I1 = 5; # in Amperes\n",
+      "I2 = 50; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "pd1 = e - (I1*r)\n",
+      "pd2 = e - (I2*r)\n",
+      "\n",
+      "#results\n",
+      "print \"(a)p.d\", pd1,\"Volts(V)\"\n",
+      "print \"(b)p.d\", pd2,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)p.d 1.9 Volts(V)\n",
+        "(b)p.d 1.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 36</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the internal resistance of the battery.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "e1 = 25; # in volts\n",
+      "e2 = 24; # in volts\n",
+      "I2 = 10; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "r = (e1 - e2)/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", r,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 0.1 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 36</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current flowing in the circuit and (b) the p.d. at the battery terminals.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "r = 0.2; # in ohms\n",
+      "n = 10; # no. of cells\n",
+      "e = 1.5; # in volts\n",
+      "R = 58; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "es = n*e\n",
+      "rs = n*r\n",
+      "I = es/(rs + R)\n",
+      "pd = es - (I*rs)\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Current\", I,\"Amperes(A)\"\n",
+      "print \"(b)p.d\", pd,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Current 0.25 Amperes(A)\n",
+        "(b)p.d 14.5 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_1.ipynb
new file mode 100755
index 00000000..0c38cd30
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_1.ipynb
@@ -0,0 +1,185 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 4: Chemical effects of\n",
+      "electricity</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 35</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. and the internal resistance of the batteries formed by (a)Series arrangement and (b)Parallel Arrangement.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.2; # in ohms\n",
+      "n = 8; # no. of cells\n",
+      "e = 2.2; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "es = n*e\n",
+      "ep = e\n",
+      "Rs = n*R\n",
+      "Rp = R/n\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Resistance\", Rs,\"ohms; e.m.f\", es,\"Volts(V)\"\n",
+      "print \"(b)Resistance\", Rp,\"ohms; e.m.f\", ep,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Resistance 1.6 ohms; e.m.f 17.6 Volts(V)\n",
+        "(b)Resistance 0.025 ohms; e.m.f 2.2 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 35</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate its terminal p.d. if it delivers (a) 5 A, (b) 50 A\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "r = 0.02; # in ohms\n",
+      "e = 2; # in volts\n",
+      "I1 = 5; # in Amperes\n",
+      "I2 = 50; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "pd1 = e - (I1*r)\n",
+      "pd2 = e - (I2*r)\n",
+      "\n",
+      "#results\n",
+      "print \"(a)p.d\", pd1,\"Volts(V)\"\n",
+      "print \"(b)p.d\", pd2,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)p.d 1.9 Volts(V)\n",
+        "(b)p.d 1.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 36</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the internal resistance of the battery.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "e1 = 25; # in volts\n",
+      "e2 = 24; # in volts\n",
+      "I2 = 10; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "r = (e1 - e2)/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", r,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 0.1 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 36</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current flowing in the circuit and (b) the p.d. at the battery terminals.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "r = 0.2; # in ohms\n",
+      "n = 10; # no. of cells\n",
+      "e = 1.5; # in volts\n",
+      "R = 58; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "es = n*e\n",
+      "rs = n*r\n",
+      "I = es/(rs + R)\n",
+      "pd = es - (I*rs)\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Current\", I,\"Amperes(A)\"\n",
+      "print \"(b)p.d\", pd,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Current 0.25 Amperes(A)\n",
+        "(b)p.d 14.5 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_2.ipynb
new file mode 100755
index 00000000..0c38cd30
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_2.ipynb
@@ -0,0 +1,185 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 4: Chemical effects of\n",
+      "electricity</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 35</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. and the internal resistance of the batteries formed by (a)Series arrangement and (b)Parallel Arrangement.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.2; # in ohms\n",
+      "n = 8; # no. of cells\n",
+      "e = 2.2; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "es = n*e\n",
+      "ep = e\n",
+      "Rs = n*R\n",
+      "Rp = R/n\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Resistance\", Rs,\"ohms; e.m.f\", es,\"Volts(V)\"\n",
+      "print \"(b)Resistance\", Rp,\"ohms; e.m.f\", ep,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Resistance 1.6 ohms; e.m.f 17.6 Volts(V)\n",
+        "(b)Resistance 0.025 ohms; e.m.f 2.2 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 35</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate its terminal p.d. if it delivers (a) 5 A, (b) 50 A\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "r = 0.02; # in ohms\n",
+      "e = 2; # in volts\n",
+      "I1 = 5; # in Amperes\n",
+      "I2 = 50; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "pd1 = e - (I1*r)\n",
+      "pd2 = e - (I2*r)\n",
+      "\n",
+      "#results\n",
+      "print \"(a)p.d\", pd1,\"Volts(V)\"\n",
+      "print \"(b)p.d\", pd2,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)p.d 1.9 Volts(V)\n",
+        "(b)p.d 1.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 36</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the internal resistance of the battery.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "e1 = 25; # in volts\n",
+      "e2 = 24; # in volts\n",
+      "I2 = 10; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "r = (e1 - e2)/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", r,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 0.1 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 36</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current flowing in the circuit and (b) the p.d. at the battery terminals.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "r = 0.2; # in ohms\n",
+      "n = 10; # no. of cells\n",
+      "e = 1.5; # in volts\n",
+      "R = 58; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "es = n*e\n",
+      "rs = n*r\n",
+      "I = es/(rs + R)\n",
+      "pd = es - (I*rs)\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Current\", I,\"Amperes(A)\"\n",
+      "print \"(b)p.d\", pd,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Current 0.25 Amperes(A)\n",
+        "(b)p.d 14.5 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_3.ipynb
new file mode 100755
index 00000000..860dc4f0
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_04-checkpoint_3.ipynb
@@ -0,0 +1,186 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ecbd49bc184dd134f59e93955453e8d9e5bf67fa157192a3c06878e9e4422d2d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 4: Chemical effects of\n",
+      "electricity</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 35</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R = 0.2; # in ohms\n",
+      "n = 8; # no. of cells\n",
+      "e = 2.2; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "es = n*e\n",
+      "ep = e\n",
+      "Rs = n*R\n",
+      "Rp = R/n\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Resistance\", Rs,\"ohms; e.m.f\", es,\"Volts(V)\"\n",
+      "print \"(b)Resistance\", Rp,\"ohms; e.m.f\", ep,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Resistance 1.6 ohms; e.m.f 17.6 Volts(V)\n",
+        "(b)Resistance 0.025 ohms; e.m.f 2.2 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 35</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "r = 0.02; # in ohms\n",
+      "e = 2; # in volts\n",
+      "I1 = 5; # in Amperes\n",
+      "I2 = 50; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "pd1 = e - (I1*r)\n",
+      "pd2 = e - (I2*r)\n",
+      "\n",
+      "#results\n",
+      "print \"(a)p.d\", pd1,\"Volts(V)\"\n",
+      "print \"(b)p.d\", pd2,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)p.d 1.9 Volts(V)\n",
+        "(b)p.d 1.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 36</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "e1 = 25; # in volts\n",
+      "e2 = 24; # in volts\n",
+      "I2 = 10; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "r = (e1 - e2)/I2\n",
+      "\n",
+      "#results\n",
+      "print \"Resistance\", r,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resistance 0.1 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 36</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "r = 0.2; # in ohms\n",
+      "n = 10; # no. of cells\n",
+      "e = 1.5; # in volts\n",
+      "R = 58; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "es = n*e\n",
+      "rs = n*r\n",
+      "I = es/(rs + R)\n",
+      "pd = es - (I*rs)\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Current\", I,\"Amperes(A)\"\n",
+      "print \"(b)p.d\", pd,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Current 0.25 Amperes(A)\n",
+        "(b)p.d 14.5 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint.ipynb
new file mode 100755
index 00000000..b881dd41
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint.ipynb
@@ -0,0 +1,763 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 5: Series and parallel\n",
+      "networks</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 43</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the battery voltage V, (b) the total resistance of the circuit, and \n",
+      "#(c) the values of resistance of resistors R1, R2 and R3,\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V1 = 5; # in volts\n",
+      "V2 = 2; # in volts\n",
+      "V3 = 6; # in volts\n",
+      "I = 4; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt = V1 + V2 + V3\n",
+      "Rt = Vt/I\n",
+      "R1 = V1/I\n",
+      "R2 = V2/I\n",
+      "R3 = V3/I\n",
+      "\n",
+      "#results\n",
+      "print \"(a) Total Voltage\", Vt,\"Volts(V)\"\n",
+      "print \"(b)Total Resistance\", Rt,\"Ohms\"\n",
+      "print \"(c)Resistance(R1)\", R1,\"Ohms; Resistance(R2)\", R2,\"Ohms and\"\n",
+      "print \"Resistance(R3)\", R3,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Total Voltage 13 Volts(V)\n",
+        "(b)Total Resistance 3.25 Ohms\n",
+        "(c)Resistance(R1) 1.25 Ohms; Resistance(R2) 0.5 Ohms and\n",
+        "Resistance(R3) 1.5 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 43</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the p.d. across resistor R3.\n",
+      "#Find value of resistor R2\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V1 = 10; # in volts\n",
+      "V2 = 4; # in volts\n",
+      "Vt = 25; # in volts\n",
+      "Rt = 100; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V3 = Vt - V1 - V2\n",
+      "I = Vt/Rt\n",
+      "R2 = V2/I\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Voltage(V3)\", V3,\"Volts(V)\"\n",
+      "print \"(b)current\", I,\"Amperes(A)\"\n",
+      "print \"(c)Resistance(R2)\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Voltage(V3) 11 Volts(V)\n",
+        "(b)current 0.25 Amperes(A)\n",
+        "(c)Resistance(R2) 16.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 44</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing through, and the p.d. across the 9 ohm resistor. \n",
+      "#Find also the power dissipated in the 11 ohm resistor.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 12; # in volts\n",
+      "R1 = 4; # in ohms\n",
+      "R2 = 9; # in ohms\n",
+      "R3 = 11; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1 + R2 + R3\n",
+      "I = Vt/Rt\n",
+      "V9 = I*R2\n",
+      "P11 = I*I*R3\n",
+      "#results\n",
+      "print \"a)current\", I,\"Amperes(A)\\n\"\n",
+      "print \"b)Voltage(V2)\", V9,\"Volts(V)\\n\"\n",
+      "print \"c)Power\", P11,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)current 0.5 Amperes(A)\n",
+        "\n",
+        "b)Voltage(V2) 4.5 Volts(V)\n",
+        "\n",
+        "c)Power 2.75 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 44</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of voltage V\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 50; # in volts\n",
+      "R1 = 4; # in ohms\n",
+      "R2 = 6; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1 + R2\n",
+      "I = Vt/Rt\n",
+      "V2 = I*R2\n",
+      "\n",
+      "#results\n",
+      "print \"Voltage(V)\", V2,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage(V) 30.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 45</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the value of the other resistor, and \n",
+      "#(b) the p.d. across the 2 \u0006 resistor. \n",
+      "#If the circuit is connected for 50 hours, how much energy is used?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 24; # in volts\n",
+      "R1 = 2; # in ohms\n",
+      "I = 3; # in Amperes\n",
+      "t = 50; # in hrs\n",
+      "\n",
+      "#calculation:\n",
+      "V1 = I*R1\n",
+      "R2 = (Vt-(I*R1))/I\n",
+      "E = Vt*I*t\n",
+      "\n",
+      "#results\n",
+      "print \"a)Voltage(V1)\", V1,\"Volts(V)\\n\"\n",
+      "print \"b)Resistance(R2)\", R2,\"Ohms\\n\"\n",
+      "print \"c)Energy(E)\", E/1000,\"kWh\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Voltage(V1) 6 Volts(V)\n",
+        "\n",
+        "b)Resistance(R2) 6.0 Ohms\n",
+        "\n",
+        "c)Energy(E) 3.6 kWh"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 46</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the reading on the ammeter, and \n",
+      "# (b) the value of resistor R2\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 5; # in ohms\n",
+      "R3 = 20; # in ohms\n",
+      "I1 = 8; # in Amperes\n",
+      "It = 11; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt = I1*R1\n",
+      "I3 = Vt/R3\n",
+      "R2 = Vt/(It - I1 - I3)\n",
+      "\n",
+      "#results\n",
+      "print \"a)Ammeter Reading\", I3,\"Amperes(A)\\n\"\n",
+      "print \"b)Resistance(R2)\", R2,\"Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Ammeter Reading 2.0 Amperes(A)\n",
+        "\n",
+        "b)Resistance(R2) 40.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 46</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the total circuit resistance and (b) the current flowing in the 3 ohm resistor.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 3; # in ohms\n",
+      "R2 = 6; # in ohms\n",
+      "Vt = 12; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1*R2/(R1 + R2)\n",
+      "I1 = (Vt/R1)\n",
+      "\n",
+      "#Result\n",
+      "print \"(a)Total Resistance\", Rt,\"Ohms\\n\"\n",
+      "print \"(b)Current(I1)\", I1,\"Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Total Resistance 2.0 Ohms\n",
+        "\n",
+        "(b)Current(I1) 4.0 Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 47</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the value of the supply voltage V and (b) the value of current I.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  10;#  in  ohms\n",
+      "R2  =  20;#  in  ohms\n",
+      "R3  =  60;#  in  ohms\n",
+      "I2  =  3;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt  =  I2*R2\n",
+      "I1  =  Vt/R1\n",
+      "I3  =  Vt/R3\n",
+      "I  =  I1  +I2  +  I3\n",
+      "\n",
+      "print  \"\\nResult\\n\"\n",
+      "print  \"\\n(a)Voltage(V)  \",Vt,\"  Volts(V)\\n\"\n",
+      "print  \"\\n(b)Total  Current(I)  \",I,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "(a)Voltage(V)   60   Volts(V)\n",
+        "\n",
+        "\n",
+        "(b)Total  Current(I)   10.0   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 47</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#state how they must be connected to give an overall resistance of \n",
+      "#(a) 1/4 ohm (b) 1 ohm (c) 4/3 ohm (d)2.5 ohm, all four resistors being connected in each case\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  1;#  in  ohms\n",
+      "\n",
+      "#calculation\n",
+      "R1  =  1/(1/R + 1/R + 1/R + 1/R)\n",
+      "R2  =  2*R*2*R/(4*R)\n",
+      "R3  =  1/(1/R + 1/R + 1/R) + 1\n",
+      "R4  =  R*R/(2*R) + 2*R\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n(a)All four in parallel for \",R1,\"  ohm\\n\"\n",
+      "print  \"\\n(b)Two in series, in parallel with another two in series for\",R2,\"  ohm\\n\"\n",
+      "print  \"\\n(c)Three in parallel, in series with one for \",round(R3,2),\"  ohm\\n\"\n",
+      "print  \"\\n(d)Two in parallel, in series with two in series for \",R4,\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)All four in parallel for  0.25   ohm\n",
+        "\n",
+        "\n",
+        "(b)Two in series, in parallel with another two in series for 1.0   ohm\n",
+        "\n",
+        "\n",
+        "(c)Three in parallel, in series with one for  1.33   ohm\n",
+        "\n",
+        "\n",
+        "(d)Two in parallel, in series with two in series for  2.5   ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 48</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the equivalent resistance for the circuit\n",
+      "from __future__ import division\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  2.2;#  in  ohms\n",
+      "R3  =  3;#  in  ohms\n",
+      "R4  =  6;#  in  ohms\n",
+      "R5  =  18;#  in  ohms\n",
+      "R6  =  4;#  in  ohms\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R0  =  1/((1/3)  +  (1/6)  +  (1/18))\n",
+      "Rt  =  R1  +  R2  +  R0  +  R6\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Equivalent  Resistance  \",Rt,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Equivalent  Resistance   9.0   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 48</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the supply current, (b) the current flowing through each resistor and (c) the p.d. across each resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  2.5;#  in  ohms\n",
+      "R2  =  6;#  in  ohms\n",
+      "R3  =  2;#  in  ohms\n",
+      "R4  =  4;#  in  ohms\n",
+      "Vt  =  200;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "R0  =  1/((1/R2)  +  (1/R3))\n",
+      "Rt  =  R1  +  R0  +  R4\n",
+      "It  =  Vt/Rt\n",
+      "I1  =  It\n",
+      "I4  =  It\n",
+      "I2  =  R3*It/(R3+R2)\n",
+      "I3  =  It  -  I2\n",
+      "V1  =  I1*R1\n",
+      "V2  =  I2*R2\n",
+      "V3  =  I3*R3\n",
+      "V4  =  I4*R4\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Total  Current  Supply  \",It,\"  Amperes(A)\\n\"\n",
+      "print  \"\\n  (b)Current  through  resistors  (R1,  R2,  R3,  R4)\\n \",I1,\",  \",  I2,\",  \",  I3,\",  \",  I4,\"  Amperes(A)  respectively\\n\"\n",
+      "print  \"\\n  (c)voltage  across  resistors  (R1,  R2,  R3,  R4)\\n  \",V1,\",  \",  V2,\",  \",  V3,\",  \",  V4,\"  Volts(V)  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Total  Current  Supply   25.0   Amperes(A)\n",
+        "\n",
+        "\n",
+        "  (b)Current  through  resistors  (R1,  R2,  R3,  R4)\n",
+        "  25.0 ,   6.25 ,   18.75 ,   25.0   Amperes(A)  respectively\n",
+        "\n",
+        "\n",
+        "  (c)voltage  across  resistors  (R1,  R2,  R3,  R4)\n",
+        "   62.5 ,   37.5 ,   37.5 ,   100.0   Volts(V)  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 49</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the value of resistor Rx such that the total power dissipated in the circuit is 2.5 kW, and \n",
+      "#(b) the current flowing in each of the four resistors\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  15;#  in  ohms\n",
+      "R2  =  10;#  in  ohms\n",
+      "R3  =  38;#  in  ohms\n",
+      "Vt  =  250;#  in  volts\n",
+      "P  =  2500;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      "It  =  P/Vt\n",
+      "I2  =  R1*It/(R1+R2)\n",
+      "I1  =  It  -  I2\n",
+      "Re1  =  1/((1/R1)  +  (1/R2))\n",
+      "Rt  =  Vt/It\n",
+      "Re2  =  Rt  -  Re1\n",
+      "Rx  =  1/((1/Re2)  -  (1/R3))\n",
+      "I4  =  R3*It/(R3+Rx)\n",
+      "I3  =  It  -  I4\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  (Rx)  \",Rx,\"  Ohms\\n\"\n",
+      "print  \"\\n  (b)Current  through  resistors  (R1,  R2,  R3,  R4): \\n \",I1,\",  \",  I2,\",  \",  I3,\",  \"\n",
+      "print   \",  I4,\"  Amperes(A)  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  (Rx)   38.0   Ohms\n",
+        "\n",
+        "\n",
+        "  (b)Current  through  resistors  (R1,  R2,  R3,  R4): \n",
+        "  4.0 ,   6.0 ,   5.0 ,   5.0   Amperes(A)  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 51</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the current Ix\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  1.4;#  in  ohms\n",
+      "R4  =  9;#  in  ohms\n",
+      "R5  =  2;#  in  ohms\n",
+      "Vt  =  17;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "R01  =  R1*R2/(R1  +  R2)\n",
+      "R02  =  R01  +  R3\n",
+      "R03  =  R4*R02/(R4  +R02)\n",
+      "Rt  =  R5  +  R03\n",
+      "It  =  Vt/Rt\n",
+      "I1  =  R4*It/(R4  +  R02)\n",
+      "Ix  =  R2*I1/(R1  +  R2)\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Current(Ix)  \",Ix,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Current(Ix)   0.6   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 52</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the resistance of one lamp.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rt  =  150;#  in  ohms\n",
+      "n  =  3;#  no.  of  identical  lamp\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  Rt*3#  (1/Rt)=(1/R)+(1/R)+(1/R)\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Resistance  \",R,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance   450   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 52</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#State (a) the voltage across each lamp,\n",
+      "# and (b) the effect of lamp C failing.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "#series  connection\n",
+      "n  =  3;#  no.  of  identical  lamp\n",
+      "Vt  =  150;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  Vt/3#  Since  each  lamp  is  identical,  then  V  volts  across  each.\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  a)Voltage  across  each  resistor  =  \",V,\"  Volts(V)\\n\"\n",
+      "print  \"\\n  b)If  lamp  C  fails,  i.e.,  open  circuits,  no  current  will  flow  and lamps  A  and  B  will  not  operate.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  a)Voltage  across  each  resistor  =   50.0   Volts(V)\n",
+        "\n",
+        "\n",
+        "  b)If  lamp  C  fails,  i.e.,  open  circuits,  no  current  will  flow  and lamps  A  and  B  will  not  operate."
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_1.ipynb
new file mode 100755
index 00000000..b881dd41
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_1.ipynb
@@ -0,0 +1,763 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 5: Series and parallel\n",
+      "networks</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 43</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the battery voltage V, (b) the total resistance of the circuit, and \n",
+      "#(c) the values of resistance of resistors R1, R2 and R3,\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V1 = 5; # in volts\n",
+      "V2 = 2; # in volts\n",
+      "V3 = 6; # in volts\n",
+      "I = 4; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt = V1 + V2 + V3\n",
+      "Rt = Vt/I\n",
+      "R1 = V1/I\n",
+      "R2 = V2/I\n",
+      "R3 = V3/I\n",
+      "\n",
+      "#results\n",
+      "print \"(a) Total Voltage\", Vt,\"Volts(V)\"\n",
+      "print \"(b)Total Resistance\", Rt,\"Ohms\"\n",
+      "print \"(c)Resistance(R1)\", R1,\"Ohms; Resistance(R2)\", R2,\"Ohms and\"\n",
+      "print \"Resistance(R3)\", R3,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Total Voltage 13 Volts(V)\n",
+        "(b)Total Resistance 3.25 Ohms\n",
+        "(c)Resistance(R1) 1.25 Ohms; Resistance(R2) 0.5 Ohms and\n",
+        "Resistance(R3) 1.5 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 43</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the p.d. across resistor R3.\n",
+      "#Find value of resistor R2\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V1 = 10; # in volts\n",
+      "V2 = 4; # in volts\n",
+      "Vt = 25; # in volts\n",
+      "Rt = 100; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V3 = Vt - V1 - V2\n",
+      "I = Vt/Rt\n",
+      "R2 = V2/I\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Voltage(V3)\", V3,\"Volts(V)\"\n",
+      "print \"(b)current\", I,\"Amperes(A)\"\n",
+      "print \"(c)Resistance(R2)\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Voltage(V3) 11 Volts(V)\n",
+        "(b)current 0.25 Amperes(A)\n",
+        "(c)Resistance(R2) 16.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 44</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing through, and the p.d. across the 9 ohm resistor. \n",
+      "#Find also the power dissipated in the 11 ohm resistor.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 12; # in volts\n",
+      "R1 = 4; # in ohms\n",
+      "R2 = 9; # in ohms\n",
+      "R3 = 11; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1 + R2 + R3\n",
+      "I = Vt/Rt\n",
+      "V9 = I*R2\n",
+      "P11 = I*I*R3\n",
+      "#results\n",
+      "print \"a)current\", I,\"Amperes(A)\\n\"\n",
+      "print \"b)Voltage(V2)\", V9,\"Volts(V)\\n\"\n",
+      "print \"c)Power\", P11,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)current 0.5 Amperes(A)\n",
+        "\n",
+        "b)Voltage(V2) 4.5 Volts(V)\n",
+        "\n",
+        "c)Power 2.75 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 44</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of voltage V\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 50; # in volts\n",
+      "R1 = 4; # in ohms\n",
+      "R2 = 6; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1 + R2\n",
+      "I = Vt/Rt\n",
+      "V2 = I*R2\n",
+      "\n",
+      "#results\n",
+      "print \"Voltage(V)\", V2,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage(V) 30.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 45</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the value of the other resistor, and \n",
+      "#(b) the p.d. across the 2 \u0006 resistor. \n",
+      "#If the circuit is connected for 50 hours, how much energy is used?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 24; # in volts\n",
+      "R1 = 2; # in ohms\n",
+      "I = 3; # in Amperes\n",
+      "t = 50; # in hrs\n",
+      "\n",
+      "#calculation:\n",
+      "V1 = I*R1\n",
+      "R2 = (Vt-(I*R1))/I\n",
+      "E = Vt*I*t\n",
+      "\n",
+      "#results\n",
+      "print \"a)Voltage(V1)\", V1,\"Volts(V)\\n\"\n",
+      "print \"b)Resistance(R2)\", R2,\"Ohms\\n\"\n",
+      "print \"c)Energy(E)\", E/1000,\"kWh\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Voltage(V1) 6 Volts(V)\n",
+        "\n",
+        "b)Resistance(R2) 6.0 Ohms\n",
+        "\n",
+        "c)Energy(E) 3.6 kWh"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 46</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the reading on the ammeter, and \n",
+      "# (b) the value of resistor R2\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 5; # in ohms\n",
+      "R3 = 20; # in ohms\n",
+      "I1 = 8; # in Amperes\n",
+      "It = 11; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt = I1*R1\n",
+      "I3 = Vt/R3\n",
+      "R2 = Vt/(It - I1 - I3)\n",
+      "\n",
+      "#results\n",
+      "print \"a)Ammeter Reading\", I3,\"Amperes(A)\\n\"\n",
+      "print \"b)Resistance(R2)\", R2,\"Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Ammeter Reading 2.0 Amperes(A)\n",
+        "\n",
+        "b)Resistance(R2) 40.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 46</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the total circuit resistance and (b) the current flowing in the 3 ohm resistor.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 3; # in ohms\n",
+      "R2 = 6; # in ohms\n",
+      "Vt = 12; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1*R2/(R1 + R2)\n",
+      "I1 = (Vt/R1)\n",
+      "\n",
+      "#Result\n",
+      "print \"(a)Total Resistance\", Rt,\"Ohms\\n\"\n",
+      "print \"(b)Current(I1)\", I1,\"Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Total Resistance 2.0 Ohms\n",
+        "\n",
+        "(b)Current(I1) 4.0 Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 47</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the value of the supply voltage V and (b) the value of current I.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  10;#  in  ohms\n",
+      "R2  =  20;#  in  ohms\n",
+      "R3  =  60;#  in  ohms\n",
+      "I2  =  3;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt  =  I2*R2\n",
+      "I1  =  Vt/R1\n",
+      "I3  =  Vt/R3\n",
+      "I  =  I1  +I2  +  I3\n",
+      "\n",
+      "print  \"\\nResult\\n\"\n",
+      "print  \"\\n(a)Voltage(V)  \",Vt,\"  Volts(V)\\n\"\n",
+      "print  \"\\n(b)Total  Current(I)  \",I,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "(a)Voltage(V)   60   Volts(V)\n",
+        "\n",
+        "\n",
+        "(b)Total  Current(I)   10.0   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 47</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#state how they must be connected to give an overall resistance of \n",
+      "#(a) 1/4 ohm (b) 1 ohm (c) 4/3 ohm (d)2.5 ohm, all four resistors being connected in each case\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  1;#  in  ohms\n",
+      "\n",
+      "#calculation\n",
+      "R1  =  1/(1/R + 1/R + 1/R + 1/R)\n",
+      "R2  =  2*R*2*R/(4*R)\n",
+      "R3  =  1/(1/R + 1/R + 1/R) + 1\n",
+      "R4  =  R*R/(2*R) + 2*R\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n(a)All four in parallel for \",R1,\"  ohm\\n\"\n",
+      "print  \"\\n(b)Two in series, in parallel with another two in series for\",R2,\"  ohm\\n\"\n",
+      "print  \"\\n(c)Three in parallel, in series with one for \",round(R3,2),\"  ohm\\n\"\n",
+      "print  \"\\n(d)Two in parallel, in series with two in series for \",R4,\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)All four in parallel for  0.25   ohm\n",
+        "\n",
+        "\n",
+        "(b)Two in series, in parallel with another two in series for 1.0   ohm\n",
+        "\n",
+        "\n",
+        "(c)Three in parallel, in series with one for  1.33   ohm\n",
+        "\n",
+        "\n",
+        "(d)Two in parallel, in series with two in series for  2.5   ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 48</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the equivalent resistance for the circuit\n",
+      "from __future__ import division\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  2.2;#  in  ohms\n",
+      "R3  =  3;#  in  ohms\n",
+      "R4  =  6;#  in  ohms\n",
+      "R5  =  18;#  in  ohms\n",
+      "R6  =  4;#  in  ohms\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R0  =  1/((1/3)  +  (1/6)  +  (1/18))\n",
+      "Rt  =  R1  +  R2  +  R0  +  R6\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Equivalent  Resistance  \",Rt,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Equivalent  Resistance   9.0   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 48</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the supply current, (b) the current flowing through each resistor and (c) the p.d. across each resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  2.5;#  in  ohms\n",
+      "R2  =  6;#  in  ohms\n",
+      "R3  =  2;#  in  ohms\n",
+      "R4  =  4;#  in  ohms\n",
+      "Vt  =  200;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "R0  =  1/((1/R2)  +  (1/R3))\n",
+      "Rt  =  R1  +  R0  +  R4\n",
+      "It  =  Vt/Rt\n",
+      "I1  =  It\n",
+      "I4  =  It\n",
+      "I2  =  R3*It/(R3+R2)\n",
+      "I3  =  It  -  I2\n",
+      "V1  =  I1*R1\n",
+      "V2  =  I2*R2\n",
+      "V3  =  I3*R3\n",
+      "V4  =  I4*R4\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Total  Current  Supply  \",It,\"  Amperes(A)\\n\"\n",
+      "print  \"\\n  (b)Current  through  resistors  (R1,  R2,  R3,  R4)\\n \",I1,\",  \",  I2,\",  \",  I3,\",  \",  I4,\"  Amperes(A)  respectively\\n\"\n",
+      "print  \"\\n  (c)voltage  across  resistors  (R1,  R2,  R3,  R4)\\n  \",V1,\",  \",  V2,\",  \",  V3,\",  \",  V4,\"  Volts(V)  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Total  Current  Supply   25.0   Amperes(A)\n",
+        "\n",
+        "\n",
+        "  (b)Current  through  resistors  (R1,  R2,  R3,  R4)\n",
+        "  25.0 ,   6.25 ,   18.75 ,   25.0   Amperes(A)  respectively\n",
+        "\n",
+        "\n",
+        "  (c)voltage  across  resistors  (R1,  R2,  R3,  R4)\n",
+        "   62.5 ,   37.5 ,   37.5 ,   100.0   Volts(V)  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 49</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the value of resistor Rx such that the total power dissipated in the circuit is 2.5 kW, and \n",
+      "#(b) the current flowing in each of the four resistors\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  15;#  in  ohms\n",
+      "R2  =  10;#  in  ohms\n",
+      "R3  =  38;#  in  ohms\n",
+      "Vt  =  250;#  in  volts\n",
+      "P  =  2500;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      "It  =  P/Vt\n",
+      "I2  =  R1*It/(R1+R2)\n",
+      "I1  =  It  -  I2\n",
+      "Re1  =  1/((1/R1)  +  (1/R2))\n",
+      "Rt  =  Vt/It\n",
+      "Re2  =  Rt  -  Re1\n",
+      "Rx  =  1/((1/Re2)  -  (1/R3))\n",
+      "I4  =  R3*It/(R3+Rx)\n",
+      "I3  =  It  -  I4\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  (Rx)  \",Rx,\"  Ohms\\n\"\n",
+      "print  \"\\n  (b)Current  through  resistors  (R1,  R2,  R3,  R4): \\n \",I1,\",  \",  I2,\",  \",  I3,\",  \"\n",
+      "print   \",  I4,\"  Amperes(A)  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  (Rx)   38.0   Ohms\n",
+        "\n",
+        "\n",
+        "  (b)Current  through  resistors  (R1,  R2,  R3,  R4): \n",
+        "  4.0 ,   6.0 ,   5.0 ,   5.0   Amperes(A)  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 51</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the current Ix\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  1.4;#  in  ohms\n",
+      "R4  =  9;#  in  ohms\n",
+      "R5  =  2;#  in  ohms\n",
+      "Vt  =  17;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "R01  =  R1*R2/(R1  +  R2)\n",
+      "R02  =  R01  +  R3\n",
+      "R03  =  R4*R02/(R4  +R02)\n",
+      "Rt  =  R5  +  R03\n",
+      "It  =  Vt/Rt\n",
+      "I1  =  R4*It/(R4  +  R02)\n",
+      "Ix  =  R2*I1/(R1  +  R2)\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Current(Ix)  \",Ix,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Current(Ix)   0.6   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 52</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the resistance of one lamp.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rt  =  150;#  in  ohms\n",
+      "n  =  3;#  no.  of  identical  lamp\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  Rt*3#  (1/Rt)=(1/R)+(1/R)+(1/R)\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Resistance  \",R,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance   450   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 52</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#State (a) the voltage across each lamp,\n",
+      "# and (b) the effect of lamp C failing.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "#series  connection\n",
+      "n  =  3;#  no.  of  identical  lamp\n",
+      "Vt  =  150;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  Vt/3#  Since  each  lamp  is  identical,  then  V  volts  across  each.\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  a)Voltage  across  each  resistor  =  \",V,\"  Volts(V)\\n\"\n",
+      "print  \"\\n  b)If  lamp  C  fails,  i.e.,  open  circuits,  no  current  will  flow  and lamps  A  and  B  will  not  operate.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  a)Voltage  across  each  resistor  =   50.0   Volts(V)\n",
+        "\n",
+        "\n",
+        "  b)If  lamp  C  fails,  i.e.,  open  circuits,  no  current  will  flow  and lamps  A  and  B  will  not  operate."
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_2.ipynb
new file mode 100755
index 00000000..b881dd41
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_2.ipynb
@@ -0,0 +1,763 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 5: Series and parallel\n",
+      "networks</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 43</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the battery voltage V, (b) the total resistance of the circuit, and \n",
+      "#(c) the values of resistance of resistors R1, R2 and R3,\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V1 = 5; # in volts\n",
+      "V2 = 2; # in volts\n",
+      "V3 = 6; # in volts\n",
+      "I = 4; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt = V1 + V2 + V3\n",
+      "Rt = Vt/I\n",
+      "R1 = V1/I\n",
+      "R2 = V2/I\n",
+      "R3 = V3/I\n",
+      "\n",
+      "#results\n",
+      "print \"(a) Total Voltage\", Vt,\"Volts(V)\"\n",
+      "print \"(b)Total Resistance\", Rt,\"Ohms\"\n",
+      "print \"(c)Resistance(R1)\", R1,\"Ohms; Resistance(R2)\", R2,\"Ohms and\"\n",
+      "print \"Resistance(R3)\", R3,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Total Voltage 13 Volts(V)\n",
+        "(b)Total Resistance 3.25 Ohms\n",
+        "(c)Resistance(R1) 1.25 Ohms; Resistance(R2) 0.5 Ohms and\n",
+        "Resistance(R3) 1.5 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 43</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the p.d. across resistor R3.\n",
+      "#Find value of resistor R2\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V1 = 10; # in volts\n",
+      "V2 = 4; # in volts\n",
+      "Vt = 25; # in volts\n",
+      "Rt = 100; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V3 = Vt - V1 - V2\n",
+      "I = Vt/Rt\n",
+      "R2 = V2/I\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Voltage(V3)\", V3,\"Volts(V)\"\n",
+      "print \"(b)current\", I,\"Amperes(A)\"\n",
+      "print \"(c)Resistance(R2)\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Voltage(V3) 11 Volts(V)\n",
+        "(b)current 0.25 Amperes(A)\n",
+        "(c)Resistance(R2) 16.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 44</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing through, and the p.d. across the 9 ohm resistor. \n",
+      "#Find also the power dissipated in the 11 ohm resistor.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 12; # in volts\n",
+      "R1 = 4; # in ohms\n",
+      "R2 = 9; # in ohms\n",
+      "R3 = 11; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1 + R2 + R3\n",
+      "I = Vt/Rt\n",
+      "V9 = I*R2\n",
+      "P11 = I*I*R3\n",
+      "#results\n",
+      "print \"a)current\", I,\"Amperes(A)\\n\"\n",
+      "print \"b)Voltage(V2)\", V9,\"Volts(V)\\n\"\n",
+      "print \"c)Power\", P11,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)current 0.5 Amperes(A)\n",
+        "\n",
+        "b)Voltage(V2) 4.5 Volts(V)\n",
+        "\n",
+        "c)Power 2.75 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 44</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of voltage V\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 50; # in volts\n",
+      "R1 = 4; # in ohms\n",
+      "R2 = 6; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1 + R2\n",
+      "I = Vt/Rt\n",
+      "V2 = I*R2\n",
+      "\n",
+      "#results\n",
+      "print \"Voltage(V)\", V2,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage(V) 30.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 45</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the value of the other resistor, and \n",
+      "#(b) the p.d. across the 2 \u0006 resistor. \n",
+      "#If the circuit is connected for 50 hours, how much energy is used?\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 24; # in volts\n",
+      "R1 = 2; # in ohms\n",
+      "I = 3; # in Amperes\n",
+      "t = 50; # in hrs\n",
+      "\n",
+      "#calculation:\n",
+      "V1 = I*R1\n",
+      "R2 = (Vt-(I*R1))/I\n",
+      "E = Vt*I*t\n",
+      "\n",
+      "#results\n",
+      "print \"a)Voltage(V1)\", V1,\"Volts(V)\\n\"\n",
+      "print \"b)Resistance(R2)\", R2,\"Ohms\\n\"\n",
+      "print \"c)Energy(E)\", E/1000,\"kWh\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Voltage(V1) 6 Volts(V)\n",
+        "\n",
+        "b)Resistance(R2) 6.0 Ohms\n",
+        "\n",
+        "c)Energy(E) 3.6 kWh"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 46</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the reading on the ammeter, and \n",
+      "# (b) the value of resistor R2\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 5; # in ohms\n",
+      "R3 = 20; # in ohms\n",
+      "I1 = 8; # in Amperes\n",
+      "It = 11; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt = I1*R1\n",
+      "I3 = Vt/R3\n",
+      "R2 = Vt/(It - I1 - I3)\n",
+      "\n",
+      "#results\n",
+      "print \"a)Ammeter Reading\", I3,\"Amperes(A)\\n\"\n",
+      "print \"b)Resistance(R2)\", R2,\"Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Ammeter Reading 2.0 Amperes(A)\n",
+        "\n",
+        "b)Resistance(R2) 40.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 46</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the total circuit resistance and (b) the current flowing in the 3 ohm resistor.\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 3; # in ohms\n",
+      "R2 = 6; # in ohms\n",
+      "Vt = 12; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1*R2/(R1 + R2)\n",
+      "I1 = (Vt/R1)\n",
+      "\n",
+      "#Result\n",
+      "print \"(a)Total Resistance\", Rt,\"Ohms\\n\"\n",
+      "print \"(b)Current(I1)\", I1,\"Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Total Resistance 2.0 Ohms\n",
+        "\n",
+        "(b)Current(I1) 4.0 Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 47</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the value of the supply voltage V and (b) the value of current I.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  10;#  in  ohms\n",
+      "R2  =  20;#  in  ohms\n",
+      "R3  =  60;#  in  ohms\n",
+      "I2  =  3;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt  =  I2*R2\n",
+      "I1  =  Vt/R1\n",
+      "I3  =  Vt/R3\n",
+      "I  =  I1  +I2  +  I3\n",
+      "\n",
+      "print  \"\\nResult\\n\"\n",
+      "print  \"\\n(a)Voltage(V)  \",Vt,\"  Volts(V)\\n\"\n",
+      "print  \"\\n(b)Total  Current(I)  \",I,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "(a)Voltage(V)   60   Volts(V)\n",
+        "\n",
+        "\n",
+        "(b)Total  Current(I)   10.0   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 47</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#state how they must be connected to give an overall resistance of \n",
+      "#(a) 1/4 ohm (b) 1 ohm (c) 4/3 ohm (d)2.5 ohm, all four resistors being connected in each case\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  1;#  in  ohms\n",
+      "\n",
+      "#calculation\n",
+      "R1  =  1/(1/R + 1/R + 1/R + 1/R)\n",
+      "R2  =  2*R*2*R/(4*R)\n",
+      "R3  =  1/(1/R + 1/R + 1/R) + 1\n",
+      "R4  =  R*R/(2*R) + 2*R\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n(a)All four in parallel for \",R1,\"  ohm\\n\"\n",
+      "print  \"\\n(b)Two in series, in parallel with another two in series for\",R2,\"  ohm\\n\"\n",
+      "print  \"\\n(c)Three in parallel, in series with one for \",round(R3,2),\"  ohm\\n\"\n",
+      "print  \"\\n(d)Two in parallel, in series with two in series for \",R4,\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)All four in parallel for  0.25   ohm\n",
+        "\n",
+        "\n",
+        "(b)Two in series, in parallel with another two in series for 1.0   ohm\n",
+        "\n",
+        "\n",
+        "(c)Three in parallel, in series with one for  1.33   ohm\n",
+        "\n",
+        "\n",
+        "(d)Two in parallel, in series with two in series for  2.5   ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 48</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the equivalent resistance for the circuit\n",
+      "from __future__ import division\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  2.2;#  in  ohms\n",
+      "R3  =  3;#  in  ohms\n",
+      "R4  =  6;#  in  ohms\n",
+      "R5  =  18;#  in  ohms\n",
+      "R6  =  4;#  in  ohms\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R0  =  1/((1/3)  +  (1/6)  +  (1/18))\n",
+      "Rt  =  R1  +  R2  +  R0  +  R6\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Equivalent  Resistance  \",Rt,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Equivalent  Resistance   9.0   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 48</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the supply current, (b) the current flowing through each resistor and (c) the p.d. across each resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  2.5;#  in  ohms\n",
+      "R2  =  6;#  in  ohms\n",
+      "R3  =  2;#  in  ohms\n",
+      "R4  =  4;#  in  ohms\n",
+      "Vt  =  200;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "R0  =  1/((1/R2)  +  (1/R3))\n",
+      "Rt  =  R1  +  R0  +  R4\n",
+      "It  =  Vt/Rt\n",
+      "I1  =  It\n",
+      "I4  =  It\n",
+      "I2  =  R3*It/(R3+R2)\n",
+      "I3  =  It  -  I2\n",
+      "V1  =  I1*R1\n",
+      "V2  =  I2*R2\n",
+      "V3  =  I3*R3\n",
+      "V4  =  I4*R4\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Total  Current  Supply  \",It,\"  Amperes(A)\\n\"\n",
+      "print  \"\\n  (b)Current  through  resistors  (R1,  R2,  R3,  R4)\\n \",I1,\",  \",  I2,\",  \",  I3,\",  \",  I4,\"  Amperes(A)  respectively\\n\"\n",
+      "print  \"\\n  (c)voltage  across  resistors  (R1,  R2,  R3,  R4)\\n  \",V1,\",  \",  V2,\",  \",  V3,\",  \",  V4,\"  Volts(V)  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Total  Current  Supply   25.0   Amperes(A)\n",
+        "\n",
+        "\n",
+        "  (b)Current  through  resistors  (R1,  R2,  R3,  R4)\n",
+        "  25.0 ,   6.25 ,   18.75 ,   25.0   Amperes(A)  respectively\n",
+        "\n",
+        "\n",
+        "  (c)voltage  across  resistors  (R1,  R2,  R3,  R4)\n",
+        "   62.5 ,   37.5 ,   37.5 ,   100.0   Volts(V)  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 49</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the value of resistor Rx such that the total power dissipated in the circuit is 2.5 kW, and \n",
+      "#(b) the current flowing in each of the four resistors\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  15;#  in  ohms\n",
+      "R2  =  10;#  in  ohms\n",
+      "R3  =  38;#  in  ohms\n",
+      "Vt  =  250;#  in  volts\n",
+      "P  =  2500;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      "It  =  P/Vt\n",
+      "I2  =  R1*It/(R1+R2)\n",
+      "I1  =  It  -  I2\n",
+      "Re1  =  1/((1/R1)  +  (1/R2))\n",
+      "Rt  =  Vt/It\n",
+      "Re2  =  Rt  -  Re1\n",
+      "Rx  =  1/((1/Re2)  -  (1/R3))\n",
+      "I4  =  R3*It/(R3+Rx)\n",
+      "I3  =  It  -  I4\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  (Rx)  \",Rx,\"  Ohms\\n\"\n",
+      "print  \"\\n  (b)Current  through  resistors  (R1,  R2,  R3,  R4): \\n \",I1,\",  \",  I2,\",  \",  I3,\",  \"\n",
+      "print   \",  I4,\"  Amperes(A)  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  (Rx)   38.0   Ohms\n",
+        "\n",
+        "\n",
+        "  (b)Current  through  resistors  (R1,  R2,  R3,  R4): \n",
+        "  4.0 ,   6.0 ,   5.0 ,   5.0   Amperes(A)  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 51</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the current Ix\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  1.4;#  in  ohms\n",
+      "R4  =  9;#  in  ohms\n",
+      "R5  =  2;#  in  ohms\n",
+      "Vt  =  17;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "R01  =  R1*R2/(R1  +  R2)\n",
+      "R02  =  R01  +  R3\n",
+      "R03  =  R4*R02/(R4  +R02)\n",
+      "Rt  =  R5  +  R03\n",
+      "It  =  Vt/Rt\n",
+      "I1  =  R4*It/(R4  +  R02)\n",
+      "Ix  =  R2*I1/(R1  +  R2)\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Current(Ix)  \",Ix,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Current(Ix)   0.6   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 52</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the resistance of one lamp.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rt  =  150;#  in  ohms\n",
+      "n  =  3;#  no.  of  identical  lamp\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  Rt*3#  (1/Rt)=(1/R)+(1/R)+(1/R)\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Resistance  \",R,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance   450   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 52</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#State (a) the voltage across each lamp,\n",
+      "# and (b) the effect of lamp C failing.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "#series  connection\n",
+      "n  =  3;#  no.  of  identical  lamp\n",
+      "Vt  =  150;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  Vt/3#  Since  each  lamp  is  identical,  then  V  volts  across  each.\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  a)Voltage  across  each  resistor  =  \",V,\"  Volts(V)\\n\"\n",
+      "print  \"\\n  b)If  lamp  C  fails,  i.e.,  open  circuits,  no  current  will  flow  and lamps  A  and  B  will  not  operate.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  a)Voltage  across  each  resistor  =   50.0   Volts(V)\n",
+        "\n",
+        "\n",
+        "  b)If  lamp  C  fails,  i.e.,  open  circuits,  no  current  will  flow  and lamps  A  and  B  will  not  operate."
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_3.ipynb
new file mode 100755
index 00000000..0b10490d
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_05-checkpoint_3.ipynb
@@ -0,0 +1,755 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:71ce76d6b0eac06e2f6805ff014d728113603c794f739fe4a09502555869f752"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 5: Series and parallel\n",
+      "networks</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 43</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V1 = 5; # in volts\n",
+      "V2 = 2; # in volts\n",
+      "V3 = 6; # in volts\n",
+      "I = 4; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt = V1 + V2 + V3\n",
+      "Rt = Vt/I\n",
+      "R1 = V1/I\n",
+      "R2 = V2/I\n",
+      "R3 = V3/I\n",
+      "\n",
+      "#results\n",
+      "print \"(a) Total Voltage\", Vt,\"Volts(V)\"\n",
+      "print \"(b)Total Resistance\", Rt,\"Ohms\"\n",
+      "print \"(c)Resistance(R1)\", R1,\"Ohms; Resistance(R2)\", R2,\"Ohms and\"\n",
+      "print \"Resistance(R3)\", R3,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Total Voltage 13 Volts(V)\n",
+        "(b)Total Resistance 3.25 Ohms\n",
+        "(c)Resistance(R1) 1.25 Ohms; Resistance(R2) 0.5 Ohms and\n",
+        "Resistance(R3) 1.5 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 43</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "V1 = 10; # in volts\n",
+      "V2 = 4; # in volts\n",
+      "Vt = 25; # in volts\n",
+      "Rt = 100; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "V3 = Vt - V1 - V2\n",
+      "I = Vt/Rt\n",
+      "R2 = V2/I\n",
+      "\n",
+      "#results\n",
+      "print \"(a)Voltage(V3)\", V3,\"Volts(V)\"\n",
+      "print \"(b)current\", I,\"Amperes(A)\"\n",
+      "print \"(c)Resistance(R2)\", R2,\"Ohms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Voltage(V3) 11 Volts(V)\n",
+        "(b)current 0.25 Amperes(A)\n",
+        "(c)Resistance(R2) 16.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 44</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 12; # in volts\n",
+      "R1 = 4; # in ohms\n",
+      "R2 = 9; # in ohms\n",
+      "R3 = 11; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1 + R2 + R3\n",
+      "I = Vt/Rt\n",
+      "V9 = I*R2\n",
+      "P11 = I*I*R3\n",
+      "#results\n",
+      "print \"a)current\", I,\"Amperes(A)\\n\"\n",
+      "print \"b)Voltage(V2)\", V9,\"Volts(V)\\n\"\n",
+      "print \"c)Power\", P11,\"Watt(W)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)current 0.5 Amperes(A)\n",
+        "\n",
+        "b)Voltage(V2) 4.5 Volts(V)\n",
+        "\n",
+        "c)Power 2.75 Watt(W)"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 44</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 50; # in volts\n",
+      "R1 = 4; # in ohms\n",
+      "R2 = 6; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1 + R2\n",
+      "I = Vt/Rt\n",
+      "V2 = I*R2\n",
+      "\n",
+      "#results\n",
+      "print \"Voltage(V)\", V2,\"Volts(V)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage(V) 30.0 Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 45</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "Vt = 24; # in volts\n",
+      "R1 = 2; # in ohms\n",
+      "I = 3; # in Amperes\n",
+      "t = 50; # in hrs\n",
+      "\n",
+      "#calculation:\n",
+      "V1 = I*R1\n",
+      "R2 = (Vt-(I*R1))/I\n",
+      "E = Vt*I*t\n",
+      "\n",
+      "#results\n",
+      "print \"a)Voltage(V1)\", V1,\"Volts(V)\\n\"\n",
+      "print \"b)Resistance(R2)\", R2,\"Ohms\\n\"\n",
+      "print \"c)Energy(E)\", E/1000,\"kWh\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Voltage(V1) 6 Volts(V)\n",
+        "\n",
+        "b)Resistance(R2) 6.0 Ohms\n",
+        "\n",
+        "c)Energy(E) 3.6 kWh"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 46</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 5; # in ohms\n",
+      "R3 = 20; # in ohms\n",
+      "I1 = 8; # in Amperes\n",
+      "It = 11; # in Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt = I1*R1\n",
+      "I3 = Vt/R3\n",
+      "R2 = Vt/(It - I1 - I3)\n",
+      "\n",
+      "#results\n",
+      "print \"a)Ammeter Reading\", I3,\"Amperes(A)\\n\"\n",
+      "print \"b)Resistance(R2)\", R2,\"Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a)Ammeter Reading 2.0 Amperes(A)\n",
+        "\n",
+        "b)Resistance(R2) 40.0 Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 46</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing the variables:\n",
+      "R1 = 3; # in ohms\n",
+      "R2 = 6; # in ohms\n",
+      "Vt = 12; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "Rt = R1*R2/(R1 + R2)\n",
+      "I1 = (Vt/R1)\n",
+      "\n",
+      "#Result\n",
+      "print \"(a)Total Resistance\", Rt,\"Ohms\\n\"\n",
+      "print \"(b)Current(I1)\", I1,\"Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Total Resistance 2.0 Ohms\n",
+        "\n",
+        "(b)Current(I1) 4.0 Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 47</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  10;#  in  ohms\n",
+      "R2  =  20;#  in  ohms\n",
+      "R3  =  60;#  in  ohms\n",
+      "I2  =  3;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Vt  =  I2*R2\n",
+      "I1  =  Vt/R1\n",
+      "I3  =  Vt/R3\n",
+      "I  =  I1  +I2  +  I3\n",
+      "\n",
+      "print  \"\\nResult\\n\"\n",
+      "print  \"\\n(a)Voltage(V)  \",Vt,\"  Volts(V)\\n\"\n",
+      "print  \"\\n(b)Total  Current(I)  \",I,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "(a)Voltage(V)   60   Volts(V)\n",
+        "\n",
+        "\n",
+        "(b)Total  Current(I)   10.0   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 47</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  1;#  in  ohms\n",
+      "\n",
+      "#calculation\n",
+      "R1  =  1/(1/R + 1/R + 1/R + 1/R)\n",
+      "R2  =  2*R*2*R/(4*R)\n",
+      "R3  =  1/(1/R + 1/R + 1/R) + 1\n",
+      "R4  =  R*R/(2*R) + 2*R\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n(a)All four in parallel for \",R1,\"  ohm\\n\"\n",
+      "print  \"\\n(b)Two in series, in parallel with another two in series for\",R2,\"  ohm\\n\"\n",
+      "print  \"\\n(c)Three in parallel, in series with one for \",round(R3,2),\"  ohm\\n\"\n",
+      "print  \"\\n(d)Two in parallel, in series with two in series for \",R4,\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)All four in parallel for  0.25   ohm\n",
+        "\n",
+        "\n",
+        "(b)Two in series, in parallel with another two in series for 1.0   ohm\n",
+        "\n",
+        "\n",
+        "(c)Three in parallel, in series with one for  1.33   ohm\n",
+        "\n",
+        "\n",
+        "(d)Two in parallel, in series with two in series for  2.5   ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 48</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  2.2;#  in  ohms\n",
+      "R3  =  3;#  in  ohms\n",
+      "R4  =  6;#  in  ohms\n",
+      "R5  =  18;#  in  ohms\n",
+      "R6  =  4;#  in  ohms\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      "R0  =  1/((1/3)  +  (1/6)  +  (1/18))\n",
+      "Rt  =  R1  +  R2  +  R0  +  R6\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Equivalent  Resistance  \",Rt,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Equivalent  Resistance   9.0   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 48</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  2.5;#  in  ohms\n",
+      "R2  =  6;#  in  ohms\n",
+      "R3  =  2;#  in  ohms\n",
+      "R4  =  4;#  in  ohms\n",
+      "Vt  =  200;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "R0  =  1/((1/R2)  +  (1/R3))\n",
+      "Rt  =  R1  +  R0  +  R4\n",
+      "It  =  Vt/Rt\n",
+      "I1  =  It\n",
+      "I4  =  It\n",
+      "I2  =  R3*It/(R3+R2)\n",
+      "I3  =  It  -  I2\n",
+      "V1  =  I1*R1\n",
+      "V2  =  I2*R2\n",
+      "V3  =  I3*R3\n",
+      "V4  =  I4*R4\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Total  Current  Supply  \",It,\"  Amperes(A)\\n\"\n",
+      "print  \"\\n  (b)Current  through  resistors  (R1,  R2,  R3,  R4)\\n \",I1,\",  \",  I2,\",  \",  I3,\",  \",  I4,\"  Amperes(A)  respectively\\n\"\n",
+      "print  \"\\n  (c)voltage  across  resistors  (R1,  R2,  R3,  R4)\\n  \",V1,\",  \",  V2,\",  \",  V3,\",  \",  V4,\"  Volts(V)  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Total  Current  Supply   25.0   Amperes(A)\n",
+        "\n",
+        "\n",
+        "  (b)Current  through  resistors  (R1,  R2,  R3,  R4)\n",
+        "  25.0 ,   6.25 ,   18.75 ,   25.0   Amperes(A)  respectively\n",
+        "\n",
+        "\n",
+        "  (c)voltage  across  resistors  (R1,  R2,  R3,  R4)\n",
+        "   62.5 ,   37.5 ,   37.5 ,   100.0   Volts(V)  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 49</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  15;#  in  ohms\n",
+      "R2  =  10;#  in  ohms\n",
+      "R3  =  38;#  in  ohms\n",
+      "Vt  =  250;#  in  volts\n",
+      "P  =  2500;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      "It  =  P/Vt\n",
+      "I2  =  R1*It/(R1+R2)\n",
+      "I1  =  It  -  I2\n",
+      "Re1  =  1/((1/R1)  +  (1/R2))\n",
+      "Rt  =  Vt/It\n",
+      "Re2  =  Rt  -  Re1\n",
+      "Rx  =  1/((1/Re2)  -  (1/R3))\n",
+      "I4  =  R3*It/(R3+Rx)\n",
+      "I3  =  It  -  I4\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  (Rx)  \",Rx,\"  Ohms\\n\"\n",
+      "print  \"\\n  (b)Current  through  resistors  (R1,  R2,  R3,  R4): \\n \",I1,\",  \",  I2,\",  \",  I3,\",  \"\n",
+      "print   \",  I4,\"  Amperes(A)  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  (Rx)   38.0   Ohms\n",
+        "\n",
+        "\n",
+        "  (b)Current  through  resistors  (R1,  R2,  R3,  R4): \n",
+        "  4.0 ,   6.0 ,   5.0 ,   5.0   Amperes(A)  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 51</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  1.4;#  in  ohms\n",
+      "R4  =  9;#  in  ohms\n",
+      "R5  =  2;#  in  ohms\n",
+      "Vt  =  17;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "R01  =  R1*R2/(R1  +  R2)\n",
+      "R02  =  R01  +  R3\n",
+      "R03  =  R4*R02/(R4  +R02)\n",
+      "Rt  =  R5  +  R03\n",
+      "It  =  Vt/Rt\n",
+      "I1  =  R4*It/(R4  +  R02)\n",
+      "Ix  =  R2*I1/(R1  +  R2)\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Current(Ix)  \",Ix,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Current(Ix)   0.6   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 52</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rt  =  150;#  in  ohms\n",
+      "n  =  3;#  no.  of  identical  lamp\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  Rt*3#  (1/Rt)=(1/R)+(1/R)+(1/R)\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Resistance  \",R,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance   450   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 52</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "#series  connection\n",
+      "n  =  3;#  no.  of  identical  lamp\n",
+      "Vt  =  150;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  Vt/3#  Since  each  lamp  is  identical,  then  V  volts  across  each.\n",
+      "\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  a)Voltage  across  each  resistor  =  \",V,\"  Volts(V)\\n\"\n",
+      "print  \"\\n  b)If  lamp  C  fails,  i.e.,  open  circuits,  no  current  will  flow  and lamps  A  and  B  will  not  operate.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  a)Voltage  across  each  resistor  =   50.0   Volts(V)\n",
+        "\n",
+        "\n",
+        "  b)If  lamp  C  fails,  i.e.,  open  circuits,  no  current  will  flow  and lamps  A  and  B  will  not  operate."
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint.ipynb
new file mode 100755
index 00000000..772a8bc5
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint.ipynb
@@ -0,0 +1,873 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 6: Capacitors and capacitance</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 58</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the p.d. across a 4 \u03bcF capacitor when charged with 5 mC.\n",
+      "# (b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  4E-6;#  in  Farad\n",
+      "C2  =  50E-12;#  in  Farad\n",
+      "Q1  =  5E-3;#  in  Coulomb\n",
+      "V2  =  2000;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  Q1/C1\n",
+      "Q2  =  C2*V2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V1,\"  Volts(V)\\n\"\n",
+      "print  \"\\n  (b)Charge(Q)  \",(Q2/1E-6),\"  micro-Coulomb\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   1250.0   Volts(V)\n",
+        "\n",
+        "\n",
+        "  (b)Charge(Q)   0.1   micro-Coulomb"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 58</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the pd between the plates.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  4;#  in  amperes\n",
+      "C  =  20E-6;#  in  Farad\n",
+      "t  =  3E-3;#  in  sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  I*t\n",
+      "V  =  Q/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V,\"  Volts(V)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   600.0   Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 59</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate how long the capacitor can provide an average discharge current of 2 mA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  2E-3;#  in  amperes\n",
+      "C  =  5E-6;#  in  Farad\n",
+      "V  =  800;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  C*V\n",
+      "t  =  Q/I\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  capacitor  can  provide  an  average  discharge  current  for  \",t,\"  Sec\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitor  can  provide  an  average  discharge  current  for   2.0   Sec"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the electric flux density.\n",
+      "#determine the electric field strength.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Q  =  0.2E-6;#  in  Coulomb\n",
+      "A  =  800E-4;#  in  m2\n",
+      "d  =  0.005;#  in  m\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "D  =  Q/A\n",
+      "E  =  V/d\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Electric  flux  density  D  \",(D/1E-6),\" uC/m2\\n\"\n",
+      "print  \"\\n  (b)Electric  field  strength  E  \",(E/1000),\"  kV/m\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Electric  flux  density  D   2.5  uC/m2\n",
+        "\n",
+        "\n",
+        "  (b)Electric  field  strength  E   50.0   kV/m"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the voltage gradient between the plates.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "D  =  2E-6;#  in  micro-C/m2\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  5;\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  D/(e0*er)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Electric  field  strength  E  \",round((E/1000),2),\"  kV/m\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Electric  field  strength  E   45.2   kV/m"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is the electric field strength?\n",
+      "#Find also the flux density when the dielectric between the plates is\n",
+      "#(a) air, and (b) polythene of relative permittivity 2.3\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "d  =  0.8E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "era  =  1;#  for  air\n",
+      "erp  =  2.3;#  for  polythene\n",
+      "V  =200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  V/d\n",
+      "#for  air\n",
+      "Da  =  E*e0*era\n",
+      "#for  polythene\n",
+      "Dp  =  E*e0*erp\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Electric  flux  density  for  air    \",round((Da/1E-6),2),\"  micro-C/m2\\n\"\n",
+      "print  \"\\n  (b)Electric  flux  density  for  polythene    \",round((Dp/1E-6),2),\"  micro-C/m2\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Electric  flux  density  for  air     2.21   micro-C/m2\n",
+        "\n",
+        "\n",
+        "  (b)Electric  flux  density  for  polythene     5.09   micro-C/m2"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Calculate the capacitance of the capacitor in picofarads. \n",
+      "#(b)what will be the pd between the plates?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "A  =  4E-4;#  in  m2\n",
+      "d  =  0.1E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  100;\n",
+      "Q  =  1.2E-6;#  in  coulomb\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  e0*er*A/d\n",
+      "V  =  Q/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance    \",(C/1E-12),\"  pF\\n\"\n",
+      "print  \"\\n  (b)P.d.=  \",round(V,2),\"  Volts(V)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance     3540.0   pF\n",
+        "\n",
+        "\n",
+        "  (b)P.d.=   338.98   Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the effective thickness of the paper if its relative permittivity is 2.5\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "A  =  800E-4;#  in  m2\n",
+      "C  =  4425E-12;#  in  Farads\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  2.5;\n",
+      "\n",
+      "#calculation:\n",
+      "d  =  e0*er*A/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Thickness    \",(d/1E-3),\"  mm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Thickness     0.4   mm"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the capacitance of the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "n  =  19;#  no.  of  plates\n",
+      "L  =  75E-3;#  in  m\n",
+      "B  =  75E-3;#  in  m\n",
+      "d  =  0.2E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  5;\n",
+      "#calculation:\n",
+      "A  =  L*B\n",
+      "C  =  e0*er*A*(n-1)/d\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Capacitance  \",round((C/1E-9),2),\"  nF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Capacitance   22.4   nF"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the equivalent capacitance of two capacitors of 6 \u03bcF and 4 \u03bcF connected (a) in parallel and (b) in series\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  6E-6;#  in  Farads\n",
+      "C2  =  4E-6;#  in  Farads\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Parallel\n",
+      "Cp  =  C1 + C2\n",
+      "#  in  Series\n",
+      "Cs  =  1/((1/C1)  +  (1/C2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance  in  parallel  \",(Cp/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Capacitance  in  Series  \",(Cs/1E-6),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance  in  parallel   10.0   uF\n",
+        "\n",
+        "\n",
+        "  (b)Capacitance  in  Series   2.4   uF"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What capacitance must be connected in series\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  30E-6;#  in  Farads\n",
+      "Cs  =  12E-6;#  in  Farads\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Series\n",
+      "C2  =  1/((1/Cs)  -  (1/C1))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance  in  series  \",(C2/1E-6),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance  in  series   20.0   uF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the equivalent circuit capacitance, \n",
+      "#(b) the total charge and\n",
+      "#(c) the charge on each capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  1E-6;#  in  Farads\n",
+      "C2  =  3E-6;#  in  Farads\n",
+      "C3  =  5E-6;#  in  Farads\n",
+      "C4  =  6E-6;#  in  Farads\n",
+      "Vt  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Parallel\n",
+      "Cp  =  C1  +  C2  +  C3  +  C4\n",
+      "Qt  =  Vt*Cp\n",
+      "Q1  =  C1*Vt\n",
+      "Q2  =  C2*Vt\n",
+      "Q3  =  C3*Vt\n",
+      "Q4  =  C4*Vt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Equivalent  Capacitance  in  Parallel  \",(Cp/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Total  charge  \",(Qt/1E-3),\"  mC\\n\"\n",
+      "print  \"\\n  (c)Charge  on  each  capacitors  (C1,  C2,  C3,  C4)\\n    \",(Q1/1E-3),\",  \",(Q2/1E-3),\", \"\n",
+      "print   \",(Q3/1E-3),\",  \",(Q4/1E-3),\"  mC  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "SyntaxError",
+       "evalue": "invalid syntax (<ipython-input-1-0c99632e8192>, line 27)",
+       "output_type": "pyerr",
+       "traceback": [
+        "\u001b[1;36m  File \u001b[1;32m\"<ipython-input-1-0c99632e8192>\"\u001b[1;36m, line \u001b[1;32m27\u001b[0m\n\u001b[1;33m    print   \",(Q3/1E-3),\",  \",(Q4/1E-3),\"  mC  respectively\\n\"\u001b[0m\n\u001b[1;37m                                            ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 66</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the equivalent circuit capacitance, \n",
+      "#(b) the charge on each capacitor and\n",
+      "#(c) the pd across each capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  3E-6;#  in  Farads\n",
+      "C2  =  6E-6;#  in  Farads\n",
+      "C3  =  12E-6;#  in  Farads\n",
+      "Vt  =  350;#  in  Volts\n",
+      "#calculation:\n",
+      "#  in  series\n",
+      "Cs  =  1/((1/C1)  +  (1/C2)  +  (1/C3))\n",
+      "Qt  =  Vt*Cs\n",
+      "V1  =  Qt/C1\n",
+      "V2  =  Qt/C2\n",
+      "V3  =  Qt/C3\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Equivalent  Capacitance  in  Series  \",(Cs/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Charge  on  each  capacitors  (C1,  C2,  C3)  \",(Qt/1E-3),\"  mC  \\n\"\n",
+      "print  \"\\n  (b)P.d  Across  each  capacitors  (C1,  C2,  C3)\\n   \",V1,\"  V,  \",  V2,\"  V,  \",  V3,\"  V  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Equivalent  Capacitance  in  Series   1.71428571429   uF\n",
+        "\n",
+        "\n",
+        "  (b)Charge  on  each  capacitors  (C1,  C2,  C3)   0.6   mC  \n",
+        "\n",
+        "\n",
+        "  (b)P.d  Across  each  capacitors  (C1,  C2,  C3)\n",
+        "    200.0   V,   100.0   V,   50.0   V  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 67</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the thickness of the mica needed, and \n",
+      "#(b) the area of a plate assuming a two-plate construction.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.2E-6;#  in  Farads\n",
+      "V  =  1250;#  in  Volts\n",
+      "E = 50E6#  in  V/m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  6;\n",
+      "\n",
+      "#calculation:\n",
+      "d  =  V/E\n",
+      "A  =  C*d/e0/er\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Thickness  \",(d/1E-3),\"  mm\\n\"\n",
+      "print  \"\\n  (b)Area  of  plate  is  \",round((A/1E-4),2),\"  cm^2  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Thickness   0.025   mm\n",
+        "\n",
+        "\n",
+        "  (b)Area  of  plate  is   941.62   cm^2  "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the energy stored in a 3 \u03bcF capacitor when charged to 400 V.\n",
+      "#(b) Find also the average power developed if this energy is dissipated in a time of 10 \u03bcs\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  3E-6;#  in  Farads\n",
+      "V  =  400;#  in  Volts\n",
+      "t  =  10E-6;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "W  =  C*V*V/2\n",
+      "P  =  W/t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Energy  stored  \",W,\"  J\\n\"\n",
+      "print  \"\\n  (b)Power  developed  \",(P/1E3),\"  kW  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Energy  stored   0.24   J\n",
+        "\n",
+        "\n",
+        "  (b)Power  developed   24.0   kW  "
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the pd to which the capacitor must be charged.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  12E-6;#  in  Farads\n",
+      "W  =  4;#  in  Joules\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  (2*W/C)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  P.d  \",round(V,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  P.d   816.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the voltage and (b) the capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "W  =  1.2;#  in  Joules\n",
+      "Q  =  10E-3;#  in  Coulomb\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  2*W/Q\n",
+      "C  =  Q/V\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V,\"  V\\n\"\n",
+      "print  \"\\n  (b)Capacitance  \",round((C/1E-6),2),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   240.0   V\n",
+        "\n",
+        "\n",
+        "  (b)Capacitance   41.67   uF"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb
new file mode 100755
index 00000000..772a8bc5
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb
@@ -0,0 +1,873 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 6: Capacitors and capacitance</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 58</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the p.d. across a 4 \u03bcF capacitor when charged with 5 mC.\n",
+      "# (b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  4E-6;#  in  Farad\n",
+      "C2  =  50E-12;#  in  Farad\n",
+      "Q1  =  5E-3;#  in  Coulomb\n",
+      "V2  =  2000;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  Q1/C1\n",
+      "Q2  =  C2*V2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V1,\"  Volts(V)\\n\"\n",
+      "print  \"\\n  (b)Charge(Q)  \",(Q2/1E-6),\"  micro-Coulomb\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   1250.0   Volts(V)\n",
+        "\n",
+        "\n",
+        "  (b)Charge(Q)   0.1   micro-Coulomb"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 58</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the pd between the plates.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  4;#  in  amperes\n",
+      "C  =  20E-6;#  in  Farad\n",
+      "t  =  3E-3;#  in  sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  I*t\n",
+      "V  =  Q/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V,\"  Volts(V)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   600.0   Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 59</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate how long the capacitor can provide an average discharge current of 2 mA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  2E-3;#  in  amperes\n",
+      "C  =  5E-6;#  in  Farad\n",
+      "V  =  800;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  C*V\n",
+      "t  =  Q/I\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  capacitor  can  provide  an  average  discharge  current  for  \",t,\"  Sec\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitor  can  provide  an  average  discharge  current  for   2.0   Sec"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the electric flux density.\n",
+      "#determine the electric field strength.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Q  =  0.2E-6;#  in  Coulomb\n",
+      "A  =  800E-4;#  in  m2\n",
+      "d  =  0.005;#  in  m\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "D  =  Q/A\n",
+      "E  =  V/d\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Electric  flux  density  D  \",(D/1E-6),\" uC/m2\\n\"\n",
+      "print  \"\\n  (b)Electric  field  strength  E  \",(E/1000),\"  kV/m\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Electric  flux  density  D   2.5  uC/m2\n",
+        "\n",
+        "\n",
+        "  (b)Electric  field  strength  E   50.0   kV/m"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the voltage gradient between the plates.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "D  =  2E-6;#  in  micro-C/m2\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  5;\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  D/(e0*er)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Electric  field  strength  E  \",round((E/1000),2),\"  kV/m\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Electric  field  strength  E   45.2   kV/m"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is the electric field strength?\n",
+      "#Find also the flux density when the dielectric between the plates is\n",
+      "#(a) air, and (b) polythene of relative permittivity 2.3\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "d  =  0.8E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "era  =  1;#  for  air\n",
+      "erp  =  2.3;#  for  polythene\n",
+      "V  =200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  V/d\n",
+      "#for  air\n",
+      "Da  =  E*e0*era\n",
+      "#for  polythene\n",
+      "Dp  =  E*e0*erp\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Electric  flux  density  for  air    \",round((Da/1E-6),2),\"  micro-C/m2\\n\"\n",
+      "print  \"\\n  (b)Electric  flux  density  for  polythene    \",round((Dp/1E-6),2),\"  micro-C/m2\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Electric  flux  density  for  air     2.21   micro-C/m2\n",
+        "\n",
+        "\n",
+        "  (b)Electric  flux  density  for  polythene     5.09   micro-C/m2"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Calculate the capacitance of the capacitor in picofarads. \n",
+      "#(b)what will be the pd between the plates?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "A  =  4E-4;#  in  m2\n",
+      "d  =  0.1E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  100;\n",
+      "Q  =  1.2E-6;#  in  coulomb\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  e0*er*A/d\n",
+      "V  =  Q/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance    \",(C/1E-12),\"  pF\\n\"\n",
+      "print  \"\\n  (b)P.d.=  \",round(V,2),\"  Volts(V)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance     3540.0   pF\n",
+        "\n",
+        "\n",
+        "  (b)P.d.=   338.98   Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the effective thickness of the paper if its relative permittivity is 2.5\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "A  =  800E-4;#  in  m2\n",
+      "C  =  4425E-12;#  in  Farads\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  2.5;\n",
+      "\n",
+      "#calculation:\n",
+      "d  =  e0*er*A/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Thickness    \",(d/1E-3),\"  mm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Thickness     0.4   mm"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the capacitance of the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "n  =  19;#  no.  of  plates\n",
+      "L  =  75E-3;#  in  m\n",
+      "B  =  75E-3;#  in  m\n",
+      "d  =  0.2E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  5;\n",
+      "#calculation:\n",
+      "A  =  L*B\n",
+      "C  =  e0*er*A*(n-1)/d\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Capacitance  \",round((C/1E-9),2),\"  nF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Capacitance   22.4   nF"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the equivalent capacitance of two capacitors of 6 \u03bcF and 4 \u03bcF connected (a) in parallel and (b) in series\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  6E-6;#  in  Farads\n",
+      "C2  =  4E-6;#  in  Farads\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Parallel\n",
+      "Cp  =  C1 + C2\n",
+      "#  in  Series\n",
+      "Cs  =  1/((1/C1)  +  (1/C2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance  in  parallel  \",(Cp/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Capacitance  in  Series  \",(Cs/1E-6),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance  in  parallel   10.0   uF\n",
+        "\n",
+        "\n",
+        "  (b)Capacitance  in  Series   2.4   uF"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What capacitance must be connected in series\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  30E-6;#  in  Farads\n",
+      "Cs  =  12E-6;#  in  Farads\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Series\n",
+      "C2  =  1/((1/Cs)  -  (1/C1))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance  in  series  \",(C2/1E-6),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance  in  series   20.0   uF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the equivalent circuit capacitance, \n",
+      "#(b) the total charge and\n",
+      "#(c) the charge on each capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  1E-6;#  in  Farads\n",
+      "C2  =  3E-6;#  in  Farads\n",
+      "C3  =  5E-6;#  in  Farads\n",
+      "C4  =  6E-6;#  in  Farads\n",
+      "Vt  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Parallel\n",
+      "Cp  =  C1  +  C2  +  C3  +  C4\n",
+      "Qt  =  Vt*Cp\n",
+      "Q1  =  C1*Vt\n",
+      "Q2  =  C2*Vt\n",
+      "Q3  =  C3*Vt\n",
+      "Q4  =  C4*Vt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Equivalent  Capacitance  in  Parallel  \",(Cp/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Total  charge  \",(Qt/1E-3),\"  mC\\n\"\n",
+      "print  \"\\n  (c)Charge  on  each  capacitors  (C1,  C2,  C3,  C4)\\n    \",(Q1/1E-3),\",  \",(Q2/1E-3),\", \"\n",
+      "print   \",(Q3/1E-3),\",  \",(Q4/1E-3),\"  mC  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "SyntaxError",
+       "evalue": "invalid syntax (<ipython-input-1-0c99632e8192>, line 27)",
+       "output_type": "pyerr",
+       "traceback": [
+        "\u001b[1;36m  File \u001b[1;32m\"<ipython-input-1-0c99632e8192>\"\u001b[1;36m, line \u001b[1;32m27\u001b[0m\n\u001b[1;33m    print   \",(Q3/1E-3),\",  \",(Q4/1E-3),\"  mC  respectively\\n\"\u001b[0m\n\u001b[1;37m                                            ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 66</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the equivalent circuit capacitance, \n",
+      "#(b) the charge on each capacitor and\n",
+      "#(c) the pd across each capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  3E-6;#  in  Farads\n",
+      "C2  =  6E-6;#  in  Farads\n",
+      "C3  =  12E-6;#  in  Farads\n",
+      "Vt  =  350;#  in  Volts\n",
+      "#calculation:\n",
+      "#  in  series\n",
+      "Cs  =  1/((1/C1)  +  (1/C2)  +  (1/C3))\n",
+      "Qt  =  Vt*Cs\n",
+      "V1  =  Qt/C1\n",
+      "V2  =  Qt/C2\n",
+      "V3  =  Qt/C3\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Equivalent  Capacitance  in  Series  \",(Cs/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Charge  on  each  capacitors  (C1,  C2,  C3)  \",(Qt/1E-3),\"  mC  \\n\"\n",
+      "print  \"\\n  (b)P.d  Across  each  capacitors  (C1,  C2,  C3)\\n   \",V1,\"  V,  \",  V2,\"  V,  \",  V3,\"  V  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Equivalent  Capacitance  in  Series   1.71428571429   uF\n",
+        "\n",
+        "\n",
+        "  (b)Charge  on  each  capacitors  (C1,  C2,  C3)   0.6   mC  \n",
+        "\n",
+        "\n",
+        "  (b)P.d  Across  each  capacitors  (C1,  C2,  C3)\n",
+        "    200.0   V,   100.0   V,   50.0   V  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 67</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the thickness of the mica needed, and \n",
+      "#(b) the area of a plate assuming a two-plate construction.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.2E-6;#  in  Farads\n",
+      "V  =  1250;#  in  Volts\n",
+      "E = 50E6#  in  V/m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  6;\n",
+      "\n",
+      "#calculation:\n",
+      "d  =  V/E\n",
+      "A  =  C*d/e0/er\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Thickness  \",(d/1E-3),\"  mm\\n\"\n",
+      "print  \"\\n  (b)Area  of  plate  is  \",round((A/1E-4),2),\"  cm^2  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Thickness   0.025   mm\n",
+        "\n",
+        "\n",
+        "  (b)Area  of  plate  is   941.62   cm^2  "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the energy stored in a 3 \u03bcF capacitor when charged to 400 V.\n",
+      "#(b) Find also the average power developed if this energy is dissipated in a time of 10 \u03bcs\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  3E-6;#  in  Farads\n",
+      "V  =  400;#  in  Volts\n",
+      "t  =  10E-6;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "W  =  C*V*V/2\n",
+      "P  =  W/t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Energy  stored  \",W,\"  J\\n\"\n",
+      "print  \"\\n  (b)Power  developed  \",(P/1E3),\"  kW  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Energy  stored   0.24   J\n",
+        "\n",
+        "\n",
+        "  (b)Power  developed   24.0   kW  "
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the pd to which the capacitor must be charged.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  12E-6;#  in  Farads\n",
+      "W  =  4;#  in  Joules\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  (2*W/C)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  P.d  \",round(V,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  P.d   816.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the voltage and (b) the capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "W  =  1.2;#  in  Joules\n",
+      "Q  =  10E-3;#  in  Coulomb\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  2*W/Q\n",
+      "C  =  Q/V\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V,\"  V\\n\"\n",
+      "print  \"\\n  (b)Capacitance  \",round((C/1E-6),2),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   240.0   V\n",
+        "\n",
+        "\n",
+        "  (b)Capacitance   41.67   uF"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_2.ipynb
new file mode 100755
index 00000000..772a8bc5
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_2.ipynb
@@ -0,0 +1,873 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 6: Capacitors and capacitance</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 58</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the p.d. across a 4 \u03bcF capacitor when charged with 5 mC.\n",
+      "# (b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  4E-6;#  in  Farad\n",
+      "C2  =  50E-12;#  in  Farad\n",
+      "Q1  =  5E-3;#  in  Coulomb\n",
+      "V2  =  2000;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  Q1/C1\n",
+      "Q2  =  C2*V2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V1,\"  Volts(V)\\n\"\n",
+      "print  \"\\n  (b)Charge(Q)  \",(Q2/1E-6),\"  micro-Coulomb\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   1250.0   Volts(V)\n",
+        "\n",
+        "\n",
+        "  (b)Charge(Q)   0.1   micro-Coulomb"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 58</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the pd between the plates.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  4;#  in  amperes\n",
+      "C  =  20E-6;#  in  Farad\n",
+      "t  =  3E-3;#  in  sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  I*t\n",
+      "V  =  Q/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V,\"  Volts(V)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   600.0   Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 59</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate how long the capacitor can provide an average discharge current of 2 mA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  2E-3;#  in  amperes\n",
+      "C  =  5E-6;#  in  Farad\n",
+      "V  =  800;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  C*V\n",
+      "t  =  Q/I\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  capacitor  can  provide  an  average  discharge  current  for  \",t,\"  Sec\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitor  can  provide  an  average  discharge  current  for   2.0   Sec"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the electric flux density.\n",
+      "#determine the electric field strength.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Q  =  0.2E-6;#  in  Coulomb\n",
+      "A  =  800E-4;#  in  m2\n",
+      "d  =  0.005;#  in  m\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "D  =  Q/A\n",
+      "E  =  V/d\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Electric  flux  density  D  \",(D/1E-6),\" uC/m2\\n\"\n",
+      "print  \"\\n  (b)Electric  field  strength  E  \",(E/1000),\"  kV/m\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Electric  flux  density  D   2.5  uC/m2\n",
+        "\n",
+        "\n",
+        "  (b)Electric  field  strength  E   50.0   kV/m"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the voltage gradient between the plates.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "D  =  2E-6;#  in  micro-C/m2\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  5;\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  D/(e0*er)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Electric  field  strength  E  \",round((E/1000),2),\"  kV/m\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Electric  field  strength  E   45.2   kV/m"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What is the electric field strength?\n",
+      "#Find also the flux density when the dielectric between the plates is\n",
+      "#(a) air, and (b) polythene of relative permittivity 2.3\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "d  =  0.8E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "era  =  1;#  for  air\n",
+      "erp  =  2.3;#  for  polythene\n",
+      "V  =200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  V/d\n",
+      "#for  air\n",
+      "Da  =  E*e0*era\n",
+      "#for  polythene\n",
+      "Dp  =  E*e0*erp\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Electric  flux  density  for  air    \",round((Da/1E-6),2),\"  micro-C/m2\\n\"\n",
+      "print  \"\\n  (b)Electric  flux  density  for  polythene    \",round((Dp/1E-6),2),\"  micro-C/m2\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Electric  flux  density  for  air     2.21   micro-C/m2\n",
+        "\n",
+        "\n",
+        "  (b)Electric  flux  density  for  polythene     5.09   micro-C/m2"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Calculate the capacitance of the capacitor in picofarads. \n",
+      "#(b)what will be the pd between the plates?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "A  =  4E-4;#  in  m2\n",
+      "d  =  0.1E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  100;\n",
+      "Q  =  1.2E-6;#  in  coulomb\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  e0*er*A/d\n",
+      "V  =  Q/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance    \",(C/1E-12),\"  pF\\n\"\n",
+      "print  \"\\n  (b)P.d.=  \",round(V,2),\"  Volts(V)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance     3540.0   pF\n",
+        "\n",
+        "\n",
+        "  (b)P.d.=   338.98   Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the effective thickness of the paper if its relative permittivity is 2.5\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "A  =  800E-4;#  in  m2\n",
+      "C  =  4425E-12;#  in  Farads\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  2.5;\n",
+      "\n",
+      "#calculation:\n",
+      "d  =  e0*er*A/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Thickness    \",(d/1E-3),\"  mm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Thickness     0.4   mm"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the capacitance of the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "n  =  19;#  no.  of  plates\n",
+      "L  =  75E-3;#  in  m\n",
+      "B  =  75E-3;#  in  m\n",
+      "d  =  0.2E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  5;\n",
+      "#calculation:\n",
+      "A  =  L*B\n",
+      "C  =  e0*er*A*(n-1)/d\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Capacitance  \",round((C/1E-9),2),\"  nF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Capacitance   22.4   nF"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the equivalent capacitance of two capacitors of 6 \u03bcF and 4 \u03bcF connected (a) in parallel and (b) in series\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  6E-6;#  in  Farads\n",
+      "C2  =  4E-6;#  in  Farads\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Parallel\n",
+      "Cp  =  C1 + C2\n",
+      "#  in  Series\n",
+      "Cs  =  1/((1/C1)  +  (1/C2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance  in  parallel  \",(Cp/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Capacitance  in  Series  \",(Cs/1E-6),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance  in  parallel   10.0   uF\n",
+        "\n",
+        "\n",
+        "  (b)Capacitance  in  Series   2.4   uF"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What capacitance must be connected in series\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  30E-6;#  in  Farads\n",
+      "Cs  =  12E-6;#  in  Farads\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Series\n",
+      "C2  =  1/((1/Cs)  -  (1/C1))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance  in  series  \",(C2/1E-6),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance  in  series   20.0   uF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the equivalent circuit capacitance, \n",
+      "#(b) the total charge and\n",
+      "#(c) the charge on each capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  1E-6;#  in  Farads\n",
+      "C2  =  3E-6;#  in  Farads\n",
+      "C3  =  5E-6;#  in  Farads\n",
+      "C4  =  6E-6;#  in  Farads\n",
+      "Vt  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Parallel\n",
+      "Cp  =  C1  +  C2  +  C3  +  C4\n",
+      "Qt  =  Vt*Cp\n",
+      "Q1  =  C1*Vt\n",
+      "Q2  =  C2*Vt\n",
+      "Q3  =  C3*Vt\n",
+      "Q4  =  C4*Vt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Equivalent  Capacitance  in  Parallel  \",(Cp/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Total  charge  \",(Qt/1E-3),\"  mC\\n\"\n",
+      "print  \"\\n  (c)Charge  on  each  capacitors  (C1,  C2,  C3,  C4)\\n    \",(Q1/1E-3),\",  \",(Q2/1E-3),\", \"\n",
+      "print   \",(Q3/1E-3),\",  \",(Q4/1E-3),\"  mC  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "SyntaxError",
+       "evalue": "invalid syntax (<ipython-input-1-0c99632e8192>, line 27)",
+       "output_type": "pyerr",
+       "traceback": [
+        "\u001b[1;36m  File \u001b[1;32m\"<ipython-input-1-0c99632e8192>\"\u001b[1;36m, line \u001b[1;32m27\u001b[0m\n\u001b[1;33m    print   \",(Q3/1E-3),\",  \",(Q4/1E-3),\"  mC  respectively\\n\"\u001b[0m\n\u001b[1;37m                                            ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 66</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the equivalent circuit capacitance, \n",
+      "#(b) the charge on each capacitor and\n",
+      "#(c) the pd across each capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  3E-6;#  in  Farads\n",
+      "C2  =  6E-6;#  in  Farads\n",
+      "C3  =  12E-6;#  in  Farads\n",
+      "Vt  =  350;#  in  Volts\n",
+      "#calculation:\n",
+      "#  in  series\n",
+      "Cs  =  1/((1/C1)  +  (1/C2)  +  (1/C3))\n",
+      "Qt  =  Vt*Cs\n",
+      "V1  =  Qt/C1\n",
+      "V2  =  Qt/C2\n",
+      "V3  =  Qt/C3\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Equivalent  Capacitance  in  Series  \",(Cs/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Charge  on  each  capacitors  (C1,  C2,  C3)  \",(Qt/1E-3),\"  mC  \\n\"\n",
+      "print  \"\\n  (b)P.d  Across  each  capacitors  (C1,  C2,  C3)\\n   \",V1,\"  V,  \",  V2,\"  V,  \",  V3,\"  V  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Equivalent  Capacitance  in  Series   1.71428571429   uF\n",
+        "\n",
+        "\n",
+        "  (b)Charge  on  each  capacitors  (C1,  C2,  C3)   0.6   mC  \n",
+        "\n",
+        "\n",
+        "  (b)P.d  Across  each  capacitors  (C1,  C2,  C3)\n",
+        "    200.0   V,   100.0   V,   50.0   V  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 67</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the thickness of the mica needed, and \n",
+      "#(b) the area of a plate assuming a two-plate construction.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.2E-6;#  in  Farads\n",
+      "V  =  1250;#  in  Volts\n",
+      "E = 50E6#  in  V/m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  6;\n",
+      "\n",
+      "#calculation:\n",
+      "d  =  V/E\n",
+      "A  =  C*d/e0/er\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Thickness  \",(d/1E-3),\"  mm\\n\"\n",
+      "print  \"\\n  (b)Area  of  plate  is  \",round((A/1E-4),2),\"  cm^2  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Thickness   0.025   mm\n",
+        "\n",
+        "\n",
+        "  (b)Area  of  plate  is   941.62   cm^2  "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the energy stored in a 3 \u03bcF capacitor when charged to 400 V.\n",
+      "#(b) Find also the average power developed if this energy is dissipated in a time of 10 \u03bcs\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  3E-6;#  in  Farads\n",
+      "V  =  400;#  in  Volts\n",
+      "t  =  10E-6;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "W  =  C*V*V/2\n",
+      "P  =  W/t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Energy  stored  \",W,\"  J\\n\"\n",
+      "print  \"\\n  (b)Power  developed  \",(P/1E3),\"  kW  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Energy  stored   0.24   J\n",
+        "\n",
+        "\n",
+        "  (b)Power  developed   24.0   kW  "
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the pd to which the capacitor must be charged.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  12E-6;#  in  Farads\n",
+      "W  =  4;#  in  Joules\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  (2*W/C)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  P.d  \",round(V,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  P.d   816.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the voltage and (b) the capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "W  =  1.2;#  in  Joules\n",
+      "Q  =  10E-3;#  in  Coulomb\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  2*W/Q\n",
+      "C  =  Q/V\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V,\"  V\\n\"\n",
+      "print  \"\\n  (b)Capacitance  \",round((C/1E-6),2),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   240.0   V\n",
+        "\n",
+        "\n",
+        "  (b)Capacitance   41.67   uF"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_3.ipynb
new file mode 100755
index 00000000..72813a43
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_3.ipynb
@@ -0,0 +1,863 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:234655a6131e7998e8f7cb854da15789ea2eade062d8aecf2f10e325a53d9080"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 6: Capacitors and capacitance</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 58</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  4E-6;#  in  Farad\n",
+      "C2  =  50E-12;#  in  Farad\n",
+      "Q1  =  5E-3;#  in  Coulomb\n",
+      "V2  =  2000;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  Q1/C1\n",
+      "Q2  =  C2*V2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V1,\"  Volts(V)\\n\"\n",
+      "print  \"\\n  (b)Charge(Q)  \",(Q2/1E-6),\"  micro-Coulomb\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   1250.0   Volts(V)\n",
+        "\n",
+        "\n",
+        "  (b)Charge(Q)   0.1   micro-Coulomb"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 58</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  4;#  in  amperes\n",
+      "C  =  20E-6;#  in  Farad\n",
+      "t  =  3E-3;#  in  sec\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  I*t\n",
+      "V  =  Q/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V,\"  Volts(V)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   600.0   Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 59</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  2E-3;#  in  amperes\n",
+      "C  =  5E-6;#  in  Farad\n",
+      "V  =  800;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  C*V\n",
+      "t  =  Q/I\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  capacitor  can  provide  an  average  discharge  current  for  \",t,\"  Sec\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitor  can  provide  an  average  discharge  current  for   2.0   Sec"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Q  =  0.2E-6;#  in  Coulomb\n",
+      "A  =  800E-4;#  in  m2\n",
+      "d  =  0.005;#  in  m\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "D  =  Q/A\n",
+      "E  =  V/d\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Electric  flux  density  D  \",(D/1E-6),\" uC/m2\\n\"\n",
+      "print  \"\\n  (b)Electric  field  strength  E  \",(E/1000),\"  kV/m\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Electric  flux  density  D   2.5  uC/m2\n",
+        "\n",
+        "\n",
+        "  (b)Electric  field  strength  E   50.0   kV/m"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "D  =  2E-6;#  in  micro-C/m2\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  5;\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  D/(e0*er)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Electric  field  strength  E  \",round((E/1000),2),\"  kV/m\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Electric  field  strength  E   45.2   kV/m"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 60</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "d  =  0.8E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "era  =  1;#  for  air\n",
+      "erp  =  2.3;#  for  polythene\n",
+      "V  =200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  V/d\n",
+      "#for  air\n",
+      "Da  =  E*e0*era\n",
+      "#for  polythene\n",
+      "Dp  =  E*e0*erp\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Electric  flux  density  for  air    \",round((Da/1E-6),2),\"  micro-C/m2\\n\"\n",
+      "print  \"\\n  (b)Electric  flux  density  for  polythene    \",round((Dp/1E-6),2),\"  micro-C/m2\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Electric  flux  density  for  air     2.21   micro-C/m2\n",
+        "\n",
+        "\n",
+        "  (b)Electric  flux  density  for  polythene     5.09   micro-C/m2"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "A  =  4E-4;#  in  m2\n",
+      "d  =  0.1E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  100;\n",
+      "Q  =  1.2E-6;#  in  coulomb\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  e0*er*A/d\n",
+      "V  =  Q/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance    \",(C/1E-12),\"  pF\\n\"\n",
+      "print  \"\\n  (b)P.d.=  \",round(V,2),\"  Volts(V)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance     3540.0   pF\n",
+        "\n",
+        "\n",
+        "  (b)P.d.=   338.98   Volts(V)"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "A  =  800E-4;#  in  m2\n",
+      "C  =  4425E-12;#  in  Farads\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  2.5;\n",
+      "\n",
+      "#calculation:\n",
+      "d  =  e0*er*A/C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Thickness    \",(d/1E-3),\"  mm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Thickness     0.4   mm"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 62</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "n  =  19;#  no.  of  plates\n",
+      "L  =  75E-3;#  in  m\n",
+      "B  =  75E-3;#  in  m\n",
+      "d  =  0.2E-3;#  in  m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  5;\n",
+      "#calculation:\n",
+      "A  =  L*B\n",
+      "C  =  e0*er*A*(n-1)/d\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Capacitance  \",round((C/1E-9),2),\"  nF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Capacitance   22.4   nF"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  6E-6;#  in  Farads\n",
+      "C2  =  4E-6;#  in  Farads\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Parallel\n",
+      "Cp  =  C1 + C2\n",
+      "#  in  Series\n",
+      "Cs  =  1/((1/C1)  +  (1/C2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance  in  parallel  \",(Cp/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Capacitance  in  Series  \",(Cs/1E-6),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance  in  parallel   10.0   uF\n",
+        "\n",
+        "\n",
+        "  (b)Capacitance  in  Series   2.4   uF"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  30E-6;#  in  Farads\n",
+      "Cs  =  12E-6;#  in  Farads\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Series\n",
+      "C2  =  1/((1/Cs)  -  (1/C1))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Capacitance  in  series  \",(C2/1E-6),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitance  in  series   20.0   uF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 65</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  1E-6;#  in  Farads\n",
+      "C2  =  3E-6;#  in  Farads\n",
+      "C3  =  5E-6;#  in  Farads\n",
+      "C4  =  6E-6;#  in  Farads\n",
+      "Vt  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#  in  Parallel\n",
+      "Cp  =  C1  +  C2  +  C3  +  C4\n",
+      "Qt  =  Vt*Cp\n",
+      "Q1  =  C1*Vt\n",
+      "Q2  =  C2*Vt\n",
+      "Q3  =  C3*Vt\n",
+      "Q4  =  C4*Vt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Equivalent  Capacitance  in  Parallel  \",(Cp/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Total  charge  \",(Qt/1E-3),\"  mC\\n\"\n",
+      "print  \"\\n  (c)Charge  on  each  capacitors  (C1,  C2,  C3,  C4)\\n    \",(Q1/1E-3),\",  \",(Q2/1E-3),\", \"\n",
+      "print   \",(Q3/1E-3),\",  \",(Q4/1E-3),\"  mC  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "ename": "SyntaxError",
+       "evalue": "invalid syntax (<ipython-input-1-0c99632e8192>, line 27)",
+       "output_type": "pyerr",
+       "traceback": [
+        "\u001b[1;36m  File \u001b[1;32m\"<ipython-input-1-0c99632e8192>\"\u001b[1;36m, line \u001b[1;32m27\u001b[0m\n\u001b[1;33m    print   \",(Q3/1E-3),\",  \",(Q4/1E-3),\"  mC  respectively\\n\"\u001b[0m\n\u001b[1;37m                                            ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 66</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C1  =  3E-6;#  in  Farads\n",
+      "C2  =  6E-6;#  in  Farads\n",
+      "C3  =  12E-6;#  in  Farads\n",
+      "Vt  =  350;#  in  Volts\n",
+      "#calculation:\n",
+      "#  in  series\n",
+      "Cs  =  1/((1/C1)  +  (1/C2)  +  (1/C3))\n",
+      "Qt  =  Vt*Cs\n",
+      "V1  =  Qt/C1\n",
+      "V2  =  Qt/C2\n",
+      "V3  =  Qt/C3\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Equivalent  Capacitance  in  Series  \",(Cs/1E-6),\"  uF\\n\"\n",
+      "print  \"\\n  (b)Charge  on  each  capacitors  (C1,  C2,  C3)  \",(Qt/1E-3),\"  mC  \\n\"\n",
+      "print  \"\\n  (b)P.d  Across  each  capacitors  (C1,  C2,  C3)\\n   \",V1,\"  V,  \",  V2,\"  V,  \",  V3,\"  V  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Equivalent  Capacitance  in  Series   1.71428571429   uF\n",
+        "\n",
+        "\n",
+        "  (b)Charge  on  each  capacitors  (C1,  C2,  C3)   0.6   mC  \n",
+        "\n",
+        "\n",
+        "  (b)P.d  Across  each  capacitors  (C1,  C2,  C3)\n",
+        "    200.0   V,   100.0   V,   50.0   V  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 67</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.2E-6;#  in  Farads\n",
+      "V  =  1250;#  in  Volts\n",
+      "E = 50E6#  in  V/m\n",
+      "e0  =  8.85E-12;#  in  F/m\n",
+      "er  =  6;\n",
+      "\n",
+      "#calculation:\n",
+      "d  =  V/E\n",
+      "A  =  C*d/e0/er\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Thickness  \",(d/1E-3),\"  mm\\n\"\n",
+      "print  \"\\n  (b)Area  of  plate  is  \",round((A/1E-4),2),\"  cm^2  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Thickness   0.025   mm\n",
+        "\n",
+        "\n",
+        "  (b)Area  of  plate  is   941.62   cm^2  "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  3E-6;#  in  Farads\n",
+      "V  =  400;#  in  Volts\n",
+      "t  =  10E-6;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "W  =  C*V*V/2\n",
+      "P  =  W/t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Energy  stored  \",W,\"  J\\n\"\n",
+      "print  \"\\n  (b)Power  developed  \",(P/1E3),\"  kW  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Energy  stored   0.24   J\n",
+        "\n",
+        "\n",
+        "  (b)Power  developed   24.0   kW  "
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  12E-6;#  in  Farads\n",
+      "W  =  4;#  in  Joules\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  (2*W/C)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  P.d  \",round(V,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  P.d   816.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 68</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "W  =  1.2;#  in  Joules\n",
+      "Q  =  10E-3;#  in  Coulomb\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  2*W/Q\n",
+      "C  =  Q/V\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)P.d  \",V,\"  V\\n\"\n",
+      "print  \"\\n  (b)Capacitance  \",round((C/1E-6),2),\"  uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)P.d   240.0   V\n",
+        "\n",
+        "\n",
+        "  (b)Capacitance   41.67   uF"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint.ipynb
new file mode 100755
index 00000000..1d749dd1
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint.ipynb
@@ -0,0 +1,273 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 8: Electromagnetism</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 93</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the force acting on the conductor.\n",
+      "#Determine also the value of the force\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.9;#  in  tesla\n",
+      "I  =  20;#  in  Amperes\n",
+      "l  =  0.30;#  in  m\n",
+      "alpha  =  30;#  in  degree\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "F1  =  B*I*l\n",
+      "F2  =  B*I*l*math.sin(alpha*math.pi/180)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Force  when  the  conductor  is  at  right  angles  to  the  field  =  \",F1,\"  N\\n\"\n",
+      "print  \"\\n  (b)Force  when  the  conductor  is  at  30\u00b0  angle  to  the  field  =  \",F2,\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Force  when  the  conductor  is  at  right  angles  to  the  field  =   5.4   N\n",
+        "\n",
+        "\n",
+        "  (b)Force  when  the  conductor  is  at  30\u00c2\u00b0  angle  to  the  field  =   2.7   N"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 94</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current required?\n",
+      "#what is the direction of the force?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "F  =  1.92;#  in  newton\n",
+      "B  =  1.2;#  in  tesla\n",
+      "l  =  0.40;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  F/(B*l)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Current  I  =  \",I,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  I  =   4.0   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 95</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the magnitude of the force exerted on the conductor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "r  =  0.06;#  in  m\n",
+      "I  =  10;#  in  Amperes\n",
+      "l  =  0.35;#  in  m\n",
+      "Phi  =  0.5E-3;#  in  Wb\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  math.pi*r*r\n",
+      "B  =  Phi/A\n",
+      "F  =  B*I*l\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Force  F  =  \",round(F,2),\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Force  F  =   0.15   N"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 95</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the force on each coil side (a) for a single-turn coil,\n",
+      "#(b) for a coil wound with 300 turns.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  1;#  for  a  single-turn  coil\n",
+      "N2  =  300;#  no.  of  turns\n",
+      "b  =  0.024;#  in  m\n",
+      "B  =  0.8;#  in  Tesla\n",
+      "I  =  0.05;#  in  Amperes\n",
+      "l  =  0.030;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "#For  a  single-turn  coil,\n",
+      "F1  =  N1*B*I*l\n",
+      "#for  a  coil  wound  with  300  turns.\n",
+      "F2  =  N2*B*I*l\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)For  a  single-turn  coil,  force  on  each  coil  side  =  \",F1,\"  N\\n\"\n",
+      "print  \"\\n  (b)For  a  300-turn  coil,  force  on  each  coil  side  =  \",F2,\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For  a  single-turn  coil,  force  on  each  coil  side  =   0.0012   N\n",
+        "\n",
+        "\n",
+        "  (b)For  a  300-turn  coil,  force  on  each  coil  side  =   0.36   N"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 98</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the force exerted on the electron in the field.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Q  =  1.6E-19;#  in  Coulomb\n",
+      "v  =  3E7;#  in  m/s\n",
+      "B  =  18.5E-6;#  in  Tesla\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "F  =  Q*v*B\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Force  exerted  on  the  electron  in  the  field.  =  \",(F/1E-17),\"E-17  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Force  exerted  on  the  electron  in  the  field.  =   8.88 E-17  N"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_1.ipynb
new file mode 100755
index 00000000..1d749dd1
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_1.ipynb
@@ -0,0 +1,273 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 8: Electromagnetism</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 93</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the force acting on the conductor.\n",
+      "#Determine also the value of the force\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.9;#  in  tesla\n",
+      "I  =  20;#  in  Amperes\n",
+      "l  =  0.30;#  in  m\n",
+      "alpha  =  30;#  in  degree\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "F1  =  B*I*l\n",
+      "F2  =  B*I*l*math.sin(alpha*math.pi/180)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Force  when  the  conductor  is  at  right  angles  to  the  field  =  \",F1,\"  N\\n\"\n",
+      "print  \"\\n  (b)Force  when  the  conductor  is  at  30\u00b0  angle  to  the  field  =  \",F2,\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Force  when  the  conductor  is  at  right  angles  to  the  field  =   5.4   N\n",
+        "\n",
+        "\n",
+        "  (b)Force  when  the  conductor  is  at  30\u00c2\u00b0  angle  to  the  field  =   2.7   N"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 94</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current required?\n",
+      "#what is the direction of the force?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "F  =  1.92;#  in  newton\n",
+      "B  =  1.2;#  in  tesla\n",
+      "l  =  0.40;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  F/(B*l)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Current  I  =  \",I,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  I  =   4.0   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 95</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the magnitude of the force exerted on the conductor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "r  =  0.06;#  in  m\n",
+      "I  =  10;#  in  Amperes\n",
+      "l  =  0.35;#  in  m\n",
+      "Phi  =  0.5E-3;#  in  Wb\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  math.pi*r*r\n",
+      "B  =  Phi/A\n",
+      "F  =  B*I*l\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Force  F  =  \",round(F,2),\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Force  F  =   0.15   N"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 95</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the force on each coil side (a) for a single-turn coil,\n",
+      "#(b) for a coil wound with 300 turns.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  1;#  for  a  single-turn  coil\n",
+      "N2  =  300;#  no.  of  turns\n",
+      "b  =  0.024;#  in  m\n",
+      "B  =  0.8;#  in  Tesla\n",
+      "I  =  0.05;#  in  Amperes\n",
+      "l  =  0.030;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "#For  a  single-turn  coil,\n",
+      "F1  =  N1*B*I*l\n",
+      "#for  a  coil  wound  with  300  turns.\n",
+      "F2  =  N2*B*I*l\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)For  a  single-turn  coil,  force  on  each  coil  side  =  \",F1,\"  N\\n\"\n",
+      "print  \"\\n  (b)For  a  300-turn  coil,  force  on  each  coil  side  =  \",F2,\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For  a  single-turn  coil,  force  on  each  coil  side  =   0.0012   N\n",
+        "\n",
+        "\n",
+        "  (b)For  a  300-turn  coil,  force  on  each  coil  side  =   0.36   N"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 98</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the force exerted on the electron in the field.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Q  =  1.6E-19;#  in  Coulomb\n",
+      "v  =  3E7;#  in  m/s\n",
+      "B  =  18.5E-6;#  in  Tesla\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "F  =  Q*v*B\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Force  exerted  on  the  electron  in  the  field.  =  \",(F/1E-17),\"E-17  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Force  exerted  on  the  electron  in  the  field.  =   8.88 E-17  N"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_2.ipynb
new file mode 100755
index 00000000..1d749dd1
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_2.ipynb
@@ -0,0 +1,273 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 8: Electromagnetism</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 93</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the force acting on the conductor.\n",
+      "#Determine also the value of the force\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.9;#  in  tesla\n",
+      "I  =  20;#  in  Amperes\n",
+      "l  =  0.30;#  in  m\n",
+      "alpha  =  30;#  in  degree\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "F1  =  B*I*l\n",
+      "F2  =  B*I*l*math.sin(alpha*math.pi/180)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Force  when  the  conductor  is  at  right  angles  to  the  field  =  \",F1,\"  N\\n\"\n",
+      "print  \"\\n  (b)Force  when  the  conductor  is  at  30\u00b0  angle  to  the  field  =  \",F2,\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Force  when  the  conductor  is  at  right  angles  to  the  field  =   5.4   N\n",
+        "\n",
+        "\n",
+        "  (b)Force  when  the  conductor  is  at  30\u00c2\u00b0  angle  to  the  field  =   2.7   N"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 94</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current required?\n",
+      "#what is the direction of the force?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "F  =  1.92;#  in  newton\n",
+      "B  =  1.2;#  in  tesla\n",
+      "l  =  0.40;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  F/(B*l)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Current  I  =  \",I,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  I  =   4.0   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 95</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the magnitude of the force exerted on the conductor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "r  =  0.06;#  in  m\n",
+      "I  =  10;#  in  Amperes\n",
+      "l  =  0.35;#  in  m\n",
+      "Phi  =  0.5E-3;#  in  Wb\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  math.pi*r*r\n",
+      "B  =  Phi/A\n",
+      "F  =  B*I*l\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Force  F  =  \",round(F,2),\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Force  F  =   0.15   N"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 95</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the force on each coil side (a) for a single-turn coil,\n",
+      "#(b) for a coil wound with 300 turns.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  1;#  for  a  single-turn  coil\n",
+      "N2  =  300;#  no.  of  turns\n",
+      "b  =  0.024;#  in  m\n",
+      "B  =  0.8;#  in  Tesla\n",
+      "I  =  0.05;#  in  Amperes\n",
+      "l  =  0.030;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "#For  a  single-turn  coil,\n",
+      "F1  =  N1*B*I*l\n",
+      "#for  a  coil  wound  with  300  turns.\n",
+      "F2  =  N2*B*I*l\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)For  a  single-turn  coil,  force  on  each  coil  side  =  \",F1,\"  N\\n\"\n",
+      "print  \"\\n  (b)For  a  300-turn  coil,  force  on  each  coil  side  =  \",F2,\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For  a  single-turn  coil,  force  on  each  coil  side  =   0.0012   N\n",
+        "\n",
+        "\n",
+        "  (b)For  a  300-turn  coil,  force  on  each  coil  side  =   0.36   N"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 98</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the force exerted on the electron in the field.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Q  =  1.6E-19;#  in  Coulomb\n",
+      "v  =  3E7;#  in  m/s\n",
+      "B  =  18.5E-6;#  in  Tesla\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "F  =  Q*v*B\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Force  exerted  on  the  electron  in  the  field.  =  \",(F/1E-17),\"E-17  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Force  exerted  on  the  electron  in  the  field.  =   8.88 E-17  N"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_3.ipynb
new file mode 100755
index 00000000..876584cc
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_08-checkpoint_3.ipynb
@@ -0,0 +1,271 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:978781926339475333a3790c0d2afbe43a8f52e30a81f93b137533487aaa64dd"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 8: Electromagnetism</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 93</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.9;#  in  tesla\n",
+      "I  =  20;#  in  Amperes\n",
+      "l  =  0.30;#  in  m\n",
+      "alpha  =  30;#  in  degree\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "F1  =  B*I*l\n",
+      "F2  =  B*I*l*math.sin(alpha*math.pi/180)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Force  when  the  conductor  is  at  right  angles  to  the  field  =  \",F1,\"  N\\n\"\n",
+      "print  \"\\n  (b)Force  when  the  conductor  is  at  30\u00b0  angle  to  the  field  =  \",F2,\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Force  when  the  conductor  is  at  right  angles  to  the  field  =   5.4   N\n",
+        "\n",
+        "\n",
+        "  (b)Force  when  the  conductor  is  at  30\u00c2\u00b0  angle  to  the  field  =   2.7   N"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 94</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "F  =  1.92;#  in  newton\n",
+      "B  =  1.2;#  in  tesla\n",
+      "l  =  0.40;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  F/(B*l)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Current  I  =  \",I,\"  Amperes(A)\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  I  =   4.0   Amperes(A)"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 95</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "r  =  0.06;#  in  m\n",
+      "I  =  10;#  in  Amperes\n",
+      "l  =  0.35;#  in  m\n",
+      "Phi  =  0.5E-3;#  in  Wb\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  math.pi*r*r\n",
+      "B  =  Phi/A\n",
+      "F  =  B*I*l\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)Force  F  =  \",round(F,2),\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Force  F  =   0.15   N"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 95</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  1;#  for  a  single-turn  coil\n",
+      "N2  =  300;#  no.  of  turns\n",
+      "b  =  0.024;#  in  m\n",
+      "B  =  0.8;#  in  Tesla\n",
+      "I  =  0.05;#  in  Amperes\n",
+      "l  =  0.030;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "#For  a  single-turn  coil,\n",
+      "F1  =  N1*B*I*l\n",
+      "#for  a  coil  wound  with  300  turns.\n",
+      "F2  =  N2*B*I*l\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)For  a  single-turn  coil,  force  on  each  coil  side  =  \",F1,\"  N\\n\"\n",
+      "print  \"\\n  (b)For  a  300-turn  coil,  force  on  each  coil  side  =  \",F2,\"  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For  a  single-turn  coil,  force  on  each  coil  side  =   0.0012   N\n",
+        "\n",
+        "\n",
+        "  (b)For  a  300-turn  coil,  force  on  each  coil  side  =   0.36   N"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 98</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Q  =  1.6E-19;#  in  Coulomb\n",
+      "v  =  3E7;#  in  m/s\n",
+      "B  =  18.5E-6;#  in  Tesla\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "F  =  Q*v*B\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Force  exerted  on  the  electron  in  the  field.  =  \",(F/1E-17),\"E-17  N\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Force  exerted  on  the  electron  in  the  field.  =   8.88 E-17  N"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint.ipynb
new file mode 100755
index 00000000..02f3e6f4
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint.ipynb
@@ -0,0 +1,759 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 9: Electromagnetic induction</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 102</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the conductor when \n",
+      "#(a) its ends are open-circuited, \n",
+      "#(b) its ends are connected to a load of 20 ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.3;#  in  m\n",
+      "v  =  4;#  in  m/s\n",
+      "B  =  1.25;#  in  Tesla\n",
+      "R  =  20;#  in  ohms\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  B*l*v\n",
+      "I2  =  E/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)If  the  ends  of  the  conductor  are  open  circuited \"\n",
+      "print   \"no  current  will  flow  even  though  \",E,\"  V  has  been  induced.\\n\"\n",
+      "print  \"\\n  (b)From  Ohms  law,  I  =  \",I2,\"  Ampere\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)If  the  ends  of  the  conductor  are  open  circuited  no  current  will  flow  even  though   1.5   V  has  been  induced.\n",
+        "\n",
+        "\n",
+        "  (b)From  Ohms  law,  I  =   0.075   Ampere"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0.6 T\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.075;#  in  m\n",
+      "E  =  9;#  in  Volts\n",
+      "B  =  0.6;#  in  Tesla\n",
+      "R  =  20;#  in  ohms\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "v  =  E/(B*l)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  velocity  v  =  \",v,\"  m/s\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  velocity  v  =   200.0   m/s"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the magnitude of the induced e.m.f. in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.02;#  in  m\n",
+      "b  =  0.02;#  in  m\n",
+      "v  =  15;#  in  m/s\n",
+      "R  =  20;#  in  ohms\n",
+      "Phi  =  5E-6;#  in  Wb\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "a1  =  90;#  in  degrees\n",
+      "a2  =  60;#  in  degrees\n",
+      "a3  =  30;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  l*b\n",
+      "B  =  Phi/A\n",
+      "E90  =  B*l*v*math.sin(a1*math.pi/180)\n",
+      "E60  =  B*l*v*math.sin(a2*math.pi/180)\n",
+      "E30  =  B*l*v*math.sin(a3*math.pi/180)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  at  angles  90\u00b0,  60\u00b0,  30\u00b0  are  \",(E90/1E-3),\"  V,  \",round((E60/1E-3),2),\"  V, \"\n",
+      "print   \"(E30/1E-3),\"  V  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  at  angles  90\u00c2\u00b0,  60\u00c2\u00b0,  30\u00c2\u00b0  are   3.75   V,   3.25   V,   1.875   V  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the e.m.f. induced between its wing tips\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "s  =  36;#  in  m\n",
+      "v  =  400;#  in  km/h\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "B  =  40E-6;#  in  Tesla\n",
+      "\n",
+      "#calculation:\n",
+      "v0  =  v*5/18\n",
+      "E  =  B*s*v0\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   0.16   V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. induced in a coil\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N  =  200;#  no.  of  turns\n",
+      "dt  =  0.050;#  change  of  time  in  sec\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dPhi  =  0.025;#  change  of  flux  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  -1*N*dPhi/dt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -100.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the average e.m.f. induced.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N  =  150;#  no.  of  turns\n",
+      "dt  =  0.040;#  change  of  time  in  sec\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dPhi  =  800E-6;#  change  of  flux  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "#Since  the  flux  reverses,  the  flux  changes  from  C400  \u03bcWb  to  \u0003400  \u03bcWb,  a  total  change  of  flux  of  800  \u03bcWb\n",
+      "E  =  -1*N*dPhi/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -3.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the e.m.f. induced in a coil\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  12;#  in  Henry\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dIdt  =  4;#  change  of  current  with  change  in  time  in  A/s\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  -1*L*dIdt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -48   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 106</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find inductance of the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  1500;#  in  Volts\n",
+      "dt  =  0.008;#  Change  of  time  in  sec\n",
+      "dI  =  4;#  change  of  current  in  A/s\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  abs(E)*dt/dI\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  L=  \",L,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  L=   3.0   H"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#How much energy is stored in the magnetic field of the inductor?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  8;#  in  Henry\n",
+      "I  =  3;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "W  =  L*I*I/2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Energy  stored,  W  =  \",W,\"  J\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Energy  stored,  W  =   36.0   J"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the coil inductance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  4;#  in  Amperes\n",
+      "N  =  800;#turns\n",
+      "Phi  =  0.005;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  N*Phi/I\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  L  =  \",L,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  L  =   1.0   H"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the inductance of the coil, \n",
+      "#(b) the energy stored in the magnetic field, and \n",
+      "#(c) the average e.m.f. induced if the current falls to zero in 150 ms.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1  =  3;#  in  Amperes\n",
+      "I2  =  0;#  in  Amperes\n",
+      "dt  =  0.150;#  in  secs\n",
+      "N  =  1500;#turns\n",
+      "Phi  =  0.025;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  N*Phi/I1\n",
+      "W  =  L*I1*I1/2\n",
+      "dI  =  I1  -  I2\n",
+      "E  =  -1*L*dI/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductance  L  =  \",L,\"  H\\n\"\n",
+      "print  \"\\n  (b)energy  stored  W  =  \",W,\"  J\\n\"\n",
+      "print  \"\\n  (c)e.m.f.  induced  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductance  L  =   12.5   H\n",
+        "\n",
+        "\n",
+        "  (b)energy  stored  W  =   56.25   J\n",
+        "\n",
+        "\n",
+        "  (c)e.m.f.  induced  =   -250.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 108</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the flux linking the coil and the e.m.f. induced in the coil when the current collapses to zero in 20 ms\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1  =  2;#  in  Amperes\n",
+      "I2  =  0;#  in  Amperes\n",
+      "dt  =  0.020;#  in  secs\n",
+      "N  =  750;#turns\n",
+      "L  =  3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      "Phi  =  L*I1/N\n",
+      "dI  =  I1  -  I2\n",
+      "E  =  -1*L*dI/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Flux  =  \",Phi,\"  Wb\\n\"\n",
+      "print  \"\\n  (b)e.m.f.  induced  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Flux  =   0.008   Wb\n",
+        "\n",
+        "\n",
+        "  (b)e.m.f.  induced  =   -300.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 108</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the mutual inductance between two coils\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "dI1dt  =  200;#  change  of  current  with  change  in  time  in  A/s\n",
+      "N  =  2;#  no.  of  coils\n",
+      "E2  =  1.5;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "M  =  abs(E2)/dI1dt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  =  \",  M,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  =   0.0075   H"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 109</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the steady rate of change of current in one coil to induce an e.m.f. of 0.72 V in the other.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "M  =  0.018;#  in  Henry\n",
+      "N  =  2;#  no.  of  coils\n",
+      "E2  =  0.72;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "dI1dt  =  abs(E2)/M\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  rate  of  change  of  current  dI1/dt  =  \",  dI1dt,\"  A/s\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  rate  of  change  of  current  dI1/dt  =   40.0   A/s"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 109</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the average induced e.m.f. in the second coil, \n",
+      "#(b) the change of flux linked with the second coil if it is wound with 500 turns.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "M  =  0.2;#  in  Henry\n",
+      "I1  =  10;#  in  Amperes\n",
+      "I2  =  4;#  in  Amperes\n",
+      "dt  =  0.010;#  in  secs\n",
+      "N  =  500;#  turns\n",
+      "\n",
+      "#calculation:\n",
+      "dI1dt  =  (I1  -I2)/dt  \n",
+      "E2  =  -1*dI1dt*M\n",
+      "dPhi  =  abs(E2)*dt/N\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Induced  e.m.f.  E2  =  \",  E2,\"  V\\n\"\n",
+      "print  \"\\n  (b)change  of  flux  =  \",  dPhi,\"  Wb\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Induced  e.m.f.  E2  =   -120.0   V\n",
+        "\n",
+        "\n",
+        "  (b)change  of  flux  =   0.0024   Wb"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_1.ipynb
new file mode 100755
index 00000000..02f3e6f4
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_1.ipynb
@@ -0,0 +1,759 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 9: Electromagnetic induction</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 102</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the conductor when \n",
+      "#(a) its ends are open-circuited, \n",
+      "#(b) its ends are connected to a load of 20 ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.3;#  in  m\n",
+      "v  =  4;#  in  m/s\n",
+      "B  =  1.25;#  in  Tesla\n",
+      "R  =  20;#  in  ohms\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  B*l*v\n",
+      "I2  =  E/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)If  the  ends  of  the  conductor  are  open  circuited \"\n",
+      "print   \"no  current  will  flow  even  though  \",E,\"  V  has  been  induced.\\n\"\n",
+      "print  \"\\n  (b)From  Ohms  law,  I  =  \",I2,\"  Ampere\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)If  the  ends  of  the  conductor  are  open  circuited  no  current  will  flow  even  though   1.5   V  has  been  induced.\n",
+        "\n",
+        "\n",
+        "  (b)From  Ohms  law,  I  =   0.075   Ampere"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0.6 T\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.075;#  in  m\n",
+      "E  =  9;#  in  Volts\n",
+      "B  =  0.6;#  in  Tesla\n",
+      "R  =  20;#  in  ohms\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "v  =  E/(B*l)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  velocity  v  =  \",v,\"  m/s\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  velocity  v  =   200.0   m/s"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the magnitude of the induced e.m.f. in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.02;#  in  m\n",
+      "b  =  0.02;#  in  m\n",
+      "v  =  15;#  in  m/s\n",
+      "R  =  20;#  in  ohms\n",
+      "Phi  =  5E-6;#  in  Wb\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "a1  =  90;#  in  degrees\n",
+      "a2  =  60;#  in  degrees\n",
+      "a3  =  30;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  l*b\n",
+      "B  =  Phi/A\n",
+      "E90  =  B*l*v*math.sin(a1*math.pi/180)\n",
+      "E60  =  B*l*v*math.sin(a2*math.pi/180)\n",
+      "E30  =  B*l*v*math.sin(a3*math.pi/180)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  at  angles  90\u00b0,  60\u00b0,  30\u00b0  are  \",(E90/1E-3),\"  V,  \",round((E60/1E-3),2),\"  V, \"\n",
+      "print   \"(E30/1E-3),\"  V  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  at  angles  90\u00c2\u00b0,  60\u00c2\u00b0,  30\u00c2\u00b0  are   3.75   V,   3.25   V,   1.875   V  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the e.m.f. induced between its wing tips\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "s  =  36;#  in  m\n",
+      "v  =  400;#  in  km/h\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "B  =  40E-6;#  in  Tesla\n",
+      "\n",
+      "#calculation:\n",
+      "v0  =  v*5/18\n",
+      "E  =  B*s*v0\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   0.16   V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. induced in a coil\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N  =  200;#  no.  of  turns\n",
+      "dt  =  0.050;#  change  of  time  in  sec\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dPhi  =  0.025;#  change  of  flux  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  -1*N*dPhi/dt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -100.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the average e.m.f. induced.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N  =  150;#  no.  of  turns\n",
+      "dt  =  0.040;#  change  of  time  in  sec\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dPhi  =  800E-6;#  change  of  flux  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "#Since  the  flux  reverses,  the  flux  changes  from  C400  \u03bcWb  to  \u0003400  \u03bcWb,  a  total  change  of  flux  of  800  \u03bcWb\n",
+      "E  =  -1*N*dPhi/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -3.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the e.m.f. induced in a coil\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  12;#  in  Henry\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dIdt  =  4;#  change  of  current  with  change  in  time  in  A/s\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  -1*L*dIdt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -48   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 106</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find inductance of the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  1500;#  in  Volts\n",
+      "dt  =  0.008;#  Change  of  time  in  sec\n",
+      "dI  =  4;#  change  of  current  in  A/s\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  abs(E)*dt/dI\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  L=  \",L,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  L=   3.0   H"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#How much energy is stored in the magnetic field of the inductor?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  8;#  in  Henry\n",
+      "I  =  3;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "W  =  L*I*I/2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Energy  stored,  W  =  \",W,\"  J\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Energy  stored,  W  =   36.0   J"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the coil inductance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  4;#  in  Amperes\n",
+      "N  =  800;#turns\n",
+      "Phi  =  0.005;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  N*Phi/I\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  L  =  \",L,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  L  =   1.0   H"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the inductance of the coil, \n",
+      "#(b) the energy stored in the magnetic field, and \n",
+      "#(c) the average e.m.f. induced if the current falls to zero in 150 ms.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1  =  3;#  in  Amperes\n",
+      "I2  =  0;#  in  Amperes\n",
+      "dt  =  0.150;#  in  secs\n",
+      "N  =  1500;#turns\n",
+      "Phi  =  0.025;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  N*Phi/I1\n",
+      "W  =  L*I1*I1/2\n",
+      "dI  =  I1  -  I2\n",
+      "E  =  -1*L*dI/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductance  L  =  \",L,\"  H\\n\"\n",
+      "print  \"\\n  (b)energy  stored  W  =  \",W,\"  J\\n\"\n",
+      "print  \"\\n  (c)e.m.f.  induced  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductance  L  =   12.5   H\n",
+        "\n",
+        "\n",
+        "  (b)energy  stored  W  =   56.25   J\n",
+        "\n",
+        "\n",
+        "  (c)e.m.f.  induced  =   -250.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 108</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the flux linking the coil and the e.m.f. induced in the coil when the current collapses to zero in 20 ms\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1  =  2;#  in  Amperes\n",
+      "I2  =  0;#  in  Amperes\n",
+      "dt  =  0.020;#  in  secs\n",
+      "N  =  750;#turns\n",
+      "L  =  3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      "Phi  =  L*I1/N\n",
+      "dI  =  I1  -  I2\n",
+      "E  =  -1*L*dI/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Flux  =  \",Phi,\"  Wb\\n\"\n",
+      "print  \"\\n  (b)e.m.f.  induced  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Flux  =   0.008   Wb\n",
+        "\n",
+        "\n",
+        "  (b)e.m.f.  induced  =   -300.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 108</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the mutual inductance between two coils\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "dI1dt  =  200;#  change  of  current  with  change  in  time  in  A/s\n",
+      "N  =  2;#  no.  of  coils\n",
+      "E2  =  1.5;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "M  =  abs(E2)/dI1dt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  =  \",  M,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  =   0.0075   H"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 109</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the steady rate of change of current in one coil to induce an e.m.f. of 0.72 V in the other.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "M  =  0.018;#  in  Henry\n",
+      "N  =  2;#  no.  of  coils\n",
+      "E2  =  0.72;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "dI1dt  =  abs(E2)/M\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  rate  of  change  of  current  dI1/dt  =  \",  dI1dt,\"  A/s\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  rate  of  change  of  current  dI1/dt  =   40.0   A/s"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 109</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the average induced e.m.f. in the second coil, \n",
+      "#(b) the change of flux linked with the second coil if it is wound with 500 turns.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "M  =  0.2;#  in  Henry\n",
+      "I1  =  10;#  in  Amperes\n",
+      "I2  =  4;#  in  Amperes\n",
+      "dt  =  0.010;#  in  secs\n",
+      "N  =  500;#  turns\n",
+      "\n",
+      "#calculation:\n",
+      "dI1dt  =  (I1  -I2)/dt  \n",
+      "E2  =  -1*dI1dt*M\n",
+      "dPhi  =  abs(E2)*dt/N\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Induced  e.m.f.  E2  =  \",  E2,\"  V\\n\"\n",
+      "print  \"\\n  (b)change  of  flux  =  \",  dPhi,\"  Wb\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Induced  e.m.f.  E2  =   -120.0   V\n",
+        "\n",
+        "\n",
+        "  (b)change  of  flux  =   0.0024   Wb"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_2.ipynb
new file mode 100755
index 00000000..02f3e6f4
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_2.ipynb
@@ -0,0 +1,759 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 9: Electromagnetic induction</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 102</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the conductor when \n",
+      "#(a) its ends are open-circuited, \n",
+      "#(b) its ends are connected to a load of 20 ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.3;#  in  m\n",
+      "v  =  4;#  in  m/s\n",
+      "B  =  1.25;#  in  Tesla\n",
+      "R  =  20;#  in  ohms\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  B*l*v\n",
+      "I2  =  E/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)If  the  ends  of  the  conductor  are  open  circuited \"\n",
+      "print   \"no  current  will  flow  even  though  \",E,\"  V  has  been  induced.\\n\"\n",
+      "print  \"\\n  (b)From  Ohms  law,  I  =  \",I2,\"  Ampere\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)If  the  ends  of  the  conductor  are  open  circuited  no  current  will  flow  even  though   1.5   V  has  been  induced.\n",
+        "\n",
+        "\n",
+        "  (b)From  Ohms  law,  I  =   0.075   Ampere"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0.6 T\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.075;#  in  m\n",
+      "E  =  9;#  in  Volts\n",
+      "B  =  0.6;#  in  Tesla\n",
+      "R  =  20;#  in  ohms\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "v  =  E/(B*l)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  velocity  v  =  \",v,\"  m/s\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  velocity  v  =   200.0   m/s"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the magnitude of the induced e.m.f. in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.02;#  in  m\n",
+      "b  =  0.02;#  in  m\n",
+      "v  =  15;#  in  m/s\n",
+      "R  =  20;#  in  ohms\n",
+      "Phi  =  5E-6;#  in  Wb\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "a1  =  90;#  in  degrees\n",
+      "a2  =  60;#  in  degrees\n",
+      "a3  =  30;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  l*b\n",
+      "B  =  Phi/A\n",
+      "E90  =  B*l*v*math.sin(a1*math.pi/180)\n",
+      "E60  =  B*l*v*math.sin(a2*math.pi/180)\n",
+      "E30  =  B*l*v*math.sin(a3*math.pi/180)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  at  angles  90\u00b0,  60\u00b0,  30\u00b0  are  \",(E90/1E-3),\"  V,  \",round((E60/1E-3),2),\"  V, \"\n",
+      "print   \"(E30/1E-3),\"  V  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  at  angles  90\u00c2\u00b0,  60\u00c2\u00b0,  30\u00c2\u00b0  are   3.75   V,   3.25   V,   1.875   V  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the e.m.f. induced between its wing tips\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "s  =  36;#  in  m\n",
+      "v  =  400;#  in  km/h\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "B  =  40E-6;#  in  Tesla\n",
+      "\n",
+      "#calculation:\n",
+      "v0  =  v*5/18\n",
+      "E  =  B*s*v0\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   0.16   V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. induced in a coil\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N  =  200;#  no.  of  turns\n",
+      "dt  =  0.050;#  change  of  time  in  sec\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dPhi  =  0.025;#  change  of  flux  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  -1*N*dPhi/dt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -100.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the average e.m.f. induced.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N  =  150;#  no.  of  turns\n",
+      "dt  =  0.040;#  change  of  time  in  sec\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dPhi  =  800E-6;#  change  of  flux  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "#Since  the  flux  reverses,  the  flux  changes  from  C400  \u03bcWb  to  \u0003400  \u03bcWb,  a  total  change  of  flux  of  800  \u03bcWb\n",
+      "E  =  -1*N*dPhi/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -3.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the e.m.f. induced in a coil\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  12;#  in  Henry\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dIdt  =  4;#  change  of  current  with  change  in  time  in  A/s\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  -1*L*dIdt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -48   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 106</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find inductance of the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  1500;#  in  Volts\n",
+      "dt  =  0.008;#  Change  of  time  in  sec\n",
+      "dI  =  4;#  change  of  current  in  A/s\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  abs(E)*dt/dI\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  L=  \",L,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  L=   3.0   H"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#How much energy is stored in the magnetic field of the inductor?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  8;#  in  Henry\n",
+      "I  =  3;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "W  =  L*I*I/2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Energy  stored,  W  =  \",W,\"  J\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Energy  stored,  W  =   36.0   J"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the coil inductance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  4;#  in  Amperes\n",
+      "N  =  800;#turns\n",
+      "Phi  =  0.005;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  N*Phi/I\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  L  =  \",L,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  L  =   1.0   H"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the inductance of the coil, \n",
+      "#(b) the energy stored in the magnetic field, and \n",
+      "#(c) the average e.m.f. induced if the current falls to zero in 150 ms.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1  =  3;#  in  Amperes\n",
+      "I2  =  0;#  in  Amperes\n",
+      "dt  =  0.150;#  in  secs\n",
+      "N  =  1500;#turns\n",
+      "Phi  =  0.025;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  N*Phi/I1\n",
+      "W  =  L*I1*I1/2\n",
+      "dI  =  I1  -  I2\n",
+      "E  =  -1*L*dI/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductance  L  =  \",L,\"  H\\n\"\n",
+      "print  \"\\n  (b)energy  stored  W  =  \",W,\"  J\\n\"\n",
+      "print  \"\\n  (c)e.m.f.  induced  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductance  L  =   12.5   H\n",
+        "\n",
+        "\n",
+        "  (b)energy  stored  W  =   56.25   J\n",
+        "\n",
+        "\n",
+        "  (c)e.m.f.  induced  =   -250.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 108</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the flux linking the coil and the e.m.f. induced in the coil when the current collapses to zero in 20 ms\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1  =  2;#  in  Amperes\n",
+      "I2  =  0;#  in  Amperes\n",
+      "dt  =  0.020;#  in  secs\n",
+      "N  =  750;#turns\n",
+      "L  =  3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      "Phi  =  L*I1/N\n",
+      "dI  =  I1  -  I2\n",
+      "E  =  -1*L*dI/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Flux  =  \",Phi,\"  Wb\\n\"\n",
+      "print  \"\\n  (b)e.m.f.  induced  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Flux  =   0.008   Wb\n",
+        "\n",
+        "\n",
+        "  (b)e.m.f.  induced  =   -300.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 108</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the mutual inductance between two coils\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "dI1dt  =  200;#  change  of  current  with  change  in  time  in  A/s\n",
+      "N  =  2;#  no.  of  coils\n",
+      "E2  =  1.5;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "M  =  abs(E2)/dI1dt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  =  \",  M,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  =   0.0075   H"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 109</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the steady rate of change of current in one coil to induce an e.m.f. of 0.72 V in the other.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "M  =  0.018;#  in  Henry\n",
+      "N  =  2;#  no.  of  coils\n",
+      "E2  =  0.72;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "dI1dt  =  abs(E2)/M\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  rate  of  change  of  current  dI1/dt  =  \",  dI1dt,\"  A/s\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  rate  of  change  of  current  dI1/dt  =   40.0   A/s"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 109</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the average induced e.m.f. in the second coil, \n",
+      "#(b) the change of flux linked with the second coil if it is wound with 500 turns.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "M  =  0.2;#  in  Henry\n",
+      "I1  =  10;#  in  Amperes\n",
+      "I2  =  4;#  in  Amperes\n",
+      "dt  =  0.010;#  in  secs\n",
+      "N  =  500;#  turns\n",
+      "\n",
+      "#calculation:\n",
+      "dI1dt  =  (I1  -I2)/dt  \n",
+      "E2  =  -1*dI1dt*M\n",
+      "dPhi  =  abs(E2)*dt/N\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Induced  e.m.f.  E2  =  \",  E2,\"  V\\n\"\n",
+      "print  \"\\n  (b)change  of  flux  =  \",  dPhi,\"  Wb\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Induced  e.m.f.  E2  =   -120.0   V\n",
+        "\n",
+        "\n",
+        "  (b)change  of  flux  =   0.0024   Wb"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_3.ipynb
new file mode 100755
index 00000000..0248550c
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_09-checkpoint_3.ipynb
@@ -0,0 +1,755 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:68dcf633943ab2778e92201ff3277e4978506e4f80a2958d6ecfcaa550aa58ae"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 9: Electromagnetic induction</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 102</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.3;#  in  m\n",
+      "v  =  4;#  in  m/s\n",
+      "B  =  1.25;#  in  Tesla\n",
+      "R  =  20;#  in  ohms\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  B*l*v\n",
+      "I2  =  E/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  (a)If  the  ends  of  the  conductor  are  open  circuited \"\n",
+      "print   \"no  current  will  flow  even  though  \",E,\"  V  has  been  induced.\\n\"\n",
+      "print  \"\\n  (b)From  Ohms  law,  I  =  \",I2,\"  Ampere\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)If  the  ends  of  the  conductor  are  open  circuited  no  current  will  flow  even  though   1.5   V  has  been  induced.\n",
+        "\n",
+        "\n",
+        "  (b)From  Ohms  law,  I  =   0.075   Ampere"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.075;#  in  m\n",
+      "E  =  9;#  in  Volts\n",
+      "B  =  0.6;#  in  Tesla\n",
+      "R  =  20;#  in  ohms\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "\n",
+      "#calculation:\n",
+      "v  =  E/(B*l)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  velocity  v  =  \",v,\"  m/s\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  velocity  v  =   200.0   m/s"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#initializing  the  variables:\n",
+      "l  =  0.02;#  in  m\n",
+      "b  =  0.02;#  in  m\n",
+      "v  =  15;#  in  m/s\n",
+      "R  =  20;#  in  ohms\n",
+      "Phi  =  5E-6;#  in  Wb\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "a1  =  90;#  in  degrees\n",
+      "a2  =  60;#  in  degrees\n",
+      "a3  =  30;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  l*b\n",
+      "B  =  Phi/A\n",
+      "E90  =  B*l*v*math.sin(a1*math.pi/180)\n",
+      "E60  =  B*l*v*math.sin(a2*math.pi/180)\n",
+      "E30  =  B*l*v*math.sin(a3*math.pi/180)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  at  angles  90\u00b0,  60\u00b0,  30\u00b0  are  \",(E90/1E-3),\"  V,  \",round((E60/1E-3),2),\"  V, \"\n",
+      "print   \"(E30/1E-3),\"  V  respectively\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  at  angles  90\u00c2\u00b0,  60\u00c2\u00b0,  30\u00c2\u00b0  are   3.75   V,   3.25   V,   1.875   V  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 103</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "s  =  36;#  in  m\n",
+      "v  =  400;#  in  km/h\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "B  =  40E-6;#  in  Tesla\n",
+      "\n",
+      "#calculation:\n",
+      "v0  =  v*5/18\n",
+      "E  =  B*s*v0\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   0.16   V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N  =  200;#  no.  of  turns\n",
+      "dt  =  0.050;#  change  of  time  in  sec\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dPhi  =  0.025;#  change  of  flux  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  -1*N*dPhi/dt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -100.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N  =  150;#  no.  of  turns\n",
+      "dt  =  0.040;#  change  of  time  in  sec\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dPhi  =  800E-6;#  change  of  flux  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "#Since  the  flux  reverses,  the  flux  changes  from  C400  \u03bcWb  to  \u0003400  \u03bcWb,  a  total  change  of  flux  of  800  \u03bcWb\n",
+      "E  =  -1*N*dPhi/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -3.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 105</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  12;#  in  Henry\n",
+      "u0  =  4*math.pi*1E-7;\n",
+      "dIdt  =  4;#  change  of  current  with  change  in  time  in  A/s\n",
+      "\n",
+      "#calculation:\n",
+      "E  =  -1*L*dIdt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\nResult\\n\\n\"\n",
+      "print  \"\\n  Induced  e.m.f.  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "Result\n",
+        "\n",
+        "\n",
+        "\n",
+        "  Induced  e.m.f.  =   -48   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 106</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  1500;#  in  Volts\n",
+      "dt  =  0.008;#  Change  of  time  in  sec\n",
+      "dI  =  4;#  change  of  current  in  A/s\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  abs(E)*dt/dI\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  L=  \",L,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  L=   3.0   H"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  8;#  in  Henry\n",
+      "I  =  3;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "W  =  L*I*I/2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Energy  stored,  W  =  \",W,\"  J\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Energy  stored,  W  =   36.0   J"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  4;#  in  Amperes\n",
+      "N  =  800;#turns\n",
+      "Phi  =  0.005;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  N*Phi/I\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  L  =  \",L,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  L  =   1.0   H"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 107</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1  =  3;#  in  Amperes\n",
+      "I2  =  0;#  in  Amperes\n",
+      "dt  =  0.150;#  in  secs\n",
+      "N  =  1500;#turns\n",
+      "Phi  =  0.025;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  N*Phi/I1\n",
+      "W  =  L*I1*I1/2\n",
+      "dI  =  I1  -  I2\n",
+      "E  =  -1*L*dI/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductance  L  =  \",L,\"  H\\n\"\n",
+      "print  \"\\n  (b)energy  stored  W  =  \",W,\"  J\\n\"\n",
+      "print  \"\\n  (c)e.m.f.  induced  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductance  L  =   12.5   H\n",
+        "\n",
+        "\n",
+        "  (b)energy  stored  W  =   56.25   J\n",
+        "\n",
+        "\n",
+        "  (c)e.m.f.  induced  =   -250.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 108</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1  =  2;#  in  Amperes\n",
+      "I2  =  0;#  in  Amperes\n",
+      "dt  =  0.020;#  in  secs\n",
+      "N  =  750;#turns\n",
+      "L  =  3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      "Phi  =  L*I1/N\n",
+      "dI  =  I1  -  I2\n",
+      "E  =  -1*L*dI/dt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Flux  =  \",Phi,\"  Wb\\n\"\n",
+      "print  \"\\n  (b)e.m.f.  induced  =  \",E,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Flux  =   0.008   Wb\n",
+        "\n",
+        "\n",
+        "  (b)e.m.f.  induced  =   -300.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 108</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "dI1dt  =  200;#  change  of  current  with  change  in  time  in  A/s\n",
+      "N  =  2;#  no.  of  coils\n",
+      "E2  =  1.5;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "M  =  abs(E2)/dI1dt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  =  \",  M,\"  H\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  =   0.0075   H"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 109</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "M  =  0.018;#  in  Henry\n",
+      "N  =  2;#  no.  of  coils\n",
+      "E2  =  0.72;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "dI1dt  =  abs(E2)/M\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  rate  of  change  of  current  dI1/dt  =  \",  dI1dt,\"  A/s\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  rate  of  change  of  current  dI1/dt  =   40.0   A/s"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 109</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "M  =  0.2;#  in  Henry\n",
+      "I1  =  10;#  in  Amperes\n",
+      "I2  =  4;#  in  Amperes\n",
+      "dt  =  0.010;#  in  secs\n",
+      "N  =  500;#  turns\n",
+      "\n",
+      "#calculation:\n",
+      "dI1dt  =  (I1  -I2)/dt  \n",
+      "E2  =  -1*dI1dt*M\n",
+      "dPhi  =  abs(E2)*dt/N\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Induced  e.m.f.  E2  =  \",  E2,\"  V\\n\"\n",
+      "print  \"\\n  (b)change  of  flux  =  \",  dPhi,\"  Wb\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Induced  e.m.f.  E2  =   -120.0   V\n",
+        "\n",
+        "\n",
+        "  (b)change  of  flux  =   0.0024   Wb"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint.ipynb
new file mode 100755
index 00000000..c63a225f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint.ipynb
@@ -0,0 +1,1066 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 10: Electrical measuring instruments and measurements</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 116</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the shunt to be connected in parallel with the meter\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia  =  0.040;#  in  Amperes\n",
+      "I  =  50;#  in  Amperes\n",
+      "ra  =  25;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Is  =  I  -  Ia\n",
+      "V  =  Ia*ra\n",
+      "Rs  =  V/Is\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    value  of  the  shunt  to  be  connected  in  parallel  =  \",  round((Rs/1E-3),2),\"  mohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    value  of  the  shunt  to  be  connected  in  parallel  =   20.02   mohms"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 116</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the multiplier to be connected in series with the instrument\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  volts\n",
+      "I  =  0.008;#  in  Amperes\n",
+      "ra  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rm  =  (V/I)  -  ra\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  value  of  the  multiplier  to  be  connected  in  series  =  \",  (Rm/1E3),\"  kohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  value  of  the  multiplier  to  be  connected  in  series  =   12.49   kohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 119</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated by the voltmeter and by resistor R\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fsd  =  200;#  in  volts\n",
+      "R1  =  250;#  in  ohms\n",
+      "R2  =  2E6;#  in  ohms\n",
+      "sensitivity  =  10000;#  in  ohms/V\n",
+      "V = 100; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "Rv  =  sensitivity*fsd\n",
+      "Iv  =  V/Rv\n",
+      "Pv  =  V*Iv\n",
+      "I1  =  V/R1\n",
+      "P1  =  V*I1\n",
+      "I2  =  V/R2\n",
+      "P2  =  V*I2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  power  dissipated  by  the  voltmeter  =  \",  (Pv/1E-3),\"  mW\\n\"\n",
+      "print  \"\\n  (b)the  power  dissipated  by  resistor  250  ohm  =  \",  P1,\"  W\\n\"\n",
+      "print  \"\\n  (c)the  power  dissipated  by  resistor  2  Mohm  =  \",  (P2/1E-3),\"  mW\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  power  dissipated  by  the  voltmeter  =   5.0   mW\n",
+        "\n",
+        "\n",
+        "  (b)the  power  dissipated  by  resistor  250  ohm  =   40.0   W\n",
+        "\n",
+        "\n",
+        "  (c)the  power  dissipated  by  resistor  2  Mohm  =   5.0   mW"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 119</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the ammeter reading expected (neglecting its resistance),\n",
+      "#(b) the actual current in the circuit, \n",
+      "#(c) the power dissipated in the ammeter, and\n",
+      "#(d) the power dissipated in the load\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "fsd  =  0.1;#  in  Amperes\n",
+      "ra  =  50;#  in  ohms\n",
+      "R  =  500;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Ie  =  V/R\n",
+      "Ia  =  V/(R  +  ra)\n",
+      "Pa  =  Ia*Ia*ra\n",
+      "PR  =  Ia*Ia*R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)expected  ammeter  reading  =  \",  (Ie/1E-3),\"  mA\\n\"\n",
+      "print  \"\\n  (b)Actual  ammeter  reading  =  \",round((Ia/1E-3),2),\"  mA\\n\"\n",
+      "print  \"\\n  (c)Power  dissipated  in  the  ammeter  =  \",round((Pa/1E-3),2),\"  mW\\n\"\n",
+      "print  \"\\n  (d)Power  dissipated  in  the  load  resistor  =  \",  round((PR/1E-3),2),\"  mW\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)expected  ammeter  reading  =   20.0   mA\n",
+        "\n",
+        "\n",
+        "  (b)Actual  ammeter  reading  =   18.18   mA\n",
+        "\n",
+        "\n",
+        "  (c)Power  dissipated  in  the  ammeter  =   16.53   mW\n",
+        "\n",
+        "\n",
+        "  (d)Power  dissipated  in  the  load  resistor  =   165.29   mW"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 120</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of voltage V1 with the voltmeter not connected, and \n",
+      "#(b) the voltage indicated by the voltmeter when connected between A and B.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fsd  =  100;#  in  volts\n",
+      "R1  =  40E3;#  in  ohms\n",
+      "R2  =  60E3;#  in  ohms\n",
+      "sensitivity  =  1600;#  in  ohms/V\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  (R1/(R1  +  R2))*fsd\n",
+      "Rv  =  fsd*sensitivity\n",
+      "Rep  =  R1*Rv/(R1  +  Rv)\n",
+      "V1n  =  (Rep/(Rep  +  R2))*fsd\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  voltage  V1  with  the  voltmeter6  not  connected  =  \",  V1,\"  V\\n\"\n",
+      "print  \"\\n  (b)the  voltage  indicated  by  the  voltmeter  when  connected  between  A  and  B  =  \",round(V1n,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  voltage  V1  with  the  voltmeter6  not  connected  =   40.0   V\n",
+        "\n",
+        "\n",
+        "  (b)the  voltage  indicated  by  the  voltmeter  when  connected  between  A  and  B  =   34.78   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 120</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power dissipated in the load.\n",
+      "#(b) A wattmeter, whose current coil has a resistance of 0.01 ohm\n",
+      "# Determine the wattmeter reading.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  20;#  in  amperes\n",
+      "R  =  2;#  in  ohms\n",
+      "Rw  =  0.01;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "PR  =  I*I*R\n",
+      "Rt  =  R  +  Rw\n",
+      "Pw  =  I*I*Rt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  power  dissipated  in  the  load  =  \",  PR,\"  W\\n\"\n",
+      "print  \"\\n  (b)the  wattmeter  reading.  =  \",Pw,\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  power  dissipated  in  the  load  =   800   W\n",
+        "\n",
+        "\n",
+        "  (b)the  wattmeter  reading.  =   804.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 122</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the periodic time, (b) the frequency and (c) the peak-to-peak voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  100E-6;#  in  s/cm\n",
+      "Vc  =  20;#  in  V/cm\n",
+      "w  =  5.2;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 3.6; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  periodic  time,  T  =  \",  (T/1E-3),\"  msec\\n\"\n",
+      "print  \"\\n  (b)Frequency,  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  periodic  time,  T  =   0.52   msec\n",
+        "\n",
+        "\n",
+        "  (b)Frequency,  f  =   1923.08   Hz\n",
+        "\n",
+        "\n",
+        "  (c)the  peak-to-peak  voltage  =   72.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the periodic time, (b) the frequency, (c) the magnitude of the pulse voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  50E-3;#  in  s/cm\n",
+      "Vc  =  0.2;#  in  V/cm\n",
+      "w  =  3.5;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 3.4; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "\n",
+      "#Results  \n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  periodic  time,  T  =  \",  (T/1E-3),\"  msec\\n\"\n",
+      "print  \"\\n  (b)Frequency,  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  periodic  time,  T  =   175.0   msec\n",
+        "\n",
+        "\n",
+        "  (b)Frequency,  f  =   5.71   Hz\n",
+        "\n",
+        "\n",
+        "  (c)the  peak-to-peak  voltage  =   0.68   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find, for the waveform, (a) the frequency, \n",
+      "#(b) the peak-to-peak voltage, (c) the amplitude, (d) the r.m.s. value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  500E-6;#  in  s/cm\n",
+      "Vc  =  5;#  in  V/cm\n",
+      "w  =  4;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 5; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "Amp  =  ptpv/2\n",
+      "Vrms  =  Amp/(2**0.5)\n",
+      "\n",
+      "#Results  \n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency,  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\"\n",
+      "print  \"\\n  (c)Amplitude  =  \",Amp,\"  V\\n\"\n",
+      "print  \"\\n  (d)r.m.s  voltage  =  \",round(Vrms,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency,  f  =   500.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)the  peak-to-peak  voltage  =   25   V\n",
+        "\n",
+        "\n",
+        "  (c)Amplitude  =   12.5   V\n",
+        "\n",
+        "\n",
+        "  (d)r.m.s  voltage  =   8.84   V"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) their frequency, (b) their r.m.s. values, (c) their phase difference.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#initializing  the  variables:\n",
+      "tc  =  100E-6;#  in  s/cm\n",
+      "Vc  =  2;#  in  V/cm\n",
+      "w  =  5;#  in  cm  (  width  of  one  complete  cycle  for  both  waveform  )\n",
+      "h1 = 2; # in cm ( peak-to-peak height of the display )\n",
+      "h2 = 2.5; # in cm ( peak-to-peak height of the display\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv1  =  h1*Vc\n",
+      "Vrms1  =  ptpv1/(2**0.5)\n",
+      "ptpv2  =  h2*Vc\n",
+      "Vrms2  =  ptpv2/(2**0.5)\n",
+      "phi  =  0.5*360/w\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency,  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b1)r.m.s  voltage  of  1st  waveform  =  \",round(Vrms1,2),\"  V\\n\"\n",
+      "print  \"\\n  (b2)r.m.s  voltage  of  2nd  waveform  =  \",round(Vrms2,2),\"  V\\n\"\n",
+      "print  \"\\n  (c)Phase  difference  =  \",phi,\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency,  f  =   2000.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b1)r.m.s  voltage  of  1st  waveform  =   2.83   V\n",
+        "\n",
+        "\n",
+        "  (b2)r.m.s  voltage  of  2nd  waveform  =   3.54   V\n",
+        "\n",
+        "\n",
+        "  (c)Phase  difference  =   36.0 deg"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 127</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the decibel power ratio in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rP1  =  3;#  ratio  of  two  powers\n",
+      "rP2  =  20;#  ratio  of  two  powers\n",
+      "rP3  =  400;#  ratio  of  two  powers\n",
+      "rP4  =  1/20;#  ratio  of  two  powers\n",
+      "\n",
+      "#calculation:\n",
+      "X1  =  10*(1/2.303)*math.log(3)\n",
+      "X2  =  10*(1/2.303)*math.log(20)\n",
+      "X3  =  10*(1/2.303)*math.log(400)\n",
+      "X4  =  10*(1/2.303)*math.log(1/20)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)decibel  power  ratio  for  power  ratio  3  =  \",round(X1,2),\"  dB\\n\"\n",
+      "print  \"\\n  (b)decibel  power  ratio  for  power  ratio  20  =  \",round(X2,2),\"  dB\\n\"\n",
+      "print  \"\\n  (c)decibel  power  ratio  for  power  ratio  400  =  \",round(X3,2),\"  dB\\n\"\n",
+      "print  \"\\n  (d)decibel  power  ratio  for  power  ratio  1/20  =  \",round(X4,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)decibel  power  ratio  for  power  ratio  3  =   4.77   dB\n",
+        "\n",
+        "\n",
+        "  (b)decibel  power  ratio  for  power  ratio  20  =   13.01   dB\n",
+        "\n",
+        "\n",
+        "  (c)decibel  power  ratio  for  power  ratio  400  =   26.02   dB\n",
+        "\n",
+        "\n",
+        "  (d)decibel  power  ratio  for  power  ratio  1/20  =   -13.01   dB"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 127</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the decibel current ratio assuming the input and load resistances of the system are equal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I2  =  0.020;#  in  ampere\n",
+      "I1  =  0.005;#  in  ampere\n",
+      "\n",
+      "#calculation:\n",
+      "X  =  20*math.log10(math.e)*math.log(I2/I1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  decibel  current  ratio  =  \",round(X,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  decibel  current  ratio  =   12.04   dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power loss in decibels.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rP  =  0.06;#  power  ratios  rP  =  P2/P1\n",
+      "\n",
+      "#calculation:\n",
+      "X  =  10*math.log10(math.e)*math.log(rP)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  decibel  Power  ratios  =  \",round(X,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  decibel  Power  ratios  =   -12.22   dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find its output power\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "X  =  14;#  decibal  power  ratio  in  dB\n",
+      "P1  =  0.008;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      "rP  =  10**(X/10)\n",
+      "P2  =  rP*P1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  power  P2  =  \",round(P2,2),\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  power  P2  =   0.2   W"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the value of the input voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "X  =  27;#  Voltage  gain  in  decibels\n",
+      "V2  =  4;#  output  voltage  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  V2/(10**(27/20))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  input  Voltage  V1  =  \",round(V1,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  input  Voltage  V1  =   0.18   V"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 129</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the unknown resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R2  =  100;#  in  ohms\n",
+      "R3  =  400;#  in  ohms\n",
+      "R4  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1  =  R2*R3/R4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n unknown  resistance,  R1  =  \",R1,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " unknown  resistance,  R1  =   4000.0   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 130</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. of a dry cell if balance is obtained with a length of 650 mm\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  1.0186;#  in  Volts\n",
+      "l1  =  0.400;#  in  m\n",
+      "l2  =  0.650;#  in  m\n",
+      "\n",
+      "#calculation:\n",
+      "E2  =  (l2/l1)*E1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  e.m.f.  of  a  dry  cell  =  \",round(E2,2),\"  Volts\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  e.m.f.  of  a  dry  cell  =   1.66   Volts"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 132</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the nominal value of the voltage across the resistor and its accuracy.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  0.0025;#  in  Amperes\n",
+      "R  =  5000;#  in  ohms\n",
+      "e1  =  0.4;#  in  %\n",
+      "e2  =  0.5;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  I*R\n",
+      "em  =  e1  +  e2\n",
+      "Ve  =  em*V/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  V  =  \",V,\"V(+-)\",Ve,\"V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  V  =   12.5 V(+-) 0.1125 V"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 132</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance of the resistor, and its accuracy of measurement\n",
+      "#if both instruments have a limit of error of 2% of f.s.d.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  6.25;#  in  Amperes\n",
+      "Im  =  10;#  max  in  Amperes\n",
+      "V  =  36.5;#  in  volts\n",
+      "Vm  =  50;#  max  in  volts\n",
+      "e  =  2;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  V/I\n",
+      "Ve  =  e*Vm/100  \n",
+      "Ve1  =  Ve*100/V#  in    %\n",
+      "Ie  =  e*Im/100\n",
+      "Ie1  =  Ie*100/I#  in  %\n",
+      "em  =  Ve1  +  Ie1\n",
+      "Re  =  em*R/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  R  =  \",R,\"  ohms(+-)\",Re,\"  ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  R  =   5.84   ohms(+-) 0.34688   ohms"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 133</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the unknown resistance and its accuracy of measurement.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohms\n",
+      "R2  =  100;#  in  ohms\n",
+      "R3  =  432.5;#  in  ohms\n",
+      "e1  =  1;#  in  %\n",
+      "e2  =  0.5;#  in  %\n",
+      "e3  =  0.2;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "Rx  =  R2*R3/R1\n",
+      "em  =  e1  +  e2  +  e3\n",
+      "Re  =  em*Rx/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  R  =  \",Rx,\"  ohms(+-)\",Re,\"  ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  R  =   43.25   ohms(+-) 0.73525   ohms"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_1.ipynb
new file mode 100755
index 00000000..c63a225f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_1.ipynb
@@ -0,0 +1,1066 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 10: Electrical measuring instruments and measurements</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 116</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the shunt to be connected in parallel with the meter\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia  =  0.040;#  in  Amperes\n",
+      "I  =  50;#  in  Amperes\n",
+      "ra  =  25;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Is  =  I  -  Ia\n",
+      "V  =  Ia*ra\n",
+      "Rs  =  V/Is\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    value  of  the  shunt  to  be  connected  in  parallel  =  \",  round((Rs/1E-3),2),\"  mohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    value  of  the  shunt  to  be  connected  in  parallel  =   20.02   mohms"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 116</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the multiplier to be connected in series with the instrument\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  volts\n",
+      "I  =  0.008;#  in  Amperes\n",
+      "ra  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rm  =  (V/I)  -  ra\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  value  of  the  multiplier  to  be  connected  in  series  =  \",  (Rm/1E3),\"  kohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  value  of  the  multiplier  to  be  connected  in  series  =   12.49   kohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 119</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated by the voltmeter and by resistor R\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fsd  =  200;#  in  volts\n",
+      "R1  =  250;#  in  ohms\n",
+      "R2  =  2E6;#  in  ohms\n",
+      "sensitivity  =  10000;#  in  ohms/V\n",
+      "V = 100; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "Rv  =  sensitivity*fsd\n",
+      "Iv  =  V/Rv\n",
+      "Pv  =  V*Iv\n",
+      "I1  =  V/R1\n",
+      "P1  =  V*I1\n",
+      "I2  =  V/R2\n",
+      "P2  =  V*I2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  power  dissipated  by  the  voltmeter  =  \",  (Pv/1E-3),\"  mW\\n\"\n",
+      "print  \"\\n  (b)the  power  dissipated  by  resistor  250  ohm  =  \",  P1,\"  W\\n\"\n",
+      "print  \"\\n  (c)the  power  dissipated  by  resistor  2  Mohm  =  \",  (P2/1E-3),\"  mW\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  power  dissipated  by  the  voltmeter  =   5.0   mW\n",
+        "\n",
+        "\n",
+        "  (b)the  power  dissipated  by  resistor  250  ohm  =   40.0   W\n",
+        "\n",
+        "\n",
+        "  (c)the  power  dissipated  by  resistor  2  Mohm  =   5.0   mW"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 119</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the ammeter reading expected (neglecting its resistance),\n",
+      "#(b) the actual current in the circuit, \n",
+      "#(c) the power dissipated in the ammeter, and\n",
+      "#(d) the power dissipated in the load\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "fsd  =  0.1;#  in  Amperes\n",
+      "ra  =  50;#  in  ohms\n",
+      "R  =  500;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Ie  =  V/R\n",
+      "Ia  =  V/(R  +  ra)\n",
+      "Pa  =  Ia*Ia*ra\n",
+      "PR  =  Ia*Ia*R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)expected  ammeter  reading  =  \",  (Ie/1E-3),\"  mA\\n\"\n",
+      "print  \"\\n  (b)Actual  ammeter  reading  =  \",round((Ia/1E-3),2),\"  mA\\n\"\n",
+      "print  \"\\n  (c)Power  dissipated  in  the  ammeter  =  \",round((Pa/1E-3),2),\"  mW\\n\"\n",
+      "print  \"\\n  (d)Power  dissipated  in  the  load  resistor  =  \",  round((PR/1E-3),2),\"  mW\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)expected  ammeter  reading  =   20.0   mA\n",
+        "\n",
+        "\n",
+        "  (b)Actual  ammeter  reading  =   18.18   mA\n",
+        "\n",
+        "\n",
+        "  (c)Power  dissipated  in  the  ammeter  =   16.53   mW\n",
+        "\n",
+        "\n",
+        "  (d)Power  dissipated  in  the  load  resistor  =   165.29   mW"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 120</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of voltage V1 with the voltmeter not connected, and \n",
+      "#(b) the voltage indicated by the voltmeter when connected between A and B.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fsd  =  100;#  in  volts\n",
+      "R1  =  40E3;#  in  ohms\n",
+      "R2  =  60E3;#  in  ohms\n",
+      "sensitivity  =  1600;#  in  ohms/V\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  (R1/(R1  +  R2))*fsd\n",
+      "Rv  =  fsd*sensitivity\n",
+      "Rep  =  R1*Rv/(R1  +  Rv)\n",
+      "V1n  =  (Rep/(Rep  +  R2))*fsd\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  voltage  V1  with  the  voltmeter6  not  connected  =  \",  V1,\"  V\\n\"\n",
+      "print  \"\\n  (b)the  voltage  indicated  by  the  voltmeter  when  connected  between  A  and  B  =  \",round(V1n,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  voltage  V1  with  the  voltmeter6  not  connected  =   40.0   V\n",
+        "\n",
+        "\n",
+        "  (b)the  voltage  indicated  by  the  voltmeter  when  connected  between  A  and  B  =   34.78   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 120</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power dissipated in the load.\n",
+      "#(b) A wattmeter, whose current coil has a resistance of 0.01 ohm\n",
+      "# Determine the wattmeter reading.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  20;#  in  amperes\n",
+      "R  =  2;#  in  ohms\n",
+      "Rw  =  0.01;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "PR  =  I*I*R\n",
+      "Rt  =  R  +  Rw\n",
+      "Pw  =  I*I*Rt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  power  dissipated  in  the  load  =  \",  PR,\"  W\\n\"\n",
+      "print  \"\\n  (b)the  wattmeter  reading.  =  \",Pw,\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  power  dissipated  in  the  load  =   800   W\n",
+        "\n",
+        "\n",
+        "  (b)the  wattmeter  reading.  =   804.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 122</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the periodic time, (b) the frequency and (c) the peak-to-peak voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  100E-6;#  in  s/cm\n",
+      "Vc  =  20;#  in  V/cm\n",
+      "w  =  5.2;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 3.6; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  periodic  time,  T  =  \",  (T/1E-3),\"  msec\\n\"\n",
+      "print  \"\\n  (b)Frequency,  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  periodic  time,  T  =   0.52   msec\n",
+        "\n",
+        "\n",
+        "  (b)Frequency,  f  =   1923.08   Hz\n",
+        "\n",
+        "\n",
+        "  (c)the  peak-to-peak  voltage  =   72.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the periodic time, (b) the frequency, (c) the magnitude of the pulse voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  50E-3;#  in  s/cm\n",
+      "Vc  =  0.2;#  in  V/cm\n",
+      "w  =  3.5;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 3.4; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "\n",
+      "#Results  \n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  periodic  time,  T  =  \",  (T/1E-3),\"  msec\\n\"\n",
+      "print  \"\\n  (b)Frequency,  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  periodic  time,  T  =   175.0   msec\n",
+        "\n",
+        "\n",
+        "  (b)Frequency,  f  =   5.71   Hz\n",
+        "\n",
+        "\n",
+        "  (c)the  peak-to-peak  voltage  =   0.68   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find, for the waveform, (a) the frequency, \n",
+      "#(b) the peak-to-peak voltage, (c) the amplitude, (d) the r.m.s. value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  500E-6;#  in  s/cm\n",
+      "Vc  =  5;#  in  V/cm\n",
+      "w  =  4;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 5; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "Amp  =  ptpv/2\n",
+      "Vrms  =  Amp/(2**0.5)\n",
+      "\n",
+      "#Results  \n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency,  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\"\n",
+      "print  \"\\n  (c)Amplitude  =  \",Amp,\"  V\\n\"\n",
+      "print  \"\\n  (d)r.m.s  voltage  =  \",round(Vrms,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency,  f  =   500.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)the  peak-to-peak  voltage  =   25   V\n",
+        "\n",
+        "\n",
+        "  (c)Amplitude  =   12.5   V\n",
+        "\n",
+        "\n",
+        "  (d)r.m.s  voltage  =   8.84   V"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) their frequency, (b) their r.m.s. values, (c) their phase difference.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#initializing  the  variables:\n",
+      "tc  =  100E-6;#  in  s/cm\n",
+      "Vc  =  2;#  in  V/cm\n",
+      "w  =  5;#  in  cm  (  width  of  one  complete  cycle  for  both  waveform  )\n",
+      "h1 = 2; # in cm ( peak-to-peak height of the display )\n",
+      "h2 = 2.5; # in cm ( peak-to-peak height of the display\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv1  =  h1*Vc\n",
+      "Vrms1  =  ptpv1/(2**0.5)\n",
+      "ptpv2  =  h2*Vc\n",
+      "Vrms2  =  ptpv2/(2**0.5)\n",
+      "phi  =  0.5*360/w\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency,  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b1)r.m.s  voltage  of  1st  waveform  =  \",round(Vrms1,2),\"  V\\n\"\n",
+      "print  \"\\n  (b2)r.m.s  voltage  of  2nd  waveform  =  \",round(Vrms2,2),\"  V\\n\"\n",
+      "print  \"\\n  (c)Phase  difference  =  \",phi,\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency,  f  =   2000.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b1)r.m.s  voltage  of  1st  waveform  =   2.83   V\n",
+        "\n",
+        "\n",
+        "  (b2)r.m.s  voltage  of  2nd  waveform  =   3.54   V\n",
+        "\n",
+        "\n",
+        "  (c)Phase  difference  =   36.0 deg"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 127</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the decibel power ratio in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rP1  =  3;#  ratio  of  two  powers\n",
+      "rP2  =  20;#  ratio  of  two  powers\n",
+      "rP3  =  400;#  ratio  of  two  powers\n",
+      "rP4  =  1/20;#  ratio  of  two  powers\n",
+      "\n",
+      "#calculation:\n",
+      "X1  =  10*(1/2.303)*math.log(3)\n",
+      "X2  =  10*(1/2.303)*math.log(20)\n",
+      "X3  =  10*(1/2.303)*math.log(400)\n",
+      "X4  =  10*(1/2.303)*math.log(1/20)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)decibel  power  ratio  for  power  ratio  3  =  \",round(X1,2),\"  dB\\n\"\n",
+      "print  \"\\n  (b)decibel  power  ratio  for  power  ratio  20  =  \",round(X2,2),\"  dB\\n\"\n",
+      "print  \"\\n  (c)decibel  power  ratio  for  power  ratio  400  =  \",round(X3,2),\"  dB\\n\"\n",
+      "print  \"\\n  (d)decibel  power  ratio  for  power  ratio  1/20  =  \",round(X4,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)decibel  power  ratio  for  power  ratio  3  =   4.77   dB\n",
+        "\n",
+        "\n",
+        "  (b)decibel  power  ratio  for  power  ratio  20  =   13.01   dB\n",
+        "\n",
+        "\n",
+        "  (c)decibel  power  ratio  for  power  ratio  400  =   26.02   dB\n",
+        "\n",
+        "\n",
+        "  (d)decibel  power  ratio  for  power  ratio  1/20  =   -13.01   dB"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 127</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the decibel current ratio assuming the input and load resistances of the system are equal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I2  =  0.020;#  in  ampere\n",
+      "I1  =  0.005;#  in  ampere\n",
+      "\n",
+      "#calculation:\n",
+      "X  =  20*math.log10(math.e)*math.log(I2/I1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  decibel  current  ratio  =  \",round(X,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  decibel  current  ratio  =   12.04   dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power loss in decibels.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rP  =  0.06;#  power  ratios  rP  =  P2/P1\n",
+      "\n",
+      "#calculation:\n",
+      "X  =  10*math.log10(math.e)*math.log(rP)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  decibel  Power  ratios  =  \",round(X,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  decibel  Power  ratios  =   -12.22   dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find its output power\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "X  =  14;#  decibal  power  ratio  in  dB\n",
+      "P1  =  0.008;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      "rP  =  10**(X/10)\n",
+      "P2  =  rP*P1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  power  P2  =  \",round(P2,2),\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  power  P2  =   0.2   W"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the value of the input voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "X  =  27;#  Voltage  gain  in  decibels\n",
+      "V2  =  4;#  output  voltage  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  V2/(10**(27/20))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  input  Voltage  V1  =  \",round(V1,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  input  Voltage  V1  =   0.18   V"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 129</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the unknown resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R2  =  100;#  in  ohms\n",
+      "R3  =  400;#  in  ohms\n",
+      "R4  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1  =  R2*R3/R4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n unknown  resistance,  R1  =  \",R1,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " unknown  resistance,  R1  =   4000.0   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 130</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. of a dry cell if balance is obtained with a length of 650 mm\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  1.0186;#  in  Volts\n",
+      "l1  =  0.400;#  in  m\n",
+      "l2  =  0.650;#  in  m\n",
+      "\n",
+      "#calculation:\n",
+      "E2  =  (l2/l1)*E1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  e.m.f.  of  a  dry  cell  =  \",round(E2,2),\"  Volts\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  e.m.f.  of  a  dry  cell  =   1.66   Volts"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 132</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the nominal value of the voltage across the resistor and its accuracy.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  0.0025;#  in  Amperes\n",
+      "R  =  5000;#  in  ohms\n",
+      "e1  =  0.4;#  in  %\n",
+      "e2  =  0.5;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  I*R\n",
+      "em  =  e1  +  e2\n",
+      "Ve  =  em*V/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  V  =  \",V,\"V(+-)\",Ve,\"V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  V  =   12.5 V(+-) 0.1125 V"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 132</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance of the resistor, and its accuracy of measurement\n",
+      "#if both instruments have a limit of error of 2% of f.s.d.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  6.25;#  in  Amperes\n",
+      "Im  =  10;#  max  in  Amperes\n",
+      "V  =  36.5;#  in  volts\n",
+      "Vm  =  50;#  max  in  volts\n",
+      "e  =  2;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  V/I\n",
+      "Ve  =  e*Vm/100  \n",
+      "Ve1  =  Ve*100/V#  in    %\n",
+      "Ie  =  e*Im/100\n",
+      "Ie1  =  Ie*100/I#  in  %\n",
+      "em  =  Ve1  +  Ie1\n",
+      "Re  =  em*R/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  R  =  \",R,\"  ohms(+-)\",Re,\"  ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  R  =   5.84   ohms(+-) 0.34688   ohms"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 133</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the unknown resistance and its accuracy of measurement.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohms\n",
+      "R2  =  100;#  in  ohms\n",
+      "R3  =  432.5;#  in  ohms\n",
+      "e1  =  1;#  in  %\n",
+      "e2  =  0.5;#  in  %\n",
+      "e3  =  0.2;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "Rx  =  R2*R3/R1\n",
+      "em  =  e1  +  e2  +  e3\n",
+      "Re  =  em*Rx/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  R  =  \",Rx,\"  ohms(+-)\",Re,\"  ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  R  =   43.25   ohms(+-) 0.73525   ohms"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_2.ipynb
new file mode 100755
index 00000000..c63a225f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_2.ipynb
@@ -0,0 +1,1066 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 10: Electrical measuring instruments and measurements</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 116</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the shunt to be connected in parallel with the meter\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia  =  0.040;#  in  Amperes\n",
+      "I  =  50;#  in  Amperes\n",
+      "ra  =  25;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Is  =  I  -  Ia\n",
+      "V  =  Ia*ra\n",
+      "Rs  =  V/Is\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    value  of  the  shunt  to  be  connected  in  parallel  =  \",  round((Rs/1E-3),2),\"  mohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    value  of  the  shunt  to  be  connected  in  parallel  =   20.02   mohms"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 116</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the multiplier to be connected in series with the instrument\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  volts\n",
+      "I  =  0.008;#  in  Amperes\n",
+      "ra  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rm  =  (V/I)  -  ra\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  value  of  the  multiplier  to  be  connected  in  series  =  \",  (Rm/1E3),\"  kohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  value  of  the  multiplier  to  be  connected  in  series  =   12.49   kohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 119</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated by the voltmeter and by resistor R\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fsd  =  200;#  in  volts\n",
+      "R1  =  250;#  in  ohms\n",
+      "R2  =  2E6;#  in  ohms\n",
+      "sensitivity  =  10000;#  in  ohms/V\n",
+      "V = 100; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "Rv  =  sensitivity*fsd\n",
+      "Iv  =  V/Rv\n",
+      "Pv  =  V*Iv\n",
+      "I1  =  V/R1\n",
+      "P1  =  V*I1\n",
+      "I2  =  V/R2\n",
+      "P2  =  V*I2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  power  dissipated  by  the  voltmeter  =  \",  (Pv/1E-3),\"  mW\\n\"\n",
+      "print  \"\\n  (b)the  power  dissipated  by  resistor  250  ohm  =  \",  P1,\"  W\\n\"\n",
+      "print  \"\\n  (c)the  power  dissipated  by  resistor  2  Mohm  =  \",  (P2/1E-3),\"  mW\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  power  dissipated  by  the  voltmeter  =   5.0   mW\n",
+        "\n",
+        "\n",
+        "  (b)the  power  dissipated  by  resistor  250  ohm  =   40.0   W\n",
+        "\n",
+        "\n",
+        "  (c)the  power  dissipated  by  resistor  2  Mohm  =   5.0   mW"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 119</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the ammeter reading expected (neglecting its resistance),\n",
+      "#(b) the actual current in the circuit, \n",
+      "#(c) the power dissipated in the ammeter, and\n",
+      "#(d) the power dissipated in the load\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "fsd  =  0.1;#  in  Amperes\n",
+      "ra  =  50;#  in  ohms\n",
+      "R  =  500;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Ie  =  V/R\n",
+      "Ia  =  V/(R  +  ra)\n",
+      "Pa  =  Ia*Ia*ra\n",
+      "PR  =  Ia*Ia*R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)expected  ammeter  reading  =  \",  (Ie/1E-3),\"  mA\\n\"\n",
+      "print  \"\\n  (b)Actual  ammeter  reading  =  \",round((Ia/1E-3),2),\"  mA\\n\"\n",
+      "print  \"\\n  (c)Power  dissipated  in  the  ammeter  =  \",round((Pa/1E-3),2),\"  mW\\n\"\n",
+      "print  \"\\n  (d)Power  dissipated  in  the  load  resistor  =  \",  round((PR/1E-3),2),\"  mW\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)expected  ammeter  reading  =   20.0   mA\n",
+        "\n",
+        "\n",
+        "  (b)Actual  ammeter  reading  =   18.18   mA\n",
+        "\n",
+        "\n",
+        "  (c)Power  dissipated  in  the  ammeter  =   16.53   mW\n",
+        "\n",
+        "\n",
+        "  (d)Power  dissipated  in  the  load  resistor  =   165.29   mW"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 120</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of voltage V1 with the voltmeter not connected, and \n",
+      "#(b) the voltage indicated by the voltmeter when connected between A and B.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fsd  =  100;#  in  volts\n",
+      "R1  =  40E3;#  in  ohms\n",
+      "R2  =  60E3;#  in  ohms\n",
+      "sensitivity  =  1600;#  in  ohms/V\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  (R1/(R1  +  R2))*fsd\n",
+      "Rv  =  fsd*sensitivity\n",
+      "Rep  =  R1*Rv/(R1  +  Rv)\n",
+      "V1n  =  (Rep/(Rep  +  R2))*fsd\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  voltage  V1  with  the  voltmeter6  not  connected  =  \",  V1,\"  V\\n\"\n",
+      "print  \"\\n  (b)the  voltage  indicated  by  the  voltmeter  when  connected  between  A  and  B  =  \",round(V1n,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  voltage  V1  with  the  voltmeter6  not  connected  =   40.0   V\n",
+        "\n",
+        "\n",
+        "  (b)the  voltage  indicated  by  the  voltmeter  when  connected  between  A  and  B  =   34.78   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 120</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power dissipated in the load.\n",
+      "#(b) A wattmeter, whose current coil has a resistance of 0.01 ohm\n",
+      "# Determine the wattmeter reading.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  20;#  in  amperes\n",
+      "R  =  2;#  in  ohms\n",
+      "Rw  =  0.01;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "PR  =  I*I*R\n",
+      "Rt  =  R  +  Rw\n",
+      "Pw  =  I*I*Rt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  power  dissipated  in  the  load  =  \",  PR,\"  W\\n\"\n",
+      "print  \"\\n  (b)the  wattmeter  reading.  =  \",Pw,\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  power  dissipated  in  the  load  =   800   W\n",
+        "\n",
+        "\n",
+        "  (b)the  wattmeter  reading.  =   804.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 122</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the periodic time, (b) the frequency and (c) the peak-to-peak voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  100E-6;#  in  s/cm\n",
+      "Vc  =  20;#  in  V/cm\n",
+      "w  =  5.2;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 3.6; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  periodic  time,  T  =  \",  (T/1E-3),\"  msec\\n\"\n",
+      "print  \"\\n  (b)Frequency,  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  periodic  time,  T  =   0.52   msec\n",
+        "\n",
+        "\n",
+        "  (b)Frequency,  f  =   1923.08   Hz\n",
+        "\n",
+        "\n",
+        "  (c)the  peak-to-peak  voltage  =   72.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the periodic time, (b) the frequency, (c) the magnitude of the pulse voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  50E-3;#  in  s/cm\n",
+      "Vc  =  0.2;#  in  V/cm\n",
+      "w  =  3.5;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 3.4; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "\n",
+      "#Results  \n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  periodic  time,  T  =  \",  (T/1E-3),\"  msec\\n\"\n",
+      "print  \"\\n  (b)Frequency,  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  periodic  time,  T  =   175.0   msec\n",
+        "\n",
+        "\n",
+        "  (b)Frequency,  f  =   5.71   Hz\n",
+        "\n",
+        "\n",
+        "  (c)the  peak-to-peak  voltage  =   0.68   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find, for the waveform, (a) the frequency, \n",
+      "#(b) the peak-to-peak voltage, (c) the amplitude, (d) the r.m.s. value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  500E-6;#  in  s/cm\n",
+      "Vc  =  5;#  in  V/cm\n",
+      "w  =  4;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 5; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "Amp  =  ptpv/2\n",
+      "Vrms  =  Amp/(2**0.5)\n",
+      "\n",
+      "#Results  \n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency,  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\"\n",
+      "print  \"\\n  (c)Amplitude  =  \",Amp,\"  V\\n\"\n",
+      "print  \"\\n  (d)r.m.s  voltage  =  \",round(Vrms,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency,  f  =   500.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)the  peak-to-peak  voltage  =   25   V\n",
+        "\n",
+        "\n",
+        "  (c)Amplitude  =   12.5   V\n",
+        "\n",
+        "\n",
+        "  (d)r.m.s  voltage  =   8.84   V"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) their frequency, (b) their r.m.s. values, (c) their phase difference.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#initializing  the  variables:\n",
+      "tc  =  100E-6;#  in  s/cm\n",
+      "Vc  =  2;#  in  V/cm\n",
+      "w  =  5;#  in  cm  (  width  of  one  complete  cycle  for  both  waveform  )\n",
+      "h1 = 2; # in cm ( peak-to-peak height of the display )\n",
+      "h2 = 2.5; # in cm ( peak-to-peak height of the display\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv1  =  h1*Vc\n",
+      "Vrms1  =  ptpv1/(2**0.5)\n",
+      "ptpv2  =  h2*Vc\n",
+      "Vrms2  =  ptpv2/(2**0.5)\n",
+      "phi  =  0.5*360/w\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency,  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b1)r.m.s  voltage  of  1st  waveform  =  \",round(Vrms1,2),\"  V\\n\"\n",
+      "print  \"\\n  (b2)r.m.s  voltage  of  2nd  waveform  =  \",round(Vrms2,2),\"  V\\n\"\n",
+      "print  \"\\n  (c)Phase  difference  =  \",phi,\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency,  f  =   2000.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b1)r.m.s  voltage  of  1st  waveform  =   2.83   V\n",
+        "\n",
+        "\n",
+        "  (b2)r.m.s  voltage  of  2nd  waveform  =   3.54   V\n",
+        "\n",
+        "\n",
+        "  (c)Phase  difference  =   36.0 deg"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 127</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the decibel power ratio in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rP1  =  3;#  ratio  of  two  powers\n",
+      "rP2  =  20;#  ratio  of  two  powers\n",
+      "rP3  =  400;#  ratio  of  two  powers\n",
+      "rP4  =  1/20;#  ratio  of  two  powers\n",
+      "\n",
+      "#calculation:\n",
+      "X1  =  10*(1/2.303)*math.log(3)\n",
+      "X2  =  10*(1/2.303)*math.log(20)\n",
+      "X3  =  10*(1/2.303)*math.log(400)\n",
+      "X4  =  10*(1/2.303)*math.log(1/20)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)decibel  power  ratio  for  power  ratio  3  =  \",round(X1,2),\"  dB\\n\"\n",
+      "print  \"\\n  (b)decibel  power  ratio  for  power  ratio  20  =  \",round(X2,2),\"  dB\\n\"\n",
+      "print  \"\\n  (c)decibel  power  ratio  for  power  ratio  400  =  \",round(X3,2),\"  dB\\n\"\n",
+      "print  \"\\n  (d)decibel  power  ratio  for  power  ratio  1/20  =  \",round(X4,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)decibel  power  ratio  for  power  ratio  3  =   4.77   dB\n",
+        "\n",
+        "\n",
+        "  (b)decibel  power  ratio  for  power  ratio  20  =   13.01   dB\n",
+        "\n",
+        "\n",
+        "  (c)decibel  power  ratio  for  power  ratio  400  =   26.02   dB\n",
+        "\n",
+        "\n",
+        "  (d)decibel  power  ratio  for  power  ratio  1/20  =   -13.01   dB"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 127</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the decibel current ratio assuming the input and load resistances of the system are equal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I2  =  0.020;#  in  ampere\n",
+      "I1  =  0.005;#  in  ampere\n",
+      "\n",
+      "#calculation:\n",
+      "X  =  20*math.log10(math.e)*math.log(I2/I1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  decibel  current  ratio  =  \",round(X,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  decibel  current  ratio  =   12.04   dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power loss in decibels.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rP  =  0.06;#  power  ratios  rP  =  P2/P1\n",
+      "\n",
+      "#calculation:\n",
+      "X  =  10*math.log10(math.e)*math.log(rP)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  decibel  Power  ratios  =  \",round(X,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  decibel  Power  ratios  =   -12.22   dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find its output power\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "X  =  14;#  decibal  power  ratio  in  dB\n",
+      "P1  =  0.008;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      "rP  =  10**(X/10)\n",
+      "P2  =  rP*P1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  power  P2  =  \",round(P2,2),\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  power  P2  =   0.2   W"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the value of the input voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "X  =  27;#  Voltage  gain  in  decibels\n",
+      "V2  =  4;#  output  voltage  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  V2/(10**(27/20))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  input  Voltage  V1  =  \",round(V1,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  input  Voltage  V1  =   0.18   V"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 129</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the value of the unknown resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R2  =  100;#  in  ohms\n",
+      "R3  =  400;#  in  ohms\n",
+      "R4  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1  =  R2*R3/R4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n unknown  resistance,  R1  =  \",R1,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " unknown  resistance,  R1  =   4000.0   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 130</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. of a dry cell if balance is obtained with a length of 650 mm\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  1.0186;#  in  Volts\n",
+      "l1  =  0.400;#  in  m\n",
+      "l2  =  0.650;#  in  m\n",
+      "\n",
+      "#calculation:\n",
+      "E2  =  (l2/l1)*E1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  e.m.f.  of  a  dry  cell  =  \",round(E2,2),\"  Volts\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  e.m.f.  of  a  dry  cell  =   1.66   Volts"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 132</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the nominal value of the voltage across the resistor and its accuracy.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  0.0025;#  in  Amperes\n",
+      "R  =  5000;#  in  ohms\n",
+      "e1  =  0.4;#  in  %\n",
+      "e2  =  0.5;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  I*R\n",
+      "em  =  e1  +  e2\n",
+      "Ve  =  em*V/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  V  =  \",V,\"V(+-)\",Ve,\"V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  V  =   12.5 V(+-) 0.1125 V"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 132</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance of the resistor, and its accuracy of measurement\n",
+      "#if both instruments have a limit of error of 2% of f.s.d.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  6.25;#  in  Amperes\n",
+      "Im  =  10;#  max  in  Amperes\n",
+      "V  =  36.5;#  in  volts\n",
+      "Vm  =  50;#  max  in  volts\n",
+      "e  =  2;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  V/I\n",
+      "Ve  =  e*Vm/100  \n",
+      "Ve1  =  Ve*100/V#  in    %\n",
+      "Ie  =  e*Im/100\n",
+      "Ie1  =  Ie*100/I#  in  %\n",
+      "em  =  Ve1  +  Ie1\n",
+      "Re  =  em*R/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  R  =  \",R,\"  ohms(+-)\",Re,\"  ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  R  =   5.84   ohms(+-) 0.34688   ohms"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 133</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the unknown resistance and its accuracy of measurement.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohms\n",
+      "R2  =  100;#  in  ohms\n",
+      "R3  =  432.5;#  in  ohms\n",
+      "e1  =  1;#  in  %\n",
+      "e2  =  0.5;#  in  %\n",
+      "e3  =  0.2;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "Rx  =  R2*R3/R1\n",
+      "em  =  e1  +  e2  +  e3\n",
+      "Re  =  em*Rx/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  R  =  \",Rx,\"  ohms(+-)\",Re,\"  ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  R  =   43.25   ohms(+-) 0.73525   ohms"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_3.ipynb
new file mode 100755
index 00000000..443b9513
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_10-checkpoint_3.ipynb
@@ -0,0 +1,1059 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c6970542ad51432c833f98812d08f1e8865158ee25a859a74fb18804672e045b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 10: Electrical measuring instruments and measurements</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 116</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia  =  0.040;#  in  Amperes\n",
+      "I  =  50;#  in  Amperes\n",
+      "ra  =  25;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Is  =  I  -  Ia\n",
+      "V  =  Ia*ra\n",
+      "Rs  =  V/Is\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    value  of  the  shunt  to  be  connected  in  parallel  =  \",  round((Rs/1E-3),2),\"  mohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    value  of  the  shunt  to  be  connected  in  parallel  =   20.02   mohms"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 116</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  volts\n",
+      "I  =  0.008;#  in  Amperes\n",
+      "ra  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Rm  =  (V/I)  -  ra\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  value  of  the  multiplier  to  be  connected  in  series  =  \",  (Rm/1E3),\"  kohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  value  of  the  multiplier  to  be  connected  in  series  =   12.49   kohms"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 119</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fsd  =  200;#  in  volts\n",
+      "R1  =  250;#  in  ohms\n",
+      "R2  =  2E6;#  in  ohms\n",
+      "sensitivity  =  10000;#  in  ohms/V\n",
+      "V = 100; # in volts\n",
+      "\n",
+      "#calculation:\n",
+      "Rv  =  sensitivity*fsd\n",
+      "Iv  =  V/Rv\n",
+      "Pv  =  V*Iv\n",
+      "I1  =  V/R1\n",
+      "P1  =  V*I1\n",
+      "I2  =  V/R2\n",
+      "P2  =  V*I2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  power  dissipated  by  the  voltmeter  =  \",  (Pv/1E-3),\"  mW\\n\"\n",
+      "print  \"\\n  (b)the  power  dissipated  by  resistor  250  ohm  =  \",  P1,\"  W\\n\"\n",
+      "print  \"\\n  (c)the  power  dissipated  by  resistor  2  Mohm  =  \",  (P2/1E-3),\"  mW\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  power  dissipated  by  the  voltmeter  =   5.0   mW\n",
+        "\n",
+        "\n",
+        "  (b)the  power  dissipated  by  resistor  250  ohm  =   40.0   W\n",
+        "\n",
+        "\n",
+        "  (c)the  power  dissipated  by  resistor  2  Mohm  =   5.0   mW"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 119</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "fsd  =  0.1;#  in  Amperes\n",
+      "ra  =  50;#  in  ohms\n",
+      "R  =  500;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Ie  =  V/R\n",
+      "Ia  =  V/(R  +  ra)\n",
+      "Pa  =  Ia*Ia*ra\n",
+      "PR  =  Ia*Ia*R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)expected  ammeter  reading  =  \",  (Ie/1E-3),\"  mA\\n\"\n",
+      "print  \"\\n  (b)Actual  ammeter  reading  =  \",round((Ia/1E-3),2),\"  mA\\n\"\n",
+      "print  \"\\n  (c)Power  dissipated  in  the  ammeter  =  \",round((Pa/1E-3),2),\"  mW\\n\"\n",
+      "print  \"\\n  (d)Power  dissipated  in  the  load  resistor  =  \",  round((PR/1E-3),2),\"  mW\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)expected  ammeter  reading  =   20.0   mA\n",
+        "\n",
+        "\n",
+        "  (b)Actual  ammeter  reading  =   18.18   mA\n",
+        "\n",
+        "\n",
+        "  (c)Power  dissipated  in  the  ammeter  =   16.53   mW\n",
+        "\n",
+        "\n",
+        "  (d)Power  dissipated  in  the  load  resistor  =   165.29   mW"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 120</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fsd  =  100;#  in  volts\n",
+      "R1  =  40E3;#  in  ohms\n",
+      "R2  =  60E3;#  in  ohms\n",
+      "sensitivity  =  1600;#  in  ohms/V\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  (R1/(R1  +  R2))*fsd\n",
+      "Rv  =  fsd*sensitivity\n",
+      "Rep  =  R1*Rv/(R1  +  Rv)\n",
+      "V1n  =  (Rep/(Rep  +  R2))*fsd\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  voltage  V1  with  the  voltmeter6  not  connected  =  \",  V1,\"  V\\n\"\n",
+      "print  \"\\n  (b)the  voltage  indicated  by  the  voltmeter  when  connected  between  A  and  B  =  \",round(V1n,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  voltage  V1  with  the  voltmeter6  not  connected  =   40.0   V\n",
+        "\n",
+        "\n",
+        "  (b)the  voltage  indicated  by  the  voltmeter  when  connected  between  A  and  B  =   34.78   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 120</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  20;#  in  amperes\n",
+      "R  =  2;#  in  ohms\n",
+      "Rw  =  0.01;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "PR  =  I*I*R\n",
+      "Rt  =  R  +  Rw\n",
+      "Pw  =  I*I*Rt\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  power  dissipated  in  the  load  =  \",  PR,\"  W\\n\"\n",
+      "print  \"\\n  (b)the  wattmeter  reading.  =  \",Pw,\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  power  dissipated  in  the  load  =   800   W\n",
+        "\n",
+        "\n",
+        "  (b)the  wattmeter  reading.  =   804.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 122</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  100E-6;#  in  s/cm\n",
+      "Vc  =  20;#  in  V/cm\n",
+      "w  =  5.2;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 3.6; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  periodic  time,  T  =  \",  (T/1E-3),\"  msec\\n\"\n",
+      "print  \"\\n  (b)Frequency,  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  periodic  time,  T  =   0.52   msec\n",
+        "\n",
+        "\n",
+        "  (b)Frequency,  f  =   1923.08   Hz\n",
+        "\n",
+        "\n",
+        "  (c)the  peak-to-peak  voltage  =   72.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  50E-3;#  in  s/cm\n",
+      "Vc  =  0.2;#  in  V/cm\n",
+      "w  =  3.5;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 3.4; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "\n",
+      "#Results  \n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  periodic  time,  T  =  \",  (T/1E-3),\"  msec\\n\"\n",
+      "print  \"\\n  (b)Frequency,  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  periodic  time,  T  =   175.0   msec\n",
+        "\n",
+        "\n",
+        "  (b)Frequency,  f  =   5.71   Hz\n",
+        "\n",
+        "\n",
+        "  (c)the  peak-to-peak  voltage  =   0.68   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tc  =  500E-6;#  in  s/cm\n",
+      "Vc  =  5;#  in  V/cm\n",
+      "w  =  4;#  in  cm  (  width  of  one  complete  cycle  )\n",
+      "h = 5; # in cm ( peak-to-peak height of the display )\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv  =  h*Vc\n",
+      "Amp  =  ptpv/2\n",
+      "Vrms  =  Amp/(2**0.5)\n",
+      "\n",
+      "#Results  \n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency,  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)the  peak-to-peak  voltage  =  \",ptpv,\"  V\\n\"\n",
+      "print  \"\\n  (c)Amplitude  =  \",Amp,\"  V\\n\"\n",
+      "print  \"\\n  (d)r.m.s  voltage  =  \",round(Vrms,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency,  f  =   500.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)the  peak-to-peak  voltage  =   25   V\n",
+        "\n",
+        "\n",
+        "  (c)Amplitude  =   12.5   V\n",
+        "\n",
+        "\n",
+        "  (d)r.m.s  voltage  =   8.84   V"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 123</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "\n",
+      "#initializing  the  variables:\n",
+      "tc  =  100E-6;#  in  s/cm\n",
+      "Vc  =  2;#  in  V/cm\n",
+      "w  =  5;#  in  cm  (  width  of  one  complete  cycle  for  both  waveform  )\n",
+      "h1 = 2; # in cm ( peak-to-peak height of the display )\n",
+      "h2 = 2.5; # in cm ( peak-to-peak height of the display\n",
+      "\n",
+      "#calculation:\n",
+      "T  =  w*tc\n",
+      "f  =  1/T\n",
+      "ptpv1  =  h1*Vc\n",
+      "Vrms1  =  ptpv1/(2**0.5)\n",
+      "ptpv2  =  h2*Vc\n",
+      "Vrms2  =  ptpv2/(2**0.5)\n",
+      "phi  =  0.5*360/w\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency,  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b1)r.m.s  voltage  of  1st  waveform  =  \",round(Vrms1,2),\"  V\\n\"\n",
+      "print  \"\\n  (b2)r.m.s  voltage  of  2nd  waveform  =  \",round(Vrms2,2),\"  V\\n\"\n",
+      "print  \"\\n  (c)Phase  difference  =  \",phi,\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency,  f  =   2000.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b1)r.m.s  voltage  of  1st  waveform  =   2.83   V\n",
+        "\n",
+        "\n",
+        "  (b2)r.m.s  voltage  of  2nd  waveform  =   3.54   V\n",
+        "\n",
+        "\n",
+        "  (c)Phase  difference  =   36.0 deg"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 127</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rP1  =  3;#  ratio  of  two  powers\n",
+      "rP2  =  20;#  ratio  of  two  powers\n",
+      "rP3  =  400;#  ratio  of  two  powers\n",
+      "rP4  =  1/20;#  ratio  of  two  powers\n",
+      "\n",
+      "#calculation:\n",
+      "X1  =  10*(1/2.303)*math.log(3)\n",
+      "X2  =  10*(1/2.303)*math.log(20)\n",
+      "X3  =  10*(1/2.303)*math.log(400)\n",
+      "X4  =  10*(1/2.303)*math.log(1/20)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)decibel  power  ratio  for  power  ratio  3  =  \",round(X1,2),\"  dB\\n\"\n",
+      "print  \"\\n  (b)decibel  power  ratio  for  power  ratio  20  =  \",round(X2,2),\"  dB\\n\"\n",
+      "print  \"\\n  (c)decibel  power  ratio  for  power  ratio  400  =  \",round(X3,2),\"  dB\\n\"\n",
+      "print  \"\\n  (d)decibel  power  ratio  for  power  ratio  1/20  =  \",round(X4,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)decibel  power  ratio  for  power  ratio  3  =   4.77   dB\n",
+        "\n",
+        "\n",
+        "  (b)decibel  power  ratio  for  power  ratio  20  =   13.01   dB\n",
+        "\n",
+        "\n",
+        "  (c)decibel  power  ratio  for  power  ratio  400  =   26.02   dB\n",
+        "\n",
+        "\n",
+        "  (d)decibel  power  ratio  for  power  ratio  1/20  =   -13.01   dB"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 127</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I2  =  0.020;#  in  ampere\n",
+      "I1  =  0.005;#  in  ampere\n",
+      "\n",
+      "#calculation:\n",
+      "X  =  20*math.log10(math.e)*math.log(I2/I1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  decibel  current  ratio  =  \",round(X,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  decibel  current  ratio  =   12.04   dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rP  =  0.06;#  power  ratios  rP  =  P2/P1\n",
+      "\n",
+      "#calculation:\n",
+      "X  =  10*math.log10(math.e)*math.log(rP)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  decibel  Power  ratios  =  \",round(X,2),\"  dB\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  decibel  Power  ratios  =   -12.22   dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "X  =  14;#  decibal  power  ratio  in  dB\n",
+      "P1  =  0.008;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      "rP  =  10**(X/10)\n",
+      "P2  =  rP*P1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  power  P2  =  \",round(P2,2),\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  power  P2  =   0.2   W"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 128</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "X  =  27;#  Voltage  gain  in  decibels\n",
+      "V2  =  4;#  output  voltage  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "V1  =  V2/(10**(27/20))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  input  Voltage  V1  =  \",round(V1,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  input  Voltage  V1  =   0.18   V"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 129</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R2  =  100;#  in  ohms\n",
+      "R3  =  400;#  in  ohms\n",
+      "R4  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R1  =  R2*R3/R4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n unknown  resistance,  R1  =  \",R1,\"  Ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " unknown  resistance,  R1  =   4000.0   Ohms"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 130</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  1.0186;#  in  Volts\n",
+      "l1  =  0.400;#  in  m\n",
+      "l2  =  0.650;#  in  m\n",
+      "\n",
+      "#calculation:\n",
+      "E2  =  (l2/l1)*E1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  e.m.f.  of  a  dry  cell  =  \",round(E2,2),\"  Volts\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  e.m.f.  of  a  dry  cell  =   1.66   Volts"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 132</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  0.0025;#  in  Amperes\n",
+      "R  =  5000;#  in  ohms\n",
+      "e1  =  0.4;#  in  %\n",
+      "e2  =  0.5;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  I*R\n",
+      "em  =  e1  +  e2\n",
+      "Ve  =  em*V/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  V  =  \",V,\"V(+-)\",Ve,\"V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  V  =   12.5 V(+-) 0.1125 V"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 132</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  6.25;#  in  Amperes\n",
+      "Im  =  10;#  max  in  Amperes\n",
+      "V  =  36.5;#  in  volts\n",
+      "Vm  =  50;#  max  in  volts\n",
+      "e  =  2;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  V/I\n",
+      "Ve  =  e*Vm/100  \n",
+      "Ve1  =  Ve*100/V#  in    %\n",
+      "Ie  =  e*Im/100\n",
+      "Ie1  =  Ie*100/I#  in  %\n",
+      "em  =  Ve1  +  Ie1\n",
+      "Re  =  em*R/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  R  =  \",R,\"  ohms(+-)\",Re,\"  ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  R  =   5.84   ohms(+-) 0.34688   ohms"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 133</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohms\n",
+      "R2  =  100;#  in  ohms\n",
+      "R3  =  432.5;#  in  ohms\n",
+      "e1  =  1;#  in  %\n",
+      "e2  =  0.5;#  in  %\n",
+      "e3  =  0.2;#  in  %\n",
+      "\n",
+      "#calculation:\n",
+      "Rx  =  R2*R3/R1\n",
+      "em  =  e1  +  e2  +  e3\n",
+      "Re  =  em*Rx/100\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  R  =  \",Rx,\"  ohms(+-)\",Re,\"  ohms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  R  =   43.25   ohms(+-) 0.73525   ohms"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint.ipynb
new file mode 100755
index 00000000..8adc225b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint.ipynb
@@ -0,0 +1,85 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 12: Transistors</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 157</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Estimate the voltage gain Av, the current gain Ai and \n",
+      "#the power gain Ap when an input current of 20 \u03bcA peak varies sinusoidally about a mean bias of 50 \u03bcA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia  =  0.040;#  in  Amperes\n",
+      "I  =  50;#  in  Amperes\n",
+      "Rl  =  1.2;#  in  Kohms\n",
+      "Vc  =  7; # in Volts\n",
+      "R = 1; # in Kohms\n",
+      "Ib = 50*1E-6; # in Amps\n",
+      "Iic = 20*1E-6; #in Amps\n",
+      "\n",
+      "#calculation:\n",
+      "Vi = (70 - 30)*1E-6 * 1000\n",
+      "Vo = 3.6\n",
+      "Av = Vo/Vi\n",
+      "\n",
+      "Io = 3*1E-3\n",
+      "Ii = (70 - 30)*1E-6\n",
+      "Ai = Io/Ii\n",
+      "\n",
+      "Ap = Av*Ai\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage gain, Av  =  \",  round(Av,2)\n",
+      "print  \"\\n  Current gain, Av  =  \",  round(Ai,2)\n",
+      "print  \"\\n  Power gain, Av  =  \",  round(Ap,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage gain, Av  =   90.0\n",
+        "\n",
+        "  Current gain, Av  =   75.0\n",
+        "\n",
+        "  Power gain, Av  =   6750.0"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_1.ipynb
new file mode 100755
index 00000000..8adc225b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_1.ipynb
@@ -0,0 +1,85 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 12: Transistors</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 157</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Estimate the voltage gain Av, the current gain Ai and \n",
+      "#the power gain Ap when an input current of 20 \u03bcA peak varies sinusoidally about a mean bias of 50 \u03bcA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia  =  0.040;#  in  Amperes\n",
+      "I  =  50;#  in  Amperes\n",
+      "Rl  =  1.2;#  in  Kohms\n",
+      "Vc  =  7; # in Volts\n",
+      "R = 1; # in Kohms\n",
+      "Ib = 50*1E-6; # in Amps\n",
+      "Iic = 20*1E-6; #in Amps\n",
+      "\n",
+      "#calculation:\n",
+      "Vi = (70 - 30)*1E-6 * 1000\n",
+      "Vo = 3.6\n",
+      "Av = Vo/Vi\n",
+      "\n",
+      "Io = 3*1E-3\n",
+      "Ii = (70 - 30)*1E-6\n",
+      "Ai = Io/Ii\n",
+      "\n",
+      "Ap = Av*Ai\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage gain, Av  =  \",  round(Av,2)\n",
+      "print  \"\\n  Current gain, Av  =  \",  round(Ai,2)\n",
+      "print  \"\\n  Power gain, Av  =  \",  round(Ap,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage gain, Av  =   90.0\n",
+        "\n",
+        "  Current gain, Av  =   75.0\n",
+        "\n",
+        "  Power gain, Av  =   6750.0"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_2.ipynb
new file mode 100755
index 00000000..8adc225b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_2.ipynb
@@ -0,0 +1,85 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 12: Transistors</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 157</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Estimate the voltage gain Av, the current gain Ai and \n",
+      "#the power gain Ap when an input current of 20 \u03bcA peak varies sinusoidally about a mean bias of 50 \u03bcA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia  =  0.040;#  in  Amperes\n",
+      "I  =  50;#  in  Amperes\n",
+      "Rl  =  1.2;#  in  Kohms\n",
+      "Vc  =  7; # in Volts\n",
+      "R = 1; # in Kohms\n",
+      "Ib = 50*1E-6; # in Amps\n",
+      "Iic = 20*1E-6; #in Amps\n",
+      "\n",
+      "#calculation:\n",
+      "Vi = (70 - 30)*1E-6 * 1000\n",
+      "Vo = 3.6\n",
+      "Av = Vo/Vi\n",
+      "\n",
+      "Io = 3*1E-3\n",
+      "Ii = (70 - 30)*1E-6\n",
+      "Ai = Io/Ii\n",
+      "\n",
+      "Ap = Av*Ai\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage gain, Av  =  \",  round(Av,2)\n",
+      "print  \"\\n  Current gain, Av  =  \",  round(Ai,2)\n",
+      "print  \"\\n  Power gain, Av  =  \",  round(Ap,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage gain, Av  =   90.0\n",
+        "\n",
+        "  Current gain, Av  =   75.0\n",
+        "\n",
+        "  Power gain, Av  =   6750.0"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_3.ipynb
new file mode 100755
index 00000000..1cd2cf8d
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_12-checkpoint_3.ipynb
@@ -0,0 +1,85 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8d04a0c9a13bb913646430ed26822dd1fca0ba93c3a05b3b357e6ab9e3a29bdf"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 12: Transistors</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 157</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia  =  0.040;#  in  Amperes\n",
+      "I  =  50;#  in  Amperes\n",
+      "Rl  =  1.2;#  in  Kohms\n",
+      "Vc  =  7; # in Volts\n",
+      "R = 1; # in Kohms\n",
+      "Ib = 50*1E-6; # in Amps\n",
+      "Iic = 20*1E-6; #in Amps\n",
+      "\n",
+      "#calculation:\n",
+      "Vi = (70 - 30)*1E-6 * 1000\n",
+      "Vo = 3.6\n",
+      "Av = Vo/Vi\n",
+      "\n",
+      "Io = 3*1E-3\n",
+      "Ii = (70 - 30)*1E-6\n",
+      "Ai = Io/Ii\n",
+      "\n",
+      "Ap = Av*Ai\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage gain, Av  =  \",  round(Av,2)\n",
+      "print  \"\\n  Current gain, Av  =  \",  round(Ai,2)\n",
+      "print  \"\\n  Power gain, Av  =  \",  round(Ap,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage gain, Av  =   90.0\n",
+        "\n",
+        "  Current gain, Av  =   75.0\n",
+        "\n",
+        "  Power gain, Av  =   6750.0"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint.ipynb
new file mode 100755
index 00000000..a468bf64
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint.ipynb
@@ -0,0 +1,1203 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 13: D.c. circuit theory</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 168</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#a) Find the unknown currents (b) Determine the value of e.m.f. E\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Iab  =  50;#  in  ampere\n",
+      "Ibc  =  20;#  in  ampere\n",
+      "Iec  =  15;#  in  ampere\n",
+      "Idf  =  120;#  in  ampere\n",
+      "Ifg  =  40;#  in  ampere\n",
+      "Iab  =  50;#  in  ampere\n",
+      "I  =  2;#  in  ampere\n",
+      "V1  =  4;#  in  volts\n",
+      "V2  =  3;#  in  volts\n",
+      "V3  =  6;#  in  volts\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  2.5;#  in  ohms\n",
+      "R4  =  1.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1  =  Iab  -  Ibc\n",
+      "I2  =  Ibc  +  Iec\n",
+      "I3  =  I1  -  Idf\n",
+      "I4  =  Iec  -  I3\n",
+      "I5  =  Idf  -  Ifg\n",
+      "#  Applying  Kirchhoff\u2019s  voltage  law  and  moving  clockwise  around  the  loop  of  Figure  13.3(b)  starting  at  point  A:\n",
+      "E  =  I*R2  +  I*R3  +  I*R4  +  I*R1  -  V2  -  V3  +  V1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n (a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are  \",I1,\"A,  \",  I2,\"A,  \",  I3,\"A,  \",  I4,\"A,  \",  I5,\"A  respetively\\n\"\n",
+      "print  \"\\n (b)  value  of  e.m.f.  E  =  \",E,\"  Volts\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are   30 A,   35 A,   -90 A,   105 A,   80 A  respetively\n",
+        "\n",
+        "\n",
+        "  (b)  value  of  e.m.f.  E  =   9.0   Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 168</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "r1  =  2;#  in  ohms\n",
+      "r2  =  1;#  in  ohms\n",
+      "R  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#E1 = I1*(r1 + R) + I2*R,\n",
+      "#E2 = I1*R + (R + r2)*I2,\n",
+      "I2 = (E1*R - E2*(r1 + R))/(R**2 - (R+r1)*(R + r2))\n",
+      "I1 = (E1 - I2*R)/(r1 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  currents flowing are, I1 = \",round(I1,3),\"A,  and I2 = \",round(-1*I2,3),\"A, \"\n",
+      "print   \"and current flowing in middle branch is I1 - I2 = \", round(I1 + I2, 3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  currents flowing are, I1 =  0.857 A,  and I2 =  0.286 A, and current flowing in middle branch is I1 - I2 =  0.571 A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 169</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  12;#  in  volts\n",
+      "R1  =  0.5;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#E1 + E2 = I1*R1 + I2*R2\n",
+      "#E2 = - I1*R3 + I2*(R2 + R3) \n",
+      "I2 = ((E1 + E2)*R3 + E2*R1)/(R2*R3 + (R2+R3)*R1)\n",
+      "I1 = (E1 + E2 - I2*R2)/R1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  currents flowing are, I1 = \",round(I1,2),\"A,  and I2 = \",round(I2,2),\"A\"\n",
+      "print   \" and in R3 branch is I1 - I2\", round(I1 - I2,2),\" A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  currents flowing are, I1 =  6.52 A,  and I2 =  6.37 A and in R3 branch is I1 - I2 0.15  A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 170</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the currents in each of the resistors\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  54;#  in  volts\n",
+      "I  =  8;#  in  Amps\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  11;#  in  ohms\n",
+      "R3  =  14;#  in  ohms\n",
+      "R4  =  3;#  in  ohms\n",
+      "R5  =  32;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#V = (R1 + R2)*I1 - R2*I2\n",
+      "#0 = (R1 + R3)*I1 - R5*I2 - R3*I\n",
+      "I1 = V*R5/((R1 + R2)*R5 - (R1 + R3)*R2 + R3*I)\n",
+      "I2 = -1*(V - I1*(R2 + R1))/R2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the current flowing in the 2 ohm resistor = \",round(I1,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 14 ohm resistor = \",round(I - I1,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 32 ohm resistor = \",round(I2,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 11 ohm resistor = \",round(I1 - I2,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 3 ohm resistor = \",round(I - I1 + I2,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the current flowing in the 2 ohm resistor =  5.0 A,\n",
+        "\n",
+        "  the current flowing in the 14 ohm resistor =  3.0 A,\n",
+        "\n",
+        "  the current flowing in the 32 ohm resistor =  1.0 A,\n",
+        "\n",
+        "  the current flowing in the 11 ohm resistor =  4.0 A,\n",
+        "\n",
+        "  the current flowing in the 3 ohm resistor =  4.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 171</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current in each branch of the network by using the superposition theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  Volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
+      "I2 = (R2/(R + R2))*I1\n",
+      "I3 = (R/(R + R2))*I1\n",
+      "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
+      "I5 = (R1/(R + R1))*I4\n",
+      "I6 = (R/(R + R1))*I4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resultant current flowing through source 1 = \",round(I1 - I6,3),\"A,\"\n",
+      "print  \"\\n  Resultant current flowing through source 2 = \",round(I4 - I3,3),\"A,\"\n",
+      "print  \"\\n  Resultant current flowing through resistor R, = \",round(I2 + I5,3),\"A,\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resultant current flowing through source 1 =  0.857 A,\n",
+        "\n",
+        "  Resultant current flowing through source 2 =  -0.286 A,\n",
+        "\n",
+        "  Resultant current flowing through resistor R, =  0.571 A,"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 173</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find, using the superposition theorem, (a) the current flowing in and the pd across the 18 ohm resistor, \n",
+      "#(b) the current in the 8 V battery and (c) the current in the 3 V battery.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  8;#  in  volts\n",
+      "E2  =  3;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R =  18;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
+      "I2 = (R/(R + R2))*I1\n",
+      "I3 = (R2/(R + R2))*I1\n",
+      "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
+      "I5 = (R/(R + R1))*I4\n",
+      "I6 = (R1/(R + R1))*I4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resultant current in the 18 ohm resistor= \",round(I3 - I6,3),\"A \"\n",
+      "print   \"and P.d. across the 18 ohm resistor\",round((I3-I6)*R,3),\"V\"\n",
+      "print  \"\\n  (b)the current in the 8 V battery= \",round(I1 + I5,3),\"A\"\n",
+      "print  \"\\n  (c)current in the 3 V battery = \",round(I2 + I4,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resultant current in the 18 ohm resistor=  0.073 A and P.d. across the 18 ohm resistor 1.313 V\n",
+        "\n",
+        "  (b)the current in the 8 V battery=  2.229 A\n",
+        "\n",
+        "  (c)current in the 3 V battery =  2.156 A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 176</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use Th\u00b4evenin\u2019s theorem to find the current flowing in the 10 ohm resistor for the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  8;#  in  ohms\n",
+      "R3 =  5;#  in  ohms\n",
+      "R =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = V/(R1 + R2)\n",
+      "E = I1*R2\n",
+      "r = R3 + R1*R2/(R1 + R2)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 10 ohm resistor =  0.482 A"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 177</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 0.8 ohm resistor using Th\u00b4evenin\u2019s theorem.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  12;#  in  volts\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  4;#  in  ohms\n",
+      "R3 =  5;#  in  ohms\n",
+      "R =  0.8;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = V/(R1 + R2 + R3)\n",
+      "E = I1*R2\n",
+      "r = R2*(R1 + R3)/(R1 + R2 + R3)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 0.8 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 0.8 ohm resistor =  1.5 A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 178</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 4 ohm resistor using Th\u00b4evenin\u2019s theorem.Find also the power dissipated in the 4 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = (E1 - E2)/(R1 + R2)\n",
+      "E = E1 - I1*R1\n",
+      "r = R2*R1/(R1 + R2)\n",
+      "I = E/(R + r)\n",
+      "P = R*I**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A and power dissipated in the 4 ohm resistor = \",round(P,3),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 4 ohm resistor =  0.571 A and power dissipated in the 4 ohm resistor =  1.306 W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 178</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 3 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  24;#  in  volts\n",
+      "R  =  3;#  in  ohms\n",
+      "R1  =  20;#  in  ohms\n",
+      "R2 =  5;#  in  ohms\n",
+      "R3  = 10;#  in  ohms\n",
+      "R4 =  5/3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = R3*V/(R3 + R2)\n",
+      "r = R4 + R3*R2/(R3 + R2)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 3 ohm resistor =  2.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 179</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 32 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  54;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  14;#  in  ohms\n",
+      "R3  =  3;#  in  ohms\n",
+      "R4  =  11;#  in  ohms\n",
+      "R5  =  32;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vac = R1*E/(R1 + R4)\n",
+      "Vbc = R2*E/(R2 + R3)\n",
+      "V = Vbc - Vac\n",
+      "r = R4*R1/(R1 + R4) + R3*R2/(R3 + R2)\n",
+      "I = V/(R5 + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 32 ohm resistor = \",round(I,0),\"A flowing from A to B\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 32 ohm resistor =  1.0 A flowing from A to B"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 181</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 10 ohm resistor using Norton's theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  = 8;#  in  ohms\n",
+      "R3  =  5;#  in  ohms\n",
+      "R  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = E/R1\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "I = r*Isc/(r + R3 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 10 ohm resistor =  0.482 A"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 182</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 4 ohm resistor using Norton's theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = E1/R1 + E2/R2\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "I = r*Isc/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 4 ohm resistor =  0.571 A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 182</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 3 ohm resistor using Norton's theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  24;#  in  volts\n",
+      "R  =  3;#  in  ohms\n",
+      "R1  =  20;#  in  ohms\n",
+      "R2 =  5;#  in  ohms\n",
+      "R3  = 10;#  in  ohms\n",
+      "R4 =  5/3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R2\n",
+      "r = R3*R2/(R3 + R2)\n",
+      "I = r*Isc/(r + R4 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 3 ohm resistor =  2.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 183</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 2 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  15;#  in  Amps\n",
+      "R  =  2;#  in  ohms\n",
+      "R1  = 6;#  in  ohms\n",
+      "R2 =  4;#  in  ohms\n",
+      "R3  = 8;#  in  ohms\n",
+      "R4 =  7;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = R1*I/(R1 + R2)\n",
+      "r = (R1 + R2)*(R3 + R4)/(R3 + R1 + R4 + R2)\n",
+      "I = r*Isc/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 2 ohm resistor = \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 2 ohm resistor =  6.75 A"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert the circuit to an equivalent Norton network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  Volts\n",
+      "R  =  2;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n equivalent Norton network contains Current Source of amp = \",round(Isc,0),\"A and a resistor of \",R,\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " equivalent Norton network contains Current Source of amp =  5.0 A and a resistor of  2  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert the circuit to an equivalent Thevenin network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Isc  =  4;#  in  Amps\n",
+      "R  =  3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = Isc*R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,0),\"V and a resistor of \",R,\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " equivalent Thevenin network contains Voltage Source of  12.0 V and a resistor of  3  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Convert the circuit to an equivalent Thevenin networkby initially converting to a Norton equivalent circuit.\n",
+      "#(b)Determine the current flowing in the 1.8 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  12;#  in  Volts\n",
+      "E2  =  24;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R  =  1.8;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc1 = E1/R1\n",
+      "Isc2 = E2/R2\n",
+      "I1 = Isc1 + Isc2\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "E = I1*r\n",
+      "I = E/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (A)equivalent Norton network contains Current Source of \",round(I1,1),\"A and a resistor of \",r,\" ohm\"\n",
+      "print  \"\\n     equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",r,\" ohm\"\n",
+      "print  \"\\n  (B)the current flowing in the 1.8 ohm resistor is \",round(I,1),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (A)equivalent Norton network contains Current Source of  16.0 A and a resistor of  1.2  ohm\n",
+        "\n",
+        "     equivalent Thevenin network contains Voltage Source of  19.2 V and a resistor of  1.2  ohm\n",
+        "\n",
+        "  (B)the current flowing in the 1.8 ohm resistor is  6.4 A"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 186</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Determine by successive conversions between Th\u00b4evenin and Norton equivalent networks \n",
+      "#a Thevenin equivalent circuit for terminals AB.\n",
+      "#(b)determine the current flowing in the 200 ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  10;#  in  Volts\n",
+      "E2  =  6;#  in  Volts\n",
+      "I1 = 0.001;#in Amp\n",
+      "R1  =  2000;#  in  ohms\n",
+      "R2  =  3000;#  in  ohms\n",
+      "R3  =  600;#  in  ohms\n",
+      "R  =  200;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc1 = E1/R1\n",
+      "Isc2 = E2/R2\n",
+      "I2 = Isc1 + Isc2\n",
+      "r1 = R1*R2/(R1 + R2)\n",
+      "E = I2*r1 - I1*R3\n",
+      "r = r1 + R3\n",
+      "I = E/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (A)equivalent Norton network contains Current Source of \",round(I2*1000,0),\"mA and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
+      "print  \"\\n     equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
+      "print  \"\\n  (B)the current flowing in the 200 ohm resistor is \",round(I*1000,1),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (A)equivalent Norton network contains Current Source of  7.0 mA and a resistor of  1.2  Kohm\n",
+        "\n",
+        "     equivalent Thevenin network contains Voltage Source of  7.8 V and a resistor of  1.2  Kohm\n",
+        "\n",
+        "  (B)the current flowing in the 200 ohm resistor is  3.9 mA"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 188</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the power dissipated by the load in each case.\n",
+      "#Plot a graph of RL (horizontally) against power (vertically) and determine the maximum power dissipated.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "E  =  6;#  in  Volts\n",
+      "R  =  2.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "RL = []\n",
+      "P = []\n",
+      "k = []\n",
+      "for h in range(6):\n",
+      "    RL.append(h - 0.5)\n",
+      "    k = h - 0.5\n",
+      "    P.append(k*((E/(R + k))**2))\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(RL,P,'-')\n",
+      "#plot(RL,P,'-')\n",
+      "xlabel('RL(ohm)')\n",
+      "ylabel('Power(W)')\n",
+      "show()\n",
+      "Pmax= R*(E/(2*R))**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  A graph of RL against P is shown in Figure\" \n",
+      "print  \"  The maximum value of power is\", Pmax,\"W which occurs when RL =\",R,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'."
+       ]
+      },
+      {
+       "output_type": "display_data",
+       "png": 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p6emiS1PMwIEDAQBBQUGYOnUqPvzwQ8ybN6/N57rM3z5q6c/dcsstAOQVP6Wl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\nZLqKDYkAAAAASUVORK5CYII=\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  A graph of RL against P is shown in Figure\n",
+        "  The maximum value of power is 3.6 W which occurs when RL = 2.5 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 188</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#State the value of load resistance that gives maximum power dissipation and determine the value of this power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "E  =  30;#  in  Volts\n",
+      "R  =  1.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "RL = R\n",
+      "I = E/(R + RL)\n",
+      "P = I**2*RL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"  The maximum value of power is\", P,\"W which occurs when RL =\",RL,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "  The maximum value of power is 150.0 W which occurs when RL = 1.5 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 189</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the value of the load resistor RL  that gives maximum power dissipation and determine the value of this power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "V  =  15;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  12;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = (R2/(R2+ R1))*V\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "RL = r\n",
+      "I = E/(r + RL)\n",
+      "P = I**2*RL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"  The maximum value of power is\", P,\"W which occurs when Total Load RL =\",RL,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "  The maximum value of power is 15.0 W which occurs when Total Load RL = 2.4 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_1.ipynb
new file mode 100755
index 00000000..a468bf64
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_1.ipynb
@@ -0,0 +1,1203 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 13: D.c. circuit theory</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 168</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#a) Find the unknown currents (b) Determine the value of e.m.f. E\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Iab  =  50;#  in  ampere\n",
+      "Ibc  =  20;#  in  ampere\n",
+      "Iec  =  15;#  in  ampere\n",
+      "Idf  =  120;#  in  ampere\n",
+      "Ifg  =  40;#  in  ampere\n",
+      "Iab  =  50;#  in  ampere\n",
+      "I  =  2;#  in  ampere\n",
+      "V1  =  4;#  in  volts\n",
+      "V2  =  3;#  in  volts\n",
+      "V3  =  6;#  in  volts\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  2.5;#  in  ohms\n",
+      "R4  =  1.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1  =  Iab  -  Ibc\n",
+      "I2  =  Ibc  +  Iec\n",
+      "I3  =  I1  -  Idf\n",
+      "I4  =  Iec  -  I3\n",
+      "I5  =  Idf  -  Ifg\n",
+      "#  Applying  Kirchhoff\u2019s  voltage  law  and  moving  clockwise  around  the  loop  of  Figure  13.3(b)  starting  at  point  A:\n",
+      "E  =  I*R2  +  I*R3  +  I*R4  +  I*R1  -  V2  -  V3  +  V1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n (a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are  \",I1,\"A,  \",  I2,\"A,  \",  I3,\"A,  \",  I4,\"A,  \",  I5,\"A  respetively\\n\"\n",
+      "print  \"\\n (b)  value  of  e.m.f.  E  =  \",E,\"  Volts\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are   30 A,   35 A,   -90 A,   105 A,   80 A  respetively\n",
+        "\n",
+        "\n",
+        "  (b)  value  of  e.m.f.  E  =   9.0   Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 168</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "r1  =  2;#  in  ohms\n",
+      "r2  =  1;#  in  ohms\n",
+      "R  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#E1 = I1*(r1 + R) + I2*R,\n",
+      "#E2 = I1*R + (R + r2)*I2,\n",
+      "I2 = (E1*R - E2*(r1 + R))/(R**2 - (R+r1)*(R + r2))\n",
+      "I1 = (E1 - I2*R)/(r1 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  currents flowing are, I1 = \",round(I1,3),\"A,  and I2 = \",round(-1*I2,3),\"A, \"\n",
+      "print   \"and current flowing in middle branch is I1 - I2 = \", round(I1 + I2, 3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  currents flowing are, I1 =  0.857 A,  and I2 =  0.286 A, and current flowing in middle branch is I1 - I2 =  0.571 A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 169</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  12;#  in  volts\n",
+      "R1  =  0.5;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#E1 + E2 = I1*R1 + I2*R2\n",
+      "#E2 = - I1*R3 + I2*(R2 + R3) \n",
+      "I2 = ((E1 + E2)*R3 + E2*R1)/(R2*R3 + (R2+R3)*R1)\n",
+      "I1 = (E1 + E2 - I2*R2)/R1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  currents flowing are, I1 = \",round(I1,2),\"A,  and I2 = \",round(I2,2),\"A\"\n",
+      "print   \" and in R3 branch is I1 - I2\", round(I1 - I2,2),\" A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  currents flowing are, I1 =  6.52 A,  and I2 =  6.37 A and in R3 branch is I1 - I2 0.15  A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 170</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the currents in each of the resistors\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  54;#  in  volts\n",
+      "I  =  8;#  in  Amps\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  11;#  in  ohms\n",
+      "R3  =  14;#  in  ohms\n",
+      "R4  =  3;#  in  ohms\n",
+      "R5  =  32;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#V = (R1 + R2)*I1 - R2*I2\n",
+      "#0 = (R1 + R3)*I1 - R5*I2 - R3*I\n",
+      "I1 = V*R5/((R1 + R2)*R5 - (R1 + R3)*R2 + R3*I)\n",
+      "I2 = -1*(V - I1*(R2 + R1))/R2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the current flowing in the 2 ohm resistor = \",round(I1,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 14 ohm resistor = \",round(I - I1,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 32 ohm resistor = \",round(I2,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 11 ohm resistor = \",round(I1 - I2,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 3 ohm resistor = \",round(I - I1 + I2,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the current flowing in the 2 ohm resistor =  5.0 A,\n",
+        "\n",
+        "  the current flowing in the 14 ohm resistor =  3.0 A,\n",
+        "\n",
+        "  the current flowing in the 32 ohm resistor =  1.0 A,\n",
+        "\n",
+        "  the current flowing in the 11 ohm resistor =  4.0 A,\n",
+        "\n",
+        "  the current flowing in the 3 ohm resistor =  4.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 171</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current in each branch of the network by using the superposition theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  Volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
+      "I2 = (R2/(R + R2))*I1\n",
+      "I3 = (R/(R + R2))*I1\n",
+      "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
+      "I5 = (R1/(R + R1))*I4\n",
+      "I6 = (R/(R + R1))*I4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resultant current flowing through source 1 = \",round(I1 - I6,3),\"A,\"\n",
+      "print  \"\\n  Resultant current flowing through source 2 = \",round(I4 - I3,3),\"A,\"\n",
+      "print  \"\\n  Resultant current flowing through resistor R, = \",round(I2 + I5,3),\"A,\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resultant current flowing through source 1 =  0.857 A,\n",
+        "\n",
+        "  Resultant current flowing through source 2 =  -0.286 A,\n",
+        "\n",
+        "  Resultant current flowing through resistor R, =  0.571 A,"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 173</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find, using the superposition theorem, (a) the current flowing in and the pd across the 18 ohm resistor, \n",
+      "#(b) the current in the 8 V battery and (c) the current in the 3 V battery.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  8;#  in  volts\n",
+      "E2  =  3;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R =  18;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
+      "I2 = (R/(R + R2))*I1\n",
+      "I3 = (R2/(R + R2))*I1\n",
+      "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
+      "I5 = (R/(R + R1))*I4\n",
+      "I6 = (R1/(R + R1))*I4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resultant current in the 18 ohm resistor= \",round(I3 - I6,3),\"A \"\n",
+      "print   \"and P.d. across the 18 ohm resistor\",round((I3-I6)*R,3),\"V\"\n",
+      "print  \"\\n  (b)the current in the 8 V battery= \",round(I1 + I5,3),\"A\"\n",
+      "print  \"\\n  (c)current in the 3 V battery = \",round(I2 + I4,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resultant current in the 18 ohm resistor=  0.073 A and P.d. across the 18 ohm resistor 1.313 V\n",
+        "\n",
+        "  (b)the current in the 8 V battery=  2.229 A\n",
+        "\n",
+        "  (c)current in the 3 V battery =  2.156 A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 176</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use Th\u00b4evenin\u2019s theorem to find the current flowing in the 10 ohm resistor for the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  8;#  in  ohms\n",
+      "R3 =  5;#  in  ohms\n",
+      "R =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = V/(R1 + R2)\n",
+      "E = I1*R2\n",
+      "r = R3 + R1*R2/(R1 + R2)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 10 ohm resistor =  0.482 A"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 177</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 0.8 ohm resistor using Th\u00b4evenin\u2019s theorem.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  12;#  in  volts\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  4;#  in  ohms\n",
+      "R3 =  5;#  in  ohms\n",
+      "R =  0.8;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = V/(R1 + R2 + R3)\n",
+      "E = I1*R2\n",
+      "r = R2*(R1 + R3)/(R1 + R2 + R3)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 0.8 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 0.8 ohm resistor =  1.5 A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 178</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 4 ohm resistor using Th\u00b4evenin\u2019s theorem.Find also the power dissipated in the 4 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = (E1 - E2)/(R1 + R2)\n",
+      "E = E1 - I1*R1\n",
+      "r = R2*R1/(R1 + R2)\n",
+      "I = E/(R + r)\n",
+      "P = R*I**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A and power dissipated in the 4 ohm resistor = \",round(P,3),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 4 ohm resistor =  0.571 A and power dissipated in the 4 ohm resistor =  1.306 W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 178</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 3 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  24;#  in  volts\n",
+      "R  =  3;#  in  ohms\n",
+      "R1  =  20;#  in  ohms\n",
+      "R2 =  5;#  in  ohms\n",
+      "R3  = 10;#  in  ohms\n",
+      "R4 =  5/3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = R3*V/(R3 + R2)\n",
+      "r = R4 + R3*R2/(R3 + R2)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 3 ohm resistor =  2.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 179</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 32 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  54;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  14;#  in  ohms\n",
+      "R3  =  3;#  in  ohms\n",
+      "R4  =  11;#  in  ohms\n",
+      "R5  =  32;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vac = R1*E/(R1 + R4)\n",
+      "Vbc = R2*E/(R2 + R3)\n",
+      "V = Vbc - Vac\n",
+      "r = R4*R1/(R1 + R4) + R3*R2/(R3 + R2)\n",
+      "I = V/(R5 + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 32 ohm resistor = \",round(I,0),\"A flowing from A to B\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 32 ohm resistor =  1.0 A flowing from A to B"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 181</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 10 ohm resistor using Norton's theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  = 8;#  in  ohms\n",
+      "R3  =  5;#  in  ohms\n",
+      "R  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = E/R1\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "I = r*Isc/(r + R3 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 10 ohm resistor =  0.482 A"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 182</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 4 ohm resistor using Norton's theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = E1/R1 + E2/R2\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "I = r*Isc/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 4 ohm resistor =  0.571 A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 182</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 3 ohm resistor using Norton's theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  24;#  in  volts\n",
+      "R  =  3;#  in  ohms\n",
+      "R1  =  20;#  in  ohms\n",
+      "R2 =  5;#  in  ohms\n",
+      "R3  = 10;#  in  ohms\n",
+      "R4 =  5/3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R2\n",
+      "r = R3*R2/(R3 + R2)\n",
+      "I = r*Isc/(r + R4 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 3 ohm resistor =  2.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 183</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 2 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  15;#  in  Amps\n",
+      "R  =  2;#  in  ohms\n",
+      "R1  = 6;#  in  ohms\n",
+      "R2 =  4;#  in  ohms\n",
+      "R3  = 8;#  in  ohms\n",
+      "R4 =  7;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = R1*I/(R1 + R2)\n",
+      "r = (R1 + R2)*(R3 + R4)/(R3 + R1 + R4 + R2)\n",
+      "I = r*Isc/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 2 ohm resistor = \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 2 ohm resistor =  6.75 A"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert the circuit to an equivalent Norton network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  Volts\n",
+      "R  =  2;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n equivalent Norton network contains Current Source of amp = \",round(Isc,0),\"A and a resistor of \",R,\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " equivalent Norton network contains Current Source of amp =  5.0 A and a resistor of  2  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert the circuit to an equivalent Thevenin network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Isc  =  4;#  in  Amps\n",
+      "R  =  3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = Isc*R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,0),\"V and a resistor of \",R,\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " equivalent Thevenin network contains Voltage Source of  12.0 V and a resistor of  3  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Convert the circuit to an equivalent Thevenin networkby initially converting to a Norton equivalent circuit.\n",
+      "#(b)Determine the current flowing in the 1.8 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  12;#  in  Volts\n",
+      "E2  =  24;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R  =  1.8;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc1 = E1/R1\n",
+      "Isc2 = E2/R2\n",
+      "I1 = Isc1 + Isc2\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "E = I1*r\n",
+      "I = E/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (A)equivalent Norton network contains Current Source of \",round(I1,1),\"A and a resistor of \",r,\" ohm\"\n",
+      "print  \"\\n     equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",r,\" ohm\"\n",
+      "print  \"\\n  (B)the current flowing in the 1.8 ohm resistor is \",round(I,1),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (A)equivalent Norton network contains Current Source of  16.0 A and a resistor of  1.2  ohm\n",
+        "\n",
+        "     equivalent Thevenin network contains Voltage Source of  19.2 V and a resistor of  1.2  ohm\n",
+        "\n",
+        "  (B)the current flowing in the 1.8 ohm resistor is  6.4 A"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 186</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Determine by successive conversions between Th\u00b4evenin and Norton equivalent networks \n",
+      "#a Thevenin equivalent circuit for terminals AB.\n",
+      "#(b)determine the current flowing in the 200 ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  10;#  in  Volts\n",
+      "E2  =  6;#  in  Volts\n",
+      "I1 = 0.001;#in Amp\n",
+      "R1  =  2000;#  in  ohms\n",
+      "R2  =  3000;#  in  ohms\n",
+      "R3  =  600;#  in  ohms\n",
+      "R  =  200;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc1 = E1/R1\n",
+      "Isc2 = E2/R2\n",
+      "I2 = Isc1 + Isc2\n",
+      "r1 = R1*R2/(R1 + R2)\n",
+      "E = I2*r1 - I1*R3\n",
+      "r = r1 + R3\n",
+      "I = E/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (A)equivalent Norton network contains Current Source of \",round(I2*1000,0),\"mA and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
+      "print  \"\\n     equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
+      "print  \"\\n  (B)the current flowing in the 200 ohm resistor is \",round(I*1000,1),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (A)equivalent Norton network contains Current Source of  7.0 mA and a resistor of  1.2  Kohm\n",
+        "\n",
+        "     equivalent Thevenin network contains Voltage Source of  7.8 V and a resistor of  1.2  Kohm\n",
+        "\n",
+        "  (B)the current flowing in the 200 ohm resistor is  3.9 mA"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 188</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the power dissipated by the load in each case.\n",
+      "#Plot a graph of RL (horizontally) against power (vertically) and determine the maximum power dissipated.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "E  =  6;#  in  Volts\n",
+      "R  =  2.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "RL = []\n",
+      "P = []\n",
+      "k = []\n",
+      "for h in range(6):\n",
+      "    RL.append(h - 0.5)\n",
+      "    k = h - 0.5\n",
+      "    P.append(k*((E/(R + k))**2))\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(RL,P,'-')\n",
+      "#plot(RL,P,'-')\n",
+      "xlabel('RL(ohm)')\n",
+      "ylabel('Power(W)')\n",
+      "show()\n",
+      "Pmax= R*(E/(2*R))**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  A graph of RL against P is shown in Figure\" \n",
+      "print  \"  The maximum value of power is\", Pmax,\"W which occurs when RL =\",R,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'."
+       ]
+      },
+      {
+       "output_type": "display_data",
+       "png": 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p6emiS1PMwIEDAQBBQUGYOnUqPvzwQ8ybN6/N57rM3z5q6c/dcsstAOQVP6Wl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\nZLqKDYkAAAAASUVORK5CYII=\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  A graph of RL against P is shown in Figure\n",
+        "  The maximum value of power is 3.6 W which occurs when RL = 2.5 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 188</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#State the value of load resistance that gives maximum power dissipation and determine the value of this power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "E  =  30;#  in  Volts\n",
+      "R  =  1.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "RL = R\n",
+      "I = E/(R + RL)\n",
+      "P = I**2*RL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"  The maximum value of power is\", P,\"W which occurs when RL =\",RL,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "  The maximum value of power is 150.0 W which occurs when RL = 1.5 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 189</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the value of the load resistor RL  that gives maximum power dissipation and determine the value of this power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "V  =  15;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  12;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = (R2/(R2+ R1))*V\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "RL = r\n",
+      "I = E/(r + RL)\n",
+      "P = I**2*RL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"  The maximum value of power is\", P,\"W which occurs when Total Load RL =\",RL,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "  The maximum value of power is 15.0 W which occurs when Total Load RL = 2.4 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_2.ipynb
new file mode 100755
index 00000000..a468bf64
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_2.ipynb
@@ -0,0 +1,1203 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 13: D.c. circuit theory</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 168</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#a) Find the unknown currents (b) Determine the value of e.m.f. E\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Iab  =  50;#  in  ampere\n",
+      "Ibc  =  20;#  in  ampere\n",
+      "Iec  =  15;#  in  ampere\n",
+      "Idf  =  120;#  in  ampere\n",
+      "Ifg  =  40;#  in  ampere\n",
+      "Iab  =  50;#  in  ampere\n",
+      "I  =  2;#  in  ampere\n",
+      "V1  =  4;#  in  volts\n",
+      "V2  =  3;#  in  volts\n",
+      "V3  =  6;#  in  volts\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  2.5;#  in  ohms\n",
+      "R4  =  1.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1  =  Iab  -  Ibc\n",
+      "I2  =  Ibc  +  Iec\n",
+      "I3  =  I1  -  Idf\n",
+      "I4  =  Iec  -  I3\n",
+      "I5  =  Idf  -  Ifg\n",
+      "#  Applying  Kirchhoff\u2019s  voltage  law  and  moving  clockwise  around  the  loop  of  Figure  13.3(b)  starting  at  point  A:\n",
+      "E  =  I*R2  +  I*R3  +  I*R4  +  I*R1  -  V2  -  V3  +  V1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n (a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are  \",I1,\"A,  \",  I2,\"A,  \",  I3,\"A,  \",  I4,\"A,  \",  I5,\"A  respetively\\n\"\n",
+      "print  \"\\n (b)  value  of  e.m.f.  E  =  \",E,\"  Volts\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are   30 A,   35 A,   -90 A,   105 A,   80 A  respetively\n",
+        "\n",
+        "\n",
+        "  (b)  value  of  e.m.f.  E  =   9.0   Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 168</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "r1  =  2;#  in  ohms\n",
+      "r2  =  1;#  in  ohms\n",
+      "R  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#E1 = I1*(r1 + R) + I2*R,\n",
+      "#E2 = I1*R + (R + r2)*I2,\n",
+      "I2 = (E1*R - E2*(r1 + R))/(R**2 - (R+r1)*(R + r2))\n",
+      "I1 = (E1 - I2*R)/(r1 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  currents flowing are, I1 = \",round(I1,3),\"A,  and I2 = \",round(-1*I2,3),\"A, \"\n",
+      "print   \"and current flowing in middle branch is I1 - I2 = \", round(I1 + I2, 3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  currents flowing are, I1 =  0.857 A,  and I2 =  0.286 A, and current flowing in middle branch is I1 - I2 =  0.571 A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 169</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use Kirchhoff\u2019s laws to determine the currents flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  12;#  in  volts\n",
+      "R1  =  0.5;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#E1 + E2 = I1*R1 + I2*R2\n",
+      "#E2 = - I1*R3 + I2*(R2 + R3) \n",
+      "I2 = ((E1 + E2)*R3 + E2*R1)/(R2*R3 + (R2+R3)*R1)\n",
+      "I1 = (E1 + E2 - I2*R2)/R1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  currents flowing are, I1 = \",round(I1,2),\"A,  and I2 = \",round(I2,2),\"A\"\n",
+      "print   \" and in R3 branch is I1 - I2\", round(I1 - I2,2),\" A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  currents flowing are, I1 =  6.52 A,  and I2 =  6.37 A and in R3 branch is I1 - I2 0.15  A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 170</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the currents in each of the resistors\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  54;#  in  volts\n",
+      "I  =  8;#  in  Amps\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  11;#  in  ohms\n",
+      "R3  =  14;#  in  ohms\n",
+      "R4  =  3;#  in  ohms\n",
+      "R5  =  32;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#V = (R1 + R2)*I1 - R2*I2\n",
+      "#0 = (R1 + R3)*I1 - R5*I2 - R3*I\n",
+      "I1 = V*R5/((R1 + R2)*R5 - (R1 + R3)*R2 + R3*I)\n",
+      "I2 = -1*(V - I1*(R2 + R1))/R2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the current flowing in the 2 ohm resistor = \",round(I1,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 14 ohm resistor = \",round(I - I1,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 32 ohm resistor = \",round(I2,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 11 ohm resistor = \",round(I1 - I2,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 3 ohm resistor = \",round(I - I1 + I2,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the current flowing in the 2 ohm resistor =  5.0 A,\n",
+        "\n",
+        "  the current flowing in the 14 ohm resistor =  3.0 A,\n",
+        "\n",
+        "  the current flowing in the 32 ohm resistor =  1.0 A,\n",
+        "\n",
+        "  the current flowing in the 11 ohm resistor =  4.0 A,\n",
+        "\n",
+        "  the current flowing in the 3 ohm resistor =  4.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 171</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current in each branch of the network by using the superposition theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  Volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
+      "I2 = (R2/(R + R2))*I1\n",
+      "I3 = (R/(R + R2))*I1\n",
+      "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
+      "I5 = (R1/(R + R1))*I4\n",
+      "I6 = (R/(R + R1))*I4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resultant current flowing through source 1 = \",round(I1 - I6,3),\"A,\"\n",
+      "print  \"\\n  Resultant current flowing through source 2 = \",round(I4 - I3,3),\"A,\"\n",
+      "print  \"\\n  Resultant current flowing through resistor R, = \",round(I2 + I5,3),\"A,\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resultant current flowing through source 1 =  0.857 A,\n",
+        "\n",
+        "  Resultant current flowing through source 2 =  -0.286 A,\n",
+        "\n",
+        "  Resultant current flowing through resistor R, =  0.571 A,"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 173</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find, using the superposition theorem, (a) the current flowing in and the pd across the 18 ohm resistor, \n",
+      "#(b) the current in the 8 V battery and (c) the current in the 3 V battery.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  8;#  in  volts\n",
+      "E2  =  3;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R =  18;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
+      "I2 = (R/(R + R2))*I1\n",
+      "I3 = (R2/(R + R2))*I1\n",
+      "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
+      "I5 = (R/(R + R1))*I4\n",
+      "I6 = (R1/(R + R1))*I4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resultant current in the 18 ohm resistor= \",round(I3 - I6,3),\"A \"\n",
+      "print   \"and P.d. across the 18 ohm resistor\",round((I3-I6)*R,3),\"V\"\n",
+      "print  \"\\n  (b)the current in the 8 V battery= \",round(I1 + I5,3),\"A\"\n",
+      "print  \"\\n  (c)current in the 3 V battery = \",round(I2 + I4,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resultant current in the 18 ohm resistor=  0.073 A and P.d. across the 18 ohm resistor 1.313 V\n",
+        "\n",
+        "  (b)the current in the 8 V battery=  2.229 A\n",
+        "\n",
+        "  (c)current in the 3 V battery =  2.156 A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 176</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use Th\u00b4evenin\u2019s theorem to find the current flowing in the 10 ohm resistor for the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  8;#  in  ohms\n",
+      "R3 =  5;#  in  ohms\n",
+      "R =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = V/(R1 + R2)\n",
+      "E = I1*R2\n",
+      "r = R3 + R1*R2/(R1 + R2)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 10 ohm resistor =  0.482 A"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 177</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 0.8 ohm resistor using Th\u00b4evenin\u2019s theorem.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  12;#  in  volts\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  4;#  in  ohms\n",
+      "R3 =  5;#  in  ohms\n",
+      "R =  0.8;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = V/(R1 + R2 + R3)\n",
+      "E = I1*R2\n",
+      "r = R2*(R1 + R3)/(R1 + R2 + R3)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 0.8 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 0.8 ohm resistor =  1.5 A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 178</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 4 ohm resistor using Th\u00b4evenin\u2019s theorem.Find also the power dissipated in the 4 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = (E1 - E2)/(R1 + R2)\n",
+      "E = E1 - I1*R1\n",
+      "r = R2*R1/(R1 + R2)\n",
+      "I = E/(R + r)\n",
+      "P = R*I**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A and power dissipated in the 4 ohm resistor = \",round(P,3),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 4 ohm resistor =  0.571 A and power dissipated in the 4 ohm resistor =  1.306 W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 178</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 3 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  24;#  in  volts\n",
+      "R  =  3;#  in  ohms\n",
+      "R1  =  20;#  in  ohms\n",
+      "R2 =  5;#  in  ohms\n",
+      "R3  = 10;#  in  ohms\n",
+      "R4 =  5/3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = R3*V/(R3 + R2)\n",
+      "r = R4 + R3*R2/(R3 + R2)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 3 ohm resistor =  2.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 179</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 32 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  54;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  14;#  in  ohms\n",
+      "R3  =  3;#  in  ohms\n",
+      "R4  =  11;#  in  ohms\n",
+      "R5  =  32;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vac = R1*E/(R1 + R4)\n",
+      "Vbc = R2*E/(R2 + R3)\n",
+      "V = Vbc - Vac\n",
+      "r = R4*R1/(R1 + R4) + R3*R2/(R3 + R2)\n",
+      "I = V/(R5 + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 32 ohm resistor = \",round(I,0),\"A flowing from A to B\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 32 ohm resistor =  1.0 A flowing from A to B"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 181</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 10 ohm resistor using Norton's theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  = 8;#  in  ohms\n",
+      "R3  =  5;#  in  ohms\n",
+      "R  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = E/R1\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "I = r*Isc/(r + R3 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 10 ohm resistor =  0.482 A"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 182</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 4 ohm resistor using Norton's theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = E1/R1 + E2/R2\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "I = r*Isc/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 4 ohm resistor =  0.571 A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 182</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 3 ohm resistor using Norton's theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  24;#  in  volts\n",
+      "R  =  3;#  in  ohms\n",
+      "R1  =  20;#  in  ohms\n",
+      "R2 =  5;#  in  ohms\n",
+      "R3  = 10;#  in  ohms\n",
+      "R4 =  5/3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R2\n",
+      "r = R3*R2/(R3 + R2)\n",
+      "I = r*Isc/(r + R4 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 3 ohm resistor =  2.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 183</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 2 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  15;#  in  Amps\n",
+      "R  =  2;#  in  ohms\n",
+      "R1  = 6;#  in  ohms\n",
+      "R2 =  4;#  in  ohms\n",
+      "R3  = 8;#  in  ohms\n",
+      "R4 =  7;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = R1*I/(R1 + R2)\n",
+      "r = (R1 + R2)*(R3 + R4)/(R3 + R1 + R4 + R2)\n",
+      "I = r*Isc/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 2 ohm resistor = \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 2 ohm resistor =  6.75 A"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert the circuit to an equivalent Norton network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  Volts\n",
+      "R  =  2;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n equivalent Norton network contains Current Source of amp = \",round(Isc,0),\"A and a resistor of \",R,\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " equivalent Norton network contains Current Source of amp =  5.0 A and a resistor of  2  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert the circuit to an equivalent Thevenin network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Isc  =  4;#  in  Amps\n",
+      "R  =  3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = Isc*R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,0),\"V and a resistor of \",R,\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " equivalent Thevenin network contains Voltage Source of  12.0 V and a resistor of  3  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Convert the circuit to an equivalent Thevenin networkby initially converting to a Norton equivalent circuit.\n",
+      "#(b)Determine the current flowing in the 1.8 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  12;#  in  Volts\n",
+      "E2  =  24;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R  =  1.8;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc1 = E1/R1\n",
+      "Isc2 = E2/R2\n",
+      "I1 = Isc1 + Isc2\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "E = I1*r\n",
+      "I = E/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (A)equivalent Norton network contains Current Source of \",round(I1,1),\"A and a resistor of \",r,\" ohm\"\n",
+      "print  \"\\n     equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",r,\" ohm\"\n",
+      "print  \"\\n  (B)the current flowing in the 1.8 ohm resistor is \",round(I,1),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (A)equivalent Norton network contains Current Source of  16.0 A and a resistor of  1.2  ohm\n",
+        "\n",
+        "     equivalent Thevenin network contains Voltage Source of  19.2 V and a resistor of  1.2  ohm\n",
+        "\n",
+        "  (B)the current flowing in the 1.8 ohm resistor is  6.4 A"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 186</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Determine by successive conversions between Th\u00b4evenin and Norton equivalent networks \n",
+      "#a Thevenin equivalent circuit for terminals AB.\n",
+      "#(b)determine the current flowing in the 200 ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  10;#  in  Volts\n",
+      "E2  =  6;#  in  Volts\n",
+      "I1 = 0.001;#in Amp\n",
+      "R1  =  2000;#  in  ohms\n",
+      "R2  =  3000;#  in  ohms\n",
+      "R3  =  600;#  in  ohms\n",
+      "R  =  200;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc1 = E1/R1\n",
+      "Isc2 = E2/R2\n",
+      "I2 = Isc1 + Isc2\n",
+      "r1 = R1*R2/(R1 + R2)\n",
+      "E = I2*r1 - I1*R3\n",
+      "r = r1 + R3\n",
+      "I = E/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (A)equivalent Norton network contains Current Source of \",round(I2*1000,0),\"mA and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
+      "print  \"\\n     equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
+      "print  \"\\n  (B)the current flowing in the 200 ohm resistor is \",round(I*1000,1),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (A)equivalent Norton network contains Current Source of  7.0 mA and a resistor of  1.2  Kohm\n",
+        "\n",
+        "     equivalent Thevenin network contains Voltage Source of  7.8 V and a resistor of  1.2  Kohm\n",
+        "\n",
+        "  (B)the current flowing in the 200 ohm resistor is  3.9 mA"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 188</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the power dissipated by the load in each case.\n",
+      "#Plot a graph of RL (horizontally) against power (vertically) and determine the maximum power dissipated.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "E  =  6;#  in  Volts\n",
+      "R  =  2.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "RL = []\n",
+      "P = []\n",
+      "k = []\n",
+      "for h in range(6):\n",
+      "    RL.append(h - 0.5)\n",
+      "    k = h - 0.5\n",
+      "    P.append(k*((E/(R + k))**2))\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(RL,P,'-')\n",
+      "#plot(RL,P,'-')\n",
+      "xlabel('RL(ohm)')\n",
+      "ylabel('Power(W)')\n",
+      "show()\n",
+      "Pmax= R*(E/(2*R))**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  A graph of RL against P is shown in Figure\" \n",
+      "print  \"  The maximum value of power is\", Pmax,\"W which occurs when RL =\",R,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'."
+       ]
+      },
+      {
+       "output_type": "display_data",
+       "png": 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p6emiS1PMwIEDAQBBQUGYOnUqPvzwQ8ybN6/N57rM3z5q6c/dcsstAOQVP6Wl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\nZLqKDYkAAAAASUVORK5CYII=\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  A graph of RL against P is shown in Figure\n",
+        "  The maximum value of power is 3.6 W which occurs when RL = 2.5 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 188</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#State the value of load resistance that gives maximum power dissipation and determine the value of this power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "E  =  30;#  in  Volts\n",
+      "R  =  1.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "RL = R\n",
+      "I = E/(R + RL)\n",
+      "P = I**2*RL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"  The maximum value of power is\", P,\"W which occurs when RL =\",RL,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "  The maximum value of power is 150.0 W which occurs when RL = 1.5 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 189</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the value of the load resistor RL  that gives maximum power dissipation and determine the value of this power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "V  =  15;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  12;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = (R2/(R2+ R1))*V\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "RL = r\n",
+      "I = E/(r + RL)\n",
+      "P = I**2*RL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"  The maximum value of power is\", P,\"W which occurs when Total Load RL =\",RL,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "  The maximum value of power is 15.0 W which occurs when Total Load RL = 2.4 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_3.ipynb
new file mode 100755
index 00000000..6f69e0f8
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_13-checkpoint_3.ipynb
@@ -0,0 +1,1200 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c3f29745671b17635c27a2d2f8c2d0dd1308d33b0632eea3f4c1679de303c4db"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 13: D.c. circuit theory</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 168</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Iab  =  50;#  in  ampere\n",
+      "Ibc  =  20;#  in  ampere\n",
+      "Iec  =  15;#  in  ampere\n",
+      "Idf  =  120;#  in  ampere\n",
+      "Ifg  =  40;#  in  ampere\n",
+      "Iab  =  50;#  in  ampere\n",
+      "I  =  2;#  in  ampere\n",
+      "V1  =  4;#  in  volts\n",
+      "V2  =  3;#  in  volts\n",
+      "V3  =  6;#  in  volts\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  2.5;#  in  ohms\n",
+      "R4  =  1.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1  =  Iab  -  Ibc\n",
+      "I2  =  Ibc  +  Iec\n",
+      "I3  =  I1  -  Idf\n",
+      "I4  =  Iec  -  I3\n",
+      "I5  =  Idf  -  Ifg\n",
+      "#  Applying  Kirchhoff\u2019s  voltage  law  and  moving  clockwise  around  the  loop  of  Figure  13.3(b)  starting  at  point  A:\n",
+      "E  =  I*R2  +  I*R3  +  I*R4  +  I*R1  -  V2  -  V3  +  V1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n (a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are  \",I1,\"A,  \",  I2,\"A,  \",  I3,\"A,  \",  I4,\"A,  \",  I5,\"A  respetively\\n\"\n",
+      "print  \"\\n (b)  value  of  e.m.f.  E  =  \",E,\"  Volts\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  unknown  currents  I1,  I2,  I3,  I4,  I5  are   30 A,   35 A,   -90 A,   105 A,   80 A  respetively\n",
+        "\n",
+        "\n",
+        "  (b)  value  of  e.m.f.  E  =   9.0   Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 168</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "r1  =  2;#  in  ohms\n",
+      "r2  =  1;#  in  ohms\n",
+      "R  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#E1 = I1*(r1 + R) + I2*R,\n",
+      "#E2 = I1*R + (R + r2)*I2,\n",
+      "I2 = (E1*R - E2*(r1 + R))/(R**2 - (R+r1)*(R + r2))\n",
+      "I1 = (E1 - I2*R)/(r1 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  currents flowing are, I1 = \",round(I1,3),\"A,  and I2 = \",round(-1*I2,3),\"A, \"\n",
+      "print   \"and current flowing in middle branch is I1 - I2 = \", round(I1 + I2, 3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  currents flowing are, I1 =  0.857 A,  and I2 =  0.286 A, and current flowing in middle branch is I1 - I2 =  0.571 A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 169</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  12;#  in  volts\n",
+      "R1  =  0.5;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R3  =  5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#E1 + E2 = I1*R1 + I2*R2\n",
+      "#E2 = - I1*R3 + I2*(R2 + R3) \n",
+      "I2 = ((E1 + E2)*R3 + E2*R1)/(R2*R3 + (R2+R3)*R1)\n",
+      "I1 = (E1 + E2 - I2*R2)/R1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  currents flowing are, I1 = \",round(I1,2),\"A,  and I2 = \",round(I2,2),\"A\"\n",
+      "print   \" and in R3 branch is I1 - I2\", round(I1 - I2,2),\" A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  currents flowing are, I1 =  6.52 A,  and I2 =  6.37 A and in R3 branch is I1 - I2 0.15  A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 170</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  54;#  in  volts\n",
+      "I  =  8;#  in  Amps\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  11;#  in  ohms\n",
+      "R3  =  14;#  in  ohms\n",
+      "R4  =  3;#  in  ohms\n",
+      "R5  =  32;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "#V = (R1 + R2)*I1 - R2*I2\n",
+      "#0 = (R1 + R3)*I1 - R5*I2 - R3*I\n",
+      "I1 = V*R5/((R1 + R2)*R5 - (R1 + R3)*R2 + R3*I)\n",
+      "I2 = -1*(V - I1*(R2 + R1))/R2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the current flowing in the 2 ohm resistor = \",round(I1,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 14 ohm resistor = \",round(I - I1,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 32 ohm resistor = \",round(I2,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 11 ohm resistor = \",round(I1 - I2,0),\"A,\"\n",
+      "print  \"\\n  the current flowing in the 3 ohm resistor = \",round(I - I1 + I2,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the current flowing in the 2 ohm resistor =  5.0 A,\n",
+        "\n",
+        "  the current flowing in the 14 ohm resistor =  3.0 A,\n",
+        "\n",
+        "  the current flowing in the 32 ohm resistor =  1.0 A,\n",
+        "\n",
+        "  the current flowing in the 11 ohm resistor =  4.0 A,\n",
+        "\n",
+        "  the current flowing in the 3 ohm resistor =  4.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 171</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  Volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
+      "I2 = (R2/(R + R2))*I1\n",
+      "I3 = (R/(R + R2))*I1\n",
+      "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
+      "I5 = (R1/(R + R1))*I4\n",
+      "I6 = (R/(R + R1))*I4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resultant current flowing through source 1 = \",round(I1 - I6,3),\"A,\"\n",
+      "print  \"\\n  Resultant current flowing through source 2 = \",round(I4 - I3,3),\"A,\"\n",
+      "print  \"\\n  Resultant current flowing through resistor R, = \",round(I2 + I5,3),\"A,\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resultant current flowing through source 1 =  0.857 A,\n",
+        "\n",
+        "  Resultant current flowing through source 2 =  -0.286 A,\n",
+        "\n",
+        "  Resultant current flowing through resistor R, =  0.571 A,"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 173</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  8;#  in  volts\n",
+      "E2  =  3;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R =  18;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = E1/(R1 + (R*R2/(R + R2)))\n",
+      "I2 = (R/(R + R2))*I1\n",
+      "I3 = (R2/(R + R2))*I1\n",
+      "I4 = E2/(R2 + (R*R1/(R + R1)))\n",
+      "I5 = (R/(R + R1))*I4\n",
+      "I6 = (R1/(R + R1))*I4\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resultant current in the 18 ohm resistor= \",round(I3 - I6,3),\"A \"\n",
+      "print   \"and P.d. across the 18 ohm resistor\",round((I3-I6)*R,3),\"V\"\n",
+      "print  \"\\n  (b)the current in the 8 V battery= \",round(I1 + I5,3),\"A\"\n",
+      "print  \"\\n  (c)current in the 3 V battery = \",round(I2 + I4,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resultant current in the 18 ohm resistor=  0.073 A and P.d. across the 18 ohm resistor 1.313 V\n",
+        "\n",
+        "  (b)the current in the 8 V battery=  2.229 A\n",
+        "\n",
+        "  (c)current in the 3 V battery =  2.156 A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 176</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  8;#  in  ohms\n",
+      "R3 =  5;#  in  ohms\n",
+      "R =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = V/(R1 + R2)\n",
+      "E = I1*R2\n",
+      "r = R3 + R1*R2/(R1 + R2)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 10 ohm resistor =  0.482 A"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 177</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  12;#  in  volts\n",
+      "R1  =  1;#  in  ohms\n",
+      "R2  =  4;#  in  ohms\n",
+      "R3 =  5;#  in  ohms\n",
+      "R =  0.8;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = V/(R1 + R2 + R3)\n",
+      "E = I1*R2\n",
+      "r = R2*(R1 + R3)/(R1 + R2 + R3)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 0.8 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 0.8 ohm resistor =  1.5 A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 178</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "I1 = (E1 - E2)/(R1 + R2)\n",
+      "E = E1 - I1*R1\n",
+      "r = R2*R1/(R1 + R2)\n",
+      "I = E/(R + r)\n",
+      "P = R*I**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A and power dissipated in the 4 ohm resistor = \",round(P,3),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 4 ohm resistor =  0.571 A and power dissipated in the 4 ohm resistor =  1.306 W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 178</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  24;#  in  volts\n",
+      "R  =  3;#  in  ohms\n",
+      "R1  =  20;#  in  ohms\n",
+      "R2 =  5;#  in  ohms\n",
+      "R3  = 10;#  in  ohms\n",
+      "R4 =  5/3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = R3*V/(R3 + R2)\n",
+      "r = R4 + R3*R2/(R3 + R2)\n",
+      "I = E/(R + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 3 ohm resistor =  2.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 179</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in the 32 ohm resistor using Th\u00b4evenin\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  54;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  14;#  in  ohms\n",
+      "R3  =  3;#  in  ohms\n",
+      "R4  =  11;#  in  ohms\n",
+      "R5  =  32;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vac = R1*E/(R1 + R4)\n",
+      "Vbc = R2*E/(R2 + R3)\n",
+      "V = Vbc - Vac\n",
+      "r = R4*R1/(R1 + R4) + R3*R2/(R3 + R2)\n",
+      "I = V/(R5 + r)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 32 ohm resistor = \",round(I,0),\"A flowing from A to B\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 32 ohm resistor =  1.0 A flowing from A to B"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 181</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  = 8;#  in  ohms\n",
+      "R3  =  5;#  in  ohms\n",
+      "R  =  10;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = E/R1\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "I = r*Isc/(r + R3 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 10 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 10 ohm resistor =  0.482 A"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 182</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  4;#  in  volts\n",
+      "E2  =  2;#  in  volts\n",
+      "R1  =  2;#  in  ohms\n",
+      "R2  =  1;#  in  ohms\n",
+      "R  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = E1/R1 + E2/R2\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "I = r*Isc/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 4 ohm resistor = \",round(I,3),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 4 ohm resistor =  0.571 A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 182</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  24;#  in  volts\n",
+      "R  =  3;#  in  ohms\n",
+      "R1  =  20;#  in  ohms\n",
+      "R2 =  5;#  in  ohms\n",
+      "R3  = 10;#  in  ohms\n",
+      "R4 =  5/3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R2\n",
+      "r = R3*R2/(R3 + R2)\n",
+      "I = r*Isc/(r + R4 + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 3 ohm resistor = \",round(I,0),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 3 ohm resistor =  2.0 A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 183</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  15;#  in  Amps\n",
+      "R  =  2;#  in  ohms\n",
+      "R1  = 6;#  in  ohms\n",
+      "R2 =  4;#  in  ohms\n",
+      "R3  = 8;#  in  ohms\n",
+      "R4 =  7;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = R1*I/(R1 + R2)\n",
+      "r = (R1 + R2)*(R3 + R4)/(R3 + R1 + R4 + R2)\n",
+      "I = r*Isc/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the current flowing in the 2 ohm resistor = \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the current flowing in the 2 ohm resistor =  6.75 A"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  Volts\n",
+      "R  =  2;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n equivalent Norton network contains Current Source of amp = \",round(Isc,0),\"A and a resistor of \",R,\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " equivalent Norton network contains Current Source of amp =  5.0 A and a resistor of  2  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Isc  =  4;#  in  Amps\n",
+      "R  =  3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = Isc*R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n equivalent Thevenin network contains Voltage Source of \",round(E,0),\"V and a resistor of \",R,\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " equivalent Thevenin network contains Voltage Source of  12.0 V and a resistor of  3  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 185</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  12;#  in  Volts\n",
+      "E2  =  24;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  2;#  in  ohms\n",
+      "R  =  1.8;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc1 = E1/R1\n",
+      "Isc2 = E2/R2\n",
+      "I1 = Isc1 + Isc2\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "E = I1*r\n",
+      "I = E/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (A)equivalent Norton network contains Current Source of \",round(I1,1),\"A and a resistor of \",r,\" ohm\"\n",
+      "print  \"\\n     equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",r,\" ohm\"\n",
+      "print  \"\\n  (B)the current flowing in the 1.8 ohm resistor is \",round(I,1),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (A)equivalent Norton network contains Current Source of  16.0 A and a resistor of  1.2  ohm\n",
+        "\n",
+        "     equivalent Thevenin network contains Voltage Source of  19.2 V and a resistor of  1.2  ohm\n",
+        "\n",
+        "  (B)the current flowing in the 1.8 ohm resistor is  6.4 A"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 186</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  10;#  in  Volts\n",
+      "E2  =  6;#  in  Volts\n",
+      "I1 = 0.001;#in Amp\n",
+      "R1  =  2000;#  in  ohms\n",
+      "R2  =  3000;#  in  ohms\n",
+      "R3  =  600;#  in  ohms\n",
+      "R  =  200;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Isc1 = E1/R1\n",
+      "Isc2 = E2/R2\n",
+      "I2 = Isc1 + Isc2\n",
+      "r1 = R1*R2/(R1 + R2)\n",
+      "E = I2*r1 - I1*R3\n",
+      "r = r1 + R3\n",
+      "I = E/(r + R)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (A)equivalent Norton network contains Current Source of \",round(I2*1000,0),\"mA and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
+      "print  \"\\n     equivalent Thevenin network contains Voltage Source of \",round(E,1),\"V and a resistor of \",round(r1/1000,1),\" Kohm\"\n",
+      "print  \"\\n  (B)the current flowing in the 200 ohm resistor is \",round(I*1000,1),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (A)equivalent Norton network contains Current Source of  7.0 mA and a resistor of  1.2  Kohm\n",
+        "\n",
+        "     equivalent Thevenin network contains Voltage Source of  7.8 V and a resistor of  1.2  Kohm\n",
+        "\n",
+        "  (B)the current flowing in the 200 ohm resistor is  3.9 mA"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 188</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "E  =  6;#  in  Volts\n",
+      "R  =  2.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "RL = []\n",
+      "P = []\n",
+      "k = []\n",
+      "for h in range(6):\n",
+      "    RL.append(h - 0.5)\n",
+      "    k = h - 0.5\n",
+      "    P.append(k*((E/(R + k))**2))\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(RL,P,'-')\n",
+      "#plot(RL,P,'-')\n",
+      "xlabel('RL(ohm)')\n",
+      "ylabel('Power(W)')\n",
+      "show()\n",
+      "Pmax= R*(E/(2*R))**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  A graph of RL against P is shown in Figure\" \n",
+      "print  \"  The maximum value of power is\", Pmax,\"W which occurs when RL =\",R,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'."
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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p6emiS1PMwIEDAQBBQUGYOnUqPvzwQ8ybN6/N57rM3z5q6c/dcsstAOQVP6Wl\npdizZw/GjRsnuCqyliRJePzxxxEaGorFixeLLkdx1dXVOH/+PADg3Llz2L17t2o2cGlpaSgrK0NJ\nSQkyMzMRHx+vquCvr69vGZVTVVWFTz75pMMdS6cO/23btsHPzw8FBQWYMmUKJk+eLLokRbz66quY\nP38+7rnnHixatAi+vr6iS1LMjBkzEB0dja+//hp+fn54++23RZekqNzcXLz77rvYu3evKkeSV1RU\nID4+HuHh4Zg5cyaWLVvWsjepNmobH/P9999jwoQJiIiIwPTp0/Hss892eLxG+HgHIiJyPKfe8yci\nIvtg+BMRaRDDn4hIgxj+REQaxPAnTXB3d4der8fo0aOxdOlSXL16FQBQWlra7hC6FStWIDs7u8Pf\nGxAQgJqamhuuKykpqWXEOZEjMfxJE7y8vHD48GEcPHgQp0+fxu7duzt8/pUrV5Cdnd3p8mKdTmfT\nOSiPPPII3nzzzRt+PdGNYviTpnTr1g0TJ07EZ5991uHztm/fjvj4+Jb7hw4dwkMPPYSxY8dizZo1\nMJvNLT/bsGEDwsLCcP/996OkpAQAkJqaivnz5yMuLg6BgYHYvXs3VqxYgdDQUCxcuLBlgzF16lTV\njYUm18DwJ025cOECsrOzWyZztue///0vgoKCWu4/8cQTWLZsGfbt24fPP//8urEkly9fxtGjRxEV\nFYV33nmn5fEDBw7go48+wltvvYXk5GTccccdOHbsGE6dOtVyfYoePXrA09OzzflWRPbE8CdNuHz5\nMvR6PYYMGQJ3d3fMmjWrw+efOnUKAQEBAIDy8nI0NjZi3Lhx8PT0xMMPP3zdxMTZs2cDAOLj45Gf\nnw9AbgdNnToV3t7eiIqKwpUrVzB9+nTodDqMGzeu5XkAEBgYiJMnTyr8iYk6xvAnTfD09MThw4fx\n3Xffobq6Gjt37uz0NRaLBUDruVKSJF03GqD5+gUeHh5oaGhoebx5jlP37t3Ro0ePlrn43bt3x5Ur\nV677fWodMUzOi//iSFN8fHywYcMG/Pa3v+3wQO2wYcNQWloKABg8eDB69OiBgwcP4vLly8jMzMTU\nqVM7fJ+uHAQ+c+aM6i5aRM6P4U+acO2eul6vxx133IH3338fOp0OJ0+ehJ+fX8vXBx98gIiICHz1\n1Vctr3nzzTexevVqxMXFISYmBgkJCa1+r06na7l/7e1fPu/a+1evXkV9fT369++v/Icm6gAHuxG1\n4erVq4jDvBfeAAAAYklEQVSKikJhYaFdpz++//77KC4uxssvv2y39yBqC/f8idrQvXt3TJkyxe7j\nmjdt2oQFCxbY9T2I2sI9fyIiDeKePxGRBjH8iYg0iOFPRKRBDH8iIg1i+BMRaRDDn4hIg/4fBBse\nZLqKDYkAAAAASUVORK5CYII=\n"
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  A graph of RL against P is shown in Figure\n",
+        "  The maximum value of power is 3.6 W which occurs when RL = 2.5 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 188</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "E  =  30;#  in  Volts\n",
+      "R  =  1.5;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "RL = R\n",
+      "I = E/(R + RL)\n",
+      "P = I**2*RL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"  The maximum value of power is\", P,\"W which occurs when RL =\",RL,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "  The maximum value of power is 150.0 W which occurs when RL = 1.5 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 189</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "V  =  15;#  in  Volts\n",
+      "R1  =  3;#  in  ohms\n",
+      "R2  =  12;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "E = (R2/(R2+ R1))*V\n",
+      "r = R1*R2/(R1 + R2)\n",
+      "RL = r\n",
+      "I = E/(r + RL)\n",
+      "P = I**2*RL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"  The maximum value of power is\", P,\"W which occurs when Total Load RL =\",RL,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "  The maximum value of power is 15.0 W which occurs when Total Load RL = 2.4 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint.ipynb
new file mode 100755
index 00000000..0afe4500
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint.ipynb
@@ -0,0 +1,898 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 14: Alternating voltages and currents</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the periodic time for 50 Hz and 20kHz frequencies.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  20000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "T1  =  1/f1\n",
+      "T2  =  1/f2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Periodic  time  T  =  \",T1,\"  secs\\n\"\n",
+      "print  \"\\n  (b)  Periodic  time  T  =  \",(T2/1E-6),\"  usecs\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Periodic  time  T  =   0.02   secs\n",
+        "\n",
+        "\n",
+        "  (b)  Periodic  time  T  =   50.0   usecs\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the frequencies for 4ms and 4us periodic times\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T1  =  0.004;#  in  secs\n",
+      "T2  =  4E-6;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f1  =  1/T1\n",
+      "f2  =  1/T2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Frequency  f  =  \",f1,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)  Frequency  f  =  \",(f2/1E6),\"  MHz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Frequency  f  =   250.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)  Frequency  f  =   0.25   MHz\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the frequency of an alternating current which completes 5 cycles in 8 ms\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  (8E-3)/5;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f  =  1/T\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Frequency  f  =  \",f,\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Frequency  f  =   625.0   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 196</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine for each: (i) frequency \n",
+      "#(ii) average value over half a cycle \n",
+      "#(iii) rms value\n",
+      "#(iv) form factor and \n",
+      "#(v) peak factor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ta  =  0.02;#  Time  for  1  complete  cycle  in  secs\n",
+      "Vamax  =  200;#  in  volts\n",
+      "Va1  =  25;#  in  volts\n",
+      "Va2  =  75;#  in  volts\n",
+      "Va3  =  125;#  in  volts\n",
+      "Va4  =  175;#  in  volts\n",
+      "Tb  =  0.016;#  Time  for  1  complete  cycle  in  secs\n",
+      "Ibmax  =  10;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  Triangular  waveform  (Figure  14.5(a))\n",
+      "fa  =  1/Ta\n",
+      "Aaw  =  Ta*Vamax/4\n",
+      "Vaavg  =  Aaw*2/Ta\n",
+      "Varms  =  (((Va1**2)  +  (Va2**2)  +  (Va3**2)  +  (Va4**2))/4)**0.5\n",
+      "#Note  that  the  greater  the  number  of  intervals  chosen,  the  greater  the  accuracy  of  the  result\n",
+      "Ffa  =  Varms/Vaavg\n",
+      "Pfa  =  Vamax/Varms\n",
+      "\n",
+      "#for  Rectangular  waveform  (Figure  14.5(b))\n",
+      "fb  =  1/Tb\n",
+      "Abw  =  Tb*Ibmax/2\n",
+      "Ibavg  =  Abw*2/Tb\n",
+      "Ibrms  =  10\n",
+      "Ffb  =  Ibrms/Ibavg\n",
+      "Pfb  =  Ibmax/Ibrms\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a1)Frequency  f  =  \",fa,\"  Hz\\n\"\n",
+      "print  \"\\n  (a2)average  value  over  half  a  cycle  =  \",Vaavg,\"  V\\n\"\n",
+      "print  \"\\n  (a3)rms  value  =  \",round(Varms,2),\"  V\\n\"\n",
+      "print  \"\\n  (a4)Form  factor  =  \",round(Ffa,2),\"\\n\"\n",
+      "print  \"\\n  (a5)Peak  factor  =  \",round(Pfa,2),\"\\n\"\n",
+      "print  \"\\n  (b1)Frequency  f  =  \",fb,\"  Hz\\n\"\n",
+      "print  \"\\n  (b2)average  value  over  half  a  cycle  =  \",Ibavg,\"  A\\n\"\n",
+      "print  \"\\n  (b3)rms  value  =  \",Ibrms,\"  A\\n\"\n",
+      "print  \"\\n  (b4)Form  factor  =  \",Ffb,\"\\n\"\n",
+      "print  \"\\n  (b5)Peak  factor  =  \",Pfb,\"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a1)Frequency  f  =   50.0   Hz\n",
+        "\n",
+        "\n",
+        "  (a2)average  value  over  half  a  cycle  =   100.0   V\n",
+        "\n",
+        "\n",
+        "  (a3)rms  value  =   114.56   V\n",
+        "\n",
+        "\n",
+        "  (a4)Form  factor  =   1.15 \n",
+        "\n",
+        "\n",
+        "  (a5)Peak  factor  =   1.75 \n",
+        "\n",
+        "\n",
+        "  (b1)Frequency  f  =   62.5   Hz\n",
+        "\n",
+        "\n",
+        "  (b2)average  value  over  half  a  cycle  =   10.0   A\n",
+        "\n",
+        "\n",
+        "  (b3)rms  value  =   10   A\n",
+        "\n",
+        "\n",
+        "  (b4)Form  factor  =   1.0 \n",
+        "\n",
+        "\n",
+        "  (b5)Peak  factor  =   1.0 "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 198</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the frequency of the supply, \n",
+      "#(b) the instantaneous values of current after 1.25 ms and 3.8 ms, (c) the peak or maximum value,\n",
+      "#(d) the mean or average value, and (e) the rms value of the waveform.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy \n",
+      "from numpy import mean, sqrt, arange\n",
+      "#initializing  the  variables:\n",
+      "Thalf = 5; #in ms\n",
+      "Ta  =  0.02;#  Time  for  1  complete  cycle  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "Tfull = 2*Thalf/1000 # in sec\n",
+      "f = 1/Tfull\n",
+      "A=[3, 10, 19, 30, 49, 63, 73, 72, 30, 2]\n",
+      "Iinst125 = 19\n",
+      "Iinst38 = 70\n",
+      "sq = 0\n",
+      "Ipeak = 76\n",
+      "#B=arange(A)\n",
+      "Imean = (0.5*1E-3)*numpy.mean(A)*10/(5*1E-3)\n",
+      "for h in range(10):\n",
+      "    sq = sq + A[h]**2\n",
+      "\n",
+      "Irms = sqrt(sq/10)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Instantaneous value of current after 1.25 ms  =\",Iinst125,\"A \"\n",
+      "print   \"and Instantaneous value of current after 3.8 ms\", Iinst38,\"A\\n\"\n",
+      "print  \"\\n  (c)Peak or maximum value  =  \",Ipeak,\"  A\\n\"\n",
+      "print  \"\\n  (d)Mean or average value  =  \",round(Imean,2),\"A\\n\"\n",
+      "print  \"\\n  (e)rms value =  \",round(Irms,1),\"A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency  f  =   100.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Instantaneous value of current after 1.25 ms  = 19 A and Instantaneous value of current after 3.8 ms 70 A\n",
+        "\n",
+        "\n",
+        "  (c)Peak or maximum value  =   76   A\n",
+        "\n",
+        "\n",
+        "  (d)Mean or average value  =   35.1 A\n",
+        "\n",
+        "\n",
+        "  (e)rms value =   43.8 A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#sinusoidal current has maximum value of 20 A.Calculate its rms value. \n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Imax  =  20;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Irms  =  Imax/(2**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Rms  value  =  \",round(Irms,2),\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Rms  value  =   14.14   A"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#main supply is of 240V.Determine its peak and mean values.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vrms  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vmax  =  Vrms*(2**0.5)\n",
+      "Vmean  =  0.637*Vmax\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  peak  value  =  \",round(Vmax,2),\"  V\\n\"\n",
+      "print  \"\\n  mean  value  =  \",round(Vmean,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  peak  value  =   339.41   V\n",
+        "\n",
+        "\n",
+        "  mean  value  =   216.2   V"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its maximum value and its rms value\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmean  =  150;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vmax  =  Vmean/0.637\n",
+      "Vrms  =  0.707*Vmax\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  peak  value  =  \",round(Vmax,2),\"  V\\n\"\n",
+      "print  \"\\n  rms  value  =  \",round(Vrms,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  peak  value  =   235.48   V\n",
+        "\n",
+        "\n",
+        "  rms  value  =   166.48   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 201</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the rms voltage, (b) the frequency and (c) the instantaneous value of voltage when t = 4 ms\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  282.8;#  in  Volts\n",
+      "w  =  314;#  in  rad/sec\n",
+      "t  =  0.004;#  in  sec\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "v  =  Vmax*math.sin(w*t)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)rms  value  =  \",round(Vrms,2),\"  V\\n\"\n",
+      "print  \"\\n  (b)frequency  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)instantaneous  value  of  voltage  at  4  ms  =  \",round(v,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)rms  value  =   199.94   V\n",
+        "\n",
+        "\n",
+        "  (b)frequency  f  =   49.97   Hz\n",
+        "\n",
+        "\n",
+        "  (c)instantaneous  value  of  voltage  at  4  ms  =   268.9   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 202</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the amplitude, (b) the peak-to-peak value, \n",
+      "#(c) the rms value, (d) the periodic time, \n",
+      "#(e) the frequency, and (f) the phase angle (in degrees and minutes) relative to 75 sin 200(pi*t)\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  75;#  in  Volts\n",
+      "w  =  200*math.pi;#  in  rad/sec\n",
+      "t  =  0.004;#  in  sec\n",
+      "phi  =  0.25;#  in  radians\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vptp  =  2*Vmax\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "T  =  1/f\n",
+      "v  =  Vmax*math.sin(w*t)\n",
+      "phid  =  phi*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Amplitude,  or  peak  value  =  \",Vmax,\"  V\\n\"\n",
+      "print  \"\\n  (b)  Peak-to-peak  value  =  \",Vptp,\"  V\\n\"\n",
+      "print  \"\\n  (c)rms  value  =  \",Vrms,\"  V\\n\"\n",
+      "print  \"\\n  (d)periodic  time,  T  =  \",T,\"  sec\\n\"\n",
+      "print  \"\\n  (e)frequency  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (f)phase  angle  =  \",round(phid,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Amplitude,  or  peak  value  =   75   V\n",
+        "\n",
+        "\n",
+        "  (b)  Peak-to-peak  value  =   150   V\n",
+        "\n",
+        "\n",
+        "  (c)rms  value  =   53.025   V\n",
+        "\n",
+        "\n",
+        "  (d)periodic  time,  T  =   0.01   sec\n",
+        "\n",
+        "\n",
+        "  (e)frequency  f  =   100.0   Hz\n",
+        "\n",
+        "\n",
+        "  (f)phase  angle  =   14.32 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 202</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Express the instantaneous voltage in the form v = Vm*sin(wt +- phi)\t\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  40;#  in  Volts\n",
+      "T  =  0.01;#  in  sec\n",
+      "v  =  -20;#  when  t  =  0sec,  in  volts\n",
+      "t  =  0;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "w  =  2*math.pi/T\n",
+      "phir  =  math.asin(v/Vmax)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    instantaneous  voltage  v  =  \",  Vmax,\"  sin(\",round(w,2),\"t\",round(phir,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    instantaneous  voltage  v  =   40   sin( 628.32 t -0.52 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 203</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find:\n",
+      "#(a) the peak value, the periodic time, the frequency and phase angle relative to 120 sin 100*pi*t\n",
+      "#(b) the value of the current when t = 0\n",
+      "#(c) the value of the current when t = 8 ms\n",
+      "#(d) the time when the current first reaches 60 A, and\n",
+      "#(e) the time when the current is first a maximum\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Imax  =  120;#  in  Amperes\n",
+      "w  =  100*math.pi;#  in  rad/sec\n",
+      "phi  =  0.36;#  in  rad\n",
+      "t1  =  0;#  in  secs\n",
+      "t2  =  0.008;#  in  secs\n",
+      "i  =  60;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "f  =  w/(2*math.pi)\n",
+      "T  =  1/f\n",
+      "phid  =  phi*180/math.pi\n",
+      "i0  =  Imax*math.sin((w*t1)+phi)\n",
+      "i8  =  Imax*math.sin((w*t2)+phi)  \n",
+      "ti  =  (math.asin(i/Imax)  -  phi)/w\n",
+      "tm1  =  (math.asin(Imax/Imax)  -  phi)/w\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Peak  value  =  \",  Imax,\"  A,  Periodic  time  T  =  \",  T,\"  sec, \"\n",
+      "print   \" Frequency,  f  =  \",  f,\"  Hz  Phase  angle  =  \",round(phid,2),\"deg leading\\n\"\n",
+      "print  \"\\n  (b)  When  t  =  0,  i  =  \",round(i0,2),\"  A\\n\"\n",
+      "print  \"\\n  (c)When  t  =  8  ms  =  \",  round(i8,2),\"  A\\n\"\n",
+      "print  \"\\n  (d)When  i  is  60  A,  then  time  t  =  \",round((ti/1E-3),2),\"  ms\\n\"\n",
+      "print  \"\\n  (e)When  the  current  is  a  maximum,  time,  t  =  \",  round((tm1/1E-3),2),\"  ms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Peak  value  =   120   A,  Periodic  time  T  =   0.02   sec, \n",
+        " Frequency,  f  =   50.0   Hz  Phase  angle  =   20.63 deg leading\n",
+        "\n",
+        "\n",
+        "  (b)  When  t  =  0,  i  =   42.27   A\n",
+        "\n",
+        "\n",
+        "  (c)When  t  =  8  ms  =   31.81   A\n",
+        "\n",
+        "\n",
+        "  (d)When  i  is  60  A,  then  time  t  =   0.52   ms\n",
+        "\n",
+        "\n",
+        "  (e)When  the  current  is  a  maximum,  time,  t  =   3.85   ms\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 204</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#obtain a sinusoidal expression for i1 + i2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "i1max  =  20;#  in  Amperes\n",
+      "i2max  =  10;#  in  Amperes\n",
+      "phi1  =  0;#  in  rad\n",
+      "phi2  =  math.pi/3;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#Ig = i1 + i2\n",
+      "Igmax  = 26.5\n",
+      "phiIg = 19*math.pi/180\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Current  Ig = i1 + i2 =\", Igmax,\"sin(wt + \",round(phiIg,3),\") Amps\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Current  Ig = i1 + i2 = 26.5 sin(wt +  0.332 ) Amps"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 205</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find, by calculation, a sinusoidal expression to represent v1 + v2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1max  =  50;#  in  volts\n",
+      "V2max  =  100;#  in  volts\n",
+      "phi2  =  -1*math.pi/6;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#vR2  =  v1**2  +  v2**2  -  2*v1*v2  cos  150\n",
+      "phidiff  =  math.pi  +  phi2\n",
+      "Vrmax  =  (V1max**2  +  V2max**2  -  2*V1max*V2max*math.cos(phidiff))**0.5\n",
+      "#Using  the  sine  rule\n",
+      "phi  =  math.asin(V2max*math.sin(phidiff)/Vrmax)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  VR  =  \",round(Vrmax,2),\"sin(wt  -  \",round(phi,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  VR  =   145.47 sin(wt  -   0.35 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 206</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find a sinusoidal expression for \u0005i1 + i2\t of Problem 13, by calculation.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1max  =  20;#  in  volts\n",
+      "I2max  =  10;#  in  volts\n",
+      "phi2  =  1*math.pi/3;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#iR2  =  i1**2  +  i2**2  -  2*i1*i2cos150\n",
+      "phidiff  =  math.pi  -  phi2\n",
+      "Irmax  =  (I1max**2  +  I2max**2  -  2*I1max*I2max*math.cos(phidiff))**0.5\n",
+      "#Using  the  sine  rule\n",
+      "phi  =  math.asin(I2max*math.sin(phidiff)/Irmax)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  IR  =  \",  round(Irmax,2),\"sin(wt  +  \",round(phi,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  IR  =   26.46 sin(wt  +   0.33 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_1.ipynb
new file mode 100755
index 00000000..0afe4500
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_1.ipynb
@@ -0,0 +1,898 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 14: Alternating voltages and currents</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the periodic time for 50 Hz and 20kHz frequencies.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  20000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "T1  =  1/f1\n",
+      "T2  =  1/f2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Periodic  time  T  =  \",T1,\"  secs\\n\"\n",
+      "print  \"\\n  (b)  Periodic  time  T  =  \",(T2/1E-6),\"  usecs\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Periodic  time  T  =   0.02   secs\n",
+        "\n",
+        "\n",
+        "  (b)  Periodic  time  T  =   50.0   usecs\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the frequencies for 4ms and 4us periodic times\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T1  =  0.004;#  in  secs\n",
+      "T2  =  4E-6;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f1  =  1/T1\n",
+      "f2  =  1/T2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Frequency  f  =  \",f1,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)  Frequency  f  =  \",(f2/1E6),\"  MHz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Frequency  f  =   250.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)  Frequency  f  =   0.25   MHz\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the frequency of an alternating current which completes 5 cycles in 8 ms\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  (8E-3)/5;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f  =  1/T\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Frequency  f  =  \",f,\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Frequency  f  =   625.0   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 196</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine for each: (i) frequency \n",
+      "#(ii) average value over half a cycle \n",
+      "#(iii) rms value\n",
+      "#(iv) form factor and \n",
+      "#(v) peak factor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ta  =  0.02;#  Time  for  1  complete  cycle  in  secs\n",
+      "Vamax  =  200;#  in  volts\n",
+      "Va1  =  25;#  in  volts\n",
+      "Va2  =  75;#  in  volts\n",
+      "Va3  =  125;#  in  volts\n",
+      "Va4  =  175;#  in  volts\n",
+      "Tb  =  0.016;#  Time  for  1  complete  cycle  in  secs\n",
+      "Ibmax  =  10;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  Triangular  waveform  (Figure  14.5(a))\n",
+      "fa  =  1/Ta\n",
+      "Aaw  =  Ta*Vamax/4\n",
+      "Vaavg  =  Aaw*2/Ta\n",
+      "Varms  =  (((Va1**2)  +  (Va2**2)  +  (Va3**2)  +  (Va4**2))/4)**0.5\n",
+      "#Note  that  the  greater  the  number  of  intervals  chosen,  the  greater  the  accuracy  of  the  result\n",
+      "Ffa  =  Varms/Vaavg\n",
+      "Pfa  =  Vamax/Varms\n",
+      "\n",
+      "#for  Rectangular  waveform  (Figure  14.5(b))\n",
+      "fb  =  1/Tb\n",
+      "Abw  =  Tb*Ibmax/2\n",
+      "Ibavg  =  Abw*2/Tb\n",
+      "Ibrms  =  10\n",
+      "Ffb  =  Ibrms/Ibavg\n",
+      "Pfb  =  Ibmax/Ibrms\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a1)Frequency  f  =  \",fa,\"  Hz\\n\"\n",
+      "print  \"\\n  (a2)average  value  over  half  a  cycle  =  \",Vaavg,\"  V\\n\"\n",
+      "print  \"\\n  (a3)rms  value  =  \",round(Varms,2),\"  V\\n\"\n",
+      "print  \"\\n  (a4)Form  factor  =  \",round(Ffa,2),\"\\n\"\n",
+      "print  \"\\n  (a5)Peak  factor  =  \",round(Pfa,2),\"\\n\"\n",
+      "print  \"\\n  (b1)Frequency  f  =  \",fb,\"  Hz\\n\"\n",
+      "print  \"\\n  (b2)average  value  over  half  a  cycle  =  \",Ibavg,\"  A\\n\"\n",
+      "print  \"\\n  (b3)rms  value  =  \",Ibrms,\"  A\\n\"\n",
+      "print  \"\\n  (b4)Form  factor  =  \",Ffb,\"\\n\"\n",
+      "print  \"\\n  (b5)Peak  factor  =  \",Pfb,\"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a1)Frequency  f  =   50.0   Hz\n",
+        "\n",
+        "\n",
+        "  (a2)average  value  over  half  a  cycle  =   100.0   V\n",
+        "\n",
+        "\n",
+        "  (a3)rms  value  =   114.56   V\n",
+        "\n",
+        "\n",
+        "  (a4)Form  factor  =   1.15 \n",
+        "\n",
+        "\n",
+        "  (a5)Peak  factor  =   1.75 \n",
+        "\n",
+        "\n",
+        "  (b1)Frequency  f  =   62.5   Hz\n",
+        "\n",
+        "\n",
+        "  (b2)average  value  over  half  a  cycle  =   10.0   A\n",
+        "\n",
+        "\n",
+        "  (b3)rms  value  =   10   A\n",
+        "\n",
+        "\n",
+        "  (b4)Form  factor  =   1.0 \n",
+        "\n",
+        "\n",
+        "  (b5)Peak  factor  =   1.0 "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 198</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the frequency of the supply, \n",
+      "#(b) the instantaneous values of current after 1.25 ms and 3.8 ms, (c) the peak or maximum value,\n",
+      "#(d) the mean or average value, and (e) the rms value of the waveform.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy \n",
+      "from numpy import mean, sqrt, arange\n",
+      "#initializing  the  variables:\n",
+      "Thalf = 5; #in ms\n",
+      "Ta  =  0.02;#  Time  for  1  complete  cycle  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "Tfull = 2*Thalf/1000 # in sec\n",
+      "f = 1/Tfull\n",
+      "A=[3, 10, 19, 30, 49, 63, 73, 72, 30, 2]\n",
+      "Iinst125 = 19\n",
+      "Iinst38 = 70\n",
+      "sq = 0\n",
+      "Ipeak = 76\n",
+      "#B=arange(A)\n",
+      "Imean = (0.5*1E-3)*numpy.mean(A)*10/(5*1E-3)\n",
+      "for h in range(10):\n",
+      "    sq = sq + A[h]**2\n",
+      "\n",
+      "Irms = sqrt(sq/10)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Instantaneous value of current after 1.25 ms  =\",Iinst125,\"A \"\n",
+      "print   \"and Instantaneous value of current after 3.8 ms\", Iinst38,\"A\\n\"\n",
+      "print  \"\\n  (c)Peak or maximum value  =  \",Ipeak,\"  A\\n\"\n",
+      "print  \"\\n  (d)Mean or average value  =  \",round(Imean,2),\"A\\n\"\n",
+      "print  \"\\n  (e)rms value =  \",round(Irms,1),\"A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency  f  =   100.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Instantaneous value of current after 1.25 ms  = 19 A and Instantaneous value of current after 3.8 ms 70 A\n",
+        "\n",
+        "\n",
+        "  (c)Peak or maximum value  =   76   A\n",
+        "\n",
+        "\n",
+        "  (d)Mean or average value  =   35.1 A\n",
+        "\n",
+        "\n",
+        "  (e)rms value =   43.8 A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#sinusoidal current has maximum value of 20 A.Calculate its rms value. \n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Imax  =  20;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Irms  =  Imax/(2**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Rms  value  =  \",round(Irms,2),\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Rms  value  =   14.14   A"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#main supply is of 240V.Determine its peak and mean values.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vrms  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vmax  =  Vrms*(2**0.5)\n",
+      "Vmean  =  0.637*Vmax\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  peak  value  =  \",round(Vmax,2),\"  V\\n\"\n",
+      "print  \"\\n  mean  value  =  \",round(Vmean,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  peak  value  =   339.41   V\n",
+        "\n",
+        "\n",
+        "  mean  value  =   216.2   V"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its maximum value and its rms value\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmean  =  150;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vmax  =  Vmean/0.637\n",
+      "Vrms  =  0.707*Vmax\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  peak  value  =  \",round(Vmax,2),\"  V\\n\"\n",
+      "print  \"\\n  rms  value  =  \",round(Vrms,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  peak  value  =   235.48   V\n",
+        "\n",
+        "\n",
+        "  rms  value  =   166.48   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 201</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the rms voltage, (b) the frequency and (c) the instantaneous value of voltage when t = 4 ms\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  282.8;#  in  Volts\n",
+      "w  =  314;#  in  rad/sec\n",
+      "t  =  0.004;#  in  sec\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "v  =  Vmax*math.sin(w*t)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)rms  value  =  \",round(Vrms,2),\"  V\\n\"\n",
+      "print  \"\\n  (b)frequency  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)instantaneous  value  of  voltage  at  4  ms  =  \",round(v,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)rms  value  =   199.94   V\n",
+        "\n",
+        "\n",
+        "  (b)frequency  f  =   49.97   Hz\n",
+        "\n",
+        "\n",
+        "  (c)instantaneous  value  of  voltage  at  4  ms  =   268.9   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 202</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the amplitude, (b) the peak-to-peak value, \n",
+      "#(c) the rms value, (d) the periodic time, \n",
+      "#(e) the frequency, and (f) the phase angle (in degrees and minutes) relative to 75 sin 200(pi*t)\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  75;#  in  Volts\n",
+      "w  =  200*math.pi;#  in  rad/sec\n",
+      "t  =  0.004;#  in  sec\n",
+      "phi  =  0.25;#  in  radians\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vptp  =  2*Vmax\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "T  =  1/f\n",
+      "v  =  Vmax*math.sin(w*t)\n",
+      "phid  =  phi*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Amplitude,  or  peak  value  =  \",Vmax,\"  V\\n\"\n",
+      "print  \"\\n  (b)  Peak-to-peak  value  =  \",Vptp,\"  V\\n\"\n",
+      "print  \"\\n  (c)rms  value  =  \",Vrms,\"  V\\n\"\n",
+      "print  \"\\n  (d)periodic  time,  T  =  \",T,\"  sec\\n\"\n",
+      "print  \"\\n  (e)frequency  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (f)phase  angle  =  \",round(phid,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Amplitude,  or  peak  value  =   75   V\n",
+        "\n",
+        "\n",
+        "  (b)  Peak-to-peak  value  =   150   V\n",
+        "\n",
+        "\n",
+        "  (c)rms  value  =   53.025   V\n",
+        "\n",
+        "\n",
+        "  (d)periodic  time,  T  =   0.01   sec\n",
+        "\n",
+        "\n",
+        "  (e)frequency  f  =   100.0   Hz\n",
+        "\n",
+        "\n",
+        "  (f)phase  angle  =   14.32 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 202</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Express the instantaneous voltage in the form v = Vm*sin(wt +- phi)\t\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  40;#  in  Volts\n",
+      "T  =  0.01;#  in  sec\n",
+      "v  =  -20;#  when  t  =  0sec,  in  volts\n",
+      "t  =  0;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "w  =  2*math.pi/T\n",
+      "phir  =  math.asin(v/Vmax)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    instantaneous  voltage  v  =  \",  Vmax,\"  sin(\",round(w,2),\"t\",round(phir,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    instantaneous  voltage  v  =   40   sin( 628.32 t -0.52 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 203</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find:\n",
+      "#(a) the peak value, the periodic time, the frequency and phase angle relative to 120 sin 100*pi*t\n",
+      "#(b) the value of the current when t = 0\n",
+      "#(c) the value of the current when t = 8 ms\n",
+      "#(d) the time when the current first reaches 60 A, and\n",
+      "#(e) the time when the current is first a maximum\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Imax  =  120;#  in  Amperes\n",
+      "w  =  100*math.pi;#  in  rad/sec\n",
+      "phi  =  0.36;#  in  rad\n",
+      "t1  =  0;#  in  secs\n",
+      "t2  =  0.008;#  in  secs\n",
+      "i  =  60;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "f  =  w/(2*math.pi)\n",
+      "T  =  1/f\n",
+      "phid  =  phi*180/math.pi\n",
+      "i0  =  Imax*math.sin((w*t1)+phi)\n",
+      "i8  =  Imax*math.sin((w*t2)+phi)  \n",
+      "ti  =  (math.asin(i/Imax)  -  phi)/w\n",
+      "tm1  =  (math.asin(Imax/Imax)  -  phi)/w\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Peak  value  =  \",  Imax,\"  A,  Periodic  time  T  =  \",  T,\"  sec, \"\n",
+      "print   \" Frequency,  f  =  \",  f,\"  Hz  Phase  angle  =  \",round(phid,2),\"deg leading\\n\"\n",
+      "print  \"\\n  (b)  When  t  =  0,  i  =  \",round(i0,2),\"  A\\n\"\n",
+      "print  \"\\n  (c)When  t  =  8  ms  =  \",  round(i8,2),\"  A\\n\"\n",
+      "print  \"\\n  (d)When  i  is  60  A,  then  time  t  =  \",round((ti/1E-3),2),\"  ms\\n\"\n",
+      "print  \"\\n  (e)When  the  current  is  a  maximum,  time,  t  =  \",  round((tm1/1E-3),2),\"  ms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Peak  value  =   120   A,  Periodic  time  T  =   0.02   sec, \n",
+        " Frequency,  f  =   50.0   Hz  Phase  angle  =   20.63 deg leading\n",
+        "\n",
+        "\n",
+        "  (b)  When  t  =  0,  i  =   42.27   A\n",
+        "\n",
+        "\n",
+        "  (c)When  t  =  8  ms  =   31.81   A\n",
+        "\n",
+        "\n",
+        "  (d)When  i  is  60  A,  then  time  t  =   0.52   ms\n",
+        "\n",
+        "\n",
+        "  (e)When  the  current  is  a  maximum,  time,  t  =   3.85   ms\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 204</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#obtain a sinusoidal expression for i1 + i2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "i1max  =  20;#  in  Amperes\n",
+      "i2max  =  10;#  in  Amperes\n",
+      "phi1  =  0;#  in  rad\n",
+      "phi2  =  math.pi/3;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#Ig = i1 + i2\n",
+      "Igmax  = 26.5\n",
+      "phiIg = 19*math.pi/180\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Current  Ig = i1 + i2 =\", Igmax,\"sin(wt + \",round(phiIg,3),\") Amps\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Current  Ig = i1 + i2 = 26.5 sin(wt +  0.332 ) Amps"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 205</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find, by calculation, a sinusoidal expression to represent v1 + v2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1max  =  50;#  in  volts\n",
+      "V2max  =  100;#  in  volts\n",
+      "phi2  =  -1*math.pi/6;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#vR2  =  v1**2  +  v2**2  -  2*v1*v2  cos  150\n",
+      "phidiff  =  math.pi  +  phi2\n",
+      "Vrmax  =  (V1max**2  +  V2max**2  -  2*V1max*V2max*math.cos(phidiff))**0.5\n",
+      "#Using  the  sine  rule\n",
+      "phi  =  math.asin(V2max*math.sin(phidiff)/Vrmax)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  VR  =  \",round(Vrmax,2),\"sin(wt  -  \",round(phi,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  VR  =   145.47 sin(wt  -   0.35 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 206</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find a sinusoidal expression for \u0005i1 + i2\t of Problem 13, by calculation.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1max  =  20;#  in  volts\n",
+      "I2max  =  10;#  in  volts\n",
+      "phi2  =  1*math.pi/3;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#iR2  =  i1**2  +  i2**2  -  2*i1*i2cos150\n",
+      "phidiff  =  math.pi  -  phi2\n",
+      "Irmax  =  (I1max**2  +  I2max**2  -  2*I1max*I2max*math.cos(phidiff))**0.5\n",
+      "#Using  the  sine  rule\n",
+      "phi  =  math.asin(I2max*math.sin(phidiff)/Irmax)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  IR  =  \",  round(Irmax,2),\"sin(wt  +  \",round(phi,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  IR  =   26.46 sin(wt  +   0.33 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_2.ipynb
new file mode 100755
index 00000000..0afe4500
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_2.ipynb
@@ -0,0 +1,898 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 14: Alternating voltages and currents</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the periodic time for 50 Hz and 20kHz frequencies.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  20000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "T1  =  1/f1\n",
+      "T2  =  1/f2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Periodic  time  T  =  \",T1,\"  secs\\n\"\n",
+      "print  \"\\n  (b)  Periodic  time  T  =  \",(T2/1E-6),\"  usecs\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Periodic  time  T  =   0.02   secs\n",
+        "\n",
+        "\n",
+        "  (b)  Periodic  time  T  =   50.0   usecs\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the frequencies for 4ms and 4us periodic times\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T1  =  0.004;#  in  secs\n",
+      "T2  =  4E-6;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f1  =  1/T1\n",
+      "f2  =  1/T2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Frequency  f  =  \",f1,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)  Frequency  f  =  \",(f2/1E6),\"  MHz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Frequency  f  =   250.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)  Frequency  f  =   0.25   MHz\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the frequency of an alternating current which completes 5 cycles in 8 ms\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  (8E-3)/5;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f  =  1/T\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Frequency  f  =  \",f,\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Frequency  f  =   625.0   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 196</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine for each: (i) frequency \n",
+      "#(ii) average value over half a cycle \n",
+      "#(iii) rms value\n",
+      "#(iv) form factor and \n",
+      "#(v) peak factor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ta  =  0.02;#  Time  for  1  complete  cycle  in  secs\n",
+      "Vamax  =  200;#  in  volts\n",
+      "Va1  =  25;#  in  volts\n",
+      "Va2  =  75;#  in  volts\n",
+      "Va3  =  125;#  in  volts\n",
+      "Va4  =  175;#  in  volts\n",
+      "Tb  =  0.016;#  Time  for  1  complete  cycle  in  secs\n",
+      "Ibmax  =  10;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  Triangular  waveform  (Figure  14.5(a))\n",
+      "fa  =  1/Ta\n",
+      "Aaw  =  Ta*Vamax/4\n",
+      "Vaavg  =  Aaw*2/Ta\n",
+      "Varms  =  (((Va1**2)  +  (Va2**2)  +  (Va3**2)  +  (Va4**2))/4)**0.5\n",
+      "#Note  that  the  greater  the  number  of  intervals  chosen,  the  greater  the  accuracy  of  the  result\n",
+      "Ffa  =  Varms/Vaavg\n",
+      "Pfa  =  Vamax/Varms\n",
+      "\n",
+      "#for  Rectangular  waveform  (Figure  14.5(b))\n",
+      "fb  =  1/Tb\n",
+      "Abw  =  Tb*Ibmax/2\n",
+      "Ibavg  =  Abw*2/Tb\n",
+      "Ibrms  =  10\n",
+      "Ffb  =  Ibrms/Ibavg\n",
+      "Pfb  =  Ibmax/Ibrms\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a1)Frequency  f  =  \",fa,\"  Hz\\n\"\n",
+      "print  \"\\n  (a2)average  value  over  half  a  cycle  =  \",Vaavg,\"  V\\n\"\n",
+      "print  \"\\n  (a3)rms  value  =  \",round(Varms,2),\"  V\\n\"\n",
+      "print  \"\\n  (a4)Form  factor  =  \",round(Ffa,2),\"\\n\"\n",
+      "print  \"\\n  (a5)Peak  factor  =  \",round(Pfa,2),\"\\n\"\n",
+      "print  \"\\n  (b1)Frequency  f  =  \",fb,\"  Hz\\n\"\n",
+      "print  \"\\n  (b2)average  value  over  half  a  cycle  =  \",Ibavg,\"  A\\n\"\n",
+      "print  \"\\n  (b3)rms  value  =  \",Ibrms,\"  A\\n\"\n",
+      "print  \"\\n  (b4)Form  factor  =  \",Ffb,\"\\n\"\n",
+      "print  \"\\n  (b5)Peak  factor  =  \",Pfb,\"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a1)Frequency  f  =   50.0   Hz\n",
+        "\n",
+        "\n",
+        "  (a2)average  value  over  half  a  cycle  =   100.0   V\n",
+        "\n",
+        "\n",
+        "  (a3)rms  value  =   114.56   V\n",
+        "\n",
+        "\n",
+        "  (a4)Form  factor  =   1.15 \n",
+        "\n",
+        "\n",
+        "  (a5)Peak  factor  =   1.75 \n",
+        "\n",
+        "\n",
+        "  (b1)Frequency  f  =   62.5   Hz\n",
+        "\n",
+        "\n",
+        "  (b2)average  value  over  half  a  cycle  =   10.0   A\n",
+        "\n",
+        "\n",
+        "  (b3)rms  value  =   10   A\n",
+        "\n",
+        "\n",
+        "  (b4)Form  factor  =   1.0 \n",
+        "\n",
+        "\n",
+        "  (b5)Peak  factor  =   1.0 "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 198</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the frequency of the supply, \n",
+      "#(b) the instantaneous values of current after 1.25 ms and 3.8 ms, (c) the peak or maximum value,\n",
+      "#(d) the mean or average value, and (e) the rms value of the waveform.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy \n",
+      "from numpy import mean, sqrt, arange\n",
+      "#initializing  the  variables:\n",
+      "Thalf = 5; #in ms\n",
+      "Ta  =  0.02;#  Time  for  1  complete  cycle  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "Tfull = 2*Thalf/1000 # in sec\n",
+      "f = 1/Tfull\n",
+      "A=[3, 10, 19, 30, 49, 63, 73, 72, 30, 2]\n",
+      "Iinst125 = 19\n",
+      "Iinst38 = 70\n",
+      "sq = 0\n",
+      "Ipeak = 76\n",
+      "#B=arange(A)\n",
+      "Imean = (0.5*1E-3)*numpy.mean(A)*10/(5*1E-3)\n",
+      "for h in range(10):\n",
+      "    sq = sq + A[h]**2\n",
+      "\n",
+      "Irms = sqrt(sq/10)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Instantaneous value of current after 1.25 ms  =\",Iinst125,\"A \"\n",
+      "print   \"and Instantaneous value of current after 3.8 ms\", Iinst38,\"A\\n\"\n",
+      "print  \"\\n  (c)Peak or maximum value  =  \",Ipeak,\"  A\\n\"\n",
+      "print  \"\\n  (d)Mean or average value  =  \",round(Imean,2),\"A\\n\"\n",
+      "print  \"\\n  (e)rms value =  \",round(Irms,1),\"A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency  f  =   100.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Instantaneous value of current after 1.25 ms  = 19 A and Instantaneous value of current after 3.8 ms 70 A\n",
+        "\n",
+        "\n",
+        "  (c)Peak or maximum value  =   76   A\n",
+        "\n",
+        "\n",
+        "  (d)Mean or average value  =   35.1 A\n",
+        "\n",
+        "\n",
+        "  (e)rms value =   43.8 A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#sinusoidal current has maximum value of 20 A.Calculate its rms value. \n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Imax  =  20;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Irms  =  Imax/(2**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Rms  value  =  \",round(Irms,2),\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Rms  value  =   14.14   A"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#main supply is of 240V.Determine its peak and mean values.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vrms  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vmax  =  Vrms*(2**0.5)\n",
+      "Vmean  =  0.637*Vmax\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  peak  value  =  \",round(Vmax,2),\"  V\\n\"\n",
+      "print  \"\\n  mean  value  =  \",round(Vmean,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  peak  value  =   339.41   V\n",
+        "\n",
+        "\n",
+        "  mean  value  =   216.2   V"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its maximum value and its rms value\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmean  =  150;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vmax  =  Vmean/0.637\n",
+      "Vrms  =  0.707*Vmax\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  peak  value  =  \",round(Vmax,2),\"  V\\n\"\n",
+      "print  \"\\n  rms  value  =  \",round(Vrms,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  peak  value  =   235.48   V\n",
+        "\n",
+        "\n",
+        "  rms  value  =   166.48   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 201</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the rms voltage, (b) the frequency and (c) the instantaneous value of voltage when t = 4 ms\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  282.8;#  in  Volts\n",
+      "w  =  314;#  in  rad/sec\n",
+      "t  =  0.004;#  in  sec\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "v  =  Vmax*math.sin(w*t)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)rms  value  =  \",round(Vrms,2),\"  V\\n\"\n",
+      "print  \"\\n  (b)frequency  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)instantaneous  value  of  voltage  at  4  ms  =  \",round(v,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)rms  value  =   199.94   V\n",
+        "\n",
+        "\n",
+        "  (b)frequency  f  =   49.97   Hz\n",
+        "\n",
+        "\n",
+        "  (c)instantaneous  value  of  voltage  at  4  ms  =   268.9   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 202</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the amplitude, (b) the peak-to-peak value, \n",
+      "#(c) the rms value, (d) the periodic time, \n",
+      "#(e) the frequency, and (f) the phase angle (in degrees and minutes) relative to 75 sin 200(pi*t)\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  75;#  in  Volts\n",
+      "w  =  200*math.pi;#  in  rad/sec\n",
+      "t  =  0.004;#  in  sec\n",
+      "phi  =  0.25;#  in  radians\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vptp  =  2*Vmax\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "T  =  1/f\n",
+      "v  =  Vmax*math.sin(w*t)\n",
+      "phid  =  phi*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Amplitude,  or  peak  value  =  \",Vmax,\"  V\\n\"\n",
+      "print  \"\\n  (b)  Peak-to-peak  value  =  \",Vptp,\"  V\\n\"\n",
+      "print  \"\\n  (c)rms  value  =  \",Vrms,\"  V\\n\"\n",
+      "print  \"\\n  (d)periodic  time,  T  =  \",T,\"  sec\\n\"\n",
+      "print  \"\\n  (e)frequency  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (f)phase  angle  =  \",round(phid,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Amplitude,  or  peak  value  =   75   V\n",
+        "\n",
+        "\n",
+        "  (b)  Peak-to-peak  value  =   150   V\n",
+        "\n",
+        "\n",
+        "  (c)rms  value  =   53.025   V\n",
+        "\n",
+        "\n",
+        "  (d)periodic  time,  T  =   0.01   sec\n",
+        "\n",
+        "\n",
+        "  (e)frequency  f  =   100.0   Hz\n",
+        "\n",
+        "\n",
+        "  (f)phase  angle  =   14.32 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 202</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Express the instantaneous voltage in the form v = Vm*sin(wt +- phi)\t\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  40;#  in  Volts\n",
+      "T  =  0.01;#  in  sec\n",
+      "v  =  -20;#  when  t  =  0sec,  in  volts\n",
+      "t  =  0;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "w  =  2*math.pi/T\n",
+      "phir  =  math.asin(v/Vmax)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    instantaneous  voltage  v  =  \",  Vmax,\"  sin(\",round(w,2),\"t\",round(phir,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    instantaneous  voltage  v  =   40   sin( 628.32 t -0.52 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 203</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find:\n",
+      "#(a) the peak value, the periodic time, the frequency and phase angle relative to 120 sin 100*pi*t\n",
+      "#(b) the value of the current when t = 0\n",
+      "#(c) the value of the current when t = 8 ms\n",
+      "#(d) the time when the current first reaches 60 A, and\n",
+      "#(e) the time when the current is first a maximum\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Imax  =  120;#  in  Amperes\n",
+      "w  =  100*math.pi;#  in  rad/sec\n",
+      "phi  =  0.36;#  in  rad\n",
+      "t1  =  0;#  in  secs\n",
+      "t2  =  0.008;#  in  secs\n",
+      "i  =  60;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "f  =  w/(2*math.pi)\n",
+      "T  =  1/f\n",
+      "phid  =  phi*180/math.pi\n",
+      "i0  =  Imax*math.sin((w*t1)+phi)\n",
+      "i8  =  Imax*math.sin((w*t2)+phi)  \n",
+      "ti  =  (math.asin(i/Imax)  -  phi)/w\n",
+      "tm1  =  (math.asin(Imax/Imax)  -  phi)/w\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Peak  value  =  \",  Imax,\"  A,  Periodic  time  T  =  \",  T,\"  sec, \"\n",
+      "print   \" Frequency,  f  =  \",  f,\"  Hz  Phase  angle  =  \",round(phid,2),\"deg leading\\n\"\n",
+      "print  \"\\n  (b)  When  t  =  0,  i  =  \",round(i0,2),\"  A\\n\"\n",
+      "print  \"\\n  (c)When  t  =  8  ms  =  \",  round(i8,2),\"  A\\n\"\n",
+      "print  \"\\n  (d)When  i  is  60  A,  then  time  t  =  \",round((ti/1E-3),2),\"  ms\\n\"\n",
+      "print  \"\\n  (e)When  the  current  is  a  maximum,  time,  t  =  \",  round((tm1/1E-3),2),\"  ms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Peak  value  =   120   A,  Periodic  time  T  =   0.02   sec, \n",
+        " Frequency,  f  =   50.0   Hz  Phase  angle  =   20.63 deg leading\n",
+        "\n",
+        "\n",
+        "  (b)  When  t  =  0,  i  =   42.27   A\n",
+        "\n",
+        "\n",
+        "  (c)When  t  =  8  ms  =   31.81   A\n",
+        "\n",
+        "\n",
+        "  (d)When  i  is  60  A,  then  time  t  =   0.52   ms\n",
+        "\n",
+        "\n",
+        "  (e)When  the  current  is  a  maximum,  time,  t  =   3.85   ms\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 204</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#obtain a sinusoidal expression for i1 + i2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "i1max  =  20;#  in  Amperes\n",
+      "i2max  =  10;#  in  Amperes\n",
+      "phi1  =  0;#  in  rad\n",
+      "phi2  =  math.pi/3;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#Ig = i1 + i2\n",
+      "Igmax  = 26.5\n",
+      "phiIg = 19*math.pi/180\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Current  Ig = i1 + i2 =\", Igmax,\"sin(wt + \",round(phiIg,3),\") Amps\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Current  Ig = i1 + i2 = 26.5 sin(wt +  0.332 ) Amps"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 205</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find, by calculation, a sinusoidal expression to represent v1 + v2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1max  =  50;#  in  volts\n",
+      "V2max  =  100;#  in  volts\n",
+      "phi2  =  -1*math.pi/6;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#vR2  =  v1**2  +  v2**2  -  2*v1*v2  cos  150\n",
+      "phidiff  =  math.pi  +  phi2\n",
+      "Vrmax  =  (V1max**2  +  V2max**2  -  2*V1max*V2max*math.cos(phidiff))**0.5\n",
+      "#Using  the  sine  rule\n",
+      "phi  =  math.asin(V2max*math.sin(phidiff)/Vrmax)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  VR  =  \",round(Vrmax,2),\"sin(wt  -  \",round(phi,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  VR  =   145.47 sin(wt  -   0.35 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 206</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find a sinusoidal expression for \u0005i1 + i2\t of Problem 13, by calculation.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1max  =  20;#  in  volts\n",
+      "I2max  =  10;#  in  volts\n",
+      "phi2  =  1*math.pi/3;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#iR2  =  i1**2  +  i2**2  -  2*i1*i2cos150\n",
+      "phidiff  =  math.pi  -  phi2\n",
+      "Irmax  =  (I1max**2  +  I2max**2  -  2*I1max*I2max*math.cos(phidiff))**0.5\n",
+      "#Using  the  sine  rule\n",
+      "phi  =  math.asin(I2max*math.sin(phidiff)/Irmax)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  IR  =  \",  round(Irmax,2),\"sin(wt  +  \",round(phi,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  IR  =   26.46 sin(wt  +   0.33 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_3.ipynb
new file mode 100755
index 00000000..d91a1cbb
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_14-checkpoint_3.ipynb
@@ -0,0 +1,886 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a70fbdcc8331977a000d59120f5c4fa11beaa27b252ae9d065173da18ee666e9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 14: Alternating voltages and currents</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  20000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "T1  =  1/f1\n",
+      "T2  =  1/f2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Periodic  time  T  =  \",T1,\"  secs\\n\"\n",
+      "print  \"\\n  (b)  Periodic  time  T  =  \",(T2/1E-6),\"  usecs\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Periodic  time  T  =   0.02   secs\n",
+        "\n",
+        "\n",
+        "  (b)  Periodic  time  T  =   50.0   usecs\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T1  =  0.004;#  in  secs\n",
+      "T2  =  4E-6;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f1  =  1/T1\n",
+      "f2  =  1/T2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Frequency  f  =  \",f1,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)  Frequency  f  =  \",(f2/1E6),\"  MHz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Frequency  f  =   250.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)  Frequency  f  =   0.25   MHz\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 195</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  (8E-3)/5;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f  =  1/T\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Frequency  f  =  \",f,\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Frequency  f  =   625.0   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 196</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ta  =  0.02;#  Time  for  1  complete  cycle  in  secs\n",
+      "Vamax  =  200;#  in  volts\n",
+      "Va1  =  25;#  in  volts\n",
+      "Va2  =  75;#  in  volts\n",
+      "Va3  =  125;#  in  volts\n",
+      "Va4  =  175;#  in  volts\n",
+      "Tb  =  0.016;#  Time  for  1  complete  cycle  in  secs\n",
+      "Ibmax  =  10;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  Triangular  waveform  (Figure  14.5(a))\n",
+      "fa  =  1/Ta\n",
+      "Aaw  =  Ta*Vamax/4\n",
+      "Vaavg  =  Aaw*2/Ta\n",
+      "Varms  =  (((Va1**2)  +  (Va2**2)  +  (Va3**2)  +  (Va4**2))/4)**0.5\n",
+      "#Note  that  the  greater  the  number  of  intervals  chosen,  the  greater  the  accuracy  of  the  result\n",
+      "Ffa  =  Varms/Vaavg\n",
+      "Pfa  =  Vamax/Varms\n",
+      "\n",
+      "#for  Rectangular  waveform  (Figure  14.5(b))\n",
+      "fb  =  1/Tb\n",
+      "Abw  =  Tb*Ibmax/2\n",
+      "Ibavg  =  Abw*2/Tb\n",
+      "Ibrms  =  10\n",
+      "Ffb  =  Ibrms/Ibavg\n",
+      "Pfb  =  Ibmax/Ibrms\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a1)Frequency  f  =  \",fa,\"  Hz\\n\"\n",
+      "print  \"\\n  (a2)average  value  over  half  a  cycle  =  \",Vaavg,\"  V\\n\"\n",
+      "print  \"\\n  (a3)rms  value  =  \",round(Varms,2),\"  V\\n\"\n",
+      "print  \"\\n  (a4)Form  factor  =  \",round(Ffa,2),\"\\n\"\n",
+      "print  \"\\n  (a5)Peak  factor  =  \",round(Pfa,2),\"\\n\"\n",
+      "print  \"\\n  (b1)Frequency  f  =  \",fb,\"  Hz\\n\"\n",
+      "print  \"\\n  (b2)average  value  over  half  a  cycle  =  \",Ibavg,\"  A\\n\"\n",
+      "print  \"\\n  (b3)rms  value  =  \",Ibrms,\"  A\\n\"\n",
+      "print  \"\\n  (b4)Form  factor  =  \",Ffb,\"\\n\"\n",
+      "print  \"\\n  (b5)Peak  factor  =  \",Pfb,\"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a1)Frequency  f  =   50.0   Hz\n",
+        "\n",
+        "\n",
+        "  (a2)average  value  over  half  a  cycle  =   100.0   V\n",
+        "\n",
+        "\n",
+        "  (a3)rms  value  =   114.56   V\n",
+        "\n",
+        "\n",
+        "  (a4)Form  factor  =   1.15 \n",
+        "\n",
+        "\n",
+        "  (a5)Peak  factor  =   1.75 \n",
+        "\n",
+        "\n",
+        "  (b1)Frequency  f  =   62.5   Hz\n",
+        "\n",
+        "\n",
+        "  (b2)average  value  over  half  a  cycle  =   10.0   A\n",
+        "\n",
+        "\n",
+        "  (b3)rms  value  =   10   A\n",
+        "\n",
+        "\n",
+        "  (b4)Form  factor  =   1.0 \n",
+        "\n",
+        "\n",
+        "  (b5)Peak  factor  =   1.0 "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 198</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy \n",
+      "from numpy import mean, sqrt, arange\n",
+      "#initializing  the  variables:\n",
+      "Thalf = 5; #in ms\n",
+      "Ta  =  0.02;#  Time  for  1  complete  cycle  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "Tfull = 2*Thalf/1000 # in sec\n",
+      "f = 1/Tfull\n",
+      "A=[3, 10, 19, 30, 49, 63, 73, 72, 30, 2]\n",
+      "Iinst125 = 19\n",
+      "Iinst38 = 70\n",
+      "sq = 0\n",
+      "Ipeak = 76\n",
+      "#B=arange(A)\n",
+      "Imean = (0.5*1E-3)*numpy.mean(A)*10/(5*1E-3)\n",
+      "for h in range(10):\n",
+      "    sq = sq + A[h]**2\n",
+      "\n",
+      "Irms = sqrt(sq/10)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Frequency  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Instantaneous value of current after 1.25 ms  =\",Iinst125,\"A \"\n",
+      "print   \"and Instantaneous value of current after 3.8 ms\", Iinst38,\"A\\n\"\n",
+      "print  \"\\n  (c)Peak or maximum value  =  \",Ipeak,\"  A\\n\"\n",
+      "print  \"\\n  (d)Mean or average value  =  \",round(Imean,2),\"A\\n\"\n",
+      "print  \"\\n  (e)rms value =  \",round(Irms,1),\"A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Frequency  f  =   100.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Instantaneous value of current after 1.25 ms  = 19 A and Instantaneous value of current after 3.8 ms 70 A\n",
+        "\n",
+        "\n",
+        "  (c)Peak or maximum value  =   76   A\n",
+        "\n",
+        "\n",
+        "  (d)Mean or average value  =   35.1 A\n",
+        "\n",
+        "\n",
+        "  (e)rms value =   43.8 A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Imax  =  20;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Irms  =  Imax/(2**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Rms  value  =  \",round(Irms,2),\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Rms  value  =   14.14   A"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vrms  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vmax  =  Vrms*(2**0.5)\n",
+      "Vmean  =  0.637*Vmax\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  peak  value  =  \",round(Vmax,2),\"  V\\n\"\n",
+      "print  \"\\n  mean  value  =  \",round(Vmean,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  peak  value  =   339.41   V\n",
+        "\n",
+        "\n",
+        "  mean  value  =   216.2   V"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 200</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmean  =  150;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vmax  =  Vmean/0.637\n",
+      "Vrms  =  0.707*Vmax\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  peak  value  =  \",round(Vmax,2),\"  V\\n\"\n",
+      "print  \"\\n  rms  value  =  \",round(Vrms,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  peak  value  =   235.48   V\n",
+        "\n",
+        "\n",
+        "  rms  value  =   166.48   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 201</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  282.8;#  in  Volts\n",
+      "w  =  314;#  in  rad/sec\n",
+      "t  =  0.004;#  in  sec\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "v  =  Vmax*math.sin(w*t)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)rms  value  =  \",round(Vrms,2),\"  V\\n\"\n",
+      "print  \"\\n  (b)frequency  f  =  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (c)instantaneous  value  of  voltage  at  4  ms  =  \",round(v,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)rms  value  =   199.94   V\n",
+        "\n",
+        "\n",
+        "  (b)frequency  f  =   49.97   Hz\n",
+        "\n",
+        "\n",
+        "  (c)instantaneous  value  of  voltage  at  4  ms  =   268.9   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 202</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  75;#  in  Volts\n",
+      "w  =  200*math.pi;#  in  rad/sec\n",
+      "t  =  0.004;#  in  sec\n",
+      "phi  =  0.25;#  in  radians\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "Vptp  =  2*Vmax\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "T  =  1/f\n",
+      "v  =  Vmax*math.sin(w*t)\n",
+      "phid  =  phi*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Amplitude,  or  peak  value  =  \",Vmax,\"  V\\n\"\n",
+      "print  \"\\n  (b)  Peak-to-peak  value  =  \",Vptp,\"  V\\n\"\n",
+      "print  \"\\n  (c)rms  value  =  \",Vrms,\"  V\\n\"\n",
+      "print  \"\\n  (d)periodic  time,  T  =  \",T,\"  sec\\n\"\n",
+      "print  \"\\n  (e)frequency  f  =  \",f,\"  Hz\\n\"\n",
+      "print  \"\\n  (f)phase  angle  =  \",round(phid,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Amplitude,  or  peak  value  =   75   V\n",
+        "\n",
+        "\n",
+        "  (b)  Peak-to-peak  value  =   150   V\n",
+        "\n",
+        "\n",
+        "  (c)rms  value  =   53.025   V\n",
+        "\n",
+        "\n",
+        "  (d)periodic  time,  T  =   0.01   sec\n",
+        "\n",
+        "\n",
+        "  (e)frequency  f  =   100.0   Hz\n",
+        "\n",
+        "\n",
+        "  (f)phase  angle  =   14.32 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 202</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vmax  =  40;#  in  Volts\n",
+      "T  =  0.01;#  in  sec\n",
+      "v  =  -20;#  when  t  =  0sec,  in  volts\n",
+      "t  =  0;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "w  =  2*math.pi/T\n",
+      "phir  =  math.asin(v/Vmax)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    instantaneous  voltage  v  =  \",  Vmax,\"  sin(\",round(w,2),\"t\",round(phir,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    instantaneous  voltage  v  =   40   sin( 628.32 t -0.52 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 203</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Imax  =  120;#  in  Amperes\n",
+      "w  =  100*math.pi;#  in  rad/sec\n",
+      "phi  =  0.36;#  in  rad\n",
+      "t1  =  0;#  in  secs\n",
+      "t2  =  0.008;#  in  secs\n",
+      "i  =  60;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "#for  a  sine  wave\n",
+      "f  =  w/(2*math.pi)\n",
+      "T  =  1/f\n",
+      "phid  =  phi*180/math.pi\n",
+      "i0  =  Imax*math.sin((w*t1)+phi)\n",
+      "i8  =  Imax*math.sin((w*t2)+phi)  \n",
+      "ti  =  (math.asin(i/Imax)  -  phi)/w\n",
+      "tm1  =  (math.asin(Imax/Imax)  -  phi)/w\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Peak  value  =  \",  Imax,\"  A,  Periodic  time  T  =  \",  T,\"  sec, \"\n",
+      "print   \" Frequency,  f  =  \",  f,\"  Hz  Phase  angle  =  \",round(phid,2),\"deg leading\\n\"\n",
+      "print  \"\\n  (b)  When  t  =  0,  i  =  \",round(i0,2),\"  A\\n\"\n",
+      "print  \"\\n  (c)When  t  =  8  ms  =  \",  round(i8,2),\"  A\\n\"\n",
+      "print  \"\\n  (d)When  i  is  60  A,  then  time  t  =  \",round((ti/1E-3),2),\"  ms\\n\"\n",
+      "print  \"\\n  (e)When  the  current  is  a  maximum,  time,  t  =  \",  round((tm1/1E-3),2),\"  ms\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Peak  value  =   120   A,  Periodic  time  T  =   0.02   sec, \n",
+        " Frequency,  f  =   50.0   Hz  Phase  angle  =   20.63 deg leading\n",
+        "\n",
+        "\n",
+        "  (b)  When  t  =  0,  i  =   42.27   A\n",
+        "\n",
+        "\n",
+        "  (c)When  t  =  8  ms  =   31.81   A\n",
+        "\n",
+        "\n",
+        "  (d)When  i  is  60  A,  then  time  t  =   0.52   ms\n",
+        "\n",
+        "\n",
+        "  (e)When  the  current  is  a  maximum,  time,  t  =   3.85   ms\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 204</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "i1max  =  20;#  in  Amperes\n",
+      "i2max  =  10;#  in  Amperes\n",
+      "phi1  =  0;#  in  rad\n",
+      "phi2  =  math.pi/3;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#Ig = i1 + i2\n",
+      "Igmax  = 26.5\n",
+      "phiIg = 19*math.pi/180\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Current  Ig = i1 + i2 =\", Igmax,\"sin(wt + \",round(phiIg,3),\") Amps\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Current  Ig = i1 + i2 = 26.5 sin(wt +  0.332 ) Amps"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 205</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1max  =  50;#  in  volts\n",
+      "V2max  =  100;#  in  volts\n",
+      "phi2  =  -1*math.pi/6;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#vR2  =  v1**2  +  v2**2  -  2*v1*v2  cos  150\n",
+      "phidiff  =  math.pi  +  phi2\n",
+      "Vrmax  =  (V1max**2  +  V2max**2  -  2*V1max*V2max*math.cos(phidiff))**0.5\n",
+      "#Using  the  sine  rule\n",
+      "phi  =  math.asin(V2max*math.sin(phidiff)/Vrmax)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  VR  =  \",round(Vrmax,2),\"sin(wt  -  \",round(phi,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  VR  =   145.47 sin(wt  -   0.35 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 206</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I1max  =  20;#  in  volts\n",
+      "I2max  =  10;#  in  volts\n",
+      "phi2  =  1*math.pi/3;#  in  rad\n",
+      "\n",
+      "#calculation:\n",
+      "#iR2  =  i1**2  +  i2**2  -  2*i1*i2cos150\n",
+      "phidiff  =  math.pi  -  phi2\n",
+      "Irmax  =  (I1max**2  +  I2max**2  -  2*I1max*I2max*math.cos(phidiff))**0.5\n",
+      "#Using  the  sine  rule\n",
+      "phi  =  math.asin(I2max*math.sin(phidiff)/Irmax)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  IR  =  \",  round(Irmax,2),\"sin(wt  +  \",round(phi,2),\")  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  IR  =   26.46 sin(wt  +   0.33 )  V"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint.ipynb
new file mode 100755
index 00000000..e7b68d8a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint.ipynb
@@ -0,0 +1,1677 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 15: Single-phase series a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 214</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Calculate the reactance of a coil \n",
+      "#(b) Determine the inductance of the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.32;#  in  Henry\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  5000;#  in  Hz\n",
+      "Z  =  124;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f1*L\n",
+      "L  =  Z/(2*math.pi*f2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms  \\n\"\n",
+      "print  \"\\n  (b)Inductance  L  =  \",round((L/1E-3),2),\"  mH  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   100.53   ohms  \n",
+        "\n",
+        "\n",
+        "  (b)Inductance  L  =   3.95   mH  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 214</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate its inductive reactance and the resulting current if connected to\n",
+      "#(a) a 240 V, 50 Hz supply, and (b) a 100 V, 1 kHz supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.040;#  in  Henry\n",
+      "V1  =  240;#  in  volts\n",
+      "V2  =  100;#  in  volts\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL1  =  2*math.pi*f1*L\n",
+      "I1  =  V1/XL1\n",
+      "XL2  =  2*math.pi*f2*L\n",
+      "I2  =  V2/XL2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(  XL1,2),\"  ohms  and  current  I  =  \",round(  I1,2),\"  A\\n\"\n",
+      "print  \"\\n  (b)Inductive  reactance,  XL  =  \",round(  XL2,2),\"  ohms  and  current  I  =  \",round(  I2,2),\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   12.57   ohms  and  current  I  =   19.1   A\n",
+        "\n",
+        "\n",
+        "  (b)Inductive  reactance,  XL  =   251.33   ohms  and  current  I  =   0.4   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitive reactance of a capacitor of 10 \u03bcF when connected to a circuit of frequency (a) 50 Hz (b) 20 kHz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  10E-6;#  in  Farads\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  20000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Xc1  =  1/(2*math.pi*f1*C)\n",
+      "Xc2  =  1/(2*math.pi*f2*C)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Capacitive  reactance,  Xc  =  \",round(  Xc1,2),\"  ohms  \"\n",
+      "print  \"\\n  (b)Capacitive  reactance,  Xc  =  \",round(  Xc2,2),\"  ohms  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitive  reactance,  Xc  =   318.31   ohms  \n",
+        "\n",
+        "  (b)Capacitive  reactance,  Xc  =   0.8   ohms  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of its capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  40;#  in  ohms\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  1/(2*math.pi*f*Z)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Capacitance,C  =  \",round((C/1E-6),2),\" uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Capacitance,C  =   79.58  uF  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the current taken by a 23 \u03bcF capacitor when connected to a 240 V, 50 Hz supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  23E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "I  =  V/Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  I  =  \",round(I,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  I  =   1.73   A  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 216</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the supply voltage and the phase angle between current and voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vr  =  12;#  in  volts\n",
+      "Vl  =  5;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  (Vr**2  +  Vl**2)**0.5\n",
+      "phi  =  math.atan(Vl/Vr)\n",
+      "phid  =  phi*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  voltage  V  =  \",V,\"  V,  phase  angle  between  current  and  voltage  is  \",  round(phid,2),\"deg lagging\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  voltage  V  =   13.0   V,  phase  angle  between  current  and  voltage  is   22.62 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 216</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the reactance, (b) the impedance, and (c) the current taken from a 240 V, 50 Hz supply. \n",
+      "#Determine also the phase angle between the supply voltage and current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  volts\n",
+      "R  =  4;#  in  ohms\n",
+      "L  =  0.00955;#  in  Henry\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  = math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (c)Current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (d)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   3.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   5.0   ohms\n",
+        "\n",
+        "  (c)Current,  I  =   48.0   A\n",
+        "\n",
+        "  (d)phase  angle  between  the  supply  voltage  and  current  is   36.87 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 217</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the resistance, impedance, inductive reactance and inductance of the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vdc  =  12;#  in  volts\n",
+      "Vac  =  240;#  in  volts\n",
+      "Iac  =  20;#  in  Amperes\n",
+      "Idc  =  2;#  in  Amperes\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  Vdc/Idc\n",
+      "Z  =  Vac/Iac\n",
+      "XL  =  (Z**2  -  R**2)**0.5\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance,  R  =  \",R,\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",Z,\"  ohms\"\n",
+      "print  \"\\n  (c)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (d)Inductance,  L  =  \",round(L,2),\"  H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance,  R  =   6.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   12.0   ohms\n",
+        "\n",
+        "  (c)Inductive  reactance,  XL  =   10.39   ohms\n",
+        "\n",
+        "  (d)Inductance,  L  =   0.03   H"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 217</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the inductive reactance of the coil, \n",
+      "#(b) the impedance of the circuit, \n",
+      "#(c) the current in the circuit, \n",
+      "#(d) the p.d. across each component, and \n",
+      "#(e) the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  200;#  in  ohms\n",
+      "L  =  0.3183;#  in  henry\n",
+      "V  =  240;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "VL  =  I*XL\n",
+      "VR  =  I*R\n",
+      "phid  =  math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (c)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (d)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V  and  p.d.  across  resistance,  VR  =  \",round(VR,2),\"  V\"\n",
+      "print  \"\\n  (e)circuit  phase  angle  is  \",round(phid,2),\" deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   100.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   223.61   ohms\n",
+        "\n",
+        "  (c)current,  I  =   1.07   A\n",
+        "\n",
+        "  (d)p.d.  across  Inductor,  VL  =   107.33   V  and  p.d.  across  resistance,  VR  =   214.66   V\n",
+        "\n",
+        "  (e)circuit  phase  angle  is   26.56  deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 218</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the circuit impedance, \n",
+      "#(b) the current flowing, \n",
+      "#(c) the p.d. across the resistance, \n",
+      "#(d) the p.d. across the inductance and \n",
+      "#(e) the phase angle between voltage and current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  100;#  in  ohms\n",
+      "L  =  0.2;#  in  henry\n",
+      "Vmax  =  200;#  in  volts\n",
+      "w  =  500;#  in  rad/sec\n",
+      "\n",
+      "#calculation:\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  Vrms/Z\n",
+      "VL  =  I*XL\n",
+      "VR  =  I*R\n",
+      "phid  =  math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "\\\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (c)p.d.  across  resistance,  VR  =  \",round(VR,2),\"  V\"\n",
+      "print  \"\\n  (d)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (e)circuit  phase  angle  is  \",phid,\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   141.42   ohms\n",
+        "\n",
+        "  (b)current,  I  =   1.0   A\n",
+        "\n",
+        "  (c)p.d.  across  resistance,  VR  =   99.98   V\n",
+        "\n",
+        "  (d)p.d.  across  Inductor,  VL  =   99.98   V\n",
+        "\n",
+        "  (e)circuit  phase  angle  is   45.0 deg"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 218</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of the supply voltage and the voltage across the 1.273 mH inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  1.2273E-3;#  in  henry\n",
+      "f  =  5000;#  in  Hz\n",
+      "VR  =  6;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "I  =VR/R\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "V  =  I*Z\n",
+      "VL  =  I*XL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)supply  voltage  =  \",round(V,2),\"  Volts\"\n",
+      "print  \"\\n  (b)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)supply  voltage  =   9.77   Volts\n",
+        "\n",
+        "  (b)p.d.  across  Inductor,  VL  =   7.71   V"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 219</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the impedance of the circuit, \n",
+      "#(b) the current in the circuit, \n",
+      "#(c) the circuit phase angle, \n",
+      "#(d) the p.d. across the 60 ohm resistor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  60;#  in  ohms\n",
+      "Rc  =  20;#  in  ohms\n",
+      "L  =  159.2E-3;#  in  henry\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Rt  =  R  +  Rc\n",
+      "Z  =  (Rt**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan(XL/Rt)*180/math.pi\n",
+      "VR  =  I*R\n",
+      "Zc  =  (Rc**2  +  XL**2)**0.5\n",
+      "Vc  =  I*Zc\n",
+      "VL  =  I*XL\n",
+      "VRc  =  I*Rc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,3),\"  A\"\n",
+      "print  \"\\n  (c)circuit  phase  angle  is  \",round(phid,0),\"deg lagging\"\n",
+      "print  \"\\n  (d)p.d.  across  resistance,  VR  =  \",round(  VR,1),\"  V\"\n",
+      "print  \"\\n  (e)p.d.  across  coil,  Vc  =  \",round(Vc,1),\"  V\"\n",
+      "print  \"\\n  (f1)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (f2)p.d.  across  coil  resistance,  VRc  =  \",round(VRc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   94.35   ohms\n",
+        "\n",
+        "  (b)current,  I  =   2.544   A\n",
+        "\n",
+        "  (c)circuit  phase  angle  is   32.0 deg lagging\n",
+        "\n",
+        "  (d)p.d.  across  resistance,  VR  =   152.6   V\n",
+        "\n",
+        "  (e)p.d.  across  coil,  Vc  =   137.0   V\n",
+        "\n",
+        "  (f1)p.d.  across  Inductor,  VL  =   127.23   V\n",
+        "\n",
+        "  (f2)p.d.  across  coil  resistance,  VRc  =   50.88   V"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 220</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the impedance, and (b) the current taken from a 240 V, 50 Hz supply. \n",
+      "#Find also the phase angle between the supply voltage and the current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohms\n",
+      "C  =  45E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Z  =  (R**2  +  Xc**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan(Xc/R)*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (c)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   75.02   ohms\n",
+        "\n",
+        "  (b)current,  I  =   3.2   A\n",
+        "\n",
+        "  (c)phase  angle  between  the  supply  voltage  and  current  is   70.54 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 221</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate: (a) the value of capacitance, C, \n",
+      "#(b) the supply voltage, \n",
+      "#(c) the phase angle between the supply voltage and current, \n",
+      "#(d) the p.d. across the resistor, and\n",
+      "#(e) the p.d. across the capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  40;#  in  ohms\n",
+      "f  =  60;#  in  Hz\n",
+      "I  =  3;#in  amperes\n",
+      "Z  =  50;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  (Z**2  -  R**2)**0.5\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "V  =  I*Z\n",
+      "phid  =  math.atan(Xc/R)*180/math.pi\n",
+      "VR  =  I*R\n",
+      "Vc  =  I*Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance,  C  =  \",round((C/1E-6),2),\"  uF\"\n",
+      "print  \"\\n  (b)Voltage,  V  =  \",V,\"  Volts\"\n",
+      "print  \"\\n  (c)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)p.d.  across  resistance,  VR  =  \",  VR,\"  V\"\n",
+      "print  \"\\n  (e)p.d.  across  Capacitor,  Vc  =  \",Vc,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance,  C  =   88.42   uF\n",
+        "\n",
+        "  (b)Voltage,  V  =   150   Volts\n",
+        "\n",
+        "  (c)phase  angle  between  the  supply  voltage  and  current  is   36.87 deg leading\n",
+        "\n",
+        "  (d)p.d.  across  resistance,  VR  =   120   V\n",
+        "\n",
+        "  (e)p.d.  across  Capacitor,  Vc  =   90.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 222</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current flowing, \n",
+      "#(b) the phase difference between the supply voltage and current, \n",
+      "#(c) the voltage across the coil and \n",
+      "#(d) the voltage across the capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  5;#  in  ohms\n",
+      "C  =  100E-6;#  in  Farads\n",
+      "L  =  0.12;#  in  Henry\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  300;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "X  =  XL  -  Xc\n",
+      " #Since  XL  is  greater  than  Xc,  the  circuit  is  inductive.\n",
+      "Z  =  (R**2  +  (XL-Xc)**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan((XL-Xc)/R)*180/math.pi\n",
+      "Zcl  =  (R**2  +  XL**2)**0.5\n",
+      "Vcl  =  I*Zcl\n",
+      "phidc  =  math.atan(XL/R)*180/math.pi\n",
+      "Vc  =  I*Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current,I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (c)Voltage  across  the  coil,  Vcoil  =  \",round(Vcl,0),\"  Volts\"\n",
+      "print  \"\\n  (d)p.d.  across  Capacitor,  Vc  =  \",round(Vc,0),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current,I  =   38.91   A\n",
+        "\n",
+        "  (b)phase  angle  between  the  supply  voltage  and  current  is   49.57 deg\n",
+        "\n",
+        "  (c)Voltage  across  the  coil,  Vcoil  =   1480.0   Volts\n",
+        "\n",
+        "  (d)p.d.  across  Capacitor,  Vc  =   1239.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 224</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the circuit current, \n",
+      "#(b) the circuit phase angle and \n",
+      "#(c) the voltage drop across each impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohms\n",
+      "C  =  0.25E-6;#  in  Farads\n",
+      "L  =  130E-6;#  in  Henry\n",
+      "Rc  =  5;#  in  ohms\n",
+      "R2  =  10;#  in  ohms\n",
+      "f  =  20000;#  in  Hz\n",
+      "V  =  40;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "X  =  Xc  -  XL\n",
+      "R  =  R1  +  R2  +  Rc\n",
+      " #Since  Xc  is  greater  than  XL,  the  circuit  is  capacitive.\n",
+      "Z  =  (R**2  +  (Xc-XL)**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid = math.atan((Xc-XL)/R)*180/math.pi\n",
+      "V1  =  I*R1\n",
+      "V2  =  I*((Rc**2  +  XL**2)**0.5)\n",
+      "V3  =  I*((R2**2  +  Xc**2)**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current,I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)circuit  phase  angle  is  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (c1)Voltage  across  the  Resistance  of  8  ohms  =  \",round(V1,2),\"  Volts\"\n",
+      "print  \"\\n  (c2)Voltage  across  the  coil,  Vcoil  =  \",round(V2,2),\"  Volts\"\n",
+      "print  \"\\n  (c3)p.d.  across  Capacitor  resistance  circuit  =  \",round(V3,2),\"  Volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current,I  =   1.44   A\n",
+        "\n",
+        "  (b)circuit  phase  angle  is   33.97 deg leading\n",
+        "\n",
+        "  (c1)Voltage  across  the  Resistance  of  8  ohms  =   11.54   Volts\n",
+        "\n",
+        "  (c2)Voltage  across  the  coil,  Vcoil  =   24.64   Volts\n",
+        "\n",
+        "  (c3)p.d.  across  Capacitor  resistance  circuit  =   48.12   Volts"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 224</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the p.d.\u2019s V1 and V2 for the circuit\n",
+      "#determine the supply voltage V and the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  4;#  in  ohms\n",
+      "C  =  1.273E-6;#  in  Farads\n",
+      "L  =  0.286E-3;#  in  Henry\n",
+      "R2  =  8;#  in  ohms\n",
+      "f  =  5000;#  in  Hz\n",
+      "I  =  5;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "phid1  =  math.atan(XL/R1)*180/math.pi\n",
+      "V1  =  I*((R1**2  +  XL**2)**0.5)\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "V2  =  I*((R2**2  +  Xc**2)**0.5)\n",
+      "phid2  =  math.atan(Xc/R2)*180/math.pi\n",
+      "Z  =  ((R1+R2)**2  +  (Xc-XL)**2)**0.5\n",
+      "V  =  I*Z\n",
+      "phid  =  math.atan((Xc-XL)/(R1+R2))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Voltage  supply,  V  =  \",round(V,2),\"  V\"\n",
+      "print  \"\\n  (b)circuit  phase  angle  is  \",round(phid,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Voltage  supply,  V  =   100.08   V\n",
+        "\n",
+        "  (b)circuit  phase  angle  is   53.16 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 226</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#At what frequency does resonance occur?\n",
+      "#Find the current flowing at the resonant frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  60E-6;#  in  Farads\n",
+      "L  =  125E-3;#  in  Henry\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*(L*C)**0.5)\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency,Fr  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency,Fr  =   58.12   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   12.0"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 226</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the circuit resistance, and (b) the circuit capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.05E-3;#  in  Henry\n",
+      "fr  =  200000;#  in  Hz\n",
+      "V  =  0.002;#  in  Volts\n",
+      "I  =  0.1E-3;#  in  amperes\n",
+      "#calculation:\n",
+      "#  L-C-R\n",
+      "#At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "R  =  V/I\n",
+      "C  =  1/(L*(2*math.pi*fr)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance,  R  =  \",round(R,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Capacitance,  C  =  \",round((C/1E-9),2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance,  R  =   20.0   ohms\n",
+        "\n",
+        "  (b)Capacitance,  C  =   12.67 nF"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 227</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, and \n",
+      "#(b) the current at resonance. \n",
+      "#How many times greater than the supply voltage is the voltage across the reactances at resonance?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  80E-3;#  in  Henry\n",
+      "C  =  0.25E-6;#  in  Farads\n",
+      "R  =  12.5;#  in  ohms\n",
+      "V  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "VL  =  I*(2*math.pi*fr*L)\n",
+      "Vc  =  I/(2*math.pi*fr*C)\n",
+      "Vm  =  VL/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  resonant  frequency  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2),\"\"\n",
+      "print  \"\\n  (b)Voltage  magnification  at  resonance  =  \",round(Vm,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  resonant  frequency  =   1125.4   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   8.0 \n",
+        "\n",
+        "  (b)Voltage  magnification  at  resonance  =   45.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 228</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the Qfactor of the circuit at resonance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  60E-3;#  in  Henry\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "R  =  2;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  ((L/C)**0.5)/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  At  resonance,  Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  At  resonance,  Q-factor  =   22.36"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 228</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, \n",
+      "#(b) the current at resonance,\n",
+      "#(c) the voltages across the coil and the capacitor at resonance, and\n",
+      "#(d) the Q-factor of the circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  100E-3;#  in  Henry\n",
+      "C  =  2E-6;#  in  Farads\n",
+      "R  =  10;#  in  ohms\n",
+      "V  =  50;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "VL  =  I*(2*math.pi*fr*L)\n",
+      "Vc  =  I/(2*math.pi*fr*C)\n",
+      "Q  =  VL/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  resonant  frequency  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2),\"\"\n",
+      "print  \"\\n  (c)Voltage  across  coil  at  resonance  is  \",round(VL,2),\"V  \"\n",
+      "print   \"and  Voltage  across  capacitance  at  resonance  is  \",round(  Vc,2),\"V\"\n",
+      "print  \"\\n  (d)At  resonance,  Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  resonant  frequency  =   355.88   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   5.0 \n",
+        "\n",
+        "  (c)Voltage  across  coil  at  resonance  is   1118.03 V  and  Voltage  across  capacitance  at  resonance  is   1118.03 V\n",
+        "\n",
+        "  (d)At  resonance,  Q-factor  =   22.36"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 230</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the bandwidth of the filter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  20E-3;#  in  Henry\n",
+      "R  =  10;#  in  ohms\n",
+      "fr  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Qr  =  (2*math.pi*fr)*L/R\n",
+      "bw  =  fr/Qr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Bandwidth,  (f2-f1)  =  \",round(bw,2),\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Bandwidth,  (f2-f1)  =   79.58   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 231</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the power dissipated in the resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  5000;#  in  ohms\n",
+      "Imax  =  0.250;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Irms  =  0.707*Imax\n",
+      "P  =  Irms*Irms*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Power,  P  =  \",round(P,2),\"  Watts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Power,  P  =   156.2   Watts"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 231</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  60;#  in  ohms\n",
+      "L  =  75E-3;#  in  Henry\n",
+      "V  =  110;#  in  Volts\n",
+      "f  =  60;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "P  =  I*I*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Power,  P  =  \",round(P,2),\"  Watts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Power,  P  =   165.02   Watts"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 232</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the value of the inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VI  =  300;#  in  VA\n",
+      "V  =  150;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  VI/V\n",
+      "XL  =  V/I\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  =  \",round(L,2),\"  H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  =   0.24   H"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 232</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the rated power output and the corresponding reactive power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VI  =  200000;#  in  VA\n",
+      "pf  =  0.8;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "P  =  VI*pf\n",
+      "Q  =  VI*math.sin(math.acos(pf))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  rated  power  output  is  \",round(P/1000,2),\"KW  and  the  corresponding  reactive  power  is  \",round(Q/1000,2),\"kvar\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  rated  power  output  is   160.0 KW  and  the  corresponding  reactive  power  is   120.0 kvar"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 233</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the resistance, \n",
+      "#(b) the impedance, \n",
+      "#(c) the reactance, \n",
+      "#(d) the power factor, and \n",
+      "#(e) the phase angle between voltage and current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  120;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "P  =  400;#  in  Watt\n",
+      "I  =  8;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  P/(I*I)\n",
+      "Z  =  V/I\n",
+      "XL  =  (Z**2  -  R**2)**0.5\n",
+      "pf  =  P/(V*I)\n",
+      "phi  =  math.acos(pf)*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  =  \",round(R,2),\"  ohm  \"\n",
+      "print  \"\\n  (b)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (c)reactance  =  \",round(XL,2),\"  ohm  \"\n",
+      "print  \"\\n  (d)Power  factor  =  \",round(pf,2),\"\"\n",
+      "print  \"\\n  (e)phase  angle  =  \",round(phi,2),\"deg lagging\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  =   6.25   ohm  \n",
+        "\n",
+        "  (b)Impedance  Z  =   15.0   Ohm  \n",
+        "\n",
+        "  (c)reactance  =   13.64   ohm  \n",
+        "\n",
+        "  (d)Power  factor  =   0.42 \n",
+        "\n",
+        "  (e)phase  angle  =   65.38 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 29, page no. 233</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the current flowing, (b) the phase angle,\n",
+      "#(c) the resistance, (d) the impedance, and (e) the capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  60;#  in  Hz\n",
+      "P  =  100;#  in  Watt\n",
+      "pf  =  0.5;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  P/(pf*V)\n",
+      "phi  =  math.acos(pf)*180/math.pi\n",
+      "R  =  P/(I*I)\n",
+      "Z  =  V/I\n",
+      "Xc  =  (Z**2  -  R**2)**0.5\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (b)phase  angle  =  \",round(phi,2),\"deg leading\"\n",
+      "print  \"\\n  (c)resistance  =  \",round(R,2),\"  ohm  \"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)capacitance  =  \",round((C/1E-6),2),\"uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  I  =   2.0   A  \n",
+        "\n",
+        "  (b)phase  angle  =   60.0 deg leading\n",
+        "\n",
+        "  (c)resistance  =   25.0   ohm  \n",
+        "\n",
+        "  (d)Impedance  Z  =   50.0   Ohm  \n",
+        "\n",
+        "  (e)capacitance  =   61.26 uF  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_1.ipynb
new file mode 100755
index 00000000..e7b68d8a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_1.ipynb
@@ -0,0 +1,1677 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 15: Single-phase series a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 214</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Calculate the reactance of a coil \n",
+      "#(b) Determine the inductance of the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.32;#  in  Henry\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  5000;#  in  Hz\n",
+      "Z  =  124;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f1*L\n",
+      "L  =  Z/(2*math.pi*f2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms  \\n\"\n",
+      "print  \"\\n  (b)Inductance  L  =  \",round((L/1E-3),2),\"  mH  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   100.53   ohms  \n",
+        "\n",
+        "\n",
+        "  (b)Inductance  L  =   3.95   mH  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 214</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate its inductive reactance and the resulting current if connected to\n",
+      "#(a) a 240 V, 50 Hz supply, and (b) a 100 V, 1 kHz supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.040;#  in  Henry\n",
+      "V1  =  240;#  in  volts\n",
+      "V2  =  100;#  in  volts\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL1  =  2*math.pi*f1*L\n",
+      "I1  =  V1/XL1\n",
+      "XL2  =  2*math.pi*f2*L\n",
+      "I2  =  V2/XL2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(  XL1,2),\"  ohms  and  current  I  =  \",round(  I1,2),\"  A\\n\"\n",
+      "print  \"\\n  (b)Inductive  reactance,  XL  =  \",round(  XL2,2),\"  ohms  and  current  I  =  \",round(  I2,2),\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   12.57   ohms  and  current  I  =   19.1   A\n",
+        "\n",
+        "\n",
+        "  (b)Inductive  reactance,  XL  =   251.33   ohms  and  current  I  =   0.4   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitive reactance of a capacitor of 10 \u03bcF when connected to a circuit of frequency (a) 50 Hz (b) 20 kHz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  10E-6;#  in  Farads\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  20000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Xc1  =  1/(2*math.pi*f1*C)\n",
+      "Xc2  =  1/(2*math.pi*f2*C)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Capacitive  reactance,  Xc  =  \",round(  Xc1,2),\"  ohms  \"\n",
+      "print  \"\\n  (b)Capacitive  reactance,  Xc  =  \",round(  Xc2,2),\"  ohms  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitive  reactance,  Xc  =   318.31   ohms  \n",
+        "\n",
+        "  (b)Capacitive  reactance,  Xc  =   0.8   ohms  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of its capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  40;#  in  ohms\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  1/(2*math.pi*f*Z)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Capacitance,C  =  \",round((C/1E-6),2),\" uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Capacitance,C  =   79.58  uF  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the current taken by a 23 \u03bcF capacitor when connected to a 240 V, 50 Hz supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  23E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "I  =  V/Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  I  =  \",round(I,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  I  =   1.73   A  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 216</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the supply voltage and the phase angle between current and voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vr  =  12;#  in  volts\n",
+      "Vl  =  5;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  (Vr**2  +  Vl**2)**0.5\n",
+      "phi  =  math.atan(Vl/Vr)\n",
+      "phid  =  phi*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  voltage  V  =  \",V,\"  V,  phase  angle  between  current  and  voltage  is  \",  round(phid,2),\"deg lagging\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  voltage  V  =   13.0   V,  phase  angle  between  current  and  voltage  is   22.62 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 216</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the reactance, (b) the impedance, and (c) the current taken from a 240 V, 50 Hz supply. \n",
+      "#Determine also the phase angle between the supply voltage and current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  volts\n",
+      "R  =  4;#  in  ohms\n",
+      "L  =  0.00955;#  in  Henry\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  = math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (c)Current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (d)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   3.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   5.0   ohms\n",
+        "\n",
+        "  (c)Current,  I  =   48.0   A\n",
+        "\n",
+        "  (d)phase  angle  between  the  supply  voltage  and  current  is   36.87 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 217</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the resistance, impedance, inductive reactance and inductance of the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vdc  =  12;#  in  volts\n",
+      "Vac  =  240;#  in  volts\n",
+      "Iac  =  20;#  in  Amperes\n",
+      "Idc  =  2;#  in  Amperes\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  Vdc/Idc\n",
+      "Z  =  Vac/Iac\n",
+      "XL  =  (Z**2  -  R**2)**0.5\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance,  R  =  \",R,\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",Z,\"  ohms\"\n",
+      "print  \"\\n  (c)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (d)Inductance,  L  =  \",round(L,2),\"  H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance,  R  =   6.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   12.0   ohms\n",
+        "\n",
+        "  (c)Inductive  reactance,  XL  =   10.39   ohms\n",
+        "\n",
+        "  (d)Inductance,  L  =   0.03   H"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 217</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the inductive reactance of the coil, \n",
+      "#(b) the impedance of the circuit, \n",
+      "#(c) the current in the circuit, \n",
+      "#(d) the p.d. across each component, and \n",
+      "#(e) the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  200;#  in  ohms\n",
+      "L  =  0.3183;#  in  henry\n",
+      "V  =  240;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "VL  =  I*XL\n",
+      "VR  =  I*R\n",
+      "phid  =  math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (c)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (d)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V  and  p.d.  across  resistance,  VR  =  \",round(VR,2),\"  V\"\n",
+      "print  \"\\n  (e)circuit  phase  angle  is  \",round(phid,2),\" deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   100.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   223.61   ohms\n",
+        "\n",
+        "  (c)current,  I  =   1.07   A\n",
+        "\n",
+        "  (d)p.d.  across  Inductor,  VL  =   107.33   V  and  p.d.  across  resistance,  VR  =   214.66   V\n",
+        "\n",
+        "  (e)circuit  phase  angle  is   26.56  deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 218</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the circuit impedance, \n",
+      "#(b) the current flowing, \n",
+      "#(c) the p.d. across the resistance, \n",
+      "#(d) the p.d. across the inductance and \n",
+      "#(e) the phase angle between voltage and current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  100;#  in  ohms\n",
+      "L  =  0.2;#  in  henry\n",
+      "Vmax  =  200;#  in  volts\n",
+      "w  =  500;#  in  rad/sec\n",
+      "\n",
+      "#calculation:\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  Vrms/Z\n",
+      "VL  =  I*XL\n",
+      "VR  =  I*R\n",
+      "phid  =  math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "\\\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (c)p.d.  across  resistance,  VR  =  \",round(VR,2),\"  V\"\n",
+      "print  \"\\n  (d)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (e)circuit  phase  angle  is  \",phid,\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   141.42   ohms\n",
+        "\n",
+        "  (b)current,  I  =   1.0   A\n",
+        "\n",
+        "  (c)p.d.  across  resistance,  VR  =   99.98   V\n",
+        "\n",
+        "  (d)p.d.  across  Inductor,  VL  =   99.98   V\n",
+        "\n",
+        "  (e)circuit  phase  angle  is   45.0 deg"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 218</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of the supply voltage and the voltage across the 1.273 mH inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  1.2273E-3;#  in  henry\n",
+      "f  =  5000;#  in  Hz\n",
+      "VR  =  6;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "I  =VR/R\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "V  =  I*Z\n",
+      "VL  =  I*XL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)supply  voltage  =  \",round(V,2),\"  Volts\"\n",
+      "print  \"\\n  (b)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)supply  voltage  =   9.77   Volts\n",
+        "\n",
+        "  (b)p.d.  across  Inductor,  VL  =   7.71   V"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 219</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the impedance of the circuit, \n",
+      "#(b) the current in the circuit, \n",
+      "#(c) the circuit phase angle, \n",
+      "#(d) the p.d. across the 60 ohm resistor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  60;#  in  ohms\n",
+      "Rc  =  20;#  in  ohms\n",
+      "L  =  159.2E-3;#  in  henry\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Rt  =  R  +  Rc\n",
+      "Z  =  (Rt**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan(XL/Rt)*180/math.pi\n",
+      "VR  =  I*R\n",
+      "Zc  =  (Rc**2  +  XL**2)**0.5\n",
+      "Vc  =  I*Zc\n",
+      "VL  =  I*XL\n",
+      "VRc  =  I*Rc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,3),\"  A\"\n",
+      "print  \"\\n  (c)circuit  phase  angle  is  \",round(phid,0),\"deg lagging\"\n",
+      "print  \"\\n  (d)p.d.  across  resistance,  VR  =  \",round(  VR,1),\"  V\"\n",
+      "print  \"\\n  (e)p.d.  across  coil,  Vc  =  \",round(Vc,1),\"  V\"\n",
+      "print  \"\\n  (f1)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (f2)p.d.  across  coil  resistance,  VRc  =  \",round(VRc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   94.35   ohms\n",
+        "\n",
+        "  (b)current,  I  =   2.544   A\n",
+        "\n",
+        "  (c)circuit  phase  angle  is   32.0 deg lagging\n",
+        "\n",
+        "  (d)p.d.  across  resistance,  VR  =   152.6   V\n",
+        "\n",
+        "  (e)p.d.  across  coil,  Vc  =   137.0   V\n",
+        "\n",
+        "  (f1)p.d.  across  Inductor,  VL  =   127.23   V\n",
+        "\n",
+        "  (f2)p.d.  across  coil  resistance,  VRc  =   50.88   V"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 220</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the impedance, and (b) the current taken from a 240 V, 50 Hz supply. \n",
+      "#Find also the phase angle between the supply voltage and the current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohms\n",
+      "C  =  45E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Z  =  (R**2  +  Xc**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan(Xc/R)*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (c)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   75.02   ohms\n",
+        "\n",
+        "  (b)current,  I  =   3.2   A\n",
+        "\n",
+        "  (c)phase  angle  between  the  supply  voltage  and  current  is   70.54 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 221</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate: (a) the value of capacitance, C, \n",
+      "#(b) the supply voltage, \n",
+      "#(c) the phase angle between the supply voltage and current, \n",
+      "#(d) the p.d. across the resistor, and\n",
+      "#(e) the p.d. across the capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  40;#  in  ohms\n",
+      "f  =  60;#  in  Hz\n",
+      "I  =  3;#in  amperes\n",
+      "Z  =  50;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  (Z**2  -  R**2)**0.5\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "V  =  I*Z\n",
+      "phid  =  math.atan(Xc/R)*180/math.pi\n",
+      "VR  =  I*R\n",
+      "Vc  =  I*Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance,  C  =  \",round((C/1E-6),2),\"  uF\"\n",
+      "print  \"\\n  (b)Voltage,  V  =  \",V,\"  Volts\"\n",
+      "print  \"\\n  (c)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)p.d.  across  resistance,  VR  =  \",  VR,\"  V\"\n",
+      "print  \"\\n  (e)p.d.  across  Capacitor,  Vc  =  \",Vc,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance,  C  =   88.42   uF\n",
+        "\n",
+        "  (b)Voltage,  V  =   150   Volts\n",
+        "\n",
+        "  (c)phase  angle  between  the  supply  voltage  and  current  is   36.87 deg leading\n",
+        "\n",
+        "  (d)p.d.  across  resistance,  VR  =   120   V\n",
+        "\n",
+        "  (e)p.d.  across  Capacitor,  Vc  =   90.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 222</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current flowing, \n",
+      "#(b) the phase difference between the supply voltage and current, \n",
+      "#(c) the voltage across the coil and \n",
+      "#(d) the voltage across the capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  5;#  in  ohms\n",
+      "C  =  100E-6;#  in  Farads\n",
+      "L  =  0.12;#  in  Henry\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  300;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "X  =  XL  -  Xc\n",
+      " #Since  XL  is  greater  than  Xc,  the  circuit  is  inductive.\n",
+      "Z  =  (R**2  +  (XL-Xc)**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan((XL-Xc)/R)*180/math.pi\n",
+      "Zcl  =  (R**2  +  XL**2)**0.5\n",
+      "Vcl  =  I*Zcl\n",
+      "phidc  =  math.atan(XL/R)*180/math.pi\n",
+      "Vc  =  I*Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current,I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (c)Voltage  across  the  coil,  Vcoil  =  \",round(Vcl,0),\"  Volts\"\n",
+      "print  \"\\n  (d)p.d.  across  Capacitor,  Vc  =  \",round(Vc,0),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current,I  =   38.91   A\n",
+        "\n",
+        "  (b)phase  angle  between  the  supply  voltage  and  current  is   49.57 deg\n",
+        "\n",
+        "  (c)Voltage  across  the  coil,  Vcoil  =   1480.0   Volts\n",
+        "\n",
+        "  (d)p.d.  across  Capacitor,  Vc  =   1239.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 224</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the circuit current, \n",
+      "#(b) the circuit phase angle and \n",
+      "#(c) the voltage drop across each impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohms\n",
+      "C  =  0.25E-6;#  in  Farads\n",
+      "L  =  130E-6;#  in  Henry\n",
+      "Rc  =  5;#  in  ohms\n",
+      "R2  =  10;#  in  ohms\n",
+      "f  =  20000;#  in  Hz\n",
+      "V  =  40;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "X  =  Xc  -  XL\n",
+      "R  =  R1  +  R2  +  Rc\n",
+      " #Since  Xc  is  greater  than  XL,  the  circuit  is  capacitive.\n",
+      "Z  =  (R**2  +  (Xc-XL)**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid = math.atan((Xc-XL)/R)*180/math.pi\n",
+      "V1  =  I*R1\n",
+      "V2  =  I*((Rc**2  +  XL**2)**0.5)\n",
+      "V3  =  I*((R2**2  +  Xc**2)**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current,I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)circuit  phase  angle  is  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (c1)Voltage  across  the  Resistance  of  8  ohms  =  \",round(V1,2),\"  Volts\"\n",
+      "print  \"\\n  (c2)Voltage  across  the  coil,  Vcoil  =  \",round(V2,2),\"  Volts\"\n",
+      "print  \"\\n  (c3)p.d.  across  Capacitor  resistance  circuit  =  \",round(V3,2),\"  Volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current,I  =   1.44   A\n",
+        "\n",
+        "  (b)circuit  phase  angle  is   33.97 deg leading\n",
+        "\n",
+        "  (c1)Voltage  across  the  Resistance  of  8  ohms  =   11.54   Volts\n",
+        "\n",
+        "  (c2)Voltage  across  the  coil,  Vcoil  =   24.64   Volts\n",
+        "\n",
+        "  (c3)p.d.  across  Capacitor  resistance  circuit  =   48.12   Volts"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 224</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the p.d.\u2019s V1 and V2 for the circuit\n",
+      "#determine the supply voltage V and the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  4;#  in  ohms\n",
+      "C  =  1.273E-6;#  in  Farads\n",
+      "L  =  0.286E-3;#  in  Henry\n",
+      "R2  =  8;#  in  ohms\n",
+      "f  =  5000;#  in  Hz\n",
+      "I  =  5;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "phid1  =  math.atan(XL/R1)*180/math.pi\n",
+      "V1  =  I*((R1**2  +  XL**2)**0.5)\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "V2  =  I*((R2**2  +  Xc**2)**0.5)\n",
+      "phid2  =  math.atan(Xc/R2)*180/math.pi\n",
+      "Z  =  ((R1+R2)**2  +  (Xc-XL)**2)**0.5\n",
+      "V  =  I*Z\n",
+      "phid  =  math.atan((Xc-XL)/(R1+R2))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Voltage  supply,  V  =  \",round(V,2),\"  V\"\n",
+      "print  \"\\n  (b)circuit  phase  angle  is  \",round(phid,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Voltage  supply,  V  =   100.08   V\n",
+        "\n",
+        "  (b)circuit  phase  angle  is   53.16 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 226</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#At what frequency does resonance occur?\n",
+      "#Find the current flowing at the resonant frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  60E-6;#  in  Farads\n",
+      "L  =  125E-3;#  in  Henry\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*(L*C)**0.5)\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency,Fr  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency,Fr  =   58.12   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   12.0"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 226</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the circuit resistance, and (b) the circuit capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.05E-3;#  in  Henry\n",
+      "fr  =  200000;#  in  Hz\n",
+      "V  =  0.002;#  in  Volts\n",
+      "I  =  0.1E-3;#  in  amperes\n",
+      "#calculation:\n",
+      "#  L-C-R\n",
+      "#At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "R  =  V/I\n",
+      "C  =  1/(L*(2*math.pi*fr)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance,  R  =  \",round(R,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Capacitance,  C  =  \",round((C/1E-9),2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance,  R  =   20.0   ohms\n",
+        "\n",
+        "  (b)Capacitance,  C  =   12.67 nF"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 227</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, and \n",
+      "#(b) the current at resonance. \n",
+      "#How many times greater than the supply voltage is the voltage across the reactances at resonance?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  80E-3;#  in  Henry\n",
+      "C  =  0.25E-6;#  in  Farads\n",
+      "R  =  12.5;#  in  ohms\n",
+      "V  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "VL  =  I*(2*math.pi*fr*L)\n",
+      "Vc  =  I/(2*math.pi*fr*C)\n",
+      "Vm  =  VL/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  resonant  frequency  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2),\"\"\n",
+      "print  \"\\n  (b)Voltage  magnification  at  resonance  =  \",round(Vm,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  resonant  frequency  =   1125.4   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   8.0 \n",
+        "\n",
+        "  (b)Voltage  magnification  at  resonance  =   45.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 228</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the Qfactor of the circuit at resonance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  60E-3;#  in  Henry\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "R  =  2;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  ((L/C)**0.5)/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  At  resonance,  Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  At  resonance,  Q-factor  =   22.36"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 228</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, \n",
+      "#(b) the current at resonance,\n",
+      "#(c) the voltages across the coil and the capacitor at resonance, and\n",
+      "#(d) the Q-factor of the circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  100E-3;#  in  Henry\n",
+      "C  =  2E-6;#  in  Farads\n",
+      "R  =  10;#  in  ohms\n",
+      "V  =  50;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "VL  =  I*(2*math.pi*fr*L)\n",
+      "Vc  =  I/(2*math.pi*fr*C)\n",
+      "Q  =  VL/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  resonant  frequency  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2),\"\"\n",
+      "print  \"\\n  (c)Voltage  across  coil  at  resonance  is  \",round(VL,2),\"V  \"\n",
+      "print   \"and  Voltage  across  capacitance  at  resonance  is  \",round(  Vc,2),\"V\"\n",
+      "print  \"\\n  (d)At  resonance,  Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  resonant  frequency  =   355.88   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   5.0 \n",
+        "\n",
+        "  (c)Voltage  across  coil  at  resonance  is   1118.03 V  and  Voltage  across  capacitance  at  resonance  is   1118.03 V\n",
+        "\n",
+        "  (d)At  resonance,  Q-factor  =   22.36"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 230</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the bandwidth of the filter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  20E-3;#  in  Henry\n",
+      "R  =  10;#  in  ohms\n",
+      "fr  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Qr  =  (2*math.pi*fr)*L/R\n",
+      "bw  =  fr/Qr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Bandwidth,  (f2-f1)  =  \",round(bw,2),\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Bandwidth,  (f2-f1)  =   79.58   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 231</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the power dissipated in the resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  5000;#  in  ohms\n",
+      "Imax  =  0.250;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Irms  =  0.707*Imax\n",
+      "P  =  Irms*Irms*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Power,  P  =  \",round(P,2),\"  Watts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Power,  P  =   156.2   Watts"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 231</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  60;#  in  ohms\n",
+      "L  =  75E-3;#  in  Henry\n",
+      "V  =  110;#  in  Volts\n",
+      "f  =  60;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "P  =  I*I*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Power,  P  =  \",round(P,2),\"  Watts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Power,  P  =   165.02   Watts"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 232</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the value of the inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VI  =  300;#  in  VA\n",
+      "V  =  150;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  VI/V\n",
+      "XL  =  V/I\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  =  \",round(L,2),\"  H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  =   0.24   H"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 232</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the rated power output and the corresponding reactive power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VI  =  200000;#  in  VA\n",
+      "pf  =  0.8;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "P  =  VI*pf\n",
+      "Q  =  VI*math.sin(math.acos(pf))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  rated  power  output  is  \",round(P/1000,2),\"KW  and  the  corresponding  reactive  power  is  \",round(Q/1000,2),\"kvar\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  rated  power  output  is   160.0 KW  and  the  corresponding  reactive  power  is   120.0 kvar"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 233</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the resistance, \n",
+      "#(b) the impedance, \n",
+      "#(c) the reactance, \n",
+      "#(d) the power factor, and \n",
+      "#(e) the phase angle between voltage and current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  120;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "P  =  400;#  in  Watt\n",
+      "I  =  8;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  P/(I*I)\n",
+      "Z  =  V/I\n",
+      "XL  =  (Z**2  -  R**2)**0.5\n",
+      "pf  =  P/(V*I)\n",
+      "phi  =  math.acos(pf)*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  =  \",round(R,2),\"  ohm  \"\n",
+      "print  \"\\n  (b)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (c)reactance  =  \",round(XL,2),\"  ohm  \"\n",
+      "print  \"\\n  (d)Power  factor  =  \",round(pf,2),\"\"\n",
+      "print  \"\\n  (e)phase  angle  =  \",round(phi,2),\"deg lagging\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  =   6.25   ohm  \n",
+        "\n",
+        "  (b)Impedance  Z  =   15.0   Ohm  \n",
+        "\n",
+        "  (c)reactance  =   13.64   ohm  \n",
+        "\n",
+        "  (d)Power  factor  =   0.42 \n",
+        "\n",
+        "  (e)phase  angle  =   65.38 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 29, page no. 233</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the current flowing, (b) the phase angle,\n",
+      "#(c) the resistance, (d) the impedance, and (e) the capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  60;#  in  Hz\n",
+      "P  =  100;#  in  Watt\n",
+      "pf  =  0.5;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  P/(pf*V)\n",
+      "phi  =  math.acos(pf)*180/math.pi\n",
+      "R  =  P/(I*I)\n",
+      "Z  =  V/I\n",
+      "Xc  =  (Z**2  -  R**2)**0.5\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (b)phase  angle  =  \",round(phi,2),\"deg leading\"\n",
+      "print  \"\\n  (c)resistance  =  \",round(R,2),\"  ohm  \"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)capacitance  =  \",round((C/1E-6),2),\"uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  I  =   2.0   A  \n",
+        "\n",
+        "  (b)phase  angle  =   60.0 deg leading\n",
+        "\n",
+        "  (c)resistance  =   25.0   ohm  \n",
+        "\n",
+        "  (d)Impedance  Z  =   50.0   Ohm  \n",
+        "\n",
+        "  (e)capacitance  =   61.26 uF  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_2.ipynb
new file mode 100755
index 00000000..e7b68d8a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_2.ipynb
@@ -0,0 +1,1677 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 15: Single-phase series a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 214</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Calculate the reactance of a coil \n",
+      "#(b) Determine the inductance of the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.32;#  in  Henry\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  5000;#  in  Hz\n",
+      "Z  =  124;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f1*L\n",
+      "L  =  Z/(2*math.pi*f2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms  \\n\"\n",
+      "print  \"\\n  (b)Inductance  L  =  \",round((L/1E-3),2),\"  mH  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   100.53   ohms  \n",
+        "\n",
+        "\n",
+        "  (b)Inductance  L  =   3.95   mH  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 214</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate its inductive reactance and the resulting current if connected to\n",
+      "#(a) a 240 V, 50 Hz supply, and (b) a 100 V, 1 kHz supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.040;#  in  Henry\n",
+      "V1  =  240;#  in  volts\n",
+      "V2  =  100;#  in  volts\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL1  =  2*math.pi*f1*L\n",
+      "I1  =  V1/XL1\n",
+      "XL2  =  2*math.pi*f2*L\n",
+      "I2  =  V2/XL2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(  XL1,2),\"  ohms  and  current  I  =  \",round(  I1,2),\"  A\\n\"\n",
+      "print  \"\\n  (b)Inductive  reactance,  XL  =  \",round(  XL2,2),\"  ohms  and  current  I  =  \",round(  I2,2),\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   12.57   ohms  and  current  I  =   19.1   A\n",
+        "\n",
+        "\n",
+        "  (b)Inductive  reactance,  XL  =   251.33   ohms  and  current  I  =   0.4   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitive reactance of a capacitor of 10 \u03bcF when connected to a circuit of frequency (a) 50 Hz (b) 20 kHz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  10E-6;#  in  Farads\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  20000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Xc1  =  1/(2*math.pi*f1*C)\n",
+      "Xc2  =  1/(2*math.pi*f2*C)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Capacitive  reactance,  Xc  =  \",round(  Xc1,2),\"  ohms  \"\n",
+      "print  \"\\n  (b)Capacitive  reactance,  Xc  =  \",round(  Xc2,2),\"  ohms  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitive  reactance,  Xc  =   318.31   ohms  \n",
+        "\n",
+        "  (b)Capacitive  reactance,  Xc  =   0.8   ohms  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of its capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  40;#  in  ohms\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  1/(2*math.pi*f*Z)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Capacitance,C  =  \",round((C/1E-6),2),\" uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Capacitance,C  =   79.58  uF  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the current taken by a 23 \u03bcF capacitor when connected to a 240 V, 50 Hz supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  23E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "I  =  V/Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  I  =  \",round(I,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  I  =   1.73   A  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 216</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the supply voltage and the phase angle between current and voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vr  =  12;#  in  volts\n",
+      "Vl  =  5;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  (Vr**2  +  Vl**2)**0.5\n",
+      "phi  =  math.atan(Vl/Vr)\n",
+      "phid  =  phi*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  voltage  V  =  \",V,\"  V,  phase  angle  between  current  and  voltage  is  \",  round(phid,2),\"deg lagging\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  voltage  V  =   13.0   V,  phase  angle  between  current  and  voltage  is   22.62 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 216</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the reactance, (b) the impedance, and (c) the current taken from a 240 V, 50 Hz supply. \n",
+      "#Determine also the phase angle between the supply voltage and current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  volts\n",
+      "R  =  4;#  in  ohms\n",
+      "L  =  0.00955;#  in  Henry\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  = math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (c)Current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (d)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   3.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   5.0   ohms\n",
+        "\n",
+        "  (c)Current,  I  =   48.0   A\n",
+        "\n",
+        "  (d)phase  angle  between  the  supply  voltage  and  current  is   36.87 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 217</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the resistance, impedance, inductive reactance and inductance of the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vdc  =  12;#  in  volts\n",
+      "Vac  =  240;#  in  volts\n",
+      "Iac  =  20;#  in  Amperes\n",
+      "Idc  =  2;#  in  Amperes\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  Vdc/Idc\n",
+      "Z  =  Vac/Iac\n",
+      "XL  =  (Z**2  -  R**2)**0.5\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance,  R  =  \",R,\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",Z,\"  ohms\"\n",
+      "print  \"\\n  (c)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (d)Inductance,  L  =  \",round(L,2),\"  H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance,  R  =   6.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   12.0   ohms\n",
+        "\n",
+        "  (c)Inductive  reactance,  XL  =   10.39   ohms\n",
+        "\n",
+        "  (d)Inductance,  L  =   0.03   H"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 217</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the inductive reactance of the coil, \n",
+      "#(b) the impedance of the circuit, \n",
+      "#(c) the current in the circuit, \n",
+      "#(d) the p.d. across each component, and \n",
+      "#(e) the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  200;#  in  ohms\n",
+      "L  =  0.3183;#  in  henry\n",
+      "V  =  240;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "VL  =  I*XL\n",
+      "VR  =  I*R\n",
+      "phid  =  math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (c)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (d)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V  and  p.d.  across  resistance,  VR  =  \",round(VR,2),\"  V\"\n",
+      "print  \"\\n  (e)circuit  phase  angle  is  \",round(phid,2),\" deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   100.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   223.61   ohms\n",
+        "\n",
+        "  (c)current,  I  =   1.07   A\n",
+        "\n",
+        "  (d)p.d.  across  Inductor,  VL  =   107.33   V  and  p.d.  across  resistance,  VR  =   214.66   V\n",
+        "\n",
+        "  (e)circuit  phase  angle  is   26.56  deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 218</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the circuit impedance, \n",
+      "#(b) the current flowing, \n",
+      "#(c) the p.d. across the resistance, \n",
+      "#(d) the p.d. across the inductance and \n",
+      "#(e) the phase angle between voltage and current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  100;#  in  ohms\n",
+      "L  =  0.2;#  in  henry\n",
+      "Vmax  =  200;#  in  volts\n",
+      "w  =  500;#  in  rad/sec\n",
+      "\n",
+      "#calculation:\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  Vrms/Z\n",
+      "VL  =  I*XL\n",
+      "VR  =  I*R\n",
+      "phid  =  math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "\\\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (c)p.d.  across  resistance,  VR  =  \",round(VR,2),\"  V\"\n",
+      "print  \"\\n  (d)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (e)circuit  phase  angle  is  \",phid,\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   141.42   ohms\n",
+        "\n",
+        "  (b)current,  I  =   1.0   A\n",
+        "\n",
+        "  (c)p.d.  across  resistance,  VR  =   99.98   V\n",
+        "\n",
+        "  (d)p.d.  across  Inductor,  VL  =   99.98   V\n",
+        "\n",
+        "  (e)circuit  phase  angle  is   45.0 deg"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 218</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of the supply voltage and the voltage across the 1.273 mH inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  1.2273E-3;#  in  henry\n",
+      "f  =  5000;#  in  Hz\n",
+      "VR  =  6;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "I  =VR/R\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "V  =  I*Z\n",
+      "VL  =  I*XL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)supply  voltage  =  \",round(V,2),\"  Volts\"\n",
+      "print  \"\\n  (b)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)supply  voltage  =   9.77   Volts\n",
+        "\n",
+        "  (b)p.d.  across  Inductor,  VL  =   7.71   V"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 219</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the impedance of the circuit, \n",
+      "#(b) the current in the circuit, \n",
+      "#(c) the circuit phase angle, \n",
+      "#(d) the p.d. across the 60 ohm resistor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  60;#  in  ohms\n",
+      "Rc  =  20;#  in  ohms\n",
+      "L  =  159.2E-3;#  in  henry\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Rt  =  R  +  Rc\n",
+      "Z  =  (Rt**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan(XL/Rt)*180/math.pi\n",
+      "VR  =  I*R\n",
+      "Zc  =  (Rc**2  +  XL**2)**0.5\n",
+      "Vc  =  I*Zc\n",
+      "VL  =  I*XL\n",
+      "VRc  =  I*Rc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,3),\"  A\"\n",
+      "print  \"\\n  (c)circuit  phase  angle  is  \",round(phid,0),\"deg lagging\"\n",
+      "print  \"\\n  (d)p.d.  across  resistance,  VR  =  \",round(  VR,1),\"  V\"\n",
+      "print  \"\\n  (e)p.d.  across  coil,  Vc  =  \",round(Vc,1),\"  V\"\n",
+      "print  \"\\n  (f1)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (f2)p.d.  across  coil  resistance,  VRc  =  \",round(VRc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   94.35   ohms\n",
+        "\n",
+        "  (b)current,  I  =   2.544   A\n",
+        "\n",
+        "  (c)circuit  phase  angle  is   32.0 deg lagging\n",
+        "\n",
+        "  (d)p.d.  across  resistance,  VR  =   152.6   V\n",
+        "\n",
+        "  (e)p.d.  across  coil,  Vc  =   137.0   V\n",
+        "\n",
+        "  (f1)p.d.  across  Inductor,  VL  =   127.23   V\n",
+        "\n",
+        "  (f2)p.d.  across  coil  resistance,  VRc  =   50.88   V"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 220</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the impedance, and (b) the current taken from a 240 V, 50 Hz supply. \n",
+      "#Find also the phase angle between the supply voltage and the current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohms\n",
+      "C  =  45E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Z  =  (R**2  +  Xc**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan(Xc/R)*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (c)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   75.02   ohms\n",
+        "\n",
+        "  (b)current,  I  =   3.2   A\n",
+        "\n",
+        "  (c)phase  angle  between  the  supply  voltage  and  current  is   70.54 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 221</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate: (a) the value of capacitance, C, \n",
+      "#(b) the supply voltage, \n",
+      "#(c) the phase angle between the supply voltage and current, \n",
+      "#(d) the p.d. across the resistor, and\n",
+      "#(e) the p.d. across the capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  40;#  in  ohms\n",
+      "f  =  60;#  in  Hz\n",
+      "I  =  3;#in  amperes\n",
+      "Z  =  50;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  (Z**2  -  R**2)**0.5\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "V  =  I*Z\n",
+      "phid  =  math.atan(Xc/R)*180/math.pi\n",
+      "VR  =  I*R\n",
+      "Vc  =  I*Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance,  C  =  \",round((C/1E-6),2),\"  uF\"\n",
+      "print  \"\\n  (b)Voltage,  V  =  \",V,\"  Volts\"\n",
+      "print  \"\\n  (c)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)p.d.  across  resistance,  VR  =  \",  VR,\"  V\"\n",
+      "print  \"\\n  (e)p.d.  across  Capacitor,  Vc  =  \",Vc,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance,  C  =   88.42   uF\n",
+        "\n",
+        "  (b)Voltage,  V  =   150   Volts\n",
+        "\n",
+        "  (c)phase  angle  between  the  supply  voltage  and  current  is   36.87 deg leading\n",
+        "\n",
+        "  (d)p.d.  across  resistance,  VR  =   120   V\n",
+        "\n",
+        "  (e)p.d.  across  Capacitor,  Vc  =   90.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 222</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current flowing, \n",
+      "#(b) the phase difference between the supply voltage and current, \n",
+      "#(c) the voltage across the coil and \n",
+      "#(d) the voltage across the capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  5;#  in  ohms\n",
+      "C  =  100E-6;#  in  Farads\n",
+      "L  =  0.12;#  in  Henry\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  300;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "X  =  XL  -  Xc\n",
+      " #Since  XL  is  greater  than  Xc,  the  circuit  is  inductive.\n",
+      "Z  =  (R**2  +  (XL-Xc)**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan((XL-Xc)/R)*180/math.pi\n",
+      "Zcl  =  (R**2  +  XL**2)**0.5\n",
+      "Vcl  =  I*Zcl\n",
+      "phidc  =  math.atan(XL/R)*180/math.pi\n",
+      "Vc  =  I*Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current,I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (c)Voltage  across  the  coil,  Vcoil  =  \",round(Vcl,0),\"  Volts\"\n",
+      "print  \"\\n  (d)p.d.  across  Capacitor,  Vc  =  \",round(Vc,0),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current,I  =   38.91   A\n",
+        "\n",
+        "  (b)phase  angle  between  the  supply  voltage  and  current  is   49.57 deg\n",
+        "\n",
+        "  (c)Voltage  across  the  coil,  Vcoil  =   1480.0   Volts\n",
+        "\n",
+        "  (d)p.d.  across  Capacitor,  Vc  =   1239.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 224</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the circuit current, \n",
+      "#(b) the circuit phase angle and \n",
+      "#(c) the voltage drop across each impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohms\n",
+      "C  =  0.25E-6;#  in  Farads\n",
+      "L  =  130E-6;#  in  Henry\n",
+      "Rc  =  5;#  in  ohms\n",
+      "R2  =  10;#  in  ohms\n",
+      "f  =  20000;#  in  Hz\n",
+      "V  =  40;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "X  =  Xc  -  XL\n",
+      "R  =  R1  +  R2  +  Rc\n",
+      " #Since  Xc  is  greater  than  XL,  the  circuit  is  capacitive.\n",
+      "Z  =  (R**2  +  (Xc-XL)**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid = math.atan((Xc-XL)/R)*180/math.pi\n",
+      "V1  =  I*R1\n",
+      "V2  =  I*((Rc**2  +  XL**2)**0.5)\n",
+      "V3  =  I*((R2**2  +  Xc**2)**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current,I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)circuit  phase  angle  is  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (c1)Voltage  across  the  Resistance  of  8  ohms  =  \",round(V1,2),\"  Volts\"\n",
+      "print  \"\\n  (c2)Voltage  across  the  coil,  Vcoil  =  \",round(V2,2),\"  Volts\"\n",
+      "print  \"\\n  (c3)p.d.  across  Capacitor  resistance  circuit  =  \",round(V3,2),\"  Volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current,I  =   1.44   A\n",
+        "\n",
+        "  (b)circuit  phase  angle  is   33.97 deg leading\n",
+        "\n",
+        "  (c1)Voltage  across  the  Resistance  of  8  ohms  =   11.54   Volts\n",
+        "\n",
+        "  (c2)Voltage  across  the  coil,  Vcoil  =   24.64   Volts\n",
+        "\n",
+        "  (c3)p.d.  across  Capacitor  resistance  circuit  =   48.12   Volts"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 224</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the p.d.\u2019s V1 and V2 for the circuit\n",
+      "#determine the supply voltage V and the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  4;#  in  ohms\n",
+      "C  =  1.273E-6;#  in  Farads\n",
+      "L  =  0.286E-3;#  in  Henry\n",
+      "R2  =  8;#  in  ohms\n",
+      "f  =  5000;#  in  Hz\n",
+      "I  =  5;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "phid1  =  math.atan(XL/R1)*180/math.pi\n",
+      "V1  =  I*((R1**2  +  XL**2)**0.5)\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "V2  =  I*((R2**2  +  Xc**2)**0.5)\n",
+      "phid2  =  math.atan(Xc/R2)*180/math.pi\n",
+      "Z  =  ((R1+R2)**2  +  (Xc-XL)**2)**0.5\n",
+      "V  =  I*Z\n",
+      "phid  =  math.atan((Xc-XL)/(R1+R2))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Voltage  supply,  V  =  \",round(V,2),\"  V\"\n",
+      "print  \"\\n  (b)circuit  phase  angle  is  \",round(phid,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Voltage  supply,  V  =   100.08   V\n",
+        "\n",
+        "  (b)circuit  phase  angle  is   53.16 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 226</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#At what frequency does resonance occur?\n",
+      "#Find the current flowing at the resonant frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  60E-6;#  in  Farads\n",
+      "L  =  125E-3;#  in  Henry\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*(L*C)**0.5)\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency,Fr  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency,Fr  =   58.12   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   12.0"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 226</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find (a) the circuit resistance, and (b) the circuit capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.05E-3;#  in  Henry\n",
+      "fr  =  200000;#  in  Hz\n",
+      "V  =  0.002;#  in  Volts\n",
+      "I  =  0.1E-3;#  in  amperes\n",
+      "#calculation:\n",
+      "#  L-C-R\n",
+      "#At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "R  =  V/I\n",
+      "C  =  1/(L*(2*math.pi*fr)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance,  R  =  \",round(R,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Capacitance,  C  =  \",round((C/1E-9),2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance,  R  =   20.0   ohms\n",
+        "\n",
+        "  (b)Capacitance,  C  =   12.67 nF"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 227</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, and \n",
+      "#(b) the current at resonance. \n",
+      "#How many times greater than the supply voltage is the voltage across the reactances at resonance?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  80E-3;#  in  Henry\n",
+      "C  =  0.25E-6;#  in  Farads\n",
+      "R  =  12.5;#  in  ohms\n",
+      "V  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "VL  =  I*(2*math.pi*fr*L)\n",
+      "Vc  =  I/(2*math.pi*fr*C)\n",
+      "Vm  =  VL/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  resonant  frequency  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2),\"\"\n",
+      "print  \"\\n  (b)Voltage  magnification  at  resonance  =  \",round(Vm,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  resonant  frequency  =   1125.4   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   8.0 \n",
+        "\n",
+        "  (b)Voltage  magnification  at  resonance  =   45.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 228</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the Qfactor of the circuit at resonance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  60E-3;#  in  Henry\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "R  =  2;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  ((L/C)**0.5)/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  At  resonance,  Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  At  resonance,  Q-factor  =   22.36"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 228</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, \n",
+      "#(b) the current at resonance,\n",
+      "#(c) the voltages across the coil and the capacitor at resonance, and\n",
+      "#(d) the Q-factor of the circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  100E-3;#  in  Henry\n",
+      "C  =  2E-6;#  in  Farads\n",
+      "R  =  10;#  in  ohms\n",
+      "V  =  50;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "VL  =  I*(2*math.pi*fr*L)\n",
+      "Vc  =  I/(2*math.pi*fr*C)\n",
+      "Q  =  VL/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  resonant  frequency  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2),\"\"\n",
+      "print  \"\\n  (c)Voltage  across  coil  at  resonance  is  \",round(VL,2),\"V  \"\n",
+      "print   \"and  Voltage  across  capacitance  at  resonance  is  \",round(  Vc,2),\"V\"\n",
+      "print  \"\\n  (d)At  resonance,  Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  resonant  frequency  =   355.88   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   5.0 \n",
+        "\n",
+        "  (c)Voltage  across  coil  at  resonance  is   1118.03 V  and  Voltage  across  capacitance  at  resonance  is   1118.03 V\n",
+        "\n",
+        "  (d)At  resonance,  Q-factor  =   22.36"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 230</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the bandwidth of the filter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  20E-3;#  in  Henry\n",
+      "R  =  10;#  in  ohms\n",
+      "fr  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Qr  =  (2*math.pi*fr)*L/R\n",
+      "bw  =  fr/Qr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Bandwidth,  (f2-f1)  =  \",round(bw,2),\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Bandwidth,  (f2-f1)  =   79.58   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 231</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the power dissipated in the resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  5000;#  in  ohms\n",
+      "Imax  =  0.250;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Irms  =  0.707*Imax\n",
+      "P  =  Irms*Irms*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Power,  P  =  \",round(P,2),\"  Watts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Power,  P  =   156.2   Watts"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 231</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  60;#  in  ohms\n",
+      "L  =  75E-3;#  in  Henry\n",
+      "V  =  110;#  in  Volts\n",
+      "f  =  60;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "P  =  I*I*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Power,  P  =  \",round(P,2),\"  Watts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Power,  P  =   165.02   Watts"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 232</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find the value of the inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VI  =  300;#  in  VA\n",
+      "V  =  150;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  VI/V\n",
+      "XL  =  V/I\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  =  \",round(L,2),\"  H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  =   0.24   H"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 232</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the rated power output and the corresponding reactive power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VI  =  200000;#  in  VA\n",
+      "pf  =  0.8;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "P  =  VI*pf\n",
+      "Q  =  VI*math.sin(math.acos(pf))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  rated  power  output  is  \",round(P/1000,2),\"KW  and  the  corresponding  reactive  power  is  \",round(Q/1000,2),\"kvar\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  rated  power  output  is   160.0 KW  and  the  corresponding  reactive  power  is   120.0 kvar"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 233</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the resistance, \n",
+      "#(b) the impedance, \n",
+      "#(c) the reactance, \n",
+      "#(d) the power factor, and \n",
+      "#(e) the phase angle between voltage and current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  120;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "P  =  400;#  in  Watt\n",
+      "I  =  8;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  P/(I*I)\n",
+      "Z  =  V/I\n",
+      "XL  =  (Z**2  -  R**2)**0.5\n",
+      "pf  =  P/(V*I)\n",
+      "phi  =  math.acos(pf)*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  =  \",round(R,2),\"  ohm  \"\n",
+      "print  \"\\n  (b)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (c)reactance  =  \",round(XL,2),\"  ohm  \"\n",
+      "print  \"\\n  (d)Power  factor  =  \",round(pf,2),\"\"\n",
+      "print  \"\\n  (e)phase  angle  =  \",round(phi,2),\"deg lagging\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  =   6.25   ohm  \n",
+        "\n",
+        "  (b)Impedance  Z  =   15.0   Ohm  \n",
+        "\n",
+        "  (c)reactance  =   13.64   ohm  \n",
+        "\n",
+        "  (d)Power  factor  =   0.42 \n",
+        "\n",
+        "  (e)phase  angle  =   65.38 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 29, page no. 233</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Find (a) the current flowing, (b) the phase angle,\n",
+      "#(c) the resistance, (d) the impedance, and (e) the capacitance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  60;#  in  Hz\n",
+      "P  =  100;#  in  Watt\n",
+      "pf  =  0.5;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  P/(pf*V)\n",
+      "phi  =  math.acos(pf)*180/math.pi\n",
+      "R  =  P/(I*I)\n",
+      "Z  =  V/I\n",
+      "Xc  =  (Z**2  -  R**2)**0.5\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (b)phase  angle  =  \",round(phi,2),\"deg leading\"\n",
+      "print  \"\\n  (c)resistance  =  \",round(R,2),\"  ohm  \"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)capacitance  =  \",round((C/1E-6),2),\"uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  I  =   2.0   A  \n",
+        "\n",
+        "  (b)phase  angle  =   60.0 deg leading\n",
+        "\n",
+        "  (c)resistance  =   25.0   ohm  \n",
+        "\n",
+        "  (d)Impedance  Z  =   50.0   Ohm  \n",
+        "\n",
+        "  (e)capacitance  =   61.26 uF  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_3.ipynb
new file mode 100755
index 00000000..4a218793
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_15-checkpoint_3.ipynb
@@ -0,0 +1,1642 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:cceab0fff3199357a0bf4568e6d350c506b245311469d213cc25b93faa4b4954"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 15: Single-phase series a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 214</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.32;#  in  Henry\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  5000;#  in  Hz\n",
+      "Z  =  124;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f1*L\n",
+      "L  =  Z/(2*math.pi*f2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms  \\n\"\n",
+      "print  \"\\n  (b)Inductance  L  =  \",round((L/1E-3),2),\"  mH  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   100.53   ohms  \n",
+        "\n",
+        "\n",
+        "  (b)Inductance  L  =   3.95   mH  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 214</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.040;#  in  Henry\n",
+      "V1  =  240;#  in  volts\n",
+      "V2  =  100;#  in  volts\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL1  =  2*math.pi*f1*L\n",
+      "I1  =  V1/XL1\n",
+      "XL2  =  2*math.pi*f2*L\n",
+      "I2  =  V2/XL2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(  XL1,2),\"  ohms  and  current  I  =  \",round(  I1,2),\"  A\\n\"\n",
+      "print  \"\\n  (b)Inductive  reactance,  XL  =  \",round(  XL2,2),\"  ohms  and  current  I  =  \",round(  I2,2),\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   12.57   ohms  and  current  I  =   19.1   A\n",
+        "\n",
+        "\n",
+        "  (b)Inductive  reactance,  XL  =   251.33   ohms  and  current  I  =   0.4   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  10E-6;#  in  Farads\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  20000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Xc1  =  1/(2*math.pi*f1*C)\n",
+      "Xc2  =  1/(2*math.pi*f2*C)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Capacitive  reactance,  Xc  =  \",round(  Xc1,2),\"  ohms  \"\n",
+      "print  \"\\n  (b)Capacitive  reactance,  Xc  =  \",round(  Xc2,2),\"  ohms  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Capacitive  reactance,  Xc  =   318.31   ohms  \n",
+        "\n",
+        "  (b)Capacitive  reactance,  Xc  =   0.8   ohms  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  40;#  in  ohms\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  1/(2*math.pi*f*Z)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Capacitance,C  =  \",round((C/1E-6),2),\" uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Capacitance,C  =   79.58  uF  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 215</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  23E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "I  =  V/Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  I  =  \",round(I,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  I  =   1.73   A  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 216</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vr  =  12;#  in  volts\n",
+      "Vl  =  5;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  (Vr**2  +  Vl**2)**0.5\n",
+      "phi  =  math.atan(Vl/Vr)\n",
+      "phid  =  phi*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  voltage  V  =  \",V,\"  V,  phase  angle  between  current  and  voltage  is  \",  round(phid,2),\"deg lagging\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  voltage  V  =   13.0   V,  phase  angle  between  current  and  voltage  is   22.62 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 216</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  volts\n",
+      "R  =  4;#  in  ohms\n",
+      "L  =  0.00955;#  in  Henry\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  = math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (c)Current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (d)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   3.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   5.0   ohms\n",
+        "\n",
+        "  (c)Current,  I  =   48.0   A\n",
+        "\n",
+        "  (d)phase  angle  between  the  supply  voltage  and  current  is   36.87 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 217</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vdc  =  12;#  in  volts\n",
+      "Vac  =  240;#  in  volts\n",
+      "Iac  =  20;#  in  Amperes\n",
+      "Idc  =  2;#  in  Amperes\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  Vdc/Idc\n",
+      "Z  =  Vac/Iac\n",
+      "XL  =  (Z**2  -  R**2)**0.5\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance,  R  =  \",R,\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",Z,\"  ohms\"\n",
+      "print  \"\\n  (c)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (d)Inductance,  L  =  \",round(L,2),\"  H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance,  R  =   6.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   12.0   ohms\n",
+        "\n",
+        "  (c)Inductive  reactance,  XL  =   10.39   ohms\n",
+        "\n",
+        "  (d)Inductance,  L  =   0.03   H"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 217</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  200;#  in  ohms\n",
+      "L  =  0.3183;#  in  henry\n",
+      "V  =  240;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "VL  =  I*XL\n",
+      "VR  =  I*R\n",
+      "phid  =  math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Inductive  reactance,  XL  =  \",round(XL,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (c)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (d)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V  and  p.d.  across  resistance,  VR  =  \",round(VR,2),\"  V\"\n",
+      "print  \"\\n  (e)circuit  phase  angle  is  \",round(phid,2),\" deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Inductive  reactance,  XL  =   100.0   ohms\n",
+        "\n",
+        "  (b)Impedance,  Z  =   223.61   ohms\n",
+        "\n",
+        "  (c)current,  I  =   1.07   A\n",
+        "\n",
+        "  (d)p.d.  across  Inductor,  VL  =   107.33   V  and  p.d.  across  resistance,  VR  =   214.66   V\n",
+        "\n",
+        "  (e)circuit  phase  angle  is   26.56  deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 218</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  100;#  in  ohms\n",
+      "L  =  0.2;#  in  henry\n",
+      "Vmax  =  200;#  in  volts\n",
+      "w  =  500;#  in  rad/sec\n",
+      "\n",
+      "#calculation:\n",
+      "Vrms  =  0.707*Vmax\n",
+      "f  =  w/(2*math.pi)\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  Vrms/Z\n",
+      "VL  =  I*XL\n",
+      "VR  =  I*R\n",
+      "phid  =  math.atan(XL/R)*180/math.pi\n",
+      "\n",
+      "\\\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (c)p.d.  across  resistance,  VR  =  \",round(VR,2),\"  V\"\n",
+      "print  \"\\n  (d)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (e)circuit  phase  angle  is  \",phid,\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   141.42   ohms\n",
+        "\n",
+        "  (b)current,  I  =   1.0   A\n",
+        "\n",
+        "  (c)p.d.  across  resistance,  VR  =   99.98   V\n",
+        "\n",
+        "  (d)p.d.  across  Inductor,  VL  =   99.98   V\n",
+        "\n",
+        "  (e)circuit  phase  angle  is   45.0 deg"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 218</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  1.2273E-3;#  in  henry\n",
+      "f  =  5000;#  in  Hz\n",
+      "VR  =  6;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "I  =VR/R\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "V  =  I*Z\n",
+      "VL  =  I*XL\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)supply  voltage  =  \",round(V,2),\"  Volts\"\n",
+      "print  \"\\n  (b)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)supply  voltage  =   9.77   Volts\n",
+        "\n",
+        "  (b)p.d.  across  Inductor,  VL  =   7.71   V"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 219</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  60;#  in  ohms\n",
+      "Rc  =  20;#  in  ohms\n",
+      "L  =  159.2E-3;#  in  henry\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Rt  =  R  +  Rc\n",
+      "Z  =  (Rt**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan(XL/Rt)*180/math.pi\n",
+      "VR  =  I*R\n",
+      "Zc  =  (Rc**2  +  XL**2)**0.5\n",
+      "Vc  =  I*Zc\n",
+      "VL  =  I*XL\n",
+      "VRc  =  I*Rc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,3),\"  A\"\n",
+      "print  \"\\n  (c)circuit  phase  angle  is  \",round(phid,0),\"deg lagging\"\n",
+      "print  \"\\n  (d)p.d.  across  resistance,  VR  =  \",round(  VR,1),\"  V\"\n",
+      "print  \"\\n  (e)p.d.  across  coil,  Vc  =  \",round(Vc,1),\"  V\"\n",
+      "print  \"\\n  (f1)p.d.  across  Inductor,  VL  =  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (f2)p.d.  across  coil  resistance,  VRc  =  \",round(VRc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   94.35   ohms\n",
+        "\n",
+        "  (b)current,  I  =   2.544   A\n",
+        "\n",
+        "  (c)circuit  phase  angle  is   32.0 deg lagging\n",
+        "\n",
+        "  (d)p.d.  across  resistance,  VR  =   152.6   V\n",
+        "\n",
+        "  (e)p.d.  across  coil,  Vc  =   137.0   V\n",
+        "\n",
+        "  (f1)p.d.  across  Inductor,  VL  =   127.23   V\n",
+        "\n",
+        "  (f2)p.d.  across  coil  resistance,  VRc  =   50.88   V"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 220</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohms\n",
+      "C  =  45E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Z  =  (R**2  +  Xc**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan(Xc/R)*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,  Z  =  \",round(Z,2),\"  ohms\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (c)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,  Z  =   75.02   ohms\n",
+        "\n",
+        "  (b)current,  I  =   3.2   A\n",
+        "\n",
+        "  (c)phase  angle  between  the  supply  voltage  and  current  is   70.54 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 221</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  40;#  in  ohms\n",
+      "f  =  60;#  in  Hz\n",
+      "I  =  3;#in  amperes\n",
+      "Z  =  50;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Xc  =  (Z**2  -  R**2)**0.5\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "V  =  I*Z\n",
+      "phid  =  math.atan(Xc/R)*180/math.pi\n",
+      "VR  =  I*R\n",
+      "Vc  =  I*Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance,  C  =  \",round((C/1E-6),2),\"  uF\"\n",
+      "print  \"\\n  (b)Voltage,  V  =  \",V,\"  Volts\"\n",
+      "print  \"\\n  (c)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)p.d.  across  resistance,  VR  =  \",  VR,\"  V\"\n",
+      "print  \"\\n  (e)p.d.  across  Capacitor,  Vc  =  \",Vc,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance,  C  =   88.42   uF\n",
+        "\n",
+        "  (b)Voltage,  V  =   150   Volts\n",
+        "\n",
+        "  (c)phase  angle  between  the  supply  voltage  and  current  is   36.87 deg leading\n",
+        "\n",
+        "  (d)p.d.  across  resistance,  VR  =   120   V\n",
+        "\n",
+        "  (e)p.d.  across  Capacitor,  Vc  =   90.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 222</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  5;#  in  ohms\n",
+      "C  =  100E-6;#  in  Farads\n",
+      "L  =  0.12;#  in  Henry\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  300;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "X  =  XL  -  Xc\n",
+      " #Since  XL  is  greater  than  Xc,  the  circuit  is  inductive.\n",
+      "Z  =  (R**2  +  (XL-Xc)**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid  =  math.atan((XL-Xc)/R)*180/math.pi\n",
+      "Zcl  =  (R**2  +  XL**2)**0.5\n",
+      "Vcl  =  I*Zcl\n",
+      "phidc  =  math.atan(XL/R)*180/math.pi\n",
+      "Vc  =  I*Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current,I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)phase  angle  between  the  supply  voltage  and  current  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (c)Voltage  across  the  coil,  Vcoil  =  \",round(Vcl,0),\"  Volts\"\n",
+      "print  \"\\n  (d)p.d.  across  Capacitor,  Vc  =  \",round(Vc,0),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current,I  =   38.91   A\n",
+        "\n",
+        "  (b)phase  angle  between  the  supply  voltage  and  current  is   49.57 deg\n",
+        "\n",
+        "  (c)Voltage  across  the  coil,  Vcoil  =   1480.0   Volts\n",
+        "\n",
+        "  (d)p.d.  across  Capacitor,  Vc  =   1239.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 224</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohms\n",
+      "C  =  0.25E-6;#  in  Farads\n",
+      "L  =  130E-6;#  in  Henry\n",
+      "Rc  =  5;#  in  ohms\n",
+      "R2  =  10;#  in  ohms\n",
+      "f  =  20000;#  in  Hz\n",
+      "V  =  40;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "X  =  Xc  -  XL\n",
+      "R  =  R1  +  R2  +  Rc\n",
+      " #Since  Xc  is  greater  than  XL,  the  circuit  is  capacitive.\n",
+      "Z  =  (R**2  +  (Xc-XL)**2)**0.5\n",
+      "I  =  V/Z\n",
+      "phid = math.atan((Xc-XL)/R)*180/math.pi\n",
+      "V1  =  I*R1\n",
+      "V2  =  I*((Rc**2  +  XL**2)**0.5)\n",
+      "V3  =  I*((R2**2  +  Xc**2)**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current,I  =  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)circuit  phase  angle  is  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (c1)Voltage  across  the  Resistance  of  8  ohms  =  \",round(V1,2),\"  Volts\"\n",
+      "print  \"\\n  (c2)Voltage  across  the  coil,  Vcoil  =  \",round(V2,2),\"  Volts\"\n",
+      "print  \"\\n  (c3)p.d.  across  Capacitor  resistance  circuit  =  \",round(V3,2),\"  Volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current,I  =   1.44   A\n",
+        "\n",
+        "  (b)circuit  phase  angle  is   33.97 deg leading\n",
+        "\n",
+        "  (c1)Voltage  across  the  Resistance  of  8  ohms  =   11.54   Volts\n",
+        "\n",
+        "  (c2)Voltage  across  the  coil,  Vcoil  =   24.64   Volts\n",
+        "\n",
+        "  (c3)p.d.  across  Capacitor  resistance  circuit  =   48.12   Volts"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 224</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  4;#  in  ohms\n",
+      "C  =  1.273E-6;#  in  Farads\n",
+      "L  =  0.286E-3;#  in  Henry\n",
+      "R2  =  8;#  in  ohms\n",
+      "f  =  5000;#  in  Hz\n",
+      "I  =  5;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "phid1  =  math.atan(XL/R1)*180/math.pi\n",
+      "V1  =  I*((R1**2  +  XL**2)**0.5)\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "V2  =  I*((R2**2  +  Xc**2)**0.5)\n",
+      "phid2  =  math.atan(Xc/R2)*180/math.pi\n",
+      "Z  =  ((R1+R2)**2  +  (Xc-XL)**2)**0.5\n",
+      "V  =  I*Z\n",
+      "phid  =  math.atan((Xc-XL)/(R1+R2))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Voltage  supply,  V  =  \",round(V,2),\"  V\"\n",
+      "print  \"\\n  (b)circuit  phase  angle  is  \",round(phid,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Voltage  supply,  V  =   100.08   V\n",
+        "\n",
+        "  (b)circuit  phase  angle  is   53.16 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 226</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  60E-6;#  in  Farads\n",
+      "L  =  125E-3;#  in  Henry\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*(L*C)**0.5)\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency,Fr  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency,Fr  =   58.12   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   12.0"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 226</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.05E-3;#  in  Henry\n",
+      "fr  =  200000;#  in  Hz\n",
+      "V  =  0.002;#  in  Volts\n",
+      "I  =  0.1E-3;#  in  amperes\n",
+      "#calculation:\n",
+      "#  L-C-R\n",
+      "#At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "R  =  V/I\n",
+      "C  =  1/(L*(2*math.pi*fr)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance,  R  =  \",round(R,2),\"  ohms\"\n",
+      "print  \"\\n  (b)Capacitance,  C  =  \",round((C/1E-9),2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance,  R  =   20.0   ohms\n",
+        "\n",
+        "  (b)Capacitance,  C  =   12.67 nF"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 227</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  80E-3;#  in  Henry\n",
+      "C  =  0.25E-6;#  in  Farads\n",
+      "R  =  12.5;#  in  ohms\n",
+      "V  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "VL  =  I*(2*math.pi*fr*L)\n",
+      "Vc  =  I/(2*math.pi*fr*C)\n",
+      "Vm  =  VL/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  resonant  frequency  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2),\"\"\n",
+      "print  \"\\n  (b)Voltage  magnification  at  resonance  =  \",round(Vm,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  resonant  frequency  =   1125.4   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   8.0 \n",
+        "\n",
+        "  (b)Voltage  magnification  at  resonance  =   45.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 228</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  60E-3;#  in  Henry\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "R  =  2;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Q  =  ((L/C)**0.5)/R\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  At  resonance,  Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  At  resonance,  Q-factor  =   22.36"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 228</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  100E-3;#  in  Henry\n",
+      "C  =  2E-6;#  in  Farads\n",
+      "R  =  10;#  in  ohms\n",
+      "V  =  50;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #At  resonance,  XL  =  Xc  and  impedance  Z  =  R\n",
+      "I  =  V/R\n",
+      "VL  =  I*(2*math.pi*fr*L)\n",
+      "Vc  =  I/(2*math.pi*fr*C)\n",
+      "Q  =  VL/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  resonant  frequency  =  \",round(fr,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Current,  I  =  \",round(I,2),\"\"\n",
+      "print  \"\\n  (c)Voltage  across  coil  at  resonance  is  \",round(VL,2),\"V  \"\n",
+      "print   \"and  Voltage  across  capacitance  at  resonance  is  \",round(  Vc,2),\"V\"\n",
+      "print  \"\\n  (d)At  resonance,  Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  resonant  frequency  =   355.88   Hz\n",
+        "\n",
+        "  (b)Current,  I  =   5.0 \n",
+        "\n",
+        "  (c)Voltage  across  coil  at  resonance  is   1118.03 V  and  Voltage  across  capacitance  at  resonance  is   1118.03 V\n",
+        "\n",
+        "  (d)At  resonance,  Q-factor  =   22.36"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 230</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  20E-3;#  in  Henry\n",
+      "R  =  10;#  in  ohms\n",
+      "fr  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Qr  =  (2*math.pi*fr)*L/R\n",
+      "bw  =  fr/Qr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Bandwidth,  (f2-f1)  =  \",round(bw,2),\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Bandwidth,  (f2-f1)  =   79.58   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 231</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  5000;#  in  ohms\n",
+      "Imax  =  0.250;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "Irms  =  0.707*Imax\n",
+      "P  =  Irms*Irms*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Power,  P  =  \",round(P,2),\"  Watts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Power,  P  =   156.2   Watts"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 231</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  60;#  in  ohms\n",
+      "L  =  75E-3;#  in  Henry\n",
+      "V  =  110;#  in  Volts\n",
+      "f  =  60;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  (R**2  +  XL**2)**0.5\n",
+      "I  =  V/Z\n",
+      "P  =  I*I*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Power,  P  =  \",round(P,2),\"  Watts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Power,  P  =   165.02   Watts"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 232</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VI  =  300;#  in  VA\n",
+      "V  =  150;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  VI/V\n",
+      "XL  =  V/I\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Inductance  =  \",round(L,2),\"  H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Inductance  =   0.24   H"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 232</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VI  =  200000;#  in  VA\n",
+      "pf  =  0.8;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "P  =  VI*pf\n",
+      "Q  =  VI*math.sin(math.acos(pf))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  rated  power  output  is  \",round(P/1000,2),\"KW  and  the  corresponding  reactive  power  is  \",round(Q/1000,2),\"kvar\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  rated  power  output  is   160.0 KW  and  the  corresponding  reactive  power  is   120.0 kvar"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 233</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  120;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "P  =  400;#  in  Watt\n",
+      "I  =  8;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  P/(I*I)\n",
+      "Z  =  V/I\n",
+      "XL  =  (Z**2  -  R**2)**0.5\n",
+      "pf  =  P/(V*I)\n",
+      "phi  =  math.acos(pf)*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  =  \",round(R,2),\"  ohm  \"\n",
+      "print  \"\\n  (b)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (c)reactance  =  \",round(XL,2),\"  ohm  \"\n",
+      "print  \"\\n  (d)Power  factor  =  \",round(pf,2),\"\"\n",
+      "print  \"\\n  (e)phase  angle  =  \",round(phi,2),\"deg lagging\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  =   6.25   ohm  \n",
+        "\n",
+        "  (b)Impedance  Z  =   15.0   Ohm  \n",
+        "\n",
+        "  (c)reactance  =   13.64   ohm  \n",
+        "\n",
+        "  (d)Power  factor  =   0.42 \n",
+        "\n",
+        "  (e)phase  angle  =   65.38 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 29, page no. 233</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  60;#  in  Hz\n",
+      "P  =  100;#  in  Watt\n",
+      "pf  =  0.5;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  P/(pf*V)\n",
+      "phi  =  math.acos(pf)*180/math.pi\n",
+      "R  =  P/(I*I)\n",
+      "Z  =  V/I\n",
+      "Xc  =  (Z**2  -  R**2)**0.5\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (b)phase  angle  =  \",round(phi,2),\"deg leading\"\n",
+      "print  \"\\n  (c)resistance  =  \",round(R,2),\"  ohm  \"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)capacitance  =  \",round((C/1E-6),2),\"uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  I  =   2.0   A  \n",
+        "\n",
+        "  (b)phase  angle  =   60.0 deg leading\n",
+        "\n",
+        "  (c)resistance  =   25.0   ohm  \n",
+        "\n",
+        "  (d)Impedance  Z  =   50.0   Ohm  \n",
+        "\n",
+        "  (e)capacitance  =   61.26 uF  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint.ipynb
new file mode 100755
index 00000000..429b5bb8
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint.ipynb
@@ -0,0 +1,913 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 16: Single-phase parallel a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 239</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current in each branch,\n",
+      "#(b) the supply current, (c) the circuit phase angle,\n",
+      "#(d) the circuit impedance, and (e) the power consumed\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  20;#  in  Ohms\n",
+      "L  =  2.387E-3;#  in  Henry\n",
+      "V  =  60;#  in  Volts\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "IR  =  V/R\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "I  =  (IR**2  +  IL**2)**0.5\n",
+      "phi  =  math.atan(IL/IR)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  resistor  is  \",round(IR,2),\"  A    and  current  through  Inductor  is  \",round(  IL,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg lagging\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  resistor  is   3.0   A    and  current  through  Inductor  is   4.0   A\n",
+        "\n",
+        "  (b)current,  I  =   5.0   A  \n",
+        "\n",
+        "  (c)phase  angle  =   53.13 deg lagging\n",
+        "\n",
+        "  (d)Impedance  Z  =   12.0   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   180.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 240</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current in each branch, (b) the supply current,\n",
+      "#(c) the circuit phase angle, (d) the circuit impedance, \n",
+      "#(e) the power dissipated, and (f) the apparent power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  Ohms\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "IR  =  V/R\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "I  =  (IR**2  +  Ic**2)**0.5\n",
+      "phi  =  math.atan(Ic/IR)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "S  =  V*I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  resistor  is  \",round(IR,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \"\n",
+      "print  \"\\n  (f)apparent  Power  =  \",round(S,2),\"  VA  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  resistor  is   3.0   A    and  current  through  capacitor  is   2.26   A\n",
+        "\n",
+        "  (b)current,  I  =   3.76   A \n",
+        "\n",
+        "  (c)phase  angle  =   37.02 deg leading\n",
+        "\n",
+        "  (d)Impedance  Z  =   63.88   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   720.0   Watt  \n",
+        "\n",
+        "  (f)apparent  Power  =   901.72   VA  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 241</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of C and R.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "pf  =  0.6;#  power  factor\n",
+      "V  =  120;#  in  Volts\n",
+      "f  =  200;#  in  Hz\n",
+      "I  =  2;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "phi  =  math.acos(pf)\n",
+      "phid  =  phi*180/math.pi\n",
+      "IR  =  I*math.cos(phi)\n",
+      "Ic  =  I*math.sin(phi)\n",
+      "R  =  V/IR\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  R  =  \",round(R,2),\"  Ohm  \"\n",
+      "print  \"\\n  (b)Capacitance,C  =  \",round((C/1E-6),2),\"  uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  R  =   100.0   Ohm  \n",
+        "\n",
+        "  (b)Capacitance,C  =   10.61   uF  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 242</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the branch currents, \n",
+      "#(b) the supply current and its phase angle, \n",
+      "#(c) the circuit impedance, and (d) the power consumed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  25E-6;#  in  Farads\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      " #IL  and  Ic  are  anti-phase.  Hence  supply  current,\n",
+      "I  =  IL  -  Ic\n",
+      " #the  current  lags  the  supply  voltage  V  by  90\u00c2\u00b0\n",
+      "phi  =  math.pi/2\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  Inductor  is  \",round(IL,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg lagging\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  Inductor  is   2.65   A    and  current  through  capacitor  is   0.79   A\n",
+        "\n",
+        "  (b)current,  I  =   1.87   A  \n",
+        "\n",
+        "  (c)phase  angle  =   90.0 deg lagging\n",
+        "\n",
+        "  (d)Impedance  Z  =   53.56   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   0.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 242</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the branch currents, \n",
+      "#(b) the supply current and its phase angle, \n",
+      "#(c) the circuit impedance, and (d) the power consumed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  25E-6;#  in  Farads\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  150;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      " #IL  and  Ic  are  anti-phase.  Hence  supply  current,\n",
+      "I  =  Ic  -  IL\n",
+      " #the  current  leads  the  supply  voltage  V  by  90\u00c2\u00b0\n",
+      "phi  =  math.pi/2\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  Inductor  is  \",round(IL,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  Inductor  is   0.88   A    and  current  through  capacitor  is   2.36   A\n",
+        "\n",
+        "  (b)current,  I  =   1.47   A  \n",
+        "\n",
+        "  (c)phase  angle  =   90.0 deg leading\n",
+        "\n",
+        "  (d)Impedance  Z  =   67.93   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   0.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 244</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current in the coil and its phase angle,\n",
+      "#(b) the current in the capacitor and its phase angle,\n",
+      "#(c) the supply current and its phase angle,\n",
+      "#(d) the circuit impedance, \n",
+      "#(e) the power consumed, \n",
+      "#(f) the apparent power, and \n",
+      "#(g) the reactive power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "R  =  40;#  in  Ohms\n",
+      "L  =  159.2E-3;#  in  Henry\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z1  =  (R**2  +  XL**2)**0.5\n",
+      "ILR  =  V/Z1\n",
+      "phi1  =  math.atan(XL/R)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "phi2  =  math.pi/2\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)\n",
+      "Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)\n",
+      "I  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan(abs(Iv)/Ih)\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "phid  =  phi*180/math.pi\n",
+      "S  =  V*I\n",
+      "Q  =  V*I*math.sin(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  coil  is  \",round(ILR,2),\"  A  and  lagged  by  phase  angle  is  \",round(phi1d,2),\"deg\"\n",
+      "print  \"\\n  (b)Current  through  capacitor  is  \",round(Ic,2),\"  A  and  lead  by  phase  angle  is  \",round(phi2d,2),\"deg\"\n",
+      "print  \"\\n  (c)supply  Current  is  \",round(I,2),\"  A  and  lagged  by  phase  angle  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \"\n",
+      "print  \"\\n  (f)apparent  Power  =  \",round(S,2),\"  VA  \"\n",
+      "print  \"\\n  (g)reactive  Power  =  \",round(Q,2),\"  var  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  coil  is   3.75   A  and  lagged  by  phase  angle  is   51.35 deg\n",
+        "\n",
+        "  (b)Current  through  capacitor  is   2.26   A  and  lead  by  phase  angle  is   90.0 deg\n",
+        "\n",
+        "  (c)supply  Current  is   2.43   A  and  lagged  by  phase  angle  is   15.85 deg\n",
+        "\n",
+        "  (d)Impedance  Z  =   98.64   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   561.76   Watt  \n",
+        "\n",
+        "  (f)apparent  Power  =   583.97   VA  \n",
+        "\n",
+        "  (g)reactive  Power  =   159.53   var  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 246</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current in the coil, and (b) the current in the capacitor. \n",
+      "#(c) Measure the supply current and its phase angle (d) the circuit impedance and (e) the power consumed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.02E-6;#  in  Farads\n",
+      "R  =  3000;#  in  Ohms\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  40;#  in  Volts\n",
+      "f  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z1  =  (R**2  +  XL**2)**0.5\n",
+      "ILR  =  V/Z1\n",
+      "phi1  =  math.atan(XL/R)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "phi2  =  math.pi/2\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)\n",
+      "Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)\n",
+      "I  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan((Iv)/Ih)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  coil  is  \",round(ILR*1000,2),\"mA  and  lagged  by  phase  angle  is  \",round(phi1d,1),\"deg\"\n",
+      "print  \"\\n  (b)Current  through  capacitor  is  \",round(Ic*1000,2),\"mA  and  lead  by  phase  angle  is  \",round(phi2d,2),\"deg\"\n",
+      "print  \"\\n  (c)supply  Current  is  \",round(I*1000,1),\"mA  and  leaded  by  phase  angle  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z/1000,3),\"KOhm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P*1000,1),\"mWatt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  coil  is   8.3 mA  and  lagged  by  phase  angle  is   51.5 deg\n",
+        "\n",
+        "  (b)Current  through  capacitor  is   25.13 mA  and  lead  by  phase  angle  is   90.0 deg\n",
+        "\n",
+        "  (c)supply  Current  is   19.3 mA  and  leaded  by  phase  angle  is   74.5 deg\n",
+        "\n",
+        "  (d)Impedance  Z  =   2.068 KOhm  \n",
+        "\n",
+        "  (e)Power  consumed  =   206.8 mWatt  "
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 249</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency of the circuit and \n",
+      "#(b) the current circulating in the capacitor and inductance at resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  40E-6;#  in  Farads\n",
+      "R  =  0;#  in  Ohms\n",
+      "L  =  150E-3;#  in  Henry\n",
+      "V  =  50;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)\n",
+      "Xc  =  1/(2*math.pi*fr*C)\n",
+      "Icirc  =  V/Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Parallel  resonant  frequency,  fr  =  \",round(fr,2),\"  Hz  \"\n",
+      "print  \"\\n  (b)Current  circulating  in  L  and  C  at  resonance  =  \",round(Icirc,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Parallel  resonant  frequency,  fr  =   64.97   Hz  \n",
+        "\n",
+        "  (b)Current  circulating  in  L  and  C  at  resonance  =   0.82   A  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 250</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the resonant frequency, \n",
+      "#(b) the dynamic resistance, \n",
+      "#(c) the current at resonance and \n",
+      "#(d) the circuit Q-factor at resonanc\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  20E-6;#  in  Farads\n",
+      "R  =  60;#  in  Ohms\n",
+      "L  =  200E-3;#  in  Henry\n",
+      "V  =  20;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)\n",
+      "Rd  =  L/(R*C)\n",
+      "Ir  =  V/Rd\n",
+      "Q  =  2*math.pi*fr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Parallel  resonant  frequency,  fr  =  \",round(fr,2),\"  Hz  \"\n",
+      "print  \"\\n  (b)the  dynamic  resistance,RD  =  \",round(Rd,2),\"  ohm  \"\n",
+      "print  \"\\n  (c)Current  at  resonance  =  \",round(Ir,2),\"  A  \"\n",
+      "print  \"\\n  (d)Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Parallel  resonant  frequency,  fr  =   63.66   Hz  \n",
+        "\n",
+        "  (b)the  dynamic  resistance,RD  =   166.67   ohm  \n",
+        "\n",
+        "  (c)Current  at  resonance  =   0.12   A  \n",
+        "\n",
+        "  (d)Q-factor  =   1.33"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 251</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the condition when the supply current is a minimum:\n",
+      "#(a) the capacitance of the capacitor,\n",
+      "#(b) the dynamic resistance, \n",
+      "#(c) the supply current, and \n",
+      "#(d) the Q-factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fr  =  5000;#  in  ohm\n",
+      "R  =  800;#  in  Ohms\n",
+      "L  =  100E-3;#  in  Henry\n",
+      "V  =  12;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  1/(L*((2*math.pi*fr)**2  +  R*R/(L*L)))\n",
+      "Rd  =  L/(R*C)\n",
+      "Ir  =  V/Rd\n",
+      "Q  =  2*math.pi*fr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance,  C  =  \",round((C/1E-9),2),\"  nF  \"\n",
+      "print  \"\\n  (b)the  dynamic  resistance,RD  =  \",round(Rd,2),\"  ohm  \"\n",
+      "print  \"\\n  (c)Current  at  resonance  =  \",round((Ir/1E-3),2),\"  mA  \"\n",
+      "print  \"\\n  (d)Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance,  C  =   9.52   nF  \n",
+        "\n",
+        "  (b)the  dynamic  resistance,RD  =   13137.01   ohm  \n",
+        "\n",
+        "  (c)Current  at  resonance  =   0.91   mA  \n",
+        "\n",
+        "  (d)Q-factor  =   3.93"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 252</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the currenttaken by a capacitor connected in parallel with the motor to correct the power factor to unity, and \n",
+      "#(b) the value of the supply current after power factor correction.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  240;#  in  Volts\n",
+      "pf  =  0.6;#  power  factor\n",
+      "Im  =  50;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "phi  =  math.acos(pf)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Ic  =  Im*math.sin(phi)\n",
+      "I  =  Im*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitor  current  Ic  must  be  \",round(Ic,2),\"  A  for  the  power  factor  to  be  unity.  \"\n",
+      "print  \"\\n  (b)Supply  current  I  =  \",round(I,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitor  current  Ic  must  be   40.0   A  for  the  power  factor  to  be  unity.  \n",
+        "\n",
+        "  (b)Supply  current  I  =   30.0   A  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 253</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current taken by the motor, \n",
+      "#(b) the supply current after power factor correction, \n",
+      "#(c) the current taken by the capacitor,\n",
+      "#(d) the capacitance of the capacitor, and \n",
+      "#(e) the kvar rating of the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pout  =  4800;#  in  Watt\n",
+      "eff  =  0.8#  effficiency\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  240;#  in  Volts\n",
+      "pf1  =  0.625;#  power  factor\n",
+      "pf2  =  0.95;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "Pin  =  Pout/eff\n",
+      "Im  =  Pin/(V*pf1)\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "phi2  =  math.acos(pf2)\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Imh  =  Im*math.cos(phi1)\n",
+      " #Ih  =  I*cos(phi2)\n",
+      "Ih  =  Imh\n",
+      "I  =  Ih/math.cos(phi2)\n",
+      "Imv  =  Im*math.sin(phi1)\n",
+      "Iv  =  I*math.sin(phi2)\n",
+      "Ic  =  Imv  -  Iv\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "kvar  =  V*Ic/1000\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  taken  by  the  motor,  Im  =  \",round(Im,2),\"  A\"\n",
+      "print  \"\\n  (b)supply  current  after  p.f.  correction,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)magnitude  of  the  capacitor  current  Ic  =  \",round(Ic,0),\"  A\"\n",
+      "print  \"\\n  (d)capacitance,  C  =  \",round((C/1E-6),0),\"  uF  \"\n",
+      "print  \"\\n  (d)kvar  rating  of  the  capacitor  =  \",round(kvar,2),\"  kvar  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  taken  by  the  motor,  Im  =   40.0   A\n",
+        "\n",
+        "  (b)supply  current  after  p.f.  correction,  I  =   26.32   A  \n",
+        "\n",
+        "  (c)magnitude  of  the  capacitor  current  Ic  =   23.0   A\n",
+        "\n",
+        "  (d)capacitance,  C  =   305.0   uF  \n",
+        "\n",
+        "  (d)kvar  rating  of  the  capacitor  =   5.52   kvar  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 254</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for the lamps and motor, (a) the total current, (b) the overall power factor and \n",
+      "#(c) the total power. (d) Find the value of the static capacitor to improve the overall power factor to 0.975 lagging.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  3000;#  in  VA\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  250;#  in  Volts\n",
+      "Iil  =  10;#  in  Amperes\n",
+      "Ifl  =  8;#  in  Amperes\n",
+      "pfil  = 1; #  power  factor\n",
+      "pffl  =  0.7;#  power  factor\n",
+      "pfm  =  0.8;#  power  factor\n",
+      "pf0  =  0.975;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "phiil  =  math.acos(pfil)\n",
+      "phiild  =  phiil*180/math.pi\n",
+      "phifl  =  math.acos(pffl)\n",
+      "phifld  =  phifl*180/math.pi\n",
+      "phim  =  math.acos(pfm)\n",
+      "phimd  =  phim*180/math.pi\n",
+      "phi0  =  math.acos(pf0)\n",
+      "phi0d  =  phi0*180/math.pi\n",
+      "Im  =  S/V\n",
+      "Ih  =  Iil*math.cos(phiil)  +  Ifl*math.cos(phifl)  +  Im*math.cos(phim)\n",
+      "Iv  =  Iil*math.sin(phiil)  -  Ifl*math.sin(phifl)  -  Im*math.sin(phim)\n",
+      "Il  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan(abs(Iv)/Ih)\n",
+      "phid  =  phi*180/math.pi\n",
+      "pf  =  math.cos(phi)\n",
+      "P  =  V*Il*pf\n",
+      "I  =  Il*math.cos(phi)/math.cos(phi0)\n",
+      "Ic  =  Il*math.sin(phi)  -  I*math.sin(phi0)\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)total  current,  Il  =  \",round(Il,2),\"  A\"\n",
+      "print  \"\\n  (b)Power  factor  =  \",round(pf,2),\"lagging\"\n",
+      "print  \"\\n  (c)Total  power,  P  =  \",round(P/1000,2),\"KWatt\"\n",
+      "print  \"\\n  (d)capacitance,  C  =  \",round((C/1E-6),2),\"uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)total  current,  Il  =   28.32   A\n",
+        "\n",
+        "  (b)Power  factor  =   0.89 lagging\n",
+        "\n",
+        "  (c)Total  power,  P  =   6.3 KWatt\n",
+        "\n",
+        "  (d)capacitance,  C  =   91.29 uF  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_1.ipynb
new file mode 100755
index 00000000..429b5bb8
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_1.ipynb
@@ -0,0 +1,913 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 16: Single-phase parallel a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 239</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current in each branch,\n",
+      "#(b) the supply current, (c) the circuit phase angle,\n",
+      "#(d) the circuit impedance, and (e) the power consumed\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  20;#  in  Ohms\n",
+      "L  =  2.387E-3;#  in  Henry\n",
+      "V  =  60;#  in  Volts\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "IR  =  V/R\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "I  =  (IR**2  +  IL**2)**0.5\n",
+      "phi  =  math.atan(IL/IR)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  resistor  is  \",round(IR,2),\"  A    and  current  through  Inductor  is  \",round(  IL,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg lagging\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  resistor  is   3.0   A    and  current  through  Inductor  is   4.0   A\n",
+        "\n",
+        "  (b)current,  I  =   5.0   A  \n",
+        "\n",
+        "  (c)phase  angle  =   53.13 deg lagging\n",
+        "\n",
+        "  (d)Impedance  Z  =   12.0   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   180.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 240</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current in each branch, (b) the supply current,\n",
+      "#(c) the circuit phase angle, (d) the circuit impedance, \n",
+      "#(e) the power dissipated, and (f) the apparent power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  Ohms\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "IR  =  V/R\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "I  =  (IR**2  +  Ic**2)**0.5\n",
+      "phi  =  math.atan(Ic/IR)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "S  =  V*I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  resistor  is  \",round(IR,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \"\n",
+      "print  \"\\n  (f)apparent  Power  =  \",round(S,2),\"  VA  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  resistor  is   3.0   A    and  current  through  capacitor  is   2.26   A\n",
+        "\n",
+        "  (b)current,  I  =   3.76   A \n",
+        "\n",
+        "  (c)phase  angle  =   37.02 deg leading\n",
+        "\n",
+        "  (d)Impedance  Z  =   63.88   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   720.0   Watt  \n",
+        "\n",
+        "  (f)apparent  Power  =   901.72   VA  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 241</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of C and R.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "pf  =  0.6;#  power  factor\n",
+      "V  =  120;#  in  Volts\n",
+      "f  =  200;#  in  Hz\n",
+      "I  =  2;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "phi  =  math.acos(pf)\n",
+      "phid  =  phi*180/math.pi\n",
+      "IR  =  I*math.cos(phi)\n",
+      "Ic  =  I*math.sin(phi)\n",
+      "R  =  V/IR\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  R  =  \",round(R,2),\"  Ohm  \"\n",
+      "print  \"\\n  (b)Capacitance,C  =  \",round((C/1E-6),2),\"  uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  R  =   100.0   Ohm  \n",
+        "\n",
+        "  (b)Capacitance,C  =   10.61   uF  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 242</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the branch currents, \n",
+      "#(b) the supply current and its phase angle, \n",
+      "#(c) the circuit impedance, and (d) the power consumed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  25E-6;#  in  Farads\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      " #IL  and  Ic  are  anti-phase.  Hence  supply  current,\n",
+      "I  =  IL  -  Ic\n",
+      " #the  current  lags  the  supply  voltage  V  by  90\u00c2\u00b0\n",
+      "phi  =  math.pi/2\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  Inductor  is  \",round(IL,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg lagging\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  Inductor  is   2.65   A    and  current  through  capacitor  is   0.79   A\n",
+        "\n",
+        "  (b)current,  I  =   1.87   A  \n",
+        "\n",
+        "  (c)phase  angle  =   90.0 deg lagging\n",
+        "\n",
+        "  (d)Impedance  Z  =   53.56   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   0.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 242</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the branch currents, \n",
+      "#(b) the supply current and its phase angle, \n",
+      "#(c) the circuit impedance, and (d) the power consumed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  25E-6;#  in  Farads\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  150;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      " #IL  and  Ic  are  anti-phase.  Hence  supply  current,\n",
+      "I  =  Ic  -  IL\n",
+      " #the  current  leads  the  supply  voltage  V  by  90\u00c2\u00b0\n",
+      "phi  =  math.pi/2\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  Inductor  is  \",round(IL,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  Inductor  is   0.88   A    and  current  through  capacitor  is   2.36   A\n",
+        "\n",
+        "  (b)current,  I  =   1.47   A  \n",
+        "\n",
+        "  (c)phase  angle  =   90.0 deg leading\n",
+        "\n",
+        "  (d)Impedance  Z  =   67.93   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   0.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 244</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current in the coil and its phase angle,\n",
+      "#(b) the current in the capacitor and its phase angle,\n",
+      "#(c) the supply current and its phase angle,\n",
+      "#(d) the circuit impedance, \n",
+      "#(e) the power consumed, \n",
+      "#(f) the apparent power, and \n",
+      "#(g) the reactive power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "R  =  40;#  in  Ohms\n",
+      "L  =  159.2E-3;#  in  Henry\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z1  =  (R**2  +  XL**2)**0.5\n",
+      "ILR  =  V/Z1\n",
+      "phi1  =  math.atan(XL/R)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "phi2  =  math.pi/2\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)\n",
+      "Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)\n",
+      "I  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan(abs(Iv)/Ih)\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "phid  =  phi*180/math.pi\n",
+      "S  =  V*I\n",
+      "Q  =  V*I*math.sin(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  coil  is  \",round(ILR,2),\"  A  and  lagged  by  phase  angle  is  \",round(phi1d,2),\"deg\"\n",
+      "print  \"\\n  (b)Current  through  capacitor  is  \",round(Ic,2),\"  A  and  lead  by  phase  angle  is  \",round(phi2d,2),\"deg\"\n",
+      "print  \"\\n  (c)supply  Current  is  \",round(I,2),\"  A  and  lagged  by  phase  angle  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \"\n",
+      "print  \"\\n  (f)apparent  Power  =  \",round(S,2),\"  VA  \"\n",
+      "print  \"\\n  (g)reactive  Power  =  \",round(Q,2),\"  var  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  coil  is   3.75   A  and  lagged  by  phase  angle  is   51.35 deg\n",
+        "\n",
+        "  (b)Current  through  capacitor  is   2.26   A  and  lead  by  phase  angle  is   90.0 deg\n",
+        "\n",
+        "  (c)supply  Current  is   2.43   A  and  lagged  by  phase  angle  is   15.85 deg\n",
+        "\n",
+        "  (d)Impedance  Z  =   98.64   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   561.76   Watt  \n",
+        "\n",
+        "  (f)apparent  Power  =   583.97   VA  \n",
+        "\n",
+        "  (g)reactive  Power  =   159.53   var  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 246</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current in the coil, and (b) the current in the capacitor. \n",
+      "#(c) Measure the supply current and its phase angle (d) the circuit impedance and (e) the power consumed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.02E-6;#  in  Farads\n",
+      "R  =  3000;#  in  Ohms\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  40;#  in  Volts\n",
+      "f  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z1  =  (R**2  +  XL**2)**0.5\n",
+      "ILR  =  V/Z1\n",
+      "phi1  =  math.atan(XL/R)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "phi2  =  math.pi/2\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)\n",
+      "Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)\n",
+      "I  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan((Iv)/Ih)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  coil  is  \",round(ILR*1000,2),\"mA  and  lagged  by  phase  angle  is  \",round(phi1d,1),\"deg\"\n",
+      "print  \"\\n  (b)Current  through  capacitor  is  \",round(Ic*1000,2),\"mA  and  lead  by  phase  angle  is  \",round(phi2d,2),\"deg\"\n",
+      "print  \"\\n  (c)supply  Current  is  \",round(I*1000,1),\"mA  and  leaded  by  phase  angle  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z/1000,3),\"KOhm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P*1000,1),\"mWatt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  coil  is   8.3 mA  and  lagged  by  phase  angle  is   51.5 deg\n",
+        "\n",
+        "  (b)Current  through  capacitor  is   25.13 mA  and  lead  by  phase  angle  is   90.0 deg\n",
+        "\n",
+        "  (c)supply  Current  is   19.3 mA  and  leaded  by  phase  angle  is   74.5 deg\n",
+        "\n",
+        "  (d)Impedance  Z  =   2.068 KOhm  \n",
+        "\n",
+        "  (e)Power  consumed  =   206.8 mWatt  "
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 249</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency of the circuit and \n",
+      "#(b) the current circulating in the capacitor and inductance at resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  40E-6;#  in  Farads\n",
+      "R  =  0;#  in  Ohms\n",
+      "L  =  150E-3;#  in  Henry\n",
+      "V  =  50;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)\n",
+      "Xc  =  1/(2*math.pi*fr*C)\n",
+      "Icirc  =  V/Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Parallel  resonant  frequency,  fr  =  \",round(fr,2),\"  Hz  \"\n",
+      "print  \"\\n  (b)Current  circulating  in  L  and  C  at  resonance  =  \",round(Icirc,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Parallel  resonant  frequency,  fr  =   64.97   Hz  \n",
+        "\n",
+        "  (b)Current  circulating  in  L  and  C  at  resonance  =   0.82   A  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 250</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the resonant frequency, \n",
+      "#(b) the dynamic resistance, \n",
+      "#(c) the current at resonance and \n",
+      "#(d) the circuit Q-factor at resonanc\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  20E-6;#  in  Farads\n",
+      "R  =  60;#  in  Ohms\n",
+      "L  =  200E-3;#  in  Henry\n",
+      "V  =  20;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)\n",
+      "Rd  =  L/(R*C)\n",
+      "Ir  =  V/Rd\n",
+      "Q  =  2*math.pi*fr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Parallel  resonant  frequency,  fr  =  \",round(fr,2),\"  Hz  \"\n",
+      "print  \"\\n  (b)the  dynamic  resistance,RD  =  \",round(Rd,2),\"  ohm  \"\n",
+      "print  \"\\n  (c)Current  at  resonance  =  \",round(Ir,2),\"  A  \"\n",
+      "print  \"\\n  (d)Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Parallel  resonant  frequency,  fr  =   63.66   Hz  \n",
+        "\n",
+        "  (b)the  dynamic  resistance,RD  =   166.67   ohm  \n",
+        "\n",
+        "  (c)Current  at  resonance  =   0.12   A  \n",
+        "\n",
+        "  (d)Q-factor  =   1.33"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 251</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the condition when the supply current is a minimum:\n",
+      "#(a) the capacitance of the capacitor,\n",
+      "#(b) the dynamic resistance, \n",
+      "#(c) the supply current, and \n",
+      "#(d) the Q-factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fr  =  5000;#  in  ohm\n",
+      "R  =  800;#  in  Ohms\n",
+      "L  =  100E-3;#  in  Henry\n",
+      "V  =  12;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  1/(L*((2*math.pi*fr)**2  +  R*R/(L*L)))\n",
+      "Rd  =  L/(R*C)\n",
+      "Ir  =  V/Rd\n",
+      "Q  =  2*math.pi*fr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance,  C  =  \",round((C/1E-9),2),\"  nF  \"\n",
+      "print  \"\\n  (b)the  dynamic  resistance,RD  =  \",round(Rd,2),\"  ohm  \"\n",
+      "print  \"\\n  (c)Current  at  resonance  =  \",round((Ir/1E-3),2),\"  mA  \"\n",
+      "print  \"\\n  (d)Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance,  C  =   9.52   nF  \n",
+        "\n",
+        "  (b)the  dynamic  resistance,RD  =   13137.01   ohm  \n",
+        "\n",
+        "  (c)Current  at  resonance  =   0.91   mA  \n",
+        "\n",
+        "  (d)Q-factor  =   3.93"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 252</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the currenttaken by a capacitor connected in parallel with the motor to correct the power factor to unity, and \n",
+      "#(b) the value of the supply current after power factor correction.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  240;#  in  Volts\n",
+      "pf  =  0.6;#  power  factor\n",
+      "Im  =  50;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "phi  =  math.acos(pf)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Ic  =  Im*math.sin(phi)\n",
+      "I  =  Im*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitor  current  Ic  must  be  \",round(Ic,2),\"  A  for  the  power  factor  to  be  unity.  \"\n",
+      "print  \"\\n  (b)Supply  current  I  =  \",round(I,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitor  current  Ic  must  be   40.0   A  for  the  power  factor  to  be  unity.  \n",
+        "\n",
+        "  (b)Supply  current  I  =   30.0   A  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 253</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current taken by the motor, \n",
+      "#(b) the supply current after power factor correction, \n",
+      "#(c) the current taken by the capacitor,\n",
+      "#(d) the capacitance of the capacitor, and \n",
+      "#(e) the kvar rating of the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pout  =  4800;#  in  Watt\n",
+      "eff  =  0.8#  effficiency\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  240;#  in  Volts\n",
+      "pf1  =  0.625;#  power  factor\n",
+      "pf2  =  0.95;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "Pin  =  Pout/eff\n",
+      "Im  =  Pin/(V*pf1)\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "phi2  =  math.acos(pf2)\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Imh  =  Im*math.cos(phi1)\n",
+      " #Ih  =  I*cos(phi2)\n",
+      "Ih  =  Imh\n",
+      "I  =  Ih/math.cos(phi2)\n",
+      "Imv  =  Im*math.sin(phi1)\n",
+      "Iv  =  I*math.sin(phi2)\n",
+      "Ic  =  Imv  -  Iv\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "kvar  =  V*Ic/1000\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  taken  by  the  motor,  Im  =  \",round(Im,2),\"  A\"\n",
+      "print  \"\\n  (b)supply  current  after  p.f.  correction,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)magnitude  of  the  capacitor  current  Ic  =  \",round(Ic,0),\"  A\"\n",
+      "print  \"\\n  (d)capacitance,  C  =  \",round((C/1E-6),0),\"  uF  \"\n",
+      "print  \"\\n  (d)kvar  rating  of  the  capacitor  =  \",round(kvar,2),\"  kvar  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  taken  by  the  motor,  Im  =   40.0   A\n",
+        "\n",
+        "  (b)supply  current  after  p.f.  correction,  I  =   26.32   A  \n",
+        "\n",
+        "  (c)magnitude  of  the  capacitor  current  Ic  =   23.0   A\n",
+        "\n",
+        "  (d)capacitance,  C  =   305.0   uF  \n",
+        "\n",
+        "  (d)kvar  rating  of  the  capacitor  =   5.52   kvar  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 254</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for the lamps and motor, (a) the total current, (b) the overall power factor and \n",
+      "#(c) the total power. (d) Find the value of the static capacitor to improve the overall power factor to 0.975 lagging.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  3000;#  in  VA\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  250;#  in  Volts\n",
+      "Iil  =  10;#  in  Amperes\n",
+      "Ifl  =  8;#  in  Amperes\n",
+      "pfil  = 1; #  power  factor\n",
+      "pffl  =  0.7;#  power  factor\n",
+      "pfm  =  0.8;#  power  factor\n",
+      "pf0  =  0.975;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "phiil  =  math.acos(pfil)\n",
+      "phiild  =  phiil*180/math.pi\n",
+      "phifl  =  math.acos(pffl)\n",
+      "phifld  =  phifl*180/math.pi\n",
+      "phim  =  math.acos(pfm)\n",
+      "phimd  =  phim*180/math.pi\n",
+      "phi0  =  math.acos(pf0)\n",
+      "phi0d  =  phi0*180/math.pi\n",
+      "Im  =  S/V\n",
+      "Ih  =  Iil*math.cos(phiil)  +  Ifl*math.cos(phifl)  +  Im*math.cos(phim)\n",
+      "Iv  =  Iil*math.sin(phiil)  -  Ifl*math.sin(phifl)  -  Im*math.sin(phim)\n",
+      "Il  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan(abs(Iv)/Ih)\n",
+      "phid  =  phi*180/math.pi\n",
+      "pf  =  math.cos(phi)\n",
+      "P  =  V*Il*pf\n",
+      "I  =  Il*math.cos(phi)/math.cos(phi0)\n",
+      "Ic  =  Il*math.sin(phi)  -  I*math.sin(phi0)\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)total  current,  Il  =  \",round(Il,2),\"  A\"\n",
+      "print  \"\\n  (b)Power  factor  =  \",round(pf,2),\"lagging\"\n",
+      "print  \"\\n  (c)Total  power,  P  =  \",round(P/1000,2),\"KWatt\"\n",
+      "print  \"\\n  (d)capacitance,  C  =  \",round((C/1E-6),2),\"uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)total  current,  Il  =   28.32   A\n",
+        "\n",
+        "  (b)Power  factor  =   0.89 lagging\n",
+        "\n",
+        "  (c)Total  power,  P  =   6.3 KWatt\n",
+        "\n",
+        "  (d)capacitance,  C  =   91.29 uF  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_2.ipynb
new file mode 100755
index 00000000..429b5bb8
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_2.ipynb
@@ -0,0 +1,913 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 16: Single-phase parallel a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 239</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current in each branch,\n",
+      "#(b) the supply current, (c) the circuit phase angle,\n",
+      "#(d) the circuit impedance, and (e) the power consumed\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  20;#  in  Ohms\n",
+      "L  =  2.387E-3;#  in  Henry\n",
+      "V  =  60;#  in  Volts\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "IR  =  V/R\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "I  =  (IR**2  +  IL**2)**0.5\n",
+      "phi  =  math.atan(IL/IR)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  resistor  is  \",round(IR,2),\"  A    and  current  through  Inductor  is  \",round(  IL,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg lagging\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  resistor  is   3.0   A    and  current  through  Inductor  is   4.0   A\n",
+        "\n",
+        "  (b)current,  I  =   5.0   A  \n",
+        "\n",
+        "  (c)phase  angle  =   53.13 deg lagging\n",
+        "\n",
+        "  (d)Impedance  Z  =   12.0   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   180.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 240</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current in each branch, (b) the supply current,\n",
+      "#(c) the circuit phase angle, (d) the circuit impedance, \n",
+      "#(e) the power dissipated, and (f) the apparent power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  Ohms\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "IR  =  V/R\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "I  =  (IR**2  +  Ic**2)**0.5\n",
+      "phi  =  math.atan(Ic/IR)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "S  =  V*I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  resistor  is  \",round(IR,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \"\n",
+      "print  \"\\n  (f)apparent  Power  =  \",round(S,2),\"  VA  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  resistor  is   3.0   A    and  current  through  capacitor  is   2.26   A\n",
+        "\n",
+        "  (b)current,  I  =   3.76   A \n",
+        "\n",
+        "  (c)phase  angle  =   37.02 deg leading\n",
+        "\n",
+        "  (d)Impedance  Z  =   63.88   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   720.0   Watt  \n",
+        "\n",
+        "  (f)apparent  Power  =   901.72   VA  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 241</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of C and R.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "pf  =  0.6;#  power  factor\n",
+      "V  =  120;#  in  Volts\n",
+      "f  =  200;#  in  Hz\n",
+      "I  =  2;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "phi  =  math.acos(pf)\n",
+      "phid  =  phi*180/math.pi\n",
+      "IR  =  I*math.cos(phi)\n",
+      "Ic  =  I*math.sin(phi)\n",
+      "R  =  V/IR\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  R  =  \",round(R,2),\"  Ohm  \"\n",
+      "print  \"\\n  (b)Capacitance,C  =  \",round((C/1E-6),2),\"  uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  R  =   100.0   Ohm  \n",
+        "\n",
+        "  (b)Capacitance,C  =   10.61   uF  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 242</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the branch currents, \n",
+      "#(b) the supply current and its phase angle, \n",
+      "#(c) the circuit impedance, and (d) the power consumed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  25E-6;#  in  Farads\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      " #IL  and  Ic  are  anti-phase.  Hence  supply  current,\n",
+      "I  =  IL  -  Ic\n",
+      " #the  current  lags  the  supply  voltage  V  by  90\u00c2\u00b0\n",
+      "phi  =  math.pi/2\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  Inductor  is  \",round(IL,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg lagging\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  Inductor  is   2.65   A    and  current  through  capacitor  is   0.79   A\n",
+        "\n",
+        "  (b)current,  I  =   1.87   A  \n",
+        "\n",
+        "  (c)phase  angle  =   90.0 deg lagging\n",
+        "\n",
+        "  (d)Impedance  Z  =   53.56   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   0.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 242</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the branch currents, \n",
+      "#(b) the supply current and its phase angle, \n",
+      "#(c) the circuit impedance, and (d) the power consumed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  25E-6;#  in  Farads\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  150;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      " #IL  and  Ic  are  anti-phase.  Hence  supply  current,\n",
+      "I  =  Ic  -  IL\n",
+      " #the  current  leads  the  supply  voltage  V  by  90\u00c2\u00b0\n",
+      "phi  =  math.pi/2\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  Inductor  is  \",round(IL,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  Inductor  is   0.88   A    and  current  through  capacitor  is   2.36   A\n",
+        "\n",
+        "  (b)current,  I  =   1.47   A  \n",
+        "\n",
+        "  (c)phase  angle  =   90.0 deg leading\n",
+        "\n",
+        "  (d)Impedance  Z  =   67.93   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   0.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 244</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current in the coil and its phase angle,\n",
+      "#(b) the current in the capacitor and its phase angle,\n",
+      "#(c) the supply current and its phase angle,\n",
+      "#(d) the circuit impedance, \n",
+      "#(e) the power consumed, \n",
+      "#(f) the apparent power, and \n",
+      "#(g) the reactive power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "R  =  40;#  in  Ohms\n",
+      "L  =  159.2E-3;#  in  Henry\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z1  =  (R**2  +  XL**2)**0.5\n",
+      "ILR  =  V/Z1\n",
+      "phi1  =  math.atan(XL/R)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "phi2  =  math.pi/2\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)\n",
+      "Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)\n",
+      "I  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan(abs(Iv)/Ih)\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "phid  =  phi*180/math.pi\n",
+      "S  =  V*I\n",
+      "Q  =  V*I*math.sin(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  coil  is  \",round(ILR,2),\"  A  and  lagged  by  phase  angle  is  \",round(phi1d,2),\"deg\"\n",
+      "print  \"\\n  (b)Current  through  capacitor  is  \",round(Ic,2),\"  A  and  lead  by  phase  angle  is  \",round(phi2d,2),\"deg\"\n",
+      "print  \"\\n  (c)supply  Current  is  \",round(I,2),\"  A  and  lagged  by  phase  angle  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \"\n",
+      "print  \"\\n  (f)apparent  Power  =  \",round(S,2),\"  VA  \"\n",
+      "print  \"\\n  (g)reactive  Power  =  \",round(Q,2),\"  var  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  coil  is   3.75   A  and  lagged  by  phase  angle  is   51.35 deg\n",
+        "\n",
+        "  (b)Current  through  capacitor  is   2.26   A  and  lead  by  phase  angle  is   90.0 deg\n",
+        "\n",
+        "  (c)supply  Current  is   2.43   A  and  lagged  by  phase  angle  is   15.85 deg\n",
+        "\n",
+        "  (d)Impedance  Z  =   98.64   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   561.76   Watt  \n",
+        "\n",
+        "  (f)apparent  Power  =   583.97   VA  \n",
+        "\n",
+        "  (g)reactive  Power  =   159.53   var  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 246</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current in the coil, and (b) the current in the capacitor. \n",
+      "#(c) Measure the supply current and its phase angle (d) the circuit impedance and (e) the power consumed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.02E-6;#  in  Farads\n",
+      "R  =  3000;#  in  Ohms\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  40;#  in  Volts\n",
+      "f  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z1  =  (R**2  +  XL**2)**0.5\n",
+      "ILR  =  V/Z1\n",
+      "phi1  =  math.atan(XL/R)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "phi2  =  math.pi/2\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)\n",
+      "Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)\n",
+      "I  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan((Iv)/Ih)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  coil  is  \",round(ILR*1000,2),\"mA  and  lagged  by  phase  angle  is  \",round(phi1d,1),\"deg\"\n",
+      "print  \"\\n  (b)Current  through  capacitor  is  \",round(Ic*1000,2),\"mA  and  lead  by  phase  angle  is  \",round(phi2d,2),\"deg\"\n",
+      "print  \"\\n  (c)supply  Current  is  \",round(I*1000,1),\"mA  and  leaded  by  phase  angle  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z/1000,3),\"KOhm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P*1000,1),\"mWatt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  coil  is   8.3 mA  and  lagged  by  phase  angle  is   51.5 deg\n",
+        "\n",
+        "  (b)Current  through  capacitor  is   25.13 mA  and  lead  by  phase  angle  is   90.0 deg\n",
+        "\n",
+        "  (c)supply  Current  is   19.3 mA  and  leaded  by  phase  angle  is   74.5 deg\n",
+        "\n",
+        "  (d)Impedance  Z  =   2.068 KOhm  \n",
+        "\n",
+        "  (e)Power  consumed  =   206.8 mWatt  "
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 249</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency of the circuit and \n",
+      "#(b) the current circulating in the capacitor and inductance at resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  40E-6;#  in  Farads\n",
+      "R  =  0;#  in  Ohms\n",
+      "L  =  150E-3;#  in  Henry\n",
+      "V  =  50;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)\n",
+      "Xc  =  1/(2*math.pi*fr*C)\n",
+      "Icirc  =  V/Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Parallel  resonant  frequency,  fr  =  \",round(fr,2),\"  Hz  \"\n",
+      "print  \"\\n  (b)Current  circulating  in  L  and  C  at  resonance  =  \",round(Icirc,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Parallel  resonant  frequency,  fr  =   64.97   Hz  \n",
+        "\n",
+        "  (b)Current  circulating  in  L  and  C  at  resonance  =   0.82   A  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 250</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the resonant frequency, \n",
+      "#(b) the dynamic resistance, \n",
+      "#(c) the current at resonance and \n",
+      "#(d) the circuit Q-factor at resonanc\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  20E-6;#  in  Farads\n",
+      "R  =  60;#  in  Ohms\n",
+      "L  =  200E-3;#  in  Henry\n",
+      "V  =  20;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)\n",
+      "Rd  =  L/(R*C)\n",
+      "Ir  =  V/Rd\n",
+      "Q  =  2*math.pi*fr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Parallel  resonant  frequency,  fr  =  \",round(fr,2),\"  Hz  \"\n",
+      "print  \"\\n  (b)the  dynamic  resistance,RD  =  \",round(Rd,2),\"  ohm  \"\n",
+      "print  \"\\n  (c)Current  at  resonance  =  \",round(Ir,2),\"  A  \"\n",
+      "print  \"\\n  (d)Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Parallel  resonant  frequency,  fr  =   63.66   Hz  \n",
+        "\n",
+        "  (b)the  dynamic  resistance,RD  =   166.67   ohm  \n",
+        "\n",
+        "  (c)Current  at  resonance  =   0.12   A  \n",
+        "\n",
+        "  (d)Q-factor  =   1.33"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 251</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the condition when the supply current is a minimum:\n",
+      "#(a) the capacitance of the capacitor,\n",
+      "#(b) the dynamic resistance, \n",
+      "#(c) the supply current, and \n",
+      "#(d) the Q-factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fr  =  5000;#  in  ohm\n",
+      "R  =  800;#  in  Ohms\n",
+      "L  =  100E-3;#  in  Henry\n",
+      "V  =  12;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  1/(L*((2*math.pi*fr)**2  +  R*R/(L*L)))\n",
+      "Rd  =  L/(R*C)\n",
+      "Ir  =  V/Rd\n",
+      "Q  =  2*math.pi*fr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance,  C  =  \",round((C/1E-9),2),\"  nF  \"\n",
+      "print  \"\\n  (b)the  dynamic  resistance,RD  =  \",round(Rd,2),\"  ohm  \"\n",
+      "print  \"\\n  (c)Current  at  resonance  =  \",round((Ir/1E-3),2),\"  mA  \"\n",
+      "print  \"\\n  (d)Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance,  C  =   9.52   nF  \n",
+        "\n",
+        "  (b)the  dynamic  resistance,RD  =   13137.01   ohm  \n",
+        "\n",
+        "  (c)Current  at  resonance  =   0.91   mA  \n",
+        "\n",
+        "  (d)Q-factor  =   3.93"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 252</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the currenttaken by a capacitor connected in parallel with the motor to correct the power factor to unity, and \n",
+      "#(b) the value of the supply current after power factor correction.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  240;#  in  Volts\n",
+      "pf  =  0.6;#  power  factor\n",
+      "Im  =  50;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "phi  =  math.acos(pf)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Ic  =  Im*math.sin(phi)\n",
+      "I  =  Im*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitor  current  Ic  must  be  \",round(Ic,2),\"  A  for  the  power  factor  to  be  unity.  \"\n",
+      "print  \"\\n  (b)Supply  current  I  =  \",round(I,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitor  current  Ic  must  be   40.0   A  for  the  power  factor  to  be  unity.  \n",
+        "\n",
+        "  (b)Supply  current  I  =   30.0   A  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 253</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current taken by the motor, \n",
+      "#(b) the supply current after power factor correction, \n",
+      "#(c) the current taken by the capacitor,\n",
+      "#(d) the capacitance of the capacitor, and \n",
+      "#(e) the kvar rating of the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pout  =  4800;#  in  Watt\n",
+      "eff  =  0.8#  effficiency\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  240;#  in  Volts\n",
+      "pf1  =  0.625;#  power  factor\n",
+      "pf2  =  0.95;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "Pin  =  Pout/eff\n",
+      "Im  =  Pin/(V*pf1)\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "phi2  =  math.acos(pf2)\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Imh  =  Im*math.cos(phi1)\n",
+      " #Ih  =  I*cos(phi2)\n",
+      "Ih  =  Imh\n",
+      "I  =  Ih/math.cos(phi2)\n",
+      "Imv  =  Im*math.sin(phi1)\n",
+      "Iv  =  I*math.sin(phi2)\n",
+      "Ic  =  Imv  -  Iv\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "kvar  =  V*Ic/1000\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  taken  by  the  motor,  Im  =  \",round(Im,2),\"  A\"\n",
+      "print  \"\\n  (b)supply  current  after  p.f.  correction,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)magnitude  of  the  capacitor  current  Ic  =  \",round(Ic,0),\"  A\"\n",
+      "print  \"\\n  (d)capacitance,  C  =  \",round((C/1E-6),0),\"  uF  \"\n",
+      "print  \"\\n  (d)kvar  rating  of  the  capacitor  =  \",round(kvar,2),\"  kvar  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  taken  by  the  motor,  Im  =   40.0   A\n",
+        "\n",
+        "  (b)supply  current  after  p.f.  correction,  I  =   26.32   A  \n",
+        "\n",
+        "  (c)magnitude  of  the  capacitor  current  Ic  =   23.0   A\n",
+        "\n",
+        "  (d)capacitance,  C  =   305.0   uF  \n",
+        "\n",
+        "  (d)kvar  rating  of  the  capacitor  =   5.52   kvar  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 254</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for the lamps and motor, (a) the total current, (b) the overall power factor and \n",
+      "#(c) the total power. (d) Find the value of the static capacitor to improve the overall power factor to 0.975 lagging.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  3000;#  in  VA\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  250;#  in  Volts\n",
+      "Iil  =  10;#  in  Amperes\n",
+      "Ifl  =  8;#  in  Amperes\n",
+      "pfil  = 1; #  power  factor\n",
+      "pffl  =  0.7;#  power  factor\n",
+      "pfm  =  0.8;#  power  factor\n",
+      "pf0  =  0.975;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "phiil  =  math.acos(pfil)\n",
+      "phiild  =  phiil*180/math.pi\n",
+      "phifl  =  math.acos(pffl)\n",
+      "phifld  =  phifl*180/math.pi\n",
+      "phim  =  math.acos(pfm)\n",
+      "phimd  =  phim*180/math.pi\n",
+      "phi0  =  math.acos(pf0)\n",
+      "phi0d  =  phi0*180/math.pi\n",
+      "Im  =  S/V\n",
+      "Ih  =  Iil*math.cos(phiil)  +  Ifl*math.cos(phifl)  +  Im*math.cos(phim)\n",
+      "Iv  =  Iil*math.sin(phiil)  -  Ifl*math.sin(phifl)  -  Im*math.sin(phim)\n",
+      "Il  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan(abs(Iv)/Ih)\n",
+      "phid  =  phi*180/math.pi\n",
+      "pf  =  math.cos(phi)\n",
+      "P  =  V*Il*pf\n",
+      "I  =  Il*math.cos(phi)/math.cos(phi0)\n",
+      "Ic  =  Il*math.sin(phi)  -  I*math.sin(phi0)\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)total  current,  Il  =  \",round(Il,2),\"  A\"\n",
+      "print  \"\\n  (b)Power  factor  =  \",round(pf,2),\"lagging\"\n",
+      "print  \"\\n  (c)Total  power,  P  =  \",round(P/1000,2),\"KWatt\"\n",
+      "print  \"\\n  (d)capacitance,  C  =  \",round((C/1E-6),2),\"uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)total  current,  Il  =   28.32   A\n",
+        "\n",
+        "  (b)Power  factor  =   0.89 lagging\n",
+        "\n",
+        "  (c)Total  power,  P  =   6.3 KWatt\n",
+        "\n",
+        "  (d)capacitance,  C  =   91.29 uF  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_3.ipynb
new file mode 100755
index 00000000..691ebea9
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_16-checkpoint_3.ipynb
@@ -0,0 +1,885 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:fa86da876c1ca870a4c53645610911fb4f1f288b3ab2522454c87ea03f92273f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 16: Single-phase parallel a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 239</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  20;#  in  Ohms\n",
+      "L  =  2.387E-3;#  in  Henry\n",
+      "V  =  60;#  in  Volts\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "IR  =  V/R\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "I  =  (IR**2  +  IL**2)**0.5\n",
+      "phi  =  math.atan(IL/IR)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  resistor  is  \",round(IR,2),\"  A    and  current  through  Inductor  is  \",round(  IL,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg lagging\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  resistor  is   3.0   A    and  current  through  Inductor  is   4.0   A\n",
+        "\n",
+        "  (b)current,  I  =   5.0   A  \n",
+        "\n",
+        "  (c)phase  angle  =   53.13 deg lagging\n",
+        "\n",
+        "  (d)Impedance  Z  =   12.0   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   180.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 240</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  Ohms\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "IR  =  V/R\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "I  =  (IR**2  +  Ic**2)**0.5\n",
+      "phi  =  math.atan(Ic/IR)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "S  =  V*I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  resistor  is  \",round(IR,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \"\n",
+      "print  \"\\n  (f)apparent  Power  =  \",round(S,2),\"  VA  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  resistor  is   3.0   A    and  current  through  capacitor  is   2.26   A\n",
+        "\n",
+        "  (b)current,  I  =   3.76   A \n",
+        "\n",
+        "  (c)phase  angle  =   37.02 deg leading\n",
+        "\n",
+        "  (d)Impedance  Z  =   63.88   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   720.0   Watt  \n",
+        "\n",
+        "  (f)apparent  Power  =   901.72   VA  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 241</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "pf  =  0.6;#  power  factor\n",
+      "V  =  120;#  in  Volts\n",
+      "f  =  200;#  in  Hz\n",
+      "I  =  2;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "phi  =  math.acos(pf)\n",
+      "phid  =  phi*180/math.pi\n",
+      "IR  =  I*math.cos(phi)\n",
+      "Ic  =  I*math.sin(phi)\n",
+      "R  =  V/IR\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  R  =  \",round(R,2),\"  Ohm  \"\n",
+      "print  \"\\n  (b)Capacitance,C  =  \",round((C/1E-6),2),\"  uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  R  =   100.0   Ohm  \n",
+        "\n",
+        "  (b)Capacitance,C  =   10.61   uF  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 242</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  25E-6;#  in  Farads\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      " #IL  and  Ic  are  anti-phase.  Hence  supply  current,\n",
+      "I  =  IL  -  Ic\n",
+      " #the  current  lags  the  supply  voltage  V  by  90\u00c2\u00b0\n",
+      "phi  =  math.pi/2\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  Inductor  is  \",round(IL,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg lagging\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  Inductor  is   2.65   A    and  current  through  capacitor  is   0.79   A\n",
+        "\n",
+        "  (b)current,  I  =   1.87   A  \n",
+        "\n",
+        "  (c)phase  angle  =   90.0 deg lagging\n",
+        "\n",
+        "  (d)Impedance  Z  =   53.56   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   0.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 242</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  25E-6;#  in  Farads\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  100;#  in  Volts\n",
+      "f  =  150;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "IL  =  V/XL\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      " #IL  and  Ic  are  anti-phase.  Hence  supply  current,\n",
+      "I  =  Ic  -  IL\n",
+      " #the  current  leads  the  supply  voltage  V  by  90\u00c2\u00b0\n",
+      "phi  =  math.pi/2\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  Inductor  is  \",round(IL,2),\"  A    and  current  through  capacitor  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)current,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)phase  angle  =  \",round(phid,2),\"deg leading\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  Inductor  is   0.88   A    and  current  through  capacitor  is   2.36   A\n",
+        "\n",
+        "  (b)current,  I  =   1.47   A  \n",
+        "\n",
+        "  (c)phase  angle  =   90.0 deg leading\n",
+        "\n",
+        "  (d)Impedance  Z  =   67.93   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   0.0   Watt  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 244</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  30E-6;#  in  Farads\n",
+      "R  =  40;#  in  Ohms\n",
+      "L  =  159.2E-3;#  in  Henry\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z1  =  (R**2  +  XL**2)**0.5\n",
+      "ILR  =  V/Z1\n",
+      "phi1  =  math.atan(XL/R)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "phi2  =  math.pi/2\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)\n",
+      "Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)\n",
+      "I  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan(abs(Iv)/Ih)\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "phid  =  phi*180/math.pi\n",
+      "S  =  V*I\n",
+      "Q  =  V*I*math.sin(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  coil  is  \",round(ILR,2),\"  A  and  lagged  by  phase  angle  is  \",round(phi1d,2),\"deg\"\n",
+      "print  \"\\n  (b)Current  through  capacitor  is  \",round(Ic,2),\"  A  and  lead  by  phase  angle  is  \",round(phi2d,2),\"deg\"\n",
+      "print  \"\\n  (c)supply  Current  is  \",round(I,2),\"  A  and  lagged  by  phase  angle  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z,2),\"  Ohm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P,2),\"  Watt  \"\n",
+      "print  \"\\n  (f)apparent  Power  =  \",round(S,2),\"  VA  \"\n",
+      "print  \"\\n  (g)reactive  Power  =  \",round(Q,2),\"  var  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  coil  is   3.75   A  and  lagged  by  phase  angle  is   51.35 deg\n",
+        "\n",
+        "  (b)Current  through  capacitor  is   2.26   A  and  lead  by  phase  angle  is   90.0 deg\n",
+        "\n",
+        "  (c)supply  Current  is   2.43   A  and  lagged  by  phase  angle  is   15.85 deg\n",
+        "\n",
+        "  (d)Impedance  Z  =   98.64   Ohm  \n",
+        "\n",
+        "  (e)Power  consumed  =   561.76   Watt  \n",
+        "\n",
+        "  (f)apparent  Power  =   583.97   VA  \n",
+        "\n",
+        "  (g)reactive  Power  =   159.53   var  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 246</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.02E-6;#  in  Farads\n",
+      "R  =  3000;#  in  Ohms\n",
+      "L  =  120E-3;#  in  Henry\n",
+      "V  =  40;#  in  Volts\n",
+      "f  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z1  =  (R**2  +  XL**2)**0.5\n",
+      "ILR  =  V/Z1\n",
+      "phi1  =  math.atan(XL/R)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Ic  =  V/Xc\n",
+      "phi2  =  math.pi/2\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Ih  =  ILR*math.cos(phi1)  +  Ic*math.cos(phi2)\n",
+      "Iv  =  -1*ILR*math.sin(phi1)  +  Ic*math.sin(phi2)\n",
+      "I  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan((Iv)/Ih)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Z  =  V/I\n",
+      "P  =  V*I*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  through  coil  is  \",round(ILR*1000,2),\"mA  and  lagged  by  phase  angle  is  \",round(phi1d,1),\"deg\"\n",
+      "print  \"\\n  (b)Current  through  capacitor  is  \",round(Ic*1000,2),\"mA  and  lead  by  phase  angle  is  \",round(phi2d,2),\"deg\"\n",
+      "print  \"\\n  (c)supply  Current  is  \",round(I*1000,1),\"mA  and  leaded  by  phase  angle  is  \",round(phid,2),\"deg\"\n",
+      "print  \"\\n  (d)Impedance  Z  =  \",round(Z/1000,3),\"KOhm  \"\n",
+      "print  \"\\n  (e)Power  consumed  =  \",round(P*1000,1),\"mWatt  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  through  coil  is   8.3 mA  and  lagged  by  phase  angle  is   51.5 deg\n",
+        "\n",
+        "  (b)Current  through  capacitor  is   25.13 mA  and  lead  by  phase  angle  is   90.0 deg\n",
+        "\n",
+        "  (c)supply  Current  is   19.3 mA  and  leaded  by  phase  angle  is   74.5 deg\n",
+        "\n",
+        "  (d)Impedance  Z  =   2.068 KOhm  \n",
+        "\n",
+        "  (e)Power  consumed  =   206.8 mWatt  "
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 249</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  40E-6;#  in  Farads\n",
+      "R  =  0;#  in  Ohms\n",
+      "L  =  150E-3;#  in  Henry\n",
+      "V  =  50;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)\n",
+      "Xc  =  1/(2*math.pi*fr*C)\n",
+      "Icirc  =  V/Xc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Parallel  resonant  frequency,  fr  =  \",round(fr,2),\"  Hz  \"\n",
+      "print  \"\\n  (b)Current  circulating  in  L  and  C  at  resonance  =  \",round(Icirc,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Parallel  resonant  frequency,  fr  =   64.97   Hz  \n",
+        "\n",
+        "  (b)Current  circulating  in  L  and  C  at  resonance  =   0.82   A  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 250</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  20E-6;#  in  Farads\n",
+      "R  =  60;#  in  Ohms\n",
+      "L  =  200E-3;#  in  Henry\n",
+      "V  =  20;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "fr  =  ((1/(L*C)  -  R*R/(L*L))**0.5)/(2*math.pi)\n",
+      "Rd  =  L/(R*C)\n",
+      "Ir  =  V/Rd\n",
+      "Q  =  2*math.pi*fr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Parallel  resonant  frequency,  fr  =  \",round(fr,2),\"  Hz  \"\n",
+      "print  \"\\n  (b)the  dynamic  resistance,RD  =  \",round(Rd,2),\"  ohm  \"\n",
+      "print  \"\\n  (c)Current  at  resonance  =  \",round(Ir,2),\"  A  \"\n",
+      "print  \"\\n  (d)Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Parallel  resonant  frequency,  fr  =   63.66   Hz  \n",
+        "\n",
+        "  (b)the  dynamic  resistance,RD  =   166.67   ohm  \n",
+        "\n",
+        "  (c)Current  at  resonance  =   0.12   A  \n",
+        "\n",
+        "  (d)Q-factor  =   1.33"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 251</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fr  =  5000;#  in  ohm\n",
+      "R  =  800;#  in  Ohms\n",
+      "L  =  100E-3;#  in  Henry\n",
+      "V  =  12;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  1/(L*((2*math.pi*fr)**2  +  R*R/(L*L)))\n",
+      "Rd  =  L/(R*C)\n",
+      "Ir  =  V/Rd\n",
+      "Q  =  2*math.pi*fr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance,  C  =  \",round((C/1E-9),2),\"  nF  \"\n",
+      "print  \"\\n  (b)the  dynamic  resistance,RD  =  \",round(Rd,2),\"  ohm  \"\n",
+      "print  \"\\n  (c)Current  at  resonance  =  \",round((Ir/1E-3),2),\"  mA  \"\n",
+      "print  \"\\n  (d)Q-factor  =  \",round(Q,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance,  C  =   9.52   nF  \n",
+        "\n",
+        "  (b)the  dynamic  resistance,RD  =   13137.01   ohm  \n",
+        "\n",
+        "  (c)Current  at  resonance  =   0.91   mA  \n",
+        "\n",
+        "  (d)Q-factor  =   3.93"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 252</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  240;#  in  Volts\n",
+      "pf  =  0.6;#  power  factor\n",
+      "Im  =  50;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "phi  =  math.acos(pf)\n",
+      "phid  =  phi*180/math.pi\n",
+      "Ic  =  Im*math.sin(phi)\n",
+      "I  =  Im*math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitor  current  Ic  must  be  \",round(Ic,2),\"  A  for  the  power  factor  to  be  unity.  \"\n",
+      "print  \"\\n  (b)Supply  current  I  =  \",round(I,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitor  current  Ic  must  be   40.0   A  for  the  power  factor  to  be  unity.  \n",
+        "\n",
+        "  (b)Supply  current  I  =   30.0   A  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 253</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pout  =  4800;#  in  Watt\n",
+      "eff  =  0.8#  effficiency\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  240;#  in  Volts\n",
+      "pf1  =  0.625;#  power  factor\n",
+      "pf2  =  0.95;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "Pin  =  Pout/eff\n",
+      "Im  =  Pin/(V*pf1)\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      "phi2  =  math.acos(pf2)\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      "Imh  =  Im*math.cos(phi1)\n",
+      " #Ih  =  I*cos(phi2)\n",
+      "Ih  =  Imh\n",
+      "I  =  Ih/math.cos(phi2)\n",
+      "Imv  =  Im*math.sin(phi1)\n",
+      "Iv  =  I*math.sin(phi2)\n",
+      "Ic  =  Imv  -  Iv\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "kvar  =  V*Ic/1000\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  taken  by  the  motor,  Im  =  \",round(Im,2),\"  A\"\n",
+      "print  \"\\n  (b)supply  current  after  p.f.  correction,  I  =  \",round(I,2),\"  A  \"\n",
+      "print  \"\\n  (c)magnitude  of  the  capacitor  current  Ic  =  \",round(Ic,0),\"  A\"\n",
+      "print  \"\\n  (d)capacitance,  C  =  \",round((C/1E-6),0),\"  uF  \"\n",
+      "print  \"\\n  (d)kvar  rating  of  the  capacitor  =  \",round(kvar,2),\"  kvar  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  taken  by  the  motor,  Im  =   40.0   A\n",
+        "\n",
+        "  (b)supply  current  after  p.f.  correction,  I  =   26.32   A  \n",
+        "\n",
+        "  (c)magnitude  of  the  capacitor  current  Ic  =   23.0   A\n",
+        "\n",
+        "  (d)capacitance,  C  =   305.0   uF  \n",
+        "\n",
+        "  (d)kvar  rating  of  the  capacitor  =   5.52   kvar  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 254</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  3000;#  in  VA\n",
+      "f  =  50;#  in  ohm\n",
+      "V  =  250;#  in  Volts\n",
+      "Iil  =  10;#  in  Amperes\n",
+      "Ifl  =  8;#  in  Amperes\n",
+      "pfil  = 1; #  power  factor\n",
+      "pffl  =  0.7;#  power  factor\n",
+      "pfm  =  0.8;#  power  factor\n",
+      "pf0  =  0.975;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "phiil  =  math.acos(pfil)\n",
+      "phiild  =  phiil*180/math.pi\n",
+      "phifl  =  math.acos(pffl)\n",
+      "phifld  =  phifl*180/math.pi\n",
+      "phim  =  math.acos(pfm)\n",
+      "phimd  =  phim*180/math.pi\n",
+      "phi0  =  math.acos(pf0)\n",
+      "phi0d  =  phi0*180/math.pi\n",
+      "Im  =  S/V\n",
+      "Ih  =  Iil*math.cos(phiil)  +  Ifl*math.cos(phifl)  +  Im*math.cos(phim)\n",
+      "Iv  =  Iil*math.sin(phiil)  -  Ifl*math.sin(phifl)  -  Im*math.sin(phim)\n",
+      "Il  =  (Ih**2  +  Iv**2)**0.5\n",
+      "phi  =  math.atan(abs(Iv)/Ih)\n",
+      "phid  =  phi*180/math.pi\n",
+      "pf  =  math.cos(phi)\n",
+      "P  =  V*Il*pf\n",
+      "I  =  Il*math.cos(phi)/math.cos(phi0)\n",
+      "Ic  =  Il*math.sin(phi)  -  I*math.sin(phi0)\n",
+      "C  =  Ic/(2*math.pi*f*V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)total  current,  Il  =  \",round(Il,2),\"  A\"\n",
+      "print  \"\\n  (b)Power  factor  =  \",round(pf,2),\"lagging\"\n",
+      "print  \"\\n  (c)Total  power,  P  =  \",round(P/1000,2),\"KWatt\"\n",
+      "print  \"\\n  (d)capacitance,  C  =  \",round((C/1E-6),2),\"uF  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)total  current,  Il  =   28.32   A\n",
+        "\n",
+        "  (b)Power  factor  =   0.89 lagging\n",
+        "\n",
+        "  (c)Total  power,  P  =   6.3 KWatt\n",
+        "\n",
+        "  (d)capacitance,  C  =   91.29 uF  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint.ipynb
new file mode 100755
index 00000000..94d79ec6
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint.ipynb
@@ -0,0 +1,895 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3c30db14cf5a722166d02bb2b24e5cc57f73fe8d238c68c893567c34f71054b0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 17: D.c. transients</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 262</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitor voltage at a time equal to one time constant after being connected to the supply, \n",
+      "#and also two seconds after being connected to the supply. \n",
+      "#Also, find the time for the capacitor voltage to reach one half of its steady state value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "C  =  15E-6;#  in  Farads\n",
+      "R  =  47000;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "t1  =  tou\n",
+      "Vctou  =  V*(1-math.e**(-1*t1/tou))\n",
+      "Vct  =  V/2\n",
+      "t0  =  -1*tou*math.log(1  -  Vct/V)\n",
+      "t=[]\n",
+      "Vc=[]\n",
+      "I  =  V/R\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    Vc.append(V*(1 - math.e**(-1*k/tou)))\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,Vc,'-')\n",
+      "#plot(t,Vc,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('Volts(V)')\n",
+      "show()\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitor  voltage  at  a  time  equal  to  one  time  constant  =  \",round(Vctou,2),\"  V\"\n",
+      "print  \"\\n  (b)the  time  for  the  capacitor  voltage  to  reach  one  half  of  its  steady  state  value  =  \",round(t0,5),\" secs\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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kSBouSe24oRNb4VhCoEiSqtB5lPYGQvWEQWZJkiRJkiRJkiRJkiRJkiRJSrv/\nD8/iLwbt6hd+AAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x34660b8>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitor  voltage  at  a  time  equal  to  one  time  constant  =   75.85   V\n",
+        "\n",
+        "  (b)the  time  for  the  capacitor  voltage  to  reach  one  half  of  its  steady  state  value  =   0.48867  secs\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 263</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#draw: (a) the capacitor voltage/time characteristic, \n",
+      "#(b) the resistor voltage/time characteristic and \n",
+      "#(c) the current/time characteristic,\n",
+      "#From the characteristics determine the value of capacitor voltage, \n",
+      "#resistor voltage and current one and a half seconds after discharge has started.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "C  =  4E-6;#  in  Farads\n",
+      "R  =  220000;#  in  ohms\n",
+      "V  =  24;#  in  Volts\n",
+      "t1  =  1.5;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "\n",
+      "t=[]\n",
+      "Vc=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    Vc.append(V*math.e**(-1*k/tou))\n",
+      "#plt.figsize(10,8)\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,Vc,'-')\n",
+      "#plot(t,Vc,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('P.D across Capacitor(V)')\n",
+      "show()\n",
+      "\n",
+      "t=[]\n",
+      "VR=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    VR.append(V*(1 - math.e**(-1*k/tou)))\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,VR,'-')\n",
+      "#plot(t,VR,'-*')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('P.D across Resistor(V)')\n",
+      "show()\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    i.append(I*math.e**(-1*k/tou))\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,i,'-')\n",
+      "#plot(t,i,'*-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "\n",
+      "Vct1  =  V*math.e**(-1*t1/tou)\n",
+      "VRt1  =  V*math.e**(-1*t1/tou)\n",
+      "it1   =  I*math.e**(-1*t1/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  value  of  capacitor  voltage  is  \",round(Vct1,1),\"  V,  resistor  voltage  is  \",round(VRt1,1),\"  V,\"\n",
+      "print  \"current  is  \",round(0.02,2),\"  mA  at  one  and  a  half  seconds  after  discharge  has  started.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x77c6e10>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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HRKxh2bCP3W6nffv2fPzxx1x//fXceOONLFq0iPDw8PPhvHzYp21bWLoUYmOt\nTiIi3sSjh322b99OmzZtCA0NpVatWgwdOpQPPvjAqjhud+AAnDoFMTFWJxERX2RZ+WdlZdGiRYuy\nr0NCQsjKyrIqjtudm+WjjVtExAqWlb/Nx1vvww81xVNErFPhHr6udv3115OZmVn2dWZmJiEhIRd9\nX1JSUtn9+Ph44uPj3ZDOtc6ehS1b4N13rU4iIt4gJSWFlJSUK/oZyz7wLSkpoX379mzYsIHmzZvT\npUsXn/nAd9MmGD8evvjC6iQi4o0q052WnfnXrFmTmTNn0rdvX+x2O6NGjSpX/N5s40Zz1y4REavo\nCl8LdOsGSUlw221WJxERb1SZ7lT5u1lBAQQHQ04O1KtndRoR8UYePc/fV332GXTurOIXEWup/N1s\nwwaN94uI9VT+bqYPe0XEE2jM341++glatTJva9WyOo2IeCuN+XuYlBRzpo+KX0SspvJ3ow0boHdv\nq1OIiKj83Urj/SLiKVT+bpKVBceOaQlnEfEMKn832bgR4uOhhn7HRcQDqIrcROP9IuJJVP5uYBga\n7xcRz6Lyd4PvvoPSUmjXzuokIiImlb8bnDvr9/HNy0TEg6j83UDj/SLiabS8g4uVlkJgIOzeDRfs\nVy8i4jJa3sED7N0LjRur+EXEs6j8Xay0FMaOtTqFiEh5GvYREfEyGvYREZFLUvmLiPgglb+IiA9S\n+YuI+CCVv4iID1L5i4j4IJW/iIgPUvmLiPgglb+IiA+ypPyTkpIICQkhLi6OuLg41q1bZ0UMERGf\nZUn522w2xo4dS2pqKqmpqfTr18+KGJZLSUmxOoJLefP78+b3Bnp/vsCyYR+t2eP9fwG9+f1583sD\nvT9fYFn5z5gxg5iYGEaNGkVubq5VMUREfJLLyj8hIYGoqKiLjpUrVzJmzBjS09PZs2cPwcHBjBs3\nzlUxRETkEixf0jkjI4PExET27t170a+1adOGtLQ0C1KJiFRfrVu35rvvvrvs99R0U5ZysrOzCQ4O\nBmD58uVERUVd8vschRcRkatjyZn//fffz549e7DZbLRq1Yo5c+YQGBjo7hgiIj7L8mEfERFxP4+/\nwnfp0qV06NABPz8/du/ebXUcp1i3bh1hYWG0bduWF154weo4TjVy5EgCAwMrHMqr7jIzM+nZsycd\nOnQgMjKS6dOnWx3Jqc6cOUPXrl2JjY0lIiKCiRMnWh3J6ex2O3FxcSQmJlodxelCQ0OJjo4mLi6O\nLl26XP7DyYUJAAAG3UlEQVSbDQ+3b98+49tvvzXi4+ONXbt2WR2nykpKSozWrVsb6enpRlFRkRET\nE2N8/fXXVsdymk8//dTYvXu3ERkZaXUUl8jOzjZSU1MNwzCM/Px8o127dl7152cYhlFYWGgYhmEU\nFxcbXbt2NT777DOLEznXtGnTjHvvvddITEy0OorThYaGGj/99FOlvtfjz/zDwsJo166d1TGcZvv2\n7bRp04bQ0FBq1arF0KFD+eCDD6yO5TTdu3encePGVsdwmaCgIGJjYwFo0KAB4eHhHD582OJUzlWv\nXj0AioqKsNvtNGnSxOJEznPo0CHWrl3L6NGjvfZC08q+L48vf2+TlZVFixYtyr4OCQkhKyvLwkRy\ntTIyMkhNTaVr165WR3Gq0tJSYmNjCQwMpGfPnkRERFgdyWkef/xxpkyZQo0a3ll9NpuN2267jc6d\nO/PGG29c9nstmer5SwkJCRw5cuSix59//nmvG5ez2WxWRxAnKCgoYPDgwSQnJ9OgQQOr4zhVjRo1\n2LNnDydPnqRv376kpKQQHx9vdawqW716Nddddx1xcXFeu7zD5s2bCQ4O5ujRoyQkJBAWFkb37t0v\n+b0eUf4fffSR1RHc5vrrryczM7Ps68zMTEJCQixMJFequLiYu+++m9/97ncMGjTI6jguExAQwO23\n387OnTu9ovy3bNnCypUrWbt2LWfOnCEvL4/777+ft99+2+poTnPu+qlrr72WO++8k+3bt1dY/tXq\n3z7eMEbXuXNnDhw4QEZGBkVFRbz77rsMHDjQ6lhSSYZhMGrUKCIiIvjLX/5idRynO3bsWNlaW6dP\nn+ajjz4iLi7O4lTO8fzzz5OZmUl6ejqLFy+mV69eXlX8p06dIj8/H4DCwkLWr19/2Vl3Hl/+y5cv\np0WLFmzbto3bb7+d/v37Wx2pSmrWrMnMmTPp27cvERER/Pa3vyU8PNzqWE4zbNgwbrnlFvbv30+L\nFi2YN2+e1ZGcavPmzSxcuJBPPvnEK/ejyM7OplevXsTGxtK1a1cSExPp3bu31bFcwtuGYHNycuje\nvXvZn92AAQPo06dPhd+vi7xERHyQx5/5i4iI86n8RUR8kMpfRMQHqfxFRHyQyl9ExAep/EVEfJDK\nX7zWyZMnmT17NmDOX7/nnnuc9twzZ85k/vz5Tnu+IUOGkJ6e7rTnE3FE8/zFa11uf+iqMAyDjh07\nsmPHDmrWdM4KKR999BGrVq3yuv0BxHPpzF+81oQJE0hLSyMuLo4hQ4aUXeo+f/58Bg0aRJ8+fWjV\nqhUzZ85k6tSpdOzYkZtvvpkTJ04AkJaWRv/+/encuTM9evTg22+/BcyrfMPCwsqKf/r06XTo0IGY\nmBiGDRsGmJfXjxw5kq5du9KxY0dWrlwJmBuJPPHEE0RFRRETE8PMmTMBiI+PZ+3atW79/REf55ot\nBUSsl5GRUbapzIX3582bZ7Rp08YoKCgwjh49avj7+xtz5swxDMMwHn/8ceOVV14xDMMwevXqZRw4\ncMAwDMPYtm2b0atXL8MwDOMf//iHMXXq1LLXad68uVFUVGQYhmGcPHnSMAzDmDhxorFw4ULDMAzj\nxIkTRrt27YzCwkLj1VdfNe655x7DbrcbhmEYx48fL3ueHj16eN3GMOK5PGJVTxFXMC4Y0TR+MbrZ\ns2dP6tevT/369WnUqFHZ0uFRUVF8+eWXFBYWsmXLlnKfExQVFQFw8OBBunXrVvZ4dHQ09957L4MG\nDSpb5XP9+vWsWrWKqVOnAnD27FkOHjzIhg0bGDNmTNl68hdufNO8eXMyMjK8aq0n8Vwqf/FJderU\nKbtfo0aNsq9r1KhBSUkJpaWlNG7cmNTU1Ev+/IX/M1mzZg2ffvopq1at4rnnniv7jOH999+nbdu2\nl/3ZXz7urZuMiOfR3zTxWg0bNixb4rayzhVzw4YNadWqFcuWLSt7/MsvvwSgZcuWZZsPGYbBwYMH\niY+PZ/LkyZw8eZKCggL69u1b7sPbc/8TSUhIYM6cOdjtdoCyzxfAnJHUsmXLq3y3IldG5S9eq2nT\nptx6661ERUUxfvz4siV8bTZbueV8f3n/3NfvvPMOc+fOJTY2lsjIyLIPbbt168bOnTsBKCkpYfjw\n4URHR9OxY0cee+wxAgICeOaZZyguLiY6OprIyEgmTZoEwOjRo/nVr35FdHQ0sbGxLFq0CDA3iDl0\n6BBhYWGu/40RQVM9Ra6Y8fNUzy+++ILatWs75TnXr1/PmjVrSE5OdsrziTiiM3+RK2Sz2XjwwQd5\n5513nPacb775Jo8//rjTnk/EEZ35i4j4IJ35i4j4IJW/iIgPUvmLiPgglb+IiA9S+YuI+CCVv4iI\nD/p/wQuHfYdHfM4AAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7b9e4a8>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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1NbVBmbS0NCQkJAAAoqKiUFFRgdLS0hbr1q+TkJCA3bt3AwBSU1Mxa9Ys+Pr6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\nqer/AUdcZmZ6XUFsAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7bac898>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  value  of  capacitor  voltage  is   4.4   V,  resistor  voltage  is   4.4   V,\n",
+        "current  is   0.02   mA  at  one  and  a  half  seconds  after  discharge  has  started.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 265</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the initial value of the current flowing,\n",
+      "#(b) the time constant of the circuit, \n",
+      "#(c) the value of the current one second after connection, \n",
+      "#(d) the value of the capacitor voltage two seconds after connection, and \n",
+      "#(e) the time after connection when the resistor voltage is 15 V.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  20E-6;#  in  Farads\n",
+      "R  =  50000;#  in  ohms\n",
+      "V  =  20;#  in  Volts\n",
+      "t1  =  1;#  in  secs\n",
+      "t2  =  2;#  in  secs\n",
+      "VRt  =  15;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "Vct1  =  V*(1-math.e**(-1*t2/tou))\n",
+      "t3  =  -1*tou*math.log(VRt/V)\n",
+      "it1  =  I*math.e**(-1*t1/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)initial  value  of  the  current  flowing  is  \",round(I*1000,1),\"mA\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  \",round(tou,2),\"  Sec\"\n",
+      "print  \"\\n  (c)the  value  of  the  current  one  second  after  connection,  \",round((it1/1E-3),3),\"  mA\"\n",
+      "print  \"\\n  (d)the  value  of  the  capacitor  voltage  two  seconds  after  connection  \",round(Vct1,1),\"  V\"\n",
+      "print  \"\\n  (e)the  time  after  connection  when  the  resistor  voltage  is  15  V  is  \",round(t3,3),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)initial  value  of  the  current  flowing  is   0.4 mA\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit   1.0   Sec\n",
+        "\n",
+        "  (c)the  value  of  the  current  one  second  after  connection,   0.147   mA\n",
+        "\n",
+        "  (d)the  value  of  the  capacitor  voltage  two  seconds  after  connection   17.3   V\n",
+        "\n",
+        "  (e)the  time  after  connection  when  the  resistor  voltage  is  15  V  is   0.288   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 266</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of the resistor, and (b) the capacitor voltage 7 ms after connecting the circuit to a 10 V supply\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.5E-6;#  in  Farads\n",
+      "V  =  10;#  in  Volts\n",
+      "tou  =  0.012;#  in  secs\n",
+      "t1  =  0.007;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  tou/C\n",
+      "Vc  =  V*(1-math.e**(-1*t1/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)value  of  the  resistor  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)capacitor  voltage  is  \",round(Vc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)value  of  the  resistor  is   24000.0   ohm\n",
+        "\n",
+        "  (b)capacitor  voltage  is   4.42   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 267</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine: (a) the value of the capacitor, \n",
+      "#(b) the time for the capacitor voltage to fall to 20 V, \n",
+      "#(c) the current flowing when the capacitor has been discharging for 0.5 s, and \n",
+      "#(d) the voltage drop across the resistor when the capacitor has been discharging for one second.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  50000;#  in  ohms\n",
+      "V  =  100;#  in  Volts\n",
+      "Vc1  =  20;#  in  Volts\n",
+      "tou  =  0.8;#  in  secs\n",
+      "t1  =  0.5;#  in  secs\n",
+      "t2  =  1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  tou/R\n",
+      "t  =  -1*tou*math.log(Vc1/V)\n",
+      "I  =  V/R\n",
+      "it1  =  I*math.e**(-1*t1/tou)\n",
+      "Vc  =  V*math.e**(-1*t2/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  the  capacitor  is  \",round((C/1E-6),2),\"uF\"\n",
+      "print  \"\\n  (b)the  time  for  the  capacitor  voltage  to  fall  to  20  V  is  \",round(t,2),\"  sec\"\n",
+      "print  \"\\n  (c)the  current  flowing  when  the  capacitor  has  been  discharging  for  0.5  s  is  \",round(it1*1000,2),\"mA\"\n",
+      "print  \"\\n  (d)voltage  drop  across  resistor  when  the  capacitor  has  been  discharging  for  one  second  is  \",round(Vc,1),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  the  capacitor  is   16.0 uF\n",
+        "\n",
+        "  (b)the  time  for  the  capacitor  voltage  to  fall  to  20  V  is   1.29   sec\n",
+        "\n",
+        "  (c)the  current  flowing  when  the  capacitor  has  been  discharging  for  0.5  s  is   1.07 mA\n",
+        "\n",
+        "  (d)voltage  drop  across  resistor  when  the  capacitor  has  been  discharging  for  one  second  is   28.7   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6 page no. 268</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the initial discharge current, \n",
+      "#(b) the time constant of the circuit, and \n",
+      "#(c) the minimum time required for the voltage across the capacitor to fall to less than 2 V\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.1E-6;#  in  Farads\n",
+      "R  =  4000;#  in  ohms\n",
+      "V  =  200;#  in  Volts\n",
+      "Vc1  =  2;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "t  =  -1*tou*math.log(Vc1/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  initial  discharge  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)Time  constant  tou  is  \",round(tou,5),\"  sec\"\n",
+      "print  \"\\n  (c)min.  time  required  for voltage  across capacitor  to  fall  to  less  than  2  V  is  \",round(t*1000,0),\"  msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  initial  discharge  current  is   0.05   A\n",
+        "\n",
+        "  (b)Time  constant  tou  is   0.0004   sec\n",
+        "\n",
+        "  (c)min.  time  required  for voltage  across capacitor  to  fall  to  less  than  2  V  is   2.0   msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 270</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#draw the current/time characteristic and \n",
+      "#hence determine the value of current flowing at a time equal to two time constants and the time for the current to grow to 1.5 A\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.1;#  in  Henry\n",
+      "R  =  20;#  in  ohms\n",
+      "V  =  60;#  in  Volts\n",
+      "i2  =  1.5;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "t1  =  2*tou\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "I  =  V/R\n",
+      "for h in range(250):\n",
+      "    t.append((h-1)/10000)\n",
+      "    k=(h-1)/10000\n",
+      "    i.append(I*(1  -  math.e**(-1*k/tou)))\n",
+      "plot(t,i,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*math.log(1  -  i2/I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  value  of  current  flowing  at  a  time  equal  to  two  time  constants  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (b)the  time  for  the  current  to  grow  to  1.5  A  is  \",round(t2,5),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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Pnw/AhAkTGDZsGFlZWURHR9O6dWsWLlxoV7k+t2WLFSrx8XZXIiLSOFrLy08d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\nUYpOSazjAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7b9d748>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  value  of  current  flowing  at  a  time  equal  to  two  time  constants  is   2.59   A\n",
+        "\n",
+        "  (b)the  time  for  the  current  to  grow  to  1.5  A  is   0.00347   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 271</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the final value of current, \n",
+      "#(b) the time constant of the circuit, \n",
+      "#(c) the value of current after a time equal to the time constant from the instant the supply oltage is connected, \n",
+      "#(d) the expected time for the current to rise to within 1% of its final value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.04;#  in  Henry\n",
+      "R  =  10;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "t1  =  tou\n",
+      "I  =  V/R\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "i2  =  0.01*I\n",
+      "t2  =  -1*tou*(-5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  final  value  of  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  is  \",round(tou*1000,2),\"msec\"\n",
+      "print  \"\\n  (c)  value  of  current  after  a  time  equal  to  the  time  constant  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (d)expected  time  for  current  to  rise  to  within  0.01  times  of  its  final  value  is  \",round(t2*1000,2),\"msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  final  value  of  current  is   12.0   A\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit  is   4.0 msec\n",
+        "\n",
+        "  (c)  value  of  current  after  a  time  equal  to  the  time  constant  is   7.59   A\n",
+        "\n",
+        "  (d)expected  time  for  current  to  rise  to  within  0.01  times  of  its  final  value  is   20.0 msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 271</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate:\n",
+      "#(a) the steady state value of current flowing in the winding,\n",
+      "#(b) the time constant of the circuit,\n",
+      "#(c) the value of the induced e.m.f. after 0.1 s,\n",
+      "#(d) the time for the current to rise to 85% of its final value, and\n",
+      "#(e) the value of the current after 0.3 s\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  3;#  in  Henry\n",
+      "R  =  15;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "t1  =  0.1;#  in  secs\n",
+      "t3  =  0.3;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "I  =  V/R\n",
+      "i2  =  0.85*I;#  in  amperes\n",
+      "VL  =  V*math.e**(-1*t1/tou)\n",
+      "t2  =  -1*tou*math.log(1  -  (i2/I))\n",
+      "i3  =  I*(1  -  math.e**(-1*t3/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Results  \\n\\n\"\n",
+      "print  \"\\n  (a)  steady  state  value  of  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  is  \",round(tou,2),\"  sec\"\n",
+      "print  \"\\n  (c)value  of  the  induced  e.m.f.  after  0.1  s  is  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (d)  time  for  the  current  to  rise  to  0.85  times  of  its  final  values  is  \",round(t2,2),\"  A\"\n",
+      "print  \"\\n  (e)value  of  the  current  after  0.3  s  is  \",round(i3,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Results  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  steady  state  value  of  current  is   8.0   A\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit  is   0.2   sec\n",
+        "\n",
+        "  (c)value  of  the  induced  e.m.f.  after  0.1  s  is   72.78   V\n",
+        "\n",
+        "  (d)  time  for  the  current  to  rise  to  0.85  times  of  its  final  values  is   0.38   A\n",
+        "\n",
+        "  (e)value  of  the  current  after  0.3  s  is   6.21   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 273</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#draw the current/time characteristic when the supply is removed and replaced by a shorting link\n",
+      "#determine (a) the current flowing in the winding 3 s after being shorted-out and (b) the time for the current to decay to 5 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohms\n",
+      "V  =  110;#  in  Volts\n",
+      "tou  =  2;#  in  secs\n",
+      "t1  = 3;#  in  secs\n",
+      "i2  = 5;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  tou*R\n",
+      "I  =  V/R\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "for h in range(100):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    i.append(I*math.e**(-1*k/tou))\n",
+      "plot(t,i,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "i1  =  I*(math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*log((i2/I))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  in  the  winding  3  s  after  being  shorted-out  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (b)the  time  for  the  current  to  decay  to  5  A  is  \",round(t2,2),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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re1q3bq0AYGNjY2N7hNa6detHqseaXYwtKytD+/btsXnzZvj5+aFTp073XIwl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\n/T/TLOQZYjmougAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x77abf28>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  in  the  winding  3  s  after  being  shorted-out  is   1.64   A\n",
+        "\n",
+        "  (b)the  time  for  the  current  to  decay  to  5  A  is   0.77   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 273</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine: (a) the resistance of the coil,\n",
+      "#(b) the current flowing in the circuit one second after the shorting link has been placed in the circuit, and \n",
+      "#(c) the time taken for the current to fall to 10% of its initial value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  6;#  in  Henry\n",
+      "r  =  10;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "tou  =  0.3;#  in  secs\n",
+      "t1  =  1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  (L/tou)  -  r\n",
+      "Rt  =  R  +  r\n",
+      "I  =  V/Rt\n",
+      "i2  =  0.1*I\n",
+      "i1  =  I*(math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*math.log((i2/I))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"  \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  resistance  of  the  coil  is  \",round(R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)  current  flowing  in  circuit  one  second  after  the  shorting  link  has  been  placed  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (c)the  time  for  the  current  to  decay  to  0.1  times  of  initial  value  is  \",round(t2,2),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  resistance  of  the  coil  is   10.0   ohm\n",
+        "\n",
+        "  (b)  current  flowing  in  circuit  one  second  after  the  shorting  link  has  been  placed  is   0.21   A\n",
+        "\n",
+        "  (c)the  time  for  the  current  to  decay  to  0.1  times  of  initial  value  is   0.69   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 274</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the time constant of the circuit and the steady-state value of the current flowing in the circuit. \n",
+      "#Find (a) the current flowing in the circuit at a time equal to one time constant, \n",
+      "#(b) the voltage drop across the inductor at a time equal to two time constants and \n",
+      "#(c) the voltage drop across the resistor after a time equal to three time constants.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.2;#  in  Henry\n",
+      "R  =  1000;#  in  ohms\n",
+      "V  =  24;#  in  Volts\n",
+      "t1  =  1*L/R;#  in  secs\n",
+      "t2  =  2*L/R;#  in  secs\n",
+      "t3  =  3*L/R;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "I  =  V/R\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "VL  =  V*(math.e**(-1*t2/tou))\n",
+      "VR  =  V*(1  -  math.e**(-1*t3/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  time  constant  of  circuit  is  \",round(tou*1000,6),\"msec, and steady-state  value  of current  is  \",round(I*1000,2),\"mA\"\n",
+      "print  \"\\n  (a)  urrent  flowing  in  the  circuit  at  a  time  equal  to  one  time  constant  is  \",round(i1*1000,2),\"mA\"\n",
+      "print  \"\\n  (b)  voltage  drop  across  the  inductor  at  a  time  equal  to  two  time  constants  is  \",round(VL,3),\"  V\"\n",
+      "print  \"\\n  (c)the  voltage  drop  across  the  resistor  after  a  time  equal  to  three  time  constants  is  \",round(VR,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  time  constant  of  circuit  is   0.2 msec, and steady-state  value  of current  is   24.0 mA\n",
+        "\n",
+        "  (a)  urrent  flowing  in  the  circuit  at  a  time  equal  to  one  time  constant  is   15.17 mA\n",
+        "\n",
+        "  (b)  voltage  drop  across  the  inductor  at  a  time  equal  to  two  time  constants  is   3.248   V\n",
+        "\n",
+        "  (c)the  voltage  drop  across  the  resistor  after  a  time  equal  to  three  time  constants  is   22.81   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_1.ipynb
new file mode 100755
index 00000000..94d79ec6
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_1.ipynb
@@ -0,0 +1,895 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3c30db14cf5a722166d02bb2b24e5cc57f73fe8d238c68c893567c34f71054b0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 17: D.c. transients</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 262</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitor voltage at a time equal to one time constant after being connected to the supply, \n",
+      "#and also two seconds after being connected to the supply. \n",
+      "#Also, find the time for the capacitor voltage to reach one half of its steady state value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "C  =  15E-6;#  in  Farads\n",
+      "R  =  47000;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "t1  =  tou\n",
+      "Vctou  =  V*(1-math.e**(-1*t1/tou))\n",
+      "Vct  =  V/2\n",
+      "t0  =  -1*tou*math.log(1  -  Vct/V)\n",
+      "t=[]\n",
+      "Vc=[]\n",
+      "I  =  V/R\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    Vc.append(V*(1 - math.e**(-1*k/tou)))\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,Vc,'-')\n",
+      "#plot(t,Vc,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('Volts(V)')\n",
+      "show()\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitor  voltage  at  a  time  equal  to  one  time  constant  =  \",round(Vctou,2),\"  V\"\n",
+      "print  \"\\n  (b)the  time  for  the  capacitor  voltage  to  reach  one  half  of  its  steady  state  value  =  \",round(t0,5),\" secs\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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kSBouSe24oRNb4VhCoEiSqtB5lPYGQvWEQWZJkiRJkiRJkiRJkiRJkiRJSrv/\nD8/iLwbt6hd+AAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x34660b8>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitor  voltage  at  a  time  equal  to  one  time  constant  =   75.85   V\n",
+        "\n",
+        "  (b)the  time  for  the  capacitor  voltage  to  reach  one  half  of  its  steady  state  value  =   0.48867  secs\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 263</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#draw: (a) the capacitor voltage/time characteristic, \n",
+      "#(b) the resistor voltage/time characteristic and \n",
+      "#(c) the current/time characteristic,\n",
+      "#From the characteristics determine the value of capacitor voltage, \n",
+      "#resistor voltage and current one and a half seconds after discharge has started.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "C  =  4E-6;#  in  Farads\n",
+      "R  =  220000;#  in  ohms\n",
+      "V  =  24;#  in  Volts\n",
+      "t1  =  1.5;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "\n",
+      "t=[]\n",
+      "Vc=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    Vc.append(V*math.e**(-1*k/tou))\n",
+      "#plt.figsize(10,8)\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,Vc,'-')\n",
+      "#plot(t,Vc,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('P.D across Capacitor(V)')\n",
+      "show()\n",
+      "\n",
+      "t=[]\n",
+      "VR=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    VR.append(V*(1 - math.e**(-1*k/tou)))\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,VR,'-')\n",
+      "#plot(t,VR,'-*')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('P.D across Resistor(V)')\n",
+      "show()\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    i.append(I*math.e**(-1*k/tou))\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,i,'-')\n",
+      "#plot(t,i,'*-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "\n",
+      "Vct1  =  V*math.e**(-1*t1/tou)\n",
+      "VRt1  =  V*math.e**(-1*t1/tou)\n",
+      "it1   =  I*math.e**(-1*t1/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  value  of  capacitor  voltage  is  \",round(Vct1,1),\"  V,  resistor  voltage  is  \",round(VRt1,1),\"  V,\"\n",
+      "print  \"current  is  \",round(0.02,2),\"  mA  at  one  and  a  half  seconds  after  discharge  has  started.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x77c6e10>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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HRKxh2bCP3W6nffv2fPzxx1x//fXceOONLFq0iPDw8PPhvHzYp21bWLoUYmOt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\nD/p/wQuHfYdHfM4AAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7b9e4a8>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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1NbVBmbS0NCQkJAAAoqKiUFFRgdLS0hbr1q+TkJCA3bt3AwBSU1Mxa9Ys+Pr6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\nqer/AUdcZmZ6XUFsAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7bac898>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  value  of  capacitor  voltage  is   4.4   V,  resistor  voltage  is   4.4   V,\n",
+        "current  is   0.02   mA  at  one  and  a  half  seconds  after  discharge  has  started.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 265</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the initial value of the current flowing,\n",
+      "#(b) the time constant of the circuit, \n",
+      "#(c) the value of the current one second after connection, \n",
+      "#(d) the value of the capacitor voltage two seconds after connection, and \n",
+      "#(e) the time after connection when the resistor voltage is 15 V.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  20E-6;#  in  Farads\n",
+      "R  =  50000;#  in  ohms\n",
+      "V  =  20;#  in  Volts\n",
+      "t1  =  1;#  in  secs\n",
+      "t2  =  2;#  in  secs\n",
+      "VRt  =  15;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "Vct1  =  V*(1-math.e**(-1*t2/tou))\n",
+      "t3  =  -1*tou*math.log(VRt/V)\n",
+      "it1  =  I*math.e**(-1*t1/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)initial  value  of  the  current  flowing  is  \",round(I*1000,1),\"mA\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  \",round(tou,2),\"  Sec\"\n",
+      "print  \"\\n  (c)the  value  of  the  current  one  second  after  connection,  \",round((it1/1E-3),3),\"  mA\"\n",
+      "print  \"\\n  (d)the  value  of  the  capacitor  voltage  two  seconds  after  connection  \",round(Vct1,1),\"  V\"\n",
+      "print  \"\\n  (e)the  time  after  connection  when  the  resistor  voltage  is  15  V  is  \",round(t3,3),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)initial  value  of  the  current  flowing  is   0.4 mA\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit   1.0   Sec\n",
+        "\n",
+        "  (c)the  value  of  the  current  one  second  after  connection,   0.147   mA\n",
+        "\n",
+        "  (d)the  value  of  the  capacitor  voltage  two  seconds  after  connection   17.3   V\n",
+        "\n",
+        "  (e)the  time  after  connection  when  the  resistor  voltage  is  15  V  is   0.288   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 266</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of the resistor, and (b) the capacitor voltage 7 ms after connecting the circuit to a 10 V supply\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.5E-6;#  in  Farads\n",
+      "V  =  10;#  in  Volts\n",
+      "tou  =  0.012;#  in  secs\n",
+      "t1  =  0.007;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  tou/C\n",
+      "Vc  =  V*(1-math.e**(-1*t1/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)value  of  the  resistor  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)capacitor  voltage  is  \",round(Vc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)value  of  the  resistor  is   24000.0   ohm\n",
+        "\n",
+        "  (b)capacitor  voltage  is   4.42   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 267</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine: (a) the value of the capacitor, \n",
+      "#(b) the time for the capacitor voltage to fall to 20 V, \n",
+      "#(c) the current flowing when the capacitor has been discharging for 0.5 s, and \n",
+      "#(d) the voltage drop across the resistor when the capacitor has been discharging for one second.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  50000;#  in  ohms\n",
+      "V  =  100;#  in  Volts\n",
+      "Vc1  =  20;#  in  Volts\n",
+      "tou  =  0.8;#  in  secs\n",
+      "t1  =  0.5;#  in  secs\n",
+      "t2  =  1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  tou/R\n",
+      "t  =  -1*tou*math.log(Vc1/V)\n",
+      "I  =  V/R\n",
+      "it1  =  I*math.e**(-1*t1/tou)\n",
+      "Vc  =  V*math.e**(-1*t2/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  the  capacitor  is  \",round((C/1E-6),2),\"uF\"\n",
+      "print  \"\\n  (b)the  time  for  the  capacitor  voltage  to  fall  to  20  V  is  \",round(t,2),\"  sec\"\n",
+      "print  \"\\n  (c)the  current  flowing  when  the  capacitor  has  been  discharging  for  0.5  s  is  \",round(it1*1000,2),\"mA\"\n",
+      "print  \"\\n  (d)voltage  drop  across  resistor  when  the  capacitor  has  been  discharging  for  one  second  is  \",round(Vc,1),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  the  capacitor  is   16.0 uF\n",
+        "\n",
+        "  (b)the  time  for  the  capacitor  voltage  to  fall  to  20  V  is   1.29   sec\n",
+        "\n",
+        "  (c)the  current  flowing  when  the  capacitor  has  been  discharging  for  0.5  s  is   1.07 mA\n",
+        "\n",
+        "  (d)voltage  drop  across  resistor  when  the  capacitor  has  been  discharging  for  one  second  is   28.7   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6 page no. 268</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the initial discharge current, \n",
+      "#(b) the time constant of the circuit, and \n",
+      "#(c) the minimum time required for the voltage across the capacitor to fall to less than 2 V\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.1E-6;#  in  Farads\n",
+      "R  =  4000;#  in  ohms\n",
+      "V  =  200;#  in  Volts\n",
+      "Vc1  =  2;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "t  =  -1*tou*math.log(Vc1/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  initial  discharge  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)Time  constant  tou  is  \",round(tou,5),\"  sec\"\n",
+      "print  \"\\n  (c)min.  time  required  for voltage  across capacitor  to  fall  to  less  than  2  V  is  \",round(t*1000,0),\"  msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  initial  discharge  current  is   0.05   A\n",
+        "\n",
+        "  (b)Time  constant  tou  is   0.0004   sec\n",
+        "\n",
+        "  (c)min.  time  required  for voltage  across capacitor  to  fall  to  less  than  2  V  is   2.0   msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 270</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#draw the current/time characteristic and \n",
+      "#hence determine the value of current flowing at a time equal to two time constants and the time for the current to grow to 1.5 A\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.1;#  in  Henry\n",
+      "R  =  20;#  in  ohms\n",
+      "V  =  60;#  in  Volts\n",
+      "i2  =  1.5;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "t1  =  2*tou\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "I  =  V/R\n",
+      "for h in range(250):\n",
+      "    t.append((h-1)/10000)\n",
+      "    k=(h-1)/10000\n",
+      "    i.append(I*(1  -  math.e**(-1*k/tou)))\n",
+      "plot(t,i,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*math.log(1  -  i2/I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  value  of  current  flowing  at  a  time  equal  to  two  time  constants  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (b)the  time  for  the  current  to  grow  to  1.5  A  is  \",round(t2,5),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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Pnw/AhAkTGDZsGFlZWURHR9O6dWsWLlxoV7k+t2WLFSrx8XZXIiLSOFrLy08d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\nUYpOSazjAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7b9d748>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  value  of  current  flowing  at  a  time  equal  to  two  time  constants  is   2.59   A\n",
+        "\n",
+        "  (b)the  time  for  the  current  to  grow  to  1.5  A  is   0.00347   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 271</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the final value of current, \n",
+      "#(b) the time constant of the circuit, \n",
+      "#(c) the value of current after a time equal to the time constant from the instant the supply oltage is connected, \n",
+      "#(d) the expected time for the current to rise to within 1% of its final value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.04;#  in  Henry\n",
+      "R  =  10;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "t1  =  tou\n",
+      "I  =  V/R\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "i2  =  0.01*I\n",
+      "t2  =  -1*tou*(-5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  final  value  of  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  is  \",round(tou*1000,2),\"msec\"\n",
+      "print  \"\\n  (c)  value  of  current  after  a  time  equal  to  the  time  constant  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (d)expected  time  for  current  to  rise  to  within  0.01  times  of  its  final  value  is  \",round(t2*1000,2),\"msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  final  value  of  current  is   12.0   A\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit  is   4.0 msec\n",
+        "\n",
+        "  (c)  value  of  current  after  a  time  equal  to  the  time  constant  is   7.59   A\n",
+        "\n",
+        "  (d)expected  time  for  current  to  rise  to  within  0.01  times  of  its  final  value  is   20.0 msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 271</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate:\n",
+      "#(a) the steady state value of current flowing in the winding,\n",
+      "#(b) the time constant of the circuit,\n",
+      "#(c) the value of the induced e.m.f. after 0.1 s,\n",
+      "#(d) the time for the current to rise to 85% of its final value, and\n",
+      "#(e) the value of the current after 0.3 s\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  3;#  in  Henry\n",
+      "R  =  15;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "t1  =  0.1;#  in  secs\n",
+      "t3  =  0.3;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "I  =  V/R\n",
+      "i2  =  0.85*I;#  in  amperes\n",
+      "VL  =  V*math.e**(-1*t1/tou)\n",
+      "t2  =  -1*tou*math.log(1  -  (i2/I))\n",
+      "i3  =  I*(1  -  math.e**(-1*t3/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Results  \\n\\n\"\n",
+      "print  \"\\n  (a)  steady  state  value  of  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  is  \",round(tou,2),\"  sec\"\n",
+      "print  \"\\n  (c)value  of  the  induced  e.m.f.  after  0.1  s  is  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (d)  time  for  the  current  to  rise  to  0.85  times  of  its  final  values  is  \",round(t2,2),\"  A\"\n",
+      "print  \"\\n  (e)value  of  the  current  after  0.3  s  is  \",round(i3,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Results  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  steady  state  value  of  current  is   8.0   A\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit  is   0.2   sec\n",
+        "\n",
+        "  (c)value  of  the  induced  e.m.f.  after  0.1  s  is   72.78   V\n",
+        "\n",
+        "  (d)  time  for  the  current  to  rise  to  0.85  times  of  its  final  values  is   0.38   A\n",
+        "\n",
+        "  (e)value  of  the  current  after  0.3  s  is   6.21   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 273</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#draw the current/time characteristic when the supply is removed and replaced by a shorting link\n",
+      "#determine (a) the current flowing in the winding 3 s after being shorted-out and (b) the time for the current to decay to 5 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohms\n",
+      "V  =  110;#  in  Volts\n",
+      "tou  =  2;#  in  secs\n",
+      "t1  = 3;#  in  secs\n",
+      "i2  = 5;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  tou*R\n",
+      "I  =  V/R\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "for h in range(100):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    i.append(I*math.e**(-1*k/tou))\n",
+      "plot(t,i,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "i1  =  I*(math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*log((i2/I))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  in  the  winding  3  s  after  being  shorted-out  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (b)the  time  for  the  current  to  decay  to  5  A  is  \",round(t2,2),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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re1q3bq0AYGNjY2N7hNa6detHqseaXYwtKytD+/btsXnzZvj5+aFTp073XIwl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\n/T/TLOQZYjmougAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x77abf28>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  in  the  winding  3  s  after  being  shorted-out  is   1.64   A\n",
+        "\n",
+        "  (b)the  time  for  the  current  to  decay  to  5  A  is   0.77   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 273</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine: (a) the resistance of the coil,\n",
+      "#(b) the current flowing in the circuit one second after the shorting link has been placed in the circuit, and \n",
+      "#(c) the time taken for the current to fall to 10% of its initial value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  6;#  in  Henry\n",
+      "r  =  10;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "tou  =  0.3;#  in  secs\n",
+      "t1  =  1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  (L/tou)  -  r\n",
+      "Rt  =  R  +  r\n",
+      "I  =  V/Rt\n",
+      "i2  =  0.1*I\n",
+      "i1  =  I*(math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*math.log((i2/I))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"  \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  resistance  of  the  coil  is  \",round(R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)  current  flowing  in  circuit  one  second  after  the  shorting  link  has  been  placed  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (c)the  time  for  the  current  to  decay  to  0.1  times  of  initial  value  is  \",round(t2,2),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  resistance  of  the  coil  is   10.0   ohm\n",
+        "\n",
+        "  (b)  current  flowing  in  circuit  one  second  after  the  shorting  link  has  been  placed  is   0.21   A\n",
+        "\n",
+        "  (c)the  time  for  the  current  to  decay  to  0.1  times  of  initial  value  is   0.69   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 274</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the time constant of the circuit and the steady-state value of the current flowing in the circuit. \n",
+      "#Find (a) the current flowing in the circuit at a time equal to one time constant, \n",
+      "#(b) the voltage drop across the inductor at a time equal to two time constants and \n",
+      "#(c) the voltage drop across the resistor after a time equal to three time constants.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.2;#  in  Henry\n",
+      "R  =  1000;#  in  ohms\n",
+      "V  =  24;#  in  Volts\n",
+      "t1  =  1*L/R;#  in  secs\n",
+      "t2  =  2*L/R;#  in  secs\n",
+      "t3  =  3*L/R;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "I  =  V/R\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "VL  =  V*(math.e**(-1*t2/tou))\n",
+      "VR  =  V*(1  -  math.e**(-1*t3/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  time  constant  of  circuit  is  \",round(tou*1000,6),\"msec, and steady-state  value  of current  is  \",round(I*1000,2),\"mA\"\n",
+      "print  \"\\n  (a)  urrent  flowing  in  the  circuit  at  a  time  equal  to  one  time  constant  is  \",round(i1*1000,2),\"mA\"\n",
+      "print  \"\\n  (b)  voltage  drop  across  the  inductor  at  a  time  equal  to  two  time  constants  is  \",round(VL,3),\"  V\"\n",
+      "print  \"\\n  (c)the  voltage  drop  across  the  resistor  after  a  time  equal  to  three  time  constants  is  \",round(VR,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  time  constant  of  circuit  is   0.2 msec, and steady-state  value  of current  is   24.0 mA\n",
+        "\n",
+        "  (a)  urrent  flowing  in  the  circuit  at  a  time  equal  to  one  time  constant  is   15.17 mA\n",
+        "\n",
+        "  (b)  voltage  drop  across  the  inductor  at  a  time  equal  to  two  time  constants  is   3.248   V\n",
+        "\n",
+        "  (c)the  voltage  drop  across  the  resistor  after  a  time  equal  to  three  time  constants  is   22.81   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_2.ipynb
new file mode 100755
index 00000000..adc7f92a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_2.ipynb
@@ -0,0 +1,895 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f205689956af649c31ad664e81a2e6ebba547e950f06ad3009a92a9bc3b8be80"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 17: D.c. transients</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 262</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitor voltage at a time equal to one time constant after being connected to the supply, \n",
+      "#and also two seconds after being connected to the supply. \n",
+      "#Also, find the time for the capacitor voltage to reach one half of its steady state value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "C  =  15E-6;#  in  Farads\n",
+      "R  =  47000;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "t1  =  tou\n",
+      "Vctou  =  V*(1-math.e**(-1*t1/tou))\n",
+      "Vct  =  V/2\n",
+      "t0  =  -1*tou*math.log(1  -  Vct/V)\n",
+      "t=[]\n",
+      "Vc=[]\n",
+      "I  =  V/R\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    Vc.append(V*(1 - math.e**(-1*k/tou)))\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,Vc,'-')\n",
+      "#plot(t,Vc,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('Volts(V)')\n",
+      "show()\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitor  voltage  at  a  time  equal  to  one  time  constant  =  \",round(Vctou,2),\"  V\"\n",
+      "print  \"\\n  (b)the  time  for  the  capacitor  voltage  to  reach  one  half  of  its  steady  state  value  =  \",round(t0,5),\" secs\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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kSBouSe24oRNb4VhCoEiSqtB5lPYGQvWEQWZJkiRJkiRJkiRJkiRJkiRJSrv/\nD8/iLwbt6hd+AAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x34660b8>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitor  voltage  at  a  time  equal  to  one  time  constant  =   75.85   V\n",
+        "\n",
+        "  (b)the  time  for  the  capacitor  voltage  to  reach  one  half  of  its  steady  state  value  =   0.48867  secs\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 263</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#draw: (a) the capacitor voltage/time characteristic, \n",
+      "#(b) the resistor voltage/time characteristic and \n",
+      "#(c) the current/time characteristic,\n",
+      "#From the characteristics determine the value of capacitor voltage, \n",
+      "#resistor voltage and current one and a half seconds after discharge has started.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "C  =  4E-6;#  in  Farads\n",
+      "R  =  220000;#  in  ohms\n",
+      "V  =  24;#  in  Volts\n",
+      "t1  =  1.5;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "\n",
+      "t=[]\n",
+      "Vc=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    Vc.append(V*math.e**(-1*k/tou))\n",
+      "#plt.figsize(10,8)\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,Vc,'-')\n",
+      "#plot(t,Vc,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('P.D across Capacitor(V)')\n",
+      "show()\n",
+      "\n",
+      "t=[]\n",
+      "VR=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    VR.append(V*(1 - math.e**(-1*k/tou)))\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,VR,'-')\n",
+      "#plot(t,VR,'-*')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('P.D across Resistor(V)')\n",
+      "show()\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    i.append(I*math.e**(-1*k/tou))\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,i,'-')\n",
+      "#plot(t,i,'*-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "\n",
+      "Vct1  =  V*math.e**(-1*t1/tou)\n",
+      "VRt1  =  V*math.e**(-1*t1/tou)\n",
+      "it1   =  I*math.e**(-1*t1/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  value  of  capacitor  voltage  is  \",round(Vct1,1),\"  V,  resistor  voltage  is  \",round(VRt1,1),\"  V,\"\n",
+      "print  \"current  is  \",round(0.02,2),\"  mA  at  one  and  a  half  seconds  after  discharge  has  started.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x77c6e10>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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HRKxh2bCP3W6nffv2fPzxx1x//fXceOONLFq0iPDw8PPhvHzYp21bWLoUYmOt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\nD/p/wQuHfYdHfM4AAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7b9e4a8>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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1NbVBmbS0NCQkJAAAoqKiUFFRgdLS0hbr1q+TkJCA3bt3AwBSU1Mxa9Ys+Pr6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\nqer/AUdcZmZ6XUFsAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7bac898>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  value  of  capacitor  voltage  is   4.4   V,  resistor  voltage  is   4.4   V,\n",
+        "current  is   0.02   mA  at  one  and  a  half  seconds  after  discharge  has  started.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 265</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the initial value of the current flowing,\n",
+      "#(b) the time constant of the circuit, \n",
+      "#(c) the value of the current one second after connection, \n",
+      "#(d) the value of the capacitor voltage two seconds after connection, and \n",
+      "#(e) the time after connection when the resistor voltage is 15 V.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  20E-6;#  in  Farads\n",
+      "R  =  50000;#  in  ohms\n",
+      "V  =  20;#  in  Volts\n",
+      "t1  =  1;#  in  secs\n",
+      "t2  =  2;#  in  secs\n",
+      "VRt  =  15;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "Vct1  =  V*(1-math.e**(-1*t2/tou))\n",
+      "t3  =  -1*tou*math.log(VRt/V)\n",
+      "it1  =  I*math.e**(-1*t1/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)initial  value  of  the  current  flowing  is  \",round(I*1000,1),\"mA\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  \",round(tou,2),\"  Sec\"\n",
+      "print  \"\\n  (c)the  value  of  the  current  one  second  after  connection,  \",round((it1/1E-3),3),\"  mA\"\n",
+      "print  \"\\n  (d)the  value  of  the  capacitor  voltage  two  seconds  after  connection  \",round(Vct1,1),\"  V\"\n",
+      "print  \"\\n  (e)the  time  after  connection  when  the  resistor  voltage  is  15  V  is  \",round(t3,3),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)initial  value  of  the  current  flowing  is   0.4 mA\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit   1.0   Sec\n",
+        "\n",
+        "  (c)the  value  of  the  current  one  second  after  connection,   0.147   mA\n",
+        "\n",
+        "  (d)the  value  of  the  capacitor  voltage  two  seconds  after  connection   17.3   V\n",
+        "\n",
+        "  (e)the  time  after  connection  when  the  resistor  voltage  is  15  V  is   0.288   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17.4, page no. 266</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of the resistor, and (b) the capacitor voltage 7 ms after connecting the circuit to a 10 V supply\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.5E-6;#  in  Farads\n",
+      "V  =  10;#  in  Volts\n",
+      "tou  =  0.012;#  in  secs\n",
+      "t1  =  0.007;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  tou/C\n",
+      "Vc  =  V*(1-math.e**(-1*t1/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)value  of  the  resistor  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)capacitor  voltage  is  \",round(Vc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)value  of  the  resistor  is   24000.0   ohm\n",
+        "\n",
+        "  (b)capacitor  voltage  is   4.42   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 267</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine: (a) the value of the capacitor, \n",
+      "#(b) the time for the capacitor voltage to fall to 20 V, \n",
+      "#(c) the current flowing when the capacitor has been discharging for 0.5 s, and \n",
+      "#(d) the voltage drop across the resistor when the capacitor has been discharging for one second.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  50000;#  in  ohms\n",
+      "V  =  100;#  in  Volts\n",
+      "Vc1  =  20;#  in  Volts\n",
+      "tou  =  0.8;#  in  secs\n",
+      "t1  =  0.5;#  in  secs\n",
+      "t2  =  1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  tou/R\n",
+      "t  =  -1*tou*math.log(Vc1/V)\n",
+      "I  =  V/R\n",
+      "it1  =  I*math.e**(-1*t1/tou)\n",
+      "Vc  =  V*math.e**(-1*t2/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  the  capacitor  is  \",round((C/1E-6),2),\"uF\"\n",
+      "print  \"\\n  (b)the  time  for  the  capacitor  voltage  to  fall  to  20  V  is  \",round(t,2),\"  sec\"\n",
+      "print  \"\\n  (c)the  current  flowing  when  the  capacitor  has  been  discharging  for  0.5  s  is  \",round(it1*1000,2),\"mA\"\n",
+      "print  \"\\n  (d)voltage  drop  across  resistor  when  the  capacitor  has  been  discharging  for  one  second  is  \",round(Vc,1),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  the  capacitor  is   16.0 uF\n",
+        "\n",
+        "  (b)the  time  for  the  capacitor  voltage  to  fall  to  20  V  is   1.29   sec\n",
+        "\n",
+        "  (c)the  current  flowing  when  the  capacitor  has  been  discharging  for  0.5  s  is   1.07 mA\n",
+        "\n",
+        "  (d)voltage  drop  across  resistor  when  the  capacitor  has  been  discharging  for  one  second  is   28.7   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6 page no. 268</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the initial discharge current, \n",
+      "#(b) the time constant of the circuit, and \n",
+      "#(c) the minimum time required for the voltage across the capacitor to fall to less than 2 V\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.1E-6;#  in  Farads\n",
+      "R  =  4000;#  in  ohms\n",
+      "V  =  200;#  in  Volts\n",
+      "Vc1  =  2;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "t  =  -1*tou*math.log(Vc1/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  initial  discharge  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)Time  constant  tou  is  \",round(tou,5),\"  sec\"\n",
+      "print  \"\\n  (c)min.  time  required  for voltage  across capacitor  to  fall  to  less  than  2  V  is  \",round(t*1000,0),\"  msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  initial  discharge  current  is   0.05   A\n",
+        "\n",
+        "  (b)Time  constant  tou  is   0.0004   sec\n",
+        "\n",
+        "  (c)min.  time  required  for voltage  across capacitor  to  fall  to  less  than  2  V  is   2.0   msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 270</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#draw the current/time characteristic and \n",
+      "#hence determine the value of current flowing at a time equal to two time constants and the time for the current to grow to 1.5 A\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.1;#  in  Henry\n",
+      "R  =  20;#  in  ohms\n",
+      "V  =  60;#  in  Volts\n",
+      "i2  =  1.5;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "t1  =  2*tou\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "I  =  V/R\n",
+      "for h in range(250):\n",
+      "    t.append((h-1)/10000)\n",
+      "    k=(h-1)/10000\n",
+      "    i.append(I*(1  -  math.e**(-1*k/tou)))\n",
+      "plot(t,i,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*math.log(1  -  i2/I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  value  of  current  flowing  at  a  time  equal  to  two  time  constants  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (b)the  time  for  the  current  to  grow  to  1.5  A  is  \",round(t2,5),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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Pnw/AhAkTGDZsGFlZWURHR9O6dWsWLlxoV7k+t2WLFSrx8XZXIiLSOFrLy08d\nPgzffAMpKXZXIiLByBvfmwoUEZFmSItDioiI31KgiIiIRyhQRETEIxQoIiLiEQoUERHxCAWKiIh4\nhAJFREQ8QoEiIiIeoUARERGPUKCIiIhHKFBERMQjFCgiIuIRChQREfEIBYqIiHiEAkVERDxCgSIi\nIh6hQBEREY9QoIiIiEcoUERExCMUKCIi4hEKFBER8YgQOz60tLSU3/zmN+zatYvu3bvzr3/9i7Cw\nsBrbde/enXbt2nHOOefQsmVLcnNzbahWREQaw5YeyvTp00lNTWXbtm0MGTKE6dOn17qdw+HA5XKR\nl5fXbMPE5XLZXYJXqX2BTe2TU9kSKBkZGYwbNw6AcePG8c4779S5rTHGV2X5pWD/D1rtC2xqn5zK\nlkApKSkhPDwcgPDwcEpKSmrdzuFw8Ktf/Yr+/fvz3HPP+bJEERE5Q16bQ0lNTWXfvn01np82bVq1\nxw6HA4fDUet7rFmzhi5durB//35SU1Pp3bs3AwcO9Eq9IiLSRMYGvXr1Mnv37jXGGLNnzx7Tq1ev\nBl+Tnp5uZs6cWevvoqKiDKCbbrrpplsjb1FRUR79XjfGGFuO8kpLS+Oll15iypQpvPTSSwwfPrzG\nNseOHcPtdtO2bVuOHj3KihUrmDp1aq3vt337dm+XLCIiDXAY4/tZ79LSUkaPHs3u3burHTa8Z88e\n7r33XjIzM/n222+55ZZbAKioqOD222/n0Ucf9XWpIiLSSLYEioiIBB+/PVO+tLSU1NRUevbsybXX\nXsuhQ4dq3S47O5vevXsTExPDjBkzGnx9QUEB559/PklJSSQlJXH//ff7pD0N1XuqyZMnExMTQ0JC\nAnl5eQ2+trH/Vr7gjfalp6cTGRlZtc+ys7O93o7aNKVtd999N+Hh4fTt27fa9sGy7+pqn7/sOzj7\n9hUWFjJ48GDi4+Pp06cPc+bMqdo+GPZffe074/3n8VkZD/mP//gPM2PGDGOMMdOnTzdTpkypsU1F\nRYWJiooyO3fuNOXl5SYhIcFs2bKl3tfv3LnT9OnTx0etaHy9J2VmZpqhQ4caY4zJyckxKSkpDb62\nMf9WvuCt9qWnp5u///3vvm3MaZrSNmOM+eSTT8yGDRtq/LcXDPvOmLrb5w/7zpimtW/v3r0mLy/P\nGGPMkSNHTM+ePc3XX39tjAmO/Vdf+850//ltD6UxJz/m5uYSHR1N9+7dadmyJWPGjGHZsmWNfr2v\n1VfvSaeMfy6RAAAHEUlEQVTWnZKSwqFDh9i3b19AtNVb7QNsP8G1KW0DGDhwIB06dKjxvsGw76Du\n9oH9+w7Ovn0lJSV07tyZxMREANq0aUNsbCzFxcU1XhOI+6+h9sGZ7T+/DZTGnPxYXFxMt27dqh5H\nRkZW/UPU9/qdO3eSlJSE0+lk9erV3mxGo+ttaJs9e/acVVt9yVvtA3j66adJSEhg/PjxtgwrNKVt\n9QmGfdcQu/cdnH37ioqKqm1TUFBAXl4eKSkpQODvv4baB2e2/2wNlNTUVPr27VvjlpGRUW27uk5+\nPP05Y0yd2518vmvXrhQWFpKXl8esWbO47bbbOHLkiAdbVbe6TuA8XWP+ImhMW33Nk+071cSJE9m5\ncydffvklXbp04eGHHz6b8prkbNt2JvsiEPddQ6/zh30HnmlfWVkZI0eOZPbs2bRp06bWzwjk/Vdb\n+850/9lyHspJH3zwQZ2/Cw8PZ9++fXTu3Jm9e/fSqVOnGttERERQWFhY9bioqIiIiIh6X9+qVSta\ntWoFQHJyMlFRUeTn55OcnOzJptXq9HoLCwuJjIysd5uioiIiIyM5ceLEGbfV1zzZvlNfe2p77rnn\nHm688UZvNaFOZ9u2k/uoLoG+7xpqnz/sO2h6+06cOMGIESO44447qp03Fyz7r672nen+89shr5Mn\nPwJ1nvzYv39/8vPzKSgooLy8nCVLlpCWllbv6w8cOIDb7Qbg22+/JT8/n0svvdQXTaq33pPS0tJ4\n+eWXAcjJySEsLIzw8PCzaquveat9e/furXr922+/XeNIIl9oStvqEwz7rj7+sO+gae0zxjB+/Hji\n4uJ48MEHa7wm0Pdffe074/13lgcVeN33339vhgwZYmJiYkxqaqo5ePCgMcaY4uJiM2zYsKrtsrKy\nTM+ePU1UVJR54oknGnz9m2++aeLj401iYqJJTk427733nk/bVVu9zz77rHn22WertnnggQdMVFSU\n6devn1m/fn29rzWm7rbawRvtGzt2rOnbt6/p16+fuemmm8y+fft816BTNKVtY8aMMV26dDGtWrUy\nkZGR5sUXXzTGBM++q6t9/rLvjDn79n366afG4XCYhIQEk5iYaBITE83y5cuNMcGx/+pr35nuP53Y\nKCIiHuG3Q14iIhJYFCgiIuIRChQREfEIBYqIiHiEAkVERDxCgSIiIh6hQJFm7/Dhw8ybNw+wTuQa\nNWqUx9577ty5LFq0yGPvN3r0aHbu3Omx9xPxJJ2HIs1eQUEBN954I5s2bfLo+xpjSE5O5vPPPyck\nxDOrHH3wwQe8++671a5ZIeIv1EORZu8///M/2bFjB0lJSYwePbpqeYlFixYxfPhwrr32Wnr06MHc\nuXOZOXMmycnJXHnllRw8eBCAHTt2MHToUPr378/VV1/N1q1bAVizZg29e/euCpM5c+YQHx9PQkIC\nt956KwBHjx7l7rvvJiUlheTk5KqFUd1uN4888gh9+/YlISGBuXPnAuB0OsnKyvLpv49Io3l+AQCR\nwFJQUFB1YahT7y9cuNBER0ebsrIys3//ftOuXTszf/58Y4wxDz30kHnqqaeMMcZcc801Jj8/3xhj\nXbjommuuMcYY89e//tXMnDmz6nO6du1qysvLjTHGHD582BhjzKOPPmpeeeUVY4wxBw8eND179jRH\njx41//jHP8yoUaOM2+02xhhTWlpa9T5XX311jYsnifgDW1cbFvEH5pRRX3PaCPDgwYNp3bo1rVu3\nJiwsrGq11b59+7Jx40aOHj3K2rVrq827lJeXA7B7924GDBhQ9Xy/fv247bbbGD58eNUigitWrODd\nd99l5syZAPz73/9m9+7dfPjhh0ycOJEWLaxBhFMvXtW1a1cKCgqIjY315D+DSJMpUETqce6551bd\nb9GiRdXjFi1aUFFRQWVlJR06dKh2ffVTnRpQmZmZfPLJJ7z77rtMmzatas7mrbfeIiYmpt7Xnv78\nyaAR8Sf6r1KavbZt257xRdZOftm3bduWHj16sHTp0qrnN27cCMAll1xSdYlcYwy7d+/G6XQyffp0\nDh8+TFlZGdddd121CfaTwZSamsr8+fOrLrVwcr4GrCPRLrnkkrNsrYj3KFCk2bvgggu46qqr6Nu3\nL3/84x+rrmJ3+hX4Tr9/8vGrr77KCy+8QGJiIn369KmaWB8wYABffPEFABUVFYwdO5Z+/fqRnJzM\n73//e9q3b89f/vIXTpw4Qb9+/ejTpw9Tp04FrIsZXXzxxfTr14/ExEQWL14MWBdCKioqonfv3t7/\nhxE5QzpsWMRLzE+HDa9bt67qKqFNtWLFCjIzM5k9e7ZH3k/Ek9RDEfESh8PBvffey6uvvuqx93z+\n+ed56KGHPPZ+Ip6kHoqIiHiEeigiIuIRChQREfEIBYqIiHiEAkVERDxCgSIiIh6hQBEREY/4f+ZL\nUYpOSazjAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7b9d748>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  value  of  current  flowing  at  a  time  equal  to  two  time  constants  is   2.59   A\n",
+        "\n",
+        "  (b)the  time  for  the  current  to  grow  to  1.5  A  is   0.00347   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 271</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the final value of current, \n",
+      "#(b) the time constant of the circuit, \n",
+      "#(c) the value of current after a time equal to the time constant from the instant the supply oltage is connected, \n",
+      "#(d) the expected time for the current to rise to within 1% of its final value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.04;#  in  Henry\n",
+      "R  =  10;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "t1  =  tou\n",
+      "I  =  V/R\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "i2  =  0.01*I\n",
+      "t2  =  -1*tou*(-5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  final  value  of  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  is  \",round(tou*1000,2),\"msec\"\n",
+      "print  \"\\n  (c)  value  of  current  after  a  time  equal  to  the  time  constant  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (d)expected  time  for  current  to  rise  to  within  0.01  times  of  its  final  value  is  \",round(t2*1000,2),\"msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  final  value  of  current  is   12.0   A\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit  is   4.0 msec\n",
+        "\n",
+        "  (c)  value  of  current  after  a  time  equal  to  the  time  constant  is   7.59   A\n",
+        "\n",
+        "  (d)expected  time  for  current  to  rise  to  within  0.01  times  of  its  final  value  is   20.0 msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 271</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate:\n",
+      "#(a) the steady state value of current flowing in the winding,\n",
+      "#(b) the time constant of the circuit,\n",
+      "#(c) the value of the induced e.m.f. after 0.1 s,\n",
+      "#(d) the time for the current to rise to 85% of its final value, and\n",
+      "#(e) the value of the current after 0.3 s\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  3;#  in  Henry\n",
+      "R  =  15;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "t1  =  0.1;#  in  secs\n",
+      "t3  =  0.3;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "I  =  V/R\n",
+      "i2  =  0.85*I;#  in  amperes\n",
+      "VL  =  V*math.e**(-1*t1/tou)\n",
+      "t2  =  -1*tou*math.log(1  -  (i2/I))\n",
+      "i3  =  I*(1  -  math.e**(-1*t3/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Results  \\n\\n\"\n",
+      "print  \"\\n  (a)  steady  state  value  of  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  is  \",round(tou,2),\"  sec\"\n",
+      "print  \"\\n  (c)value  of  the  induced  e.m.f.  after  0.1  s  is  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (d)  time  for  the  current  to  rise  to  0.85  times  of  its  final  values  is  \",round(t2,2),\"  A\"\n",
+      "print  \"\\n  (e)value  of  the  current  after  0.3  s  is  \",round(i3,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Results  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  steady  state  value  of  current  is   8.0   A\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit  is   0.2   sec\n",
+        "\n",
+        "  (c)value  of  the  induced  e.m.f.  after  0.1  s  is   72.78   V\n",
+        "\n",
+        "  (d)  time  for  the  current  to  rise  to  0.85  times  of  its  final  values  is   0.38   A\n",
+        "\n",
+        "  (e)value  of  the  current  after  0.3  s  is   6.21   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 273</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#draw the current/time characteristic when the supply is removed and replaced by a shorting link\n",
+      "#determine (a) the current flowing in the winding 3 s after being shorted-out and (b) the time for the current to decay to 5 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohms\n",
+      "V  =  110;#  in  Volts\n",
+      "tou  =  2;#  in  secs\n",
+      "t1  = 3;#  in  secs\n",
+      "i2  = 5;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  tou*R\n",
+      "I  =  V/R\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "for h in range(100):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    i.append(I*math.e**(-1*k/tou))\n",
+      "plot(t,i,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "i1  =  I*(math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*log((i2/I))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  in  the  winding  3  s  after  being  shorted-out  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (b)the  time  for  the  current  to  decay  to  5  A  is  \",round(t2,2),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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re1q3bq0AYGNjY2N7hNa6detHqseaXYwtKytD+/btsXnzZvj5+aFTp073XIwl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\n/T/TLOQZYjmougAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x77abf28>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  in  the  winding  3  s  after  being  shorted-out  is   1.64   A\n",
+        "\n",
+        "  (b)the  time  for  the  current  to  decay  to  5  A  is   0.77   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 273</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine: (a) the resistance of the coil,\n",
+      "#(b) the current flowing in the circuit one second after the shorting link has been placed in the circuit, and \n",
+      "#(c) the time taken for the current to fall to 10% of its initial value.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  6;#  in  Henry\n",
+      "r  =  10;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "tou  =  0.3;#  in  secs\n",
+      "t1  =  1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  (L/tou)  -  r\n",
+      "Rt  =  R  +  r\n",
+      "I  =  V/Rt\n",
+      "i2  =  0.1*I\n",
+      "i1  =  I*(math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*math.log((i2/I))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"  \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  resistance  of  the  coil  is  \",round(R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)  current  flowing  in  circuit  one  second  after  the  shorting  link  has  been  placed  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (c)the  time  for  the  current  to  decay  to  0.1  times  of  initial  value  is  \",round(t2,2),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  resistance  of  the  coil  is   10.0   ohm\n",
+        "\n",
+        "  (b)  current  flowing  in  circuit  one  second  after  the  shorting  link  has  been  placed  is   0.21   A\n",
+        "\n",
+        "  (c)the  time  for  the  current  to  decay  to  0.1  times  of  initial  value  is   0.69   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 274</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the time constant of the circuit and the steady-state value of the current flowing in the circuit. \n",
+      "#Find (a) the current flowing in the circuit at a time equal to one time constant, \n",
+      "#(b) the voltage drop across the inductor at a time equal to two time constants and \n",
+      "#(c) the voltage drop across the resistor after a time equal to three time constants.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.2;#  in  Henry\n",
+      "R  =  1000;#  in  ohms\n",
+      "V  =  24;#  in  Volts\n",
+      "t1  =  1*L/R;#  in  secs\n",
+      "t2  =  2*L/R;#  in  secs\n",
+      "t3  =  3*L/R;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "I  =  V/R\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "VL  =  V*(math.e**(-1*t2/tou))\n",
+      "VR  =  V*(1  -  math.e**(-1*t3/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  time  constant  of  circuit  is  \",round(tou*1000,6),\"msec, and steady-state  value  of current  is  \",round(I*1000,2),\"mA\"\n",
+      "print  \"\\n  (a)  urrent  flowing  in  the  circuit  at  a  time  equal  to  one  time  constant  is  \",round(i1*1000,2),\"mA\"\n",
+      "print  \"\\n  (b)  voltage  drop  across  the  inductor  at  a  time  equal  to  two  time  constants  is  \",round(VL,3),\"  V\"\n",
+      "print  \"\\n  (c)the  voltage  drop  across  the  resistor  after  a  time  equal  to  three  time  constants  is  \",round(VR,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  time  constant  of  circuit  is   0.2 msec, and steady-state  value  of current  is   24.0 mA\n",
+        "\n",
+        "  (a)  urrent  flowing  in  the  circuit  at  a  time  equal  to  one  time  constant  is   15.17 mA\n",
+        "\n",
+        "  (b)  voltage  drop  across  the  inductor  at  a  time  equal  to  two  time  constants  is   3.248   V\n",
+        "\n",
+        "  (c)the  voltage  drop  across  the  resistor  after  a  time  equal  to  three  time  constants  is   22.81   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_3.ipynb
new file mode 100755
index 00000000..81f680ae
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_17-checkpoint_3.ipynb
@@ -0,0 +1,865 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ca1c7401f018e12f4456a542a7349262c450bbcce53eee5003079be7be1331af"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 17: D.c. transients</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 262</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "C  =  15E-6;#  in  Farads\n",
+      "R  =  47000;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "t1  =  tou\n",
+      "Vctou  =  V*(1-math.e**(-1*t1/tou))\n",
+      "Vct  =  V/2\n",
+      "t0  =  -1*tou*math.log(1  -  Vct/V)\n",
+      "t=[]\n",
+      "Vc=[]\n",
+      "I  =  V/R\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    Vc.append(V*(1 - math.e**(-1*k/tou)))\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,Vc,'-')\n",
+      "#plot(t,Vc,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('Volts(V)')\n",
+      "show()\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitor  voltage  at  a  time  equal  to  one  time  constant  =  \",round(Vctou,2),\"  V\"\n",
+      "print  \"\\n  (b)the  time  for  the  capacitor  voltage  to  reach  one  half  of  its  steady  state  value  =  \",round(t0,5),\" secs\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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kSBouSe24oRNb4VhCoEiSqtB5lPYGQvWEQWZJkiRJkiRJkiRJkiRJkiRJSrv/\nD8/iLwbt6hd+AAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x34660b8>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitor  voltage  at  a  time  equal  to  one  time  constant  =   75.85   V\n",
+        "\n",
+        "  (b)the  time  for  the  capacitor  voltage  to  reach  one  half  of  its  steady  state  value  =   0.48867  secs\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 263</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "C  =  4E-6;#  in  Farads\n",
+      "R  =  220000;#  in  ohms\n",
+      "V  =  24;#  in  Volts\n",
+      "t1  =  1.5;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "\n",
+      "t=[]\n",
+      "Vc=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    Vc.append(V*math.e**(-1*k/tou))\n",
+      "#plt.figsize(10,8)\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,Vc,'-')\n",
+      "#plot(t,Vc,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('P.D across Capacitor(V)')\n",
+      "show()\n",
+      "\n",
+      "t=[]\n",
+      "VR=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    VR.append(V*(1 - math.e**(-1*k/tou)))\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,VR,'-')\n",
+      "#plot(t,VR,'-*')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('P.D across Resistor(V)')\n",
+      "show()\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "for h in range(50):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    i.append(I*math.e**(-1*k/tou))\n",
+      "fig  = plt.figure()\n",
+      "#canvas = fc(fig)\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(t,i,'-')\n",
+      "#plot(t,i,'*-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "\n",
+      "Vct1  =  V*math.e**(-1*t1/tou)\n",
+      "VRt1  =  V*math.e**(-1*t1/tou)\n",
+      "it1   =  I*math.e**(-1*t1/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  value  of  capacitor  voltage  is  \",round(Vct1,1),\"  V,  resistor  voltage  is  \",round(VRt1,1),\"  V,\"\n",
+      "print  \"current  is  \",round(0.02,2),\"  mA  at  one  and  a  half  seconds  after  discharge  has  started.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+       "text": [
+        "<matplotlib.figure.Figure at 0x77c6e10>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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HRKxh2bCP3W6nffv2fPzxx1x//fXceOONLFq0iPDw8PPhvHzYp21bWLoUYmOt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\nD/p/wQuHfYdHfM4AAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7b9e4a8>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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1NbVBmbS0NCQkJAAAoqKiUFFRgdLS0hbr1q+TkJCA3bt3AwBSU1Mxa9Ys+Pr6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\nqer/AUdcZmZ6XUFsAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7bac898>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  value  of  capacitor  voltage  is   4.4   V,  resistor  voltage  is   4.4   V,\n",
+        "current  is   0.02   mA  at  one  and  a  half  seconds  after  discharge  has  started.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 265</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  20E-6;#  in  Farads\n",
+      "R  =  50000;#  in  ohms\n",
+      "V  =  20;#  in  Volts\n",
+      "t1  =  1;#  in  secs\n",
+      "t2  =  2;#  in  secs\n",
+      "VRt  =  15;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "Vct1  =  V*(1-math.e**(-1*t2/tou))\n",
+      "t3  =  -1*tou*math.log(VRt/V)\n",
+      "it1  =  I*math.e**(-1*t1/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)initial  value  of  the  current  flowing  is  \",round(I*1000,1),\"mA\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  \",round(tou,2),\"  Sec\"\n",
+      "print  \"\\n  (c)the  value  of  the  current  one  second  after  connection,  \",round((it1/1E-3),3),\"  mA\"\n",
+      "print  \"\\n  (d)the  value  of  the  capacitor  voltage  two  seconds  after  connection  \",round(Vct1,1),\"  V\"\n",
+      "print  \"\\n  (e)the  time  after  connection  when  the  resistor  voltage  is  15  V  is  \",round(t3,3),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)initial  value  of  the  current  flowing  is   0.4 mA\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit   1.0   Sec\n",
+        "\n",
+        "  (c)the  value  of  the  current  one  second  after  connection,   0.147   mA\n",
+        "\n",
+        "  (d)the  value  of  the  capacitor  voltage  two  seconds  after  connection   17.3   V\n",
+        "\n",
+        "  (e)the  time  after  connection  when  the  resistor  voltage  is  15  V  is   0.288   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17.4, page no. 266</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.5E-6;#  in  Farads\n",
+      "V  =  10;#  in  Volts\n",
+      "tou  =  0.012;#  in  secs\n",
+      "t1  =  0.007;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  tou/C\n",
+      "Vc  =  V*(1-math.e**(-1*t1/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)value  of  the  resistor  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)capacitor  voltage  is  \",round(Vc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)value  of  the  resistor  is   24000.0   ohm\n",
+        "\n",
+        "  (b)capacitor  voltage  is   4.42   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 267</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  50000;#  in  ohms\n",
+      "V  =  100;#  in  Volts\n",
+      "Vc1  =  20;#  in  Volts\n",
+      "tou  =  0.8;#  in  secs\n",
+      "t1  =  0.5;#  in  secs\n",
+      "t2  =  1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "C  =  tou/R\n",
+      "t  =  -1*tou*math.log(Vc1/V)\n",
+      "I  =  V/R\n",
+      "it1  =  I*math.e**(-1*t1/tou)\n",
+      "Vc  =  V*math.e**(-1*t2/tou)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  the  capacitor  is  \",round((C/1E-6),2),\"uF\"\n",
+      "print  \"\\n  (b)the  time  for  the  capacitor  voltage  to  fall  to  20  V  is  \",round(t,2),\"  sec\"\n",
+      "print  \"\\n  (c)the  current  flowing  when  the  capacitor  has  been  discharging  for  0.5  s  is  \",round(it1*1000,2),\"mA\"\n",
+      "print  \"\\n  (d)voltage  drop  across  resistor  when  the  capacitor  has  been  discharging  for  one  second  is  \",round(Vc,1),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  the  capacitor  is   16.0 uF\n",
+        "\n",
+        "  (b)the  time  for  the  capacitor  voltage  to  fall  to  20  V  is   1.29   sec\n",
+        "\n",
+        "  (c)the  current  flowing  when  the  capacitor  has  been  discharging  for  0.5  s  is   1.07 mA\n",
+        "\n",
+        "  (d)voltage  drop  across  resistor  when  the  capacitor  has  been  discharging  for  one  second  is   28.7   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6 page no. 268</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  0.1E-6;#  in  Farads\n",
+      "R  =  4000;#  in  ohms\n",
+      "V  =  200;#  in  Volts\n",
+      "Vc1  =  2;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  R*C\n",
+      "I  =  V/R\n",
+      "t  =  -1*tou*math.log(Vc1/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  initial  discharge  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)Time  constant  tou  is  \",round(tou,5),\"  sec\"\n",
+      "print  \"\\n  (c)min.  time  required  for voltage  across capacitor  to  fall  to  less  than  2  V  is  \",round(t*1000,0),\"  msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  initial  discharge  current  is   0.05   A\n",
+        "\n",
+        "  (b)Time  constant  tou  is   0.0004   sec\n",
+        "\n",
+        "  (c)min.  time  required  for voltage  across capacitor  to  fall  to  less  than  2  V  is   2.0   msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 270</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.1;#  in  Henry\n",
+      "R  =  20;#  in  ohms\n",
+      "V  =  60;#  in  Volts\n",
+      "i2  =  1.5;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "t1  =  2*tou\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "I  =  V/R\n",
+      "for h in range(250):\n",
+      "    t.append((h-1)/10000)\n",
+      "    k=(h-1)/10000\n",
+      "    i.append(I*(1  -  math.e**(-1*k/tou)))\n",
+      "plot(t,i,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*math.log(1  -  i2/I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  value  of  current  flowing  at  a  time  equal  to  two  time  constants  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (b)the  time  for  the  current  to  grow  to  1.5  A  is  \",round(t2,5),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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Pnw/AhAkTGDZsGFlZWURHR9O6dWsWLlxoV7k+t2WLFSrx8XZXIiLSOFrLy08d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\nUYpOSazjAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7b9d748>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  value  of  current  flowing  at  a  time  equal  to  two  time  constants  is   2.59   A\n",
+        "\n",
+        "  (b)the  time  for  the  current  to  grow  to  1.5  A  is   0.00347   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 271</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.04;#  in  Henry\n",
+      "R  =  10;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "t1  =  tou\n",
+      "I  =  V/R\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "i2  =  0.01*I\n",
+      "t2  =  -1*tou*(-5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  final  value  of  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  is  \",round(tou*1000,2),\"msec\"\n",
+      "print  \"\\n  (c)  value  of  current  after  a  time  equal  to  the  time  constant  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (d)expected  time  for  current  to  rise  to  within  0.01  times  of  its  final  value  is  \",round(t2*1000,2),\"msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  final  value  of  current  is   12.0   A\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit  is   4.0 msec\n",
+        "\n",
+        "  (c)  value  of  current  after  a  time  equal  to  the  time  constant  is   7.59   A\n",
+        "\n",
+        "  (d)expected  time  for  current  to  rise  to  within  0.01  times  of  its  final  value  is   20.0 msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 271</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  3;#  in  Henry\n",
+      "R  =  15;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "t1  =  0.1;#  in  secs\n",
+      "t3  =  0.3;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "I  =  V/R\n",
+      "i2  =  0.85*I;#  in  amperes\n",
+      "VL  =  V*math.e**(-1*t1/tou)\n",
+      "t2  =  -1*tou*math.log(1  -  (i2/I))\n",
+      "i3  =  I*(1  -  math.e**(-1*t3/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Results  \\n\\n\"\n",
+      "print  \"\\n  (a)  steady  state  value  of  current  is  \",round(I,2),\"  A\"\n",
+      "print  \"\\n  (b)time  constant  of  the  circuit  is  \",round(tou,2),\"  sec\"\n",
+      "print  \"\\n  (c)value  of  the  induced  e.m.f.  after  0.1  s  is  \",round(VL,2),\"  V\"\n",
+      "print  \"\\n  (d)  time  for  the  current  to  rise  to  0.85  times  of  its  final  values  is  \",round(t2,2),\"  A\"\n",
+      "print  \"\\n  (e)value  of  the  current  after  0.3  s  is  \",round(i3,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Results  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  steady  state  value  of  current  is   8.0   A\n",
+        "\n",
+        "  (b)time  constant  of  the  circuit  is   0.2   sec\n",
+        "\n",
+        "  (c)value  of  the  induced  e.m.f.  after  0.1  s  is   72.78   V\n",
+        "\n",
+        "  (d)  time  for  the  current  to  rise  to  0.85  times  of  its  final  values  is   0.38   A\n",
+        "\n",
+        "  (e)value  of  the  current  after  0.3  s  is   6.21   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 273</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "from pylab import *\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohms\n",
+      "V  =  110;#  in  Volts\n",
+      "tou  =  2;#  in  secs\n",
+      "t1  = 3;#  in  secs\n",
+      "i2  = 5;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      "L  =  tou*R\n",
+      "I  =  V/R\n",
+      "\n",
+      "t=[]\n",
+      "i=[]\n",
+      "for h in range(100):\n",
+      "    t.append((h-1)/10)\n",
+      "    k=(h-1)/10\n",
+      "    i.append(I*math.e**(-1*k/tou))\n",
+      "plot(t,i,'-')\n",
+      "xlabel('time(sec)')\n",
+      "ylabel('current(A)')\n",
+      "show()\n",
+      "i1  =  I*(math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*log((i2/I))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \" \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  in  the  winding  3  s  after  being  shorted-out  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (b)the  time  for  the  current  to  decay  to  5  A  is  \",round(t2,2),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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re1q3bq0AYGNjY2N7hNa6detHqseaXYwtKytD+/btsXnzZvj5+aFTp073XIwl\nIiLtaTZ14+zsjEWLFqFPnz4wm82YOHEiizwRkQRSb5giIiLtST+L5tVXX0VQUBDCw8MxePBgFBQU\nyI6kirS0NAQGBqJt27b44IMPZMdR1fnz59GjRw8EBwcjJCQECxculB1JdWazGZGRkejfv7/sKKrL\nz8/H0KFDERQUBL1ej4yMDNmRVDVnzhwEBwcjNDQUCQkJuHnzpuxINTJhwgR4e3sjNDTU8rm8vDzE\nx8ejXbt26N27N/Lz8x/6GtILfe/evXHkyBEcPHgQ7dq1w5w5c2RHqjGz2YypU6ciLS0NR48exapV\nq3Ds2DHZsVTj4uKCv//97zhy5AgyMjLw4YcfOtT7A4CkpCTo9XqHXD32yiuv4JlnnsGxY8dw6NAh\nh5pSNZlMWLJkCTIzM3H48GGYzWYkJyfLjlUj48ePR1pa2l2fmzt3LuLj43Hy5En06tULc+fOfehr\nSC/08fHxqPOfQy5jYmJw4cIFyYlq7s6bxVxcXCw3izkKHx8fREREAADc3d0RFBSE7OxsyanUc+HC\nBaSmpmLSpEkOd+d2QUEBtm3bhgkTJgAQ19IaNGggOZV6PD094eLiguLiYpSVlaG4uBjNmjWTHatG\nunbtikaNGt31uXXr1mHcuHEAgHHjxmHNmjUPfQ3phf5OS5cuxTPPPCM7Ro3d72axixcvSkykHZPJ\nhP379yMmJkZ2FNVMnz4d8+bNswxAHMnZs2fRpEkTjB8/HlFRUZg8eTKKi4tlx1KNl5cXZsyYgYCA\nAPj5+aFhw4aIi4uTHUt1ly5dgre3NwDA29sbly5deuj3W+W/5Pj4eISGht7TUlJSLN8ze/Zs1K1b\nFwkJCdaIpClH/HP/fgoLCzF06FAkJSXB3d1ddhxVrF+/Hk2bNkVkZKTDjeYBsew5MzMTL730EjIz\nM+Hm5lbpn/32JCsrCwsWLIDJZEJ2djYKCwuxcuVK2bE0pdPpKq05VtmKZ9OmTQ/9+vLly5GamorN\nmzdbI47mmjVrhvPnz1s+Pn/+PPz9/SUmUt+tW7cwZMgQjBkzBgMHDpQdRzU7d+7EunXrkJqaihs3\nbuDatWsYO3YsVqxYITuaKvz9/eHv74+OHTsCAIYOHepQhX7v3r3o0qULGjduDAAYPHgwdu7cidGj\nR0tOpi5vb2/k5ubCx8cHOTk5aNq06UO/X/rfpmlpaZg3bx7Wrl2Lxx57THYcVURHR+PUqVMwmUwo\nLS3FF1+m33ZKAAAE20lEQVR8gQEDBsiOpRpFUTBx4kTo9XpMmzZNdhxVvf/++zh//jzOnj2L5ORk\n9OzZ02GKPCCurzRv3hwnT54EAKSnpyO4phup2JDAwEBkZGSgpKQEiqIgPT0der1edizVDRgwAJ9+\n+ikA4NNPP618sFWjfQ5U0KZNGyUgIECJiIhQIiIilBdffFF2JFWkpqYq7dq1U1q3bq28//77suOo\natu2bYpOp1PCw8Mt/96+++472bFUZzQalf79+8uOoboDBw4o0dHRSlhYmDJo0CAlPz9fdiRVffDB\nB4per1dCQkKUsWPHKqWlpbIj1cjIkSMVX19fxcXFRfH391eWLl2q/Pbbb0qvXr2Utm3bKvHx8crV\nq1cf+hq8YYqIyMFJn7ohIiJtsdATETk4FnoiIgfHQk9E5OBY6ImIHBwLPRGRg2OhJ7tWUFCAf/7z\nnwCAnJwcDBs2TLXXXrRoEZYvX67a6w0fPhxnz55V7fWIqorr6MmumUwm9O/fH4cPH1b1dRVFQVRU\nFPbs2QNnZ3V2Ctm0aRNSUlIccv9+sm0c0ZNde+ONN5CVlYXIyEgMHz7ccjjD8uXLMXDgQPTu3Rut\nWrXCokWLMH/+fERFRaFz5864evUqALEJVt++fREdHY1u3brhxIkTAIAdO3YgMDDQUuQXLlyI4OBg\nhIeHY9SoUQCAoqIiTJgwATExMYiKisK6desAiPMIZs6cidDQUISHh2PRokUAAIPBgNTUVKv+8yEC\nIH8LBKKaMJlMSkhIyD3Ply1bprRp00YpLCxULl++rHh6eiqLFy9WFEVRpk+frixYsEBRFEXp2bOn\ncurUKUVRFCUjI0Pp2bOnoiiKMmfOHGX+/PmWfvz8/Cy30hcUFCiKoihvvvmm8tlnnymKoihXr15V\n2rVrpxQVFSkfffSRMmzYMMVsNiuKoih5eXmW1+nWrZty9OhRbf5hED2AVXavJNKKcsfMo/K7Wcge\nPXrAzc0Nbm5uaNiwoeVYwNDQUBw6dAhFRUXYuXPnXfP6paWlAIBz584hNjbW8vmwsDAkJCRg4MCB\nlg2kNm7ciJSUFMyfPx8AcPPmTZw7dw6bN2/Giy++aNnP/s5DI/z8/GAymRzqVCeyfSz05LDq1atn\neV6nTh3Lx3Xq1EFZWRnKy8vRqFEj7N+//74/f+cvjg0bNuDHH39ESkoKZs+ebbkm8M0336Bt27YP\n/dnff94RDzQh28b/4siueXh44Pr164/0M7eLsIeHB1q1aoXVq1dbPn/o0CEAQIsWLZCbm2v5/Llz\n52AwGDB37lwUFBSgsLAQffr0uevC6u1fGPHx8Vi8eDHMZjMAWK4HAGJlUIsWLar5bomqh4We7Frj\nxo3x1FNPITQ0FK+99prlpJ3fn7rz++e3P165ciX+9a9/ISIiAiEhIZYLqrGxsdi7dy8AcSrT888/\nj7CwMERFReGVV15BgwYN8Pbbb+PWrVsICwtDSEgIZs2aBQCYNGkSAgICEBYWhoiICKxatQqAOKzl\nwoULCAwM1P4fDNEduLyS6D6U/yyv3LVrF+rWravKa27cuBEbNmxAUlKSKq9HVFUc0RPdh06nw+TJ\nk1U9b/STTz7B9OnTVXs9oqriiJ6IyMFxRE9E5OBY6ImIHBwLPRGRg2OhJyJycCz0REQOjoWeiMjB\n/T/TLOQZYjmougAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x77abf28>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  in  the  winding  3  s  after  being  shorted-out  is   1.64   A\n",
+        "\n",
+        "  (b)the  time  for  the  current  to  decay  to  5  A  is   0.77   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 273</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  6;#  in  Henry\n",
+      "r  =  10;#  in  ohms\n",
+      "V  =  120;#  in  Volts\n",
+      "tou  =  0.3;#  in  secs\n",
+      "t1  =  1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "R  =  (L/tou)  -  r\n",
+      "Rt  =  R  +  r\n",
+      "I  =  V/Rt\n",
+      "i2  =  0.1*I\n",
+      "i1  =  I*(math.e**(-1*t1/tou))\n",
+      "t2  =  -1*tou*math.log((i2/I))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"  \\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  resistance  of  the  coil  is  \",round(R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)  current  flowing  in  circuit  one  second  after  the  shorting  link  has  been  placed  is  \",round(i1,2),\"  A\"\n",
+      "print  \"\\n  (c)the  time  for  the  current  to  decay  to  0.1  times  of  initial  value  is  \",round(t2,2),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  \n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  resistance  of  the  coil  is   10.0   ohm\n",
+        "\n",
+        "  (b)  current  flowing  in  circuit  one  second  after  the  shorting  link  has  been  placed  is   0.21   A\n",
+        "\n",
+        "  (c)the  time  for  the  current  to  decay  to  0.1  times  of  initial  value  is   0.69   sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 274</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.2;#  in  Henry\n",
+      "R  =  1000;#  in  ohms\n",
+      "V  =  24;#  in  Volts\n",
+      "t1  =  1*L/R;#  in  secs\n",
+      "t2  =  2*L/R;#  in  secs\n",
+      "t3  =  3*L/R;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "tou  =  L/R\n",
+      "I  =  V/R\n",
+      "i1  =  I*(1  -  math.e**(-1*t1/tou))\n",
+      "VL  =  V*(math.e**(-1*t2/tou))\n",
+      "VR  =  V*(1  -  math.e**(-1*t3/tou))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  time  constant  of  circuit  is  \",round(tou*1000,6),\"msec, and steady-state  value  of current  is  \",round(I*1000,2),\"mA\"\n",
+      "print  \"\\n  (a)  urrent  flowing  in  the  circuit  at  a  time  equal  to  one  time  constant  is  \",round(i1*1000,2),\"mA\"\n",
+      "print  \"\\n  (b)  voltage  drop  across  the  inductor  at  a  time  equal  to  two  time  constants  is  \",round(VL,3),\"  V\"\n",
+      "print  \"\\n  (c)the  voltage  drop  across  the  resistor  after  a  time  equal  to  three  time  constants  is  \",round(VR,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  time  constant  of  circuit  is   0.2 msec, and steady-state  value  of current  is   24.0 mA\n",
+        "\n",
+        "  (a)  urrent  flowing  in  the  circuit  at  a  time  equal  to  one  time  constant  is   15.17 mA\n",
+        "\n",
+        "  (b)  voltage  drop  across  the  inductor  at  a  time  equal  to  two  time  constants  is   3.248   V\n",
+        "\n",
+        "  (c)the  voltage  drop  across  the  resistor  after  a  time  equal  to  three  time  constants  is   22.81   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint.ipynb
new file mode 100755
index 00000000..f3f727b4
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint.ipynb
@@ -0,0 +1,535 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d16259f8430b68bdc5f61e2c9d378821ed7dc4c5b9f7144ffa85cfd5e017e63d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 18: Operational amplifiers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 279</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the output voltage of the amplifier.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vi2  =  2.45;#  in  Volts\n",
+      "Vi1  =  2.35;#  in  Volts\n",
+      "A0  =  120;#  open-loop  voltage  gain\n",
+      "\n",
+      "#calculation:\n",
+      "Vo  =  A0*(Vi2  -  Vi1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  output  voltage  is  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  output  voltage  is   12.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 281</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the common-mode gain of an op amp\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  150E3;#  differential  voltage  gain  \n",
+      "CMRR  =  90;#  in  dB\n",
+      "\n",
+      "#calculation:\n",
+      "CMG  =  Vg/(10**(CMRR/20))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  common-mode  gain  is  \",round(CMG,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  common-mode  gain  is   4.74"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 282</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the common-mode gain and the CMRR.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  120;#  differential  voltage  gain  \n",
+      "Vi  =  3;#  in  Volts\n",
+      "Vo  =  0.024;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "CMG  =  Vo/Vi\n",
+      "CMRR  =  20*(1/2.303)*math.log(Vg/CMG)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  common-mode  gain  is  \",round(CMG,3),\"  and  CMRR  is  \",round(CMRR,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  common-mode  gain  is   0.008   and  CMRR  is   83.51   dB"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 283</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the output voltage when the input voltage is: (a) +0.4 V (b) -1.2 V\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rf  =  2000;#  in  ohms\n",
+      "Ri  =  1000;#  in  ohms\n",
+      "Vi1  =  0.4;#  in  Volts\n",
+      "Vi2  =  -1.2;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "Vo1  =  -1*Rf*Vi1/Ri\n",
+      "Vo2  =  -1*Rf*Vi2/Ri\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  when  the  input  voltage  is  0.4V  is  \",round(Vo1,2),\"  V \"\n",
+      "print   \" and  when  the  input  voltage  is  -1.2V  is  \",round(Vo2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage  when  the  input  voltage  is  0.4V  is   -0.8   V \n",
+        " and  when  the  input  voltage  is  -1.2V  is   2.4   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 283</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the voltage gain, and\n",
+      "#(b) the output offset voltage due to the input bias current. \n",
+      "#(c) How can the effect of input bias current be minimised?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ii  =  100E-9;#  in  Amperes\n",
+      "T  =  20;#  in  \u00b0C\n",
+      "Rf  =  1E6;#  in  ohms\n",
+      "Ri  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  -1*Rf/Ri\n",
+      "Vos  =  Ii*Ri*Rf/(Ri+Rf)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  voltage  gain  is  \",round(A,2),\"\"\n",
+      "print  \"\\n  (b)output  offset  voltage  is  \",round(Vos*1000,2),\"  mV\"\n",
+      "print  \"\\n  (c)The  effect  of  input  bias  current  can  be  minimised  by  ensuring  \"\n",
+      "print  \"that  both  inputs have the  same  driving  resistance.\" \n",
+      "print  \"This  means  that  a  resistance  of  value  of  9.9  kohm  (from  part  (b))  \"\n",
+      "print   \"should  be  placed  between  the  non-inverting  (+)  terminal  and  earth.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  voltage  gain  is   -100.0 \n",
+        "\n",
+        "  (b)output  offset  voltage  is   0.99   mV\n",
+        "\n",
+        "  (c)The  effect  of  input  bias  current  can  be  minimised  by  ensuring  \n",
+        "that  both  inputs have the  same  driving  resistance.\n",
+        "This  means  that  a  resistance  of  value  of  9.9  kohm  (from  part  (b))  \n",
+        "should  be  placed  between  the  non-inverting  (+)  terminal  and  earth.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 284</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design an inverting amplifier\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  40;#  in  dB\n",
+      "bf  =  5000;#  in  Hz\n",
+      "Ri  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  10**(Vg/20)\n",
+      "Rf  =  A*Ri\n",
+      "f  =  A*bf\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  voltage  gain  is  \",round(A,2),\",  Rf  =  \",round(Rf/1000,2),\"kohm  and  frequency  =  \",round(f/1000,2),\"  kHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  voltage  gain  is   100.0 ,  Rf  =   1000.0 kohm  and  frequency  =   500.0   kHz"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 286</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the voltage gain (b) the output voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vi  =  -0.4;#  in  Volts\n",
+      "R1  =  4700;#  in  ohms\n",
+      "R2  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  1  +  (R2/R1)\n",
+      "Vo  =  A*Vi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  the  voltage  gain  is  \",round(A,2),\"\"\n",
+      "print  \"\\n(b)  output  voltageis  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  the  voltage  gain  is   3.13 \n",
+        "\n",
+        "(b)  output  voltageis   -1.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 287</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the output voltage, Vo\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  0.5;#  in  Volts\n",
+      "V2  =  0.8;#  in  Volts\n",
+      "V3  =  1.2;#  in  Volts\n",
+      "R1  =  10000;#  in  ohms\n",
+      "R2  =  20000;#  in  ohms\n",
+      "R3  =  30000;#  in  ohms\n",
+      "Rf  =  50000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vo  =  -1*Rf*(V1/R1  +  V2/R2  +  V3/R3)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltageis  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltageis   -6.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 289</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from scipy import integrate\n",
+      "#initializing  the  variables:\n",
+      "Vs  =  -0.75;#  in  Volts\n",
+      "R  =  200000;#  in  ohms\n",
+      "C  =  2.5E-6;#  in  Farads\n",
+      "t  =  0.1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f = lambda x,a : a*1\n",
+      "y, err = integrate.quad(f, 0, 0.1, args=(-0.75,))\n",
+      "Vo  =  (-1/(C*R))*y\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  is  \",Vo,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage  is   0.15   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 290</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the output voltage Vo\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1a  =  0.005;#  in  Volts\n",
+      "V2a  =  0;#  in  Volts\n",
+      "V1b  =  0;#  in  Volts\n",
+      "V2b  =  0.005;#  in  Volts\n",
+      "V1c  =  0.05;#  in  Volts\n",
+      "V2c  =  0.025;#  in  Volts\n",
+      "V1d  =  0.025;#  in  Volts\n",
+      "V2d  =  0.05;#  in  Volts\n",
+      "R1  =  10000;#  in  ohms\n",
+      "R2  =  10000;#  in  ohms\n",
+      "R3  =  100000;#  in  ohms\n",
+      "Rf  =  100000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vo1  =  -1*Rf*V1a/R1\n",
+      "Vo2  =  (R3/(R2+R3))*(1  +  (Rf/R1))*V2b\n",
+      "Vo3  =  -1*Rf*(V1c-V2c)/R1\n",
+      "Vo4  =  (R3/(R2+R3))*(1  +  (Rf/R1))*(V2d-V1d)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)output  voltage  is  \",round(Vo1,2),\"  V\"\n",
+      "print  \"\\n  (b)output  voltage  is  \",round(Vo2,2),\"  V\"\n",
+      "print  \"\\n  (c)output  voltage  is  \",round(Vo3,2),\"  V\"\n",
+      "print  \"\\n  (d)output  voltage  is  \",round(Vo4,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)output  voltage  is   -0.05   V\n",
+        "\n",
+        "  (b)output  voltage  is   0.05   V\n",
+        "\n",
+        "  (c)output  voltage  is   -0.25   V\n",
+        "\n",
+        "  (d)output  voltage  is   0.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_1.ipynb
new file mode 100755
index 00000000..f3f727b4
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_1.ipynb
@@ -0,0 +1,535 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d16259f8430b68bdc5f61e2c9d378821ed7dc4c5b9f7144ffa85cfd5e017e63d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 18: Operational amplifiers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 279</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the output voltage of the amplifier.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vi2  =  2.45;#  in  Volts\n",
+      "Vi1  =  2.35;#  in  Volts\n",
+      "A0  =  120;#  open-loop  voltage  gain\n",
+      "\n",
+      "#calculation:\n",
+      "Vo  =  A0*(Vi2  -  Vi1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  output  voltage  is  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  output  voltage  is   12.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 281</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the common-mode gain of an op amp\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  150E3;#  differential  voltage  gain  \n",
+      "CMRR  =  90;#  in  dB\n",
+      "\n",
+      "#calculation:\n",
+      "CMG  =  Vg/(10**(CMRR/20))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  common-mode  gain  is  \",round(CMG,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  common-mode  gain  is   4.74"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 282</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the common-mode gain and the CMRR.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  120;#  differential  voltage  gain  \n",
+      "Vi  =  3;#  in  Volts\n",
+      "Vo  =  0.024;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "CMG  =  Vo/Vi\n",
+      "CMRR  =  20*(1/2.303)*math.log(Vg/CMG)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  common-mode  gain  is  \",round(CMG,3),\"  and  CMRR  is  \",round(CMRR,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  common-mode  gain  is   0.008   and  CMRR  is   83.51   dB"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 283</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the output voltage when the input voltage is: (a) +0.4 V (b) -1.2 V\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rf  =  2000;#  in  ohms\n",
+      "Ri  =  1000;#  in  ohms\n",
+      "Vi1  =  0.4;#  in  Volts\n",
+      "Vi2  =  -1.2;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "Vo1  =  -1*Rf*Vi1/Ri\n",
+      "Vo2  =  -1*Rf*Vi2/Ri\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  when  the  input  voltage  is  0.4V  is  \",round(Vo1,2),\"  V \"\n",
+      "print   \" and  when  the  input  voltage  is  -1.2V  is  \",round(Vo2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage  when  the  input  voltage  is  0.4V  is   -0.8   V \n",
+        " and  when  the  input  voltage  is  -1.2V  is   2.4   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 283</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the voltage gain, and\n",
+      "#(b) the output offset voltage due to the input bias current. \n",
+      "#(c) How can the effect of input bias current be minimised?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ii  =  100E-9;#  in  Amperes\n",
+      "T  =  20;#  in  \u00b0C\n",
+      "Rf  =  1E6;#  in  ohms\n",
+      "Ri  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  -1*Rf/Ri\n",
+      "Vos  =  Ii*Ri*Rf/(Ri+Rf)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  voltage  gain  is  \",round(A,2),\"\"\n",
+      "print  \"\\n  (b)output  offset  voltage  is  \",round(Vos*1000,2),\"  mV\"\n",
+      "print  \"\\n  (c)The  effect  of  input  bias  current  can  be  minimised  by  ensuring  \"\n",
+      "print  \"that  both  inputs have the  same  driving  resistance.\" \n",
+      "print  \"This  means  that  a  resistance  of  value  of  9.9  kohm  (from  part  (b))  \"\n",
+      "print   \"should  be  placed  between  the  non-inverting  (+)  terminal  and  earth.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  voltage  gain  is   -100.0 \n",
+        "\n",
+        "  (b)output  offset  voltage  is   0.99   mV\n",
+        "\n",
+        "  (c)The  effect  of  input  bias  current  can  be  minimised  by  ensuring  \n",
+        "that  both  inputs have the  same  driving  resistance.\n",
+        "This  means  that  a  resistance  of  value  of  9.9  kohm  (from  part  (b))  \n",
+        "should  be  placed  between  the  non-inverting  (+)  terminal  and  earth.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 284</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design an inverting amplifier\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  40;#  in  dB\n",
+      "bf  =  5000;#  in  Hz\n",
+      "Ri  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  10**(Vg/20)\n",
+      "Rf  =  A*Ri\n",
+      "f  =  A*bf\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  voltage  gain  is  \",round(A,2),\",  Rf  =  \",round(Rf/1000,2),\"kohm  and  frequency  =  \",round(f/1000,2),\"  kHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  voltage  gain  is   100.0 ,  Rf  =   1000.0 kohm  and  frequency  =   500.0   kHz"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 286</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the voltage gain (b) the output voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vi  =  -0.4;#  in  Volts\n",
+      "R1  =  4700;#  in  ohms\n",
+      "R2  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  1  +  (R2/R1)\n",
+      "Vo  =  A*Vi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  the  voltage  gain  is  \",round(A,2),\"\"\n",
+      "print  \"\\n(b)  output  voltageis  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  the  voltage  gain  is   3.13 \n",
+        "\n",
+        "(b)  output  voltageis   -1.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 287</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the output voltage, Vo\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  0.5;#  in  Volts\n",
+      "V2  =  0.8;#  in  Volts\n",
+      "V3  =  1.2;#  in  Volts\n",
+      "R1  =  10000;#  in  ohms\n",
+      "R2  =  20000;#  in  ohms\n",
+      "R3  =  30000;#  in  ohms\n",
+      "Rf  =  50000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vo  =  -1*Rf*(V1/R1  +  V2/R2  +  V3/R3)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltageis  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltageis   -6.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 289</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from scipy import integrate\n",
+      "#initializing  the  variables:\n",
+      "Vs  =  -0.75;#  in  Volts\n",
+      "R  =  200000;#  in  ohms\n",
+      "C  =  2.5E-6;#  in  Farads\n",
+      "t  =  0.1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f = lambda x,a : a*1\n",
+      "y, err = integrate.quad(f, 0, 0.1, args=(-0.75,))\n",
+      "Vo  =  (-1/(C*R))*y\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  is  \",Vo,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage  is   0.15   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 290</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the output voltage Vo\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1a  =  0.005;#  in  Volts\n",
+      "V2a  =  0;#  in  Volts\n",
+      "V1b  =  0;#  in  Volts\n",
+      "V2b  =  0.005;#  in  Volts\n",
+      "V1c  =  0.05;#  in  Volts\n",
+      "V2c  =  0.025;#  in  Volts\n",
+      "V1d  =  0.025;#  in  Volts\n",
+      "V2d  =  0.05;#  in  Volts\n",
+      "R1  =  10000;#  in  ohms\n",
+      "R2  =  10000;#  in  ohms\n",
+      "R3  =  100000;#  in  ohms\n",
+      "Rf  =  100000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vo1  =  -1*Rf*V1a/R1\n",
+      "Vo2  =  (R3/(R2+R3))*(1  +  (Rf/R1))*V2b\n",
+      "Vo3  =  -1*Rf*(V1c-V2c)/R1\n",
+      "Vo4  =  (R3/(R2+R3))*(1  +  (Rf/R1))*(V2d-V1d)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)output  voltage  is  \",round(Vo1,2),\"  V\"\n",
+      "print  \"\\n  (b)output  voltage  is  \",round(Vo2,2),\"  V\"\n",
+      "print  \"\\n  (c)output  voltage  is  \",round(Vo3,2),\"  V\"\n",
+      "print  \"\\n  (d)output  voltage  is  \",round(Vo4,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)output  voltage  is   -0.05   V\n",
+        "\n",
+        "  (b)output  voltage  is   0.05   V\n",
+        "\n",
+        "  (c)output  voltage  is   -0.25   V\n",
+        "\n",
+        "  (d)output  voltage  is   0.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb
new file mode 100755
index 00000000..f3f727b4
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb
@@ -0,0 +1,535 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d16259f8430b68bdc5f61e2c9d378821ed7dc4c5b9f7144ffa85cfd5e017e63d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 18: Operational amplifiers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 279</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the output voltage of the amplifier.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vi2  =  2.45;#  in  Volts\n",
+      "Vi1  =  2.35;#  in  Volts\n",
+      "A0  =  120;#  open-loop  voltage  gain\n",
+      "\n",
+      "#calculation:\n",
+      "Vo  =  A0*(Vi2  -  Vi1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  output  voltage  is  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  output  voltage  is   12.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 281</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the common-mode gain of an op amp\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  150E3;#  differential  voltage  gain  \n",
+      "CMRR  =  90;#  in  dB\n",
+      "\n",
+      "#calculation:\n",
+      "CMG  =  Vg/(10**(CMRR/20))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  common-mode  gain  is  \",round(CMG,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  common-mode  gain  is   4.74"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 282</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the common-mode gain and the CMRR.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  120;#  differential  voltage  gain  \n",
+      "Vi  =  3;#  in  Volts\n",
+      "Vo  =  0.024;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "CMG  =  Vo/Vi\n",
+      "CMRR  =  20*(1/2.303)*math.log(Vg/CMG)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  common-mode  gain  is  \",round(CMG,3),\"  and  CMRR  is  \",round(CMRR,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  common-mode  gain  is   0.008   and  CMRR  is   83.51   dB"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 283</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the output voltage when the input voltage is: (a) +0.4 V (b) -1.2 V\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rf  =  2000;#  in  ohms\n",
+      "Ri  =  1000;#  in  ohms\n",
+      "Vi1  =  0.4;#  in  Volts\n",
+      "Vi2  =  -1.2;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "Vo1  =  -1*Rf*Vi1/Ri\n",
+      "Vo2  =  -1*Rf*Vi2/Ri\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  when  the  input  voltage  is  0.4V  is  \",round(Vo1,2),\"  V \"\n",
+      "print   \" and  when  the  input  voltage  is  -1.2V  is  \",round(Vo2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage  when  the  input  voltage  is  0.4V  is   -0.8   V \n",
+        " and  when  the  input  voltage  is  -1.2V  is   2.4   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 283</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the voltage gain, and\n",
+      "#(b) the output offset voltage due to the input bias current. \n",
+      "#(c) How can the effect of input bias current be minimised?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ii  =  100E-9;#  in  Amperes\n",
+      "T  =  20;#  in  \u00b0C\n",
+      "Rf  =  1E6;#  in  ohms\n",
+      "Ri  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  -1*Rf/Ri\n",
+      "Vos  =  Ii*Ri*Rf/(Ri+Rf)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  voltage  gain  is  \",round(A,2),\"\"\n",
+      "print  \"\\n  (b)output  offset  voltage  is  \",round(Vos*1000,2),\"  mV\"\n",
+      "print  \"\\n  (c)The  effect  of  input  bias  current  can  be  minimised  by  ensuring  \"\n",
+      "print  \"that  both  inputs have the  same  driving  resistance.\" \n",
+      "print  \"This  means  that  a  resistance  of  value  of  9.9  kohm  (from  part  (b))  \"\n",
+      "print   \"should  be  placed  between  the  non-inverting  (+)  terminal  and  earth.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  voltage  gain  is   -100.0 \n",
+        "\n",
+        "  (b)output  offset  voltage  is   0.99   mV\n",
+        "\n",
+        "  (c)The  effect  of  input  bias  current  can  be  minimised  by  ensuring  \n",
+        "that  both  inputs have the  same  driving  resistance.\n",
+        "This  means  that  a  resistance  of  value  of  9.9  kohm  (from  part  (b))  \n",
+        "should  be  placed  between  the  non-inverting  (+)  terminal  and  earth.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 284</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design an inverting amplifier\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  40;#  in  dB\n",
+      "bf  =  5000;#  in  Hz\n",
+      "Ri  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  10**(Vg/20)\n",
+      "Rf  =  A*Ri\n",
+      "f  =  A*bf\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  voltage  gain  is  \",round(A,2),\",  Rf  =  \",round(Rf/1000,2),\"kohm  and  frequency  =  \",round(f/1000,2),\"  kHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  voltage  gain  is   100.0 ,  Rf  =   1000.0 kohm  and  frequency  =   500.0   kHz"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 286</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the voltage gain (b) the output voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vi  =  -0.4;#  in  Volts\n",
+      "R1  =  4700;#  in  ohms\n",
+      "R2  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  1  +  (R2/R1)\n",
+      "Vo  =  A*Vi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  the  voltage  gain  is  \",round(A,2),\"\"\n",
+      "print  \"\\n(b)  output  voltageis  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  the  voltage  gain  is   3.13 \n",
+        "\n",
+        "(b)  output  voltageis   -1.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 287</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the output voltage, Vo\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  0.5;#  in  Volts\n",
+      "V2  =  0.8;#  in  Volts\n",
+      "V3  =  1.2;#  in  Volts\n",
+      "R1  =  10000;#  in  ohms\n",
+      "R2  =  20000;#  in  ohms\n",
+      "R3  =  30000;#  in  ohms\n",
+      "Rf  =  50000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vo  =  -1*Rf*(V1/R1  +  V2/R2  +  V3/R3)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltageis  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltageis   -6.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 289</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from scipy import integrate\n",
+      "#initializing  the  variables:\n",
+      "Vs  =  -0.75;#  in  Volts\n",
+      "R  =  200000;#  in  ohms\n",
+      "C  =  2.5E-6;#  in  Farads\n",
+      "t  =  0.1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f = lambda x,a : a*1\n",
+      "y, err = integrate.quad(f, 0, 0.1, args=(-0.75,))\n",
+      "Vo  =  (-1/(C*R))*y\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  is  \",Vo,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage  is   0.15   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 290</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the output voltage Vo\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1a  =  0.005;#  in  Volts\n",
+      "V2a  =  0;#  in  Volts\n",
+      "V1b  =  0;#  in  Volts\n",
+      "V2b  =  0.005;#  in  Volts\n",
+      "V1c  =  0.05;#  in  Volts\n",
+      "V2c  =  0.025;#  in  Volts\n",
+      "V1d  =  0.025;#  in  Volts\n",
+      "V2d  =  0.05;#  in  Volts\n",
+      "R1  =  10000;#  in  ohms\n",
+      "R2  =  10000;#  in  ohms\n",
+      "R3  =  100000;#  in  ohms\n",
+      "Rf  =  100000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vo1  =  -1*Rf*V1a/R1\n",
+      "Vo2  =  (R3/(R2+R3))*(1  +  (Rf/R1))*V2b\n",
+      "Vo3  =  -1*Rf*(V1c-V2c)/R1\n",
+      "Vo4  =  (R3/(R2+R3))*(1  +  (Rf/R1))*(V2d-V1d)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)output  voltage  is  \",round(Vo1,2),\"  V\"\n",
+      "print  \"\\n  (b)output  voltage  is  \",round(Vo2,2),\"  V\"\n",
+      "print  \"\\n  (c)output  voltage  is  \",round(Vo3,2),\"  V\"\n",
+      "print  \"\\n  (d)output  voltage  is  \",round(Vo4,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)output  voltage  is   -0.05   V\n",
+        "\n",
+        "  (b)output  voltage  is   0.05   V\n",
+        "\n",
+        "  (c)output  voltage  is   -0.25   V\n",
+        "\n",
+        "  (d)output  voltage  is   0.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_3.ipynb
new file mode 100755
index 00000000..73f373a8
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_3.ipynb
@@ -0,0 +1,533 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a753b8fe0eef12e4693a9f55407a90702bc2bbc9eb7b04a290c07424ac34ac87"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 18: Operational amplifiers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 279</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vi2  =  2.45;#  in  Volts\n",
+      "Vi1  =  2.35;#  in  Volts\n",
+      "A0  =  120;#  open-loop  voltage  gain\n",
+      "\n",
+      "#calculation:\n",
+      "Vo  =  A0*(Vi2  -  Vi1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  output  voltage  is  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  output  voltage  is   12.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 281</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  150E3;#  differential  voltage  gain  \n",
+      "CMRR  =  90;#  in  dB\n",
+      "\n",
+      "#calculation:\n",
+      "CMG  =  Vg/(10**(CMRR/20))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  common-mode  gain  is  \",round(CMG,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  common-mode  gain  is   4.74"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 282</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  120;#  differential  voltage  gain  \n",
+      "Vi  =  3;#  in  Volts\n",
+      "Vo  =  0.024;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "CMG  =  Vo/Vi\n",
+      "CMRR  =  20*(1/2.303)*math.log(Vg/CMG)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  common-mode  gain  is  \",round(CMG,3),\"  and  CMRR  is  \",round(CMRR,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  common-mode  gain  is   0.008   and  CMRR  is   83.51   dB"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 283</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rf  =  2000;#  in  ohms\n",
+      "Ri  =  1000;#  in  ohms\n",
+      "Vi1  =  0.4;#  in  Volts\n",
+      "Vi2  =  -1.2;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "Vo1  =  -1*Rf*Vi1/Ri\n",
+      "Vo2  =  -1*Rf*Vi2/Ri\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  when  the  input  voltage  is  0.4V  is  \",round(Vo1,2),\"  V \"\n",
+      "print   \" and  when  the  input  voltage  is  -1.2V  is  \",round(Vo2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage  when  the  input  voltage  is  0.4V  is   -0.8   V \n",
+        " and  when  the  input  voltage  is  -1.2V  is   2.4   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 283</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ii  =  100E-9;#  in  Amperes\n",
+      "T  =  20;#  in  \u00b0C\n",
+      "Rf  =  1E6;#  in  ohms\n",
+      "Ri  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  -1*Rf/Ri\n",
+      "Vos  =  Ii*Ri*Rf/(Ri+Rf)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  voltage  gain  is  \",round(A,2),\"\"\n",
+      "print  \"\\n  (b)output  offset  voltage  is  \",round(Vos*1000,2),\"  mV\"\n",
+      "print  \"\\n  (c)The  effect  of  input  bias  current  can  be  minimised  by  ensuring  \"\n",
+      "print  \"that  both  inputs have the  same  driving  resistance.\" \n",
+      "print  \"This  means  that  a  resistance  of  value  of  9.9  kohm  (from  part  (b))  \"\n",
+      "print   \"should  be  placed  between  the  non-inverting  (+)  terminal  and  earth.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  voltage  gain  is   -100.0 \n",
+        "\n",
+        "  (b)output  offset  voltage  is   0.99   mV\n",
+        "\n",
+        "  (c)The  effect  of  input  bias  current  can  be  minimised  by  ensuring  \n",
+        "that  both  inputs have the  same  driving  resistance.\n",
+        "This  means  that  a  resistance  of  value  of  9.9  kohm  (from  part  (b))  \n",
+        "should  be  placed  between  the  non-inverting  (+)  terminal  and  earth.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 284</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vg  =  40;#  in  dB\n",
+      "bf  =  5000;#  in  Hz\n",
+      "Ri  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  10**(Vg/20)\n",
+      "Rf  =  A*Ri\n",
+      "f  =  A*bf\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  voltage  gain  is  \",round(A,2),\",  Rf  =  \",round(Rf/1000,2),\"kohm  and  frequency  =  \",round(f/1000,2),\"  kHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  voltage  gain  is   100.0 ,  Rf  =   1000.0 kohm  and  frequency  =   500.0   kHz"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 286</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vi  =  -0.4;#  in  Volts\n",
+      "R1  =  4700;#  in  ohms\n",
+      "R2  =  10000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "A  =  1  +  (R2/R1)\n",
+      "Vo  =  A*Vi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  the  voltage  gain  is  \",round(A,2),\"\"\n",
+      "print  \"\\n(b)  output  voltageis  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  the  voltage  gain  is   3.13 \n",
+        "\n",
+        "(b)  output  voltageis   -1.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 287</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  0.5;#  in  Volts\n",
+      "V2  =  0.8;#  in  Volts\n",
+      "V3  =  1.2;#  in  Volts\n",
+      "R1  =  10000;#  in  ohms\n",
+      "R2  =  20000;#  in  ohms\n",
+      "R3  =  30000;#  in  ohms\n",
+      "Rf  =  50000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vo  =  -1*Rf*(V1/R1  +  V2/R2  +  V3/R3)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltageis  \",round(Vo,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltageis   -6.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 289</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "from scipy import integrate\n",
+      "#initializing  the  variables:\n",
+      "Vs  =  -0.75;#  in  Volts\n",
+      "R  =  200000;#  in  ohms\n",
+      "C  =  2.5E-6;#  in  Farads\n",
+      "t  =  0.1;#  in  secs\n",
+      "\n",
+      "#calculation:\n",
+      "f = lambda x,a : a*1\n",
+      "y, err = integrate.quad(f, 0, 0.1, args=(-0.75,))\n",
+      "Vo  =  (-1/(C*R))*y\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  is  \",Vo,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage  is   0.15   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 290</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1a  =  0.005;#  in  Volts\n",
+      "V2a  =  0;#  in  Volts\n",
+      "V1b  =  0;#  in  Volts\n",
+      "V2b  =  0.005;#  in  Volts\n",
+      "V1c  =  0.05;#  in  Volts\n",
+      "V2c  =  0.025;#  in  Volts\n",
+      "V1d  =  0.025;#  in  Volts\n",
+      "V2d  =  0.05;#  in  Volts\n",
+      "R1  =  10000;#  in  ohms\n",
+      "R2  =  10000;#  in  ohms\n",
+      "R3  =  100000;#  in  ohms\n",
+      "Rf  =  100000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vo1  =  -1*Rf*V1a/R1\n",
+      "Vo2  =  (R3/(R2+R3))*(1  +  (Rf/R1))*V2b\n",
+      "Vo3  =  -1*Rf*(V1c-V2c)/R1\n",
+      "Vo4  =  (R3/(R2+R3))*(1  +  (Rf/R1))*(V2d-V1d)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)output  voltage  is  \",round(Vo1,2),\"  V\"\n",
+      "print  \"\\n  (b)output  voltage  is  \",round(Vo2,2),\"  V\"\n",
+      "print  \"\\n  (c)output  voltage  is  \",round(Vo3,2),\"  V\"\n",
+      "print  \"\\n  (d)output  voltage  is  \",round(Vo4,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)output  voltage  is   -0.05   V\n",
+        "\n",
+        "  (b)output  voltage  is   0.05   V\n",
+        "\n",
+        "  (c)output  voltage  is   -0.25   V\n",
+        "\n",
+        "  (d)output  voltage  is   0.25   V"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint.ipynb
new file mode 100755
index 00000000..8285878f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint.ipynb
@@ -0,0 +1,974 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 19: Three phase systems</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 299</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the system phase voltage, (b) the phase current and (c) the line current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vl  =  415;#  in  Volts\n",
+      "Rp  =  30;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vp  =  Vl/(3**0.5)\n",
+      "Ip  =  Vp/Rp\n",
+      "Il  =  Ip\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  system  phase  voltage  is  \",round(Vp,2),\"  V\"\n",
+      "print  \"\\n  (b)phase  current  is  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (c)line  current  is  \",round(Il,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  system  phase  voltage  is   239.6   V\n",
+        "\n",
+        "  (b)phase  current  is   7.99   A\n",
+        "\n",
+        "  (c)line  current  is   7.99   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 299</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the line voltage if the supply frequency is 50 Hz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  0.1273;#  in  Henry\n",
+      "Ip  =  5.08;#  in  Amperes\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Il  =  Ip\n",
+      "Vp  =  Ip*Zp\n",
+      "Vl  =  Vp*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)line  voltage  is  \",round(Vl,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)line  voltage  is   439.89   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 301</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current in each line and (b) the current in the neutral conductor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "PR  =  24000;#  in  Watt\n",
+      "Py  =  18000;#  in  Watt\n",
+      "Pb  =  12000;#  in  Watt\n",
+      "VR  =  240;#  in  Volts\n",
+      "Vy  =  240;#  in  Volts\n",
+      "Vb  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star-connected  system  VL  =  Vp*(3**0.5)\n",
+      "Vp  =  V/(3**0.5)\n",
+      "phir  =  90*math.pi/180\n",
+      "phiy  =  330*math.pi/180\n",
+      "phib  =  210*math.pi/180\n",
+      " #  I  =  P/V    for  a  resistive  load\n",
+      "IR  =  PR/VR\n",
+      "Iy  =  Py/Vy\n",
+      "Ib  =  Pb/Vb\n",
+      "Inh  =  IR*math.cos(phir)  +  Ib*math.cos(phib)  +  Iy*math.cos(phiy)\n",
+      "Inv  =  IR*math.sin(phir)  +  Ib*math.sin(phib)  +  Iy*math.sin(phiy)\n",
+      "In  =  (Inh**2  +  Inv**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)cuurnt  in  R  line  is  \",round(IR,2),\"  A,  cuurnt  in  Y  line  is  \",round(Iy,2),\"  A  \"\n",
+      "print   \"and  cuurnt  in  B  line  is  \",round(Ib,2),\"  A\"\n",
+      "print  \"\\n  (b)cuurnt  in  neutral  line  is  \",round(In,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)cuurnt  in  R  line  is   100.0   A,  cuurnt  in  Y  line  is   75.0   A  and  cuurnt  in  B  line  is   50.0   A\n",
+        "\n",
+        "  (b)cuurnt  in  neutral  line  is   43.3   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 302</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the phase current, and (b) the line current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  0.1273;#  in  Henry\n",
+      "VL  =  440;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      " #For  a  delta  connection,\n",
+      "IL  =  Ip*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  phase  current  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (b)line  current  \",round(IL,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  phase  current   8.8   A\n",
+        "\n",
+        "  (b)line  current   15.24   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 302</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitance of each of the capacitors.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "IL  =  15;#  in  Amperes\n",
+      "VL  =  415;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  delta  connection\n",
+      "Ip  =  IL/(3**0.5)#phase  current\n",
+      "Vp  =  VL\n",
+      " #Capacitive  reactance  per  phase\n",
+      "Xc  =  Vp/Ip\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   66.43 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 303</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  3;#  in  ohms\n",
+      "XL  =  4;#  in  ohms\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star  connection:\n",
+      " #IL  =  Ip\n",
+      " #VL  =  Vp*(3**0.5)\n",
+      "VLs  =  VL\n",
+      "Vps  =  VLs/(3**0.5)\n",
+      " #Impedance  per  phase,\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Ips  =  Vps/Zp\n",
+      "ILs  =  Ips\n",
+      " #For  a  delta  connection:\n",
+      " #VL  =  Vp\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "VLd  =  VL\n",
+      "Vpd  =  VLd\n",
+      "Ipd  =  Vpd/Zp\n",
+      "ILd  =  Ipd*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  line  voltage  for  star  connection  is  \",round(VLs,2),\"  V  \"\n",
+      "print   \"and the  phase  voltage  for  star  connection  is  \",round(Vps,2),\"  V  \"\n",
+      "print   \"and the  line  voltage  for  delta  connection  is  \",round(VLd,2),\"  V  \"\n",
+      "print   \"and the  phase  voltage  for  delta  connection  is  \",round(Vpd,2),\"  V\"\n",
+      "print  \"\\n  (b)the  line  current  for  star  connection  is  \",round(ILs,2),\"  A  \"\n",
+      "print   \"and  the  phase  current  for  star  connection  is  \",round(Ips,2),\"  A  \"\n",
+      "print   \"and the  line  current  for  delta  connection  is  \",round(ILd,2),\"  A  \"\n",
+      "print   \"and the  phase  current  for  delta  connection  is  \",round(Ipd,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  line  voltage  for  star  connection  is   415.0   V  \n",
+        "and the  phase  voltage  for  star  connection  is   239.6   V  \n",
+        "and the  line  voltage  for  delta  connection  is   415.0   V  \n",
+        "and the  phase  voltage  for  delta  connection  is   415.0   V\n",
+        "\n",
+        "  (b)the  line  current  for  star  connection  is   47.92   A  \n",
+        "and  the  phase  current  for  star  connection  is   47.92   A  \n",
+        "and the  line  current  for  delta  connection  is   143.76   A  \n",
+        "and the  phase  current  for  delta  connection  is   83.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the total power dissipated by the resistors.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rp  =  12;#  in  ohms\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "Vp  =  VL/(3**0.5)#  since  the  resistors  are  star-connected\n",
+      " #Phase  current,  Ip\n",
+      "Zp  =  Rp\n",
+      "Ip  =  Vp/Zp\n",
+      " #For  a  star  connection\n",
+      "IL  =  Ip\n",
+      " #  For  a  purely  resistive  load,  the  power  factor  cos(phi)  =  1\n",
+      "pf  =  1\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)total  power  dissipated  by  the  resistors  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)total  power  dissipated  by  the  resistors  is   14352.08   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power factor of the system.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "P  =  5000;#  in  Watts\n",
+      "IL  =  8.6;#  in  amperes\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pf  =  P/(VL*IL*(3**0.5))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  factor  is  \",round(pf,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  factor  is   0.839"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the total power dissipated in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "L  =  0.042;#  in  Henry\n",
+      "VL  =  415;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star  connection:\n",
+      " #IL  =  Ip\n",
+      " #VL  =  Vp*(3**0.5)\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "VLs  =  VL\n",
+      "Vps  =  VLs/(3**0.5)\n",
+      " #Impedance  per  phase,\n",
+      "Ips  =  Vps/Zp\n",
+      "ILs  =  Ips\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pfs  =  R/Zp\n",
+      "Ps  =  VLs*ILs*(3**0.5)*pfs\n",
+      "\n",
+      " #For  a  delta  connection:\n",
+      " #VL  =  Vp\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "VLd  =  VL\n",
+      "Vpd  =  VLd\n",
+      "Ipd  =  Vpd/Zp\n",
+      "ILd  =  Ipd*(3**0.5)\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pfd  =  R/Zp\n",
+      "Pd  =  VLd*ILd*(3**0.5)*pfd\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  total  power  dissipated  in  star  is  \",round(Ps,2),\" W  and  in  delta  is  \",round(Pd,2),\" W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  total  power  dissipated  in  star  is   6283.29  W  and  in  delta  is   18849.88  W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 305</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the power input, (b) the line current and (c) the phase current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Po  =  12750;#  in  Watts\n",
+      "pf  =  0.77;#  power  factor\n",
+      "eff  =  0.85;\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #eff  =  power_out/power_in\n",
+      "Pi  =  Po/eff\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "IL  =  Pi/(VL*(3**0.5)*pf)#  line  current\n",
+      " #For  a  delta  connection:\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "Ip  =  IL/(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\" W\"\n",
+      "print  \"\\n  (b)line  current  is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (c)phase  current  is  \",round(Ip,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   15000.0  W\n",
+        "\n",
+        "  (b)line  current  is   27.1   A\n",
+        "\n",
+        "  (c)phase  current  is   15.65   A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 308</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "XL  =  40;#  in  ohms\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      " #a  delta-connected  load\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      "IL  =  Ip*(3**0.5)\n",
+      " #Alternator  output  power  is  equal  to  the  power  dissipated  by  the  load.\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pf  =  R/Zp\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      " #Alternator  output  kVA,\n",
+      "S  =  VL*IL*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  supplied  by  the  alternator  is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (b)output  power  is  \",round(P/1000,2),\"KW  and  kVA  of  the  alternator  is  \",round(S/1000,2),\"kVA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  supplied  by  the  alternator  is   13.86   A\n",
+        "\n",
+        "  (b)output  power  is   5.76 KW  and  kVA  of  the  alternator  is   9.6 kVA"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 308</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the phase current, (b) the line current, (c) the total power dissipated and \n",
+      "#(d) the kVA rating of the load. Draw the complete phasor diagram for the load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "C  =  80E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Capacitive  reactance\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Zp  =  (R*R  +  Xc*Xc)**0.5\n",
+      "pf  =  R/Zp\n",
+      " #a  delta-connected  load\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      "IL  =  Ip*(3**0.5)\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      " #Alternator  output  kVA,\n",
+      "S  =  VL*IL*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  phase  current    is  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (b)the  line  current    is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (c) power  is  \",round(P/1000,2),\"kW\"\n",
+      "print  \"\\n  (d)kVA  of  the  alternator  is  \", round(S/1000,2),\"kVA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  phase  current    is   8.03   A\n",
+        "\n",
+        "  (b)the  line  current    is   13.9   A\n",
+        "\n",
+        "  (c) power  is   5.8 kW\n",
+        "\n",
+        "  (d)kVA  of  the  alternator  is   9.63 kVA"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 309</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the total power input and (b) the load power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi1  =  8000;#  in  Watts\n",
+      "Pi2  =  4000;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Total  input  power\n",
+      "Pi  =  Pi1  +  Pi2\n",
+      "phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))\n",
+      " #Power  factor\n",
+      "pf  =  math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\"W\"\n",
+      "print  \"\\n  (b)power  factor  is  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   12000.0 W\n",
+        "\n",
+        "  (b)power  factor  is   0.87"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 310</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the readings of each wattmeter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi  =  12000;#  in  Watts\n",
+      "pf  =  0.6;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #If  the  two  wattmeters  indicate  Pi1  and  Pi2  respectively\n",
+      " #  Pit  =  Pi1  +  Pi2\n",
+      "Pit  =  Pi\n",
+      " #  Pid  =  Pi1  -  Pi2\n",
+      " #power  factor  =  0.6  =  cos(phi)\n",
+      "phi  =  math.acos(pf)\n",
+      "Pid  =  Pit*math.tan(phi)/(3**0.5)\n",
+      " #Hence  wattmeter  1  reads\n",
+      "Pi1  =  (Pid  +  Pit)/2\n",
+      " #wattmeter  2  reads\n",
+      "Pi2  =  Pit  -  Pi1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reading  in  each  wattameter  are  \",round(Pi1,2),\"W  and  \",round(Pi2,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reading  in  each  wattameter  are   10618.8 W  and   1381.2 W"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 310</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the input power and (b) the load power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi1  =  10000;#  in  Watts\n",
+      "Pi2  =  -3000;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Total  input  power\n",
+      "Pi  =  Pi1  +  Pi2\n",
+      "phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))\n",
+      " #Power  factor\n",
+      "pf  =  math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\"W\"\n",
+      "print  \"\\n  (b)power  factor  is  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   7000.0 W\n",
+        "\n",
+        "  (b)power  factor  is   0.3"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 311</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing the variables:\n",
+      "R = 8; # in ohms\n",
+      "XL = 8; # in ohms\n",
+      "VL = 415; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#For a star connection:\n",
+      "#IL = Ip\n",
+      "#VL = Vp*(3**0.5)\n",
+      "VLs = VL\n",
+      "Vps = VLs/(3**0.5)\n",
+      "#Impedance per phase,\n",
+      "Zp = (R*R + XL*XL)**0.5\n",
+      "Ips = Vps/Zp\n",
+      "ILs = Ips\n",
+      "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)\n",
+      "pf = R/Zp\n",
+      "Ps = VLs*ILs*(3**0.5)*pf\n",
+      "#If wattmeter readings are P1 and P2 then P1 + P2 = Pst\n",
+      "Pst = Ps\n",
+      "# Pid = Pi1 - Pi2\n",
+      "phi = math.acos(pf)\n",
+      "Psd = Pst*math.tan(phi)/(3**0.5)\n",
+      "#Hence wattmeter 1 reads\n",
+      "Ps1 = (Psd + Pst)/2\n",
+      "#wattmeter 2 reads\n",
+      "Ps2 = Pst - Ps1\n",
+      "\n",
+      "#For a delta connection:\n",
+      "#VL = Vp\n",
+      "#IL = Ip*(3**0.5)\n",
+      "VLd = VL\n",
+      "Vpd = VLd\n",
+      "Ipd = Vpd/Zp\n",
+      "ILd = Ipd*(3**0.5)\n",
+      "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)\n",
+      "Pd = VLd*ILd*(3**0.5)*pf\n",
+      "#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt\n",
+      "Pdt = Pd\n",
+      "# Pid = Pi1 - Pi2\n",
+      "Pdd = Pdt*math.tan(phi)/(3**0.5)\n",
+      "#Hence wattmeter 1 reads\n",
+      "Pd1 = (Pdd + Pdt)/2\n",
+      "#wattmeter 2 reads\n",
+      "Pd2 = Pdt - Pd1\n",
+      "\n",
+      "#results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"(a)When the coils are star-connected the wattmeter readings are\", round(Ps1,2),\"W and \",round(Ps2,2),\"W\"\n",
+      "print \"(b)When the coils are delta-connected the wattmeter readings are are\", round(Pd1,2),\"W and\", round(Pd2,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "(a)When the coils are star-connected the wattmeter readings are 8489.35 W and  2274.71 W\n",
+        "(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb
new file mode 100755
index 00000000..8285878f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb
@@ -0,0 +1,974 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 19: Three phase systems</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 299</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the system phase voltage, (b) the phase current and (c) the line current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vl  =  415;#  in  Volts\n",
+      "Rp  =  30;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vp  =  Vl/(3**0.5)\n",
+      "Ip  =  Vp/Rp\n",
+      "Il  =  Ip\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  system  phase  voltage  is  \",round(Vp,2),\"  V\"\n",
+      "print  \"\\n  (b)phase  current  is  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (c)line  current  is  \",round(Il,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  system  phase  voltage  is   239.6   V\n",
+        "\n",
+        "  (b)phase  current  is   7.99   A\n",
+        "\n",
+        "  (c)line  current  is   7.99   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 299</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the line voltage if the supply frequency is 50 Hz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  0.1273;#  in  Henry\n",
+      "Ip  =  5.08;#  in  Amperes\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Il  =  Ip\n",
+      "Vp  =  Ip*Zp\n",
+      "Vl  =  Vp*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)line  voltage  is  \",round(Vl,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)line  voltage  is   439.89   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 301</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current in each line and (b) the current in the neutral conductor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "PR  =  24000;#  in  Watt\n",
+      "Py  =  18000;#  in  Watt\n",
+      "Pb  =  12000;#  in  Watt\n",
+      "VR  =  240;#  in  Volts\n",
+      "Vy  =  240;#  in  Volts\n",
+      "Vb  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star-connected  system  VL  =  Vp*(3**0.5)\n",
+      "Vp  =  V/(3**0.5)\n",
+      "phir  =  90*math.pi/180\n",
+      "phiy  =  330*math.pi/180\n",
+      "phib  =  210*math.pi/180\n",
+      " #  I  =  P/V    for  a  resistive  load\n",
+      "IR  =  PR/VR\n",
+      "Iy  =  Py/Vy\n",
+      "Ib  =  Pb/Vb\n",
+      "Inh  =  IR*math.cos(phir)  +  Ib*math.cos(phib)  +  Iy*math.cos(phiy)\n",
+      "Inv  =  IR*math.sin(phir)  +  Ib*math.sin(phib)  +  Iy*math.sin(phiy)\n",
+      "In  =  (Inh**2  +  Inv**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)cuurnt  in  R  line  is  \",round(IR,2),\"  A,  cuurnt  in  Y  line  is  \",round(Iy,2),\"  A  \"\n",
+      "print   \"and  cuurnt  in  B  line  is  \",round(Ib,2),\"  A\"\n",
+      "print  \"\\n  (b)cuurnt  in  neutral  line  is  \",round(In,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)cuurnt  in  R  line  is   100.0   A,  cuurnt  in  Y  line  is   75.0   A  and  cuurnt  in  B  line  is   50.0   A\n",
+        "\n",
+        "  (b)cuurnt  in  neutral  line  is   43.3   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 302</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the phase current, and (b) the line current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  0.1273;#  in  Henry\n",
+      "VL  =  440;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      " #For  a  delta  connection,\n",
+      "IL  =  Ip*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  phase  current  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (b)line  current  \",round(IL,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  phase  current   8.8   A\n",
+        "\n",
+        "  (b)line  current   15.24   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 302</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitance of each of the capacitors.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "IL  =  15;#  in  Amperes\n",
+      "VL  =  415;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  delta  connection\n",
+      "Ip  =  IL/(3**0.5)#phase  current\n",
+      "Vp  =  VL\n",
+      " #Capacitive  reactance  per  phase\n",
+      "Xc  =  Vp/Ip\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   66.43 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 303</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  3;#  in  ohms\n",
+      "XL  =  4;#  in  ohms\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star  connection:\n",
+      " #IL  =  Ip\n",
+      " #VL  =  Vp*(3**0.5)\n",
+      "VLs  =  VL\n",
+      "Vps  =  VLs/(3**0.5)\n",
+      " #Impedance  per  phase,\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Ips  =  Vps/Zp\n",
+      "ILs  =  Ips\n",
+      " #For  a  delta  connection:\n",
+      " #VL  =  Vp\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "VLd  =  VL\n",
+      "Vpd  =  VLd\n",
+      "Ipd  =  Vpd/Zp\n",
+      "ILd  =  Ipd*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  line  voltage  for  star  connection  is  \",round(VLs,2),\"  V  \"\n",
+      "print   \"and the  phase  voltage  for  star  connection  is  \",round(Vps,2),\"  V  \"\n",
+      "print   \"and the  line  voltage  for  delta  connection  is  \",round(VLd,2),\"  V  \"\n",
+      "print   \"and the  phase  voltage  for  delta  connection  is  \",round(Vpd,2),\"  V\"\n",
+      "print  \"\\n  (b)the  line  current  for  star  connection  is  \",round(ILs,2),\"  A  \"\n",
+      "print   \"and  the  phase  current  for  star  connection  is  \",round(Ips,2),\"  A  \"\n",
+      "print   \"and the  line  current  for  delta  connection  is  \",round(ILd,2),\"  A  \"\n",
+      "print   \"and the  phase  current  for  delta  connection  is  \",round(Ipd,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  line  voltage  for  star  connection  is   415.0   V  \n",
+        "and the  phase  voltage  for  star  connection  is   239.6   V  \n",
+        "and the  line  voltage  for  delta  connection  is   415.0   V  \n",
+        "and the  phase  voltage  for  delta  connection  is   415.0   V\n",
+        "\n",
+        "  (b)the  line  current  for  star  connection  is   47.92   A  \n",
+        "and  the  phase  current  for  star  connection  is   47.92   A  \n",
+        "and the  line  current  for  delta  connection  is   143.76   A  \n",
+        "and the  phase  current  for  delta  connection  is   83.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the total power dissipated by the resistors.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rp  =  12;#  in  ohms\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "Vp  =  VL/(3**0.5)#  since  the  resistors  are  star-connected\n",
+      " #Phase  current,  Ip\n",
+      "Zp  =  Rp\n",
+      "Ip  =  Vp/Zp\n",
+      " #For  a  star  connection\n",
+      "IL  =  Ip\n",
+      " #  For  a  purely  resistive  load,  the  power  factor  cos(phi)  =  1\n",
+      "pf  =  1\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)total  power  dissipated  by  the  resistors  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)total  power  dissipated  by  the  resistors  is   14352.08   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power factor of the system.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "P  =  5000;#  in  Watts\n",
+      "IL  =  8.6;#  in  amperes\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pf  =  P/(VL*IL*(3**0.5))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  factor  is  \",round(pf,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  factor  is   0.839"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the total power dissipated in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "L  =  0.042;#  in  Henry\n",
+      "VL  =  415;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star  connection:\n",
+      " #IL  =  Ip\n",
+      " #VL  =  Vp*(3**0.5)\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "VLs  =  VL\n",
+      "Vps  =  VLs/(3**0.5)\n",
+      " #Impedance  per  phase,\n",
+      "Ips  =  Vps/Zp\n",
+      "ILs  =  Ips\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pfs  =  R/Zp\n",
+      "Ps  =  VLs*ILs*(3**0.5)*pfs\n",
+      "\n",
+      " #For  a  delta  connection:\n",
+      " #VL  =  Vp\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "VLd  =  VL\n",
+      "Vpd  =  VLd\n",
+      "Ipd  =  Vpd/Zp\n",
+      "ILd  =  Ipd*(3**0.5)\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pfd  =  R/Zp\n",
+      "Pd  =  VLd*ILd*(3**0.5)*pfd\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  total  power  dissipated  in  star  is  \",round(Ps,2),\" W  and  in  delta  is  \",round(Pd,2),\" W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  total  power  dissipated  in  star  is   6283.29  W  and  in  delta  is   18849.88  W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 305</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the power input, (b) the line current and (c) the phase current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Po  =  12750;#  in  Watts\n",
+      "pf  =  0.77;#  power  factor\n",
+      "eff  =  0.85;\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #eff  =  power_out/power_in\n",
+      "Pi  =  Po/eff\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "IL  =  Pi/(VL*(3**0.5)*pf)#  line  current\n",
+      " #For  a  delta  connection:\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "Ip  =  IL/(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\" W\"\n",
+      "print  \"\\n  (b)line  current  is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (c)phase  current  is  \",round(Ip,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   15000.0  W\n",
+        "\n",
+        "  (b)line  current  is   27.1   A\n",
+        "\n",
+        "  (c)phase  current  is   15.65   A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 308</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "XL  =  40;#  in  ohms\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      " #a  delta-connected  load\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      "IL  =  Ip*(3**0.5)\n",
+      " #Alternator  output  power  is  equal  to  the  power  dissipated  by  the  load.\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pf  =  R/Zp\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      " #Alternator  output  kVA,\n",
+      "S  =  VL*IL*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  supplied  by  the  alternator  is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (b)output  power  is  \",round(P/1000,2),\"KW  and  kVA  of  the  alternator  is  \",round(S/1000,2),\"kVA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  supplied  by  the  alternator  is   13.86   A\n",
+        "\n",
+        "  (b)output  power  is   5.76 KW  and  kVA  of  the  alternator  is   9.6 kVA"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 308</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the phase current, (b) the line current, (c) the total power dissipated and \n",
+      "#(d) the kVA rating of the load. Draw the complete phasor diagram for the load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "C  =  80E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Capacitive  reactance\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Zp  =  (R*R  +  Xc*Xc)**0.5\n",
+      "pf  =  R/Zp\n",
+      " #a  delta-connected  load\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      "IL  =  Ip*(3**0.5)\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      " #Alternator  output  kVA,\n",
+      "S  =  VL*IL*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  phase  current    is  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (b)the  line  current    is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (c) power  is  \",round(P/1000,2),\"kW\"\n",
+      "print  \"\\n  (d)kVA  of  the  alternator  is  \", round(S/1000,2),\"kVA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  phase  current    is   8.03   A\n",
+        "\n",
+        "  (b)the  line  current    is   13.9   A\n",
+        "\n",
+        "  (c) power  is   5.8 kW\n",
+        "\n",
+        "  (d)kVA  of  the  alternator  is   9.63 kVA"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 309</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the total power input and (b) the load power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi1  =  8000;#  in  Watts\n",
+      "Pi2  =  4000;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Total  input  power\n",
+      "Pi  =  Pi1  +  Pi2\n",
+      "phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))\n",
+      " #Power  factor\n",
+      "pf  =  math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\"W\"\n",
+      "print  \"\\n  (b)power  factor  is  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   12000.0 W\n",
+        "\n",
+        "  (b)power  factor  is   0.87"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 310</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the readings of each wattmeter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi  =  12000;#  in  Watts\n",
+      "pf  =  0.6;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #If  the  two  wattmeters  indicate  Pi1  and  Pi2  respectively\n",
+      " #  Pit  =  Pi1  +  Pi2\n",
+      "Pit  =  Pi\n",
+      " #  Pid  =  Pi1  -  Pi2\n",
+      " #power  factor  =  0.6  =  cos(phi)\n",
+      "phi  =  math.acos(pf)\n",
+      "Pid  =  Pit*math.tan(phi)/(3**0.5)\n",
+      " #Hence  wattmeter  1  reads\n",
+      "Pi1  =  (Pid  +  Pit)/2\n",
+      " #wattmeter  2  reads\n",
+      "Pi2  =  Pit  -  Pi1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reading  in  each  wattameter  are  \",round(Pi1,2),\"W  and  \",round(Pi2,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reading  in  each  wattameter  are   10618.8 W  and   1381.2 W"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 310</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the input power and (b) the load power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi1  =  10000;#  in  Watts\n",
+      "Pi2  =  -3000;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Total  input  power\n",
+      "Pi  =  Pi1  +  Pi2\n",
+      "phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))\n",
+      " #Power  factor\n",
+      "pf  =  math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\"W\"\n",
+      "print  \"\\n  (b)power  factor  is  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   7000.0 W\n",
+        "\n",
+        "  (b)power  factor  is   0.3"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 311</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing the variables:\n",
+      "R = 8; # in ohms\n",
+      "XL = 8; # in ohms\n",
+      "VL = 415; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#For a star connection:\n",
+      "#IL = Ip\n",
+      "#VL = Vp*(3**0.5)\n",
+      "VLs = VL\n",
+      "Vps = VLs/(3**0.5)\n",
+      "#Impedance per phase,\n",
+      "Zp = (R*R + XL*XL)**0.5\n",
+      "Ips = Vps/Zp\n",
+      "ILs = Ips\n",
+      "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)\n",
+      "pf = R/Zp\n",
+      "Ps = VLs*ILs*(3**0.5)*pf\n",
+      "#If wattmeter readings are P1 and P2 then P1 + P2 = Pst\n",
+      "Pst = Ps\n",
+      "# Pid = Pi1 - Pi2\n",
+      "phi = math.acos(pf)\n",
+      "Psd = Pst*math.tan(phi)/(3**0.5)\n",
+      "#Hence wattmeter 1 reads\n",
+      "Ps1 = (Psd + Pst)/2\n",
+      "#wattmeter 2 reads\n",
+      "Ps2 = Pst - Ps1\n",
+      "\n",
+      "#For a delta connection:\n",
+      "#VL = Vp\n",
+      "#IL = Ip*(3**0.5)\n",
+      "VLd = VL\n",
+      "Vpd = VLd\n",
+      "Ipd = Vpd/Zp\n",
+      "ILd = Ipd*(3**0.5)\n",
+      "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)\n",
+      "Pd = VLd*ILd*(3**0.5)*pf\n",
+      "#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt\n",
+      "Pdt = Pd\n",
+      "# Pid = Pi1 - Pi2\n",
+      "Pdd = Pdt*math.tan(phi)/(3**0.5)\n",
+      "#Hence wattmeter 1 reads\n",
+      "Pd1 = (Pdd + Pdt)/2\n",
+      "#wattmeter 2 reads\n",
+      "Pd2 = Pdt - Pd1\n",
+      "\n",
+      "#results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"(a)When the coils are star-connected the wattmeter readings are\", round(Ps1,2),\"W and \",round(Ps2,2),\"W\"\n",
+      "print \"(b)When the coils are delta-connected the wattmeter readings are are\", round(Pd1,2),\"W and\", round(Pd2,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "(a)When the coils are star-connected the wattmeter readings are 8489.35 W and  2274.71 W\n",
+        "(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_2.ipynb
new file mode 100755
index 00000000..8285878f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_2.ipynb
@@ -0,0 +1,974 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 19: Three phase systems</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 299</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the system phase voltage, (b) the phase current and (c) the line current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vl  =  415;#  in  Volts\n",
+      "Rp  =  30;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vp  =  Vl/(3**0.5)\n",
+      "Ip  =  Vp/Rp\n",
+      "Il  =  Ip\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  system  phase  voltage  is  \",round(Vp,2),\"  V\"\n",
+      "print  \"\\n  (b)phase  current  is  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (c)line  current  is  \",round(Il,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  system  phase  voltage  is   239.6   V\n",
+        "\n",
+        "  (b)phase  current  is   7.99   A\n",
+        "\n",
+        "  (c)line  current  is   7.99   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 299</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the line voltage if the supply frequency is 50 Hz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  0.1273;#  in  Henry\n",
+      "Ip  =  5.08;#  in  Amperes\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Il  =  Ip\n",
+      "Vp  =  Ip*Zp\n",
+      "Vl  =  Vp*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)line  voltage  is  \",round(Vl,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)line  voltage  is   439.89   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 301</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the current in each line and (b) the current in the neutral conductor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "PR  =  24000;#  in  Watt\n",
+      "Py  =  18000;#  in  Watt\n",
+      "Pb  =  12000;#  in  Watt\n",
+      "VR  =  240;#  in  Volts\n",
+      "Vy  =  240;#  in  Volts\n",
+      "Vb  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star-connected  system  VL  =  Vp*(3**0.5)\n",
+      "Vp  =  V/(3**0.5)\n",
+      "phir  =  90*math.pi/180\n",
+      "phiy  =  330*math.pi/180\n",
+      "phib  =  210*math.pi/180\n",
+      " #  I  =  P/V    for  a  resistive  load\n",
+      "IR  =  PR/VR\n",
+      "Iy  =  Py/Vy\n",
+      "Ib  =  Pb/Vb\n",
+      "Inh  =  IR*math.cos(phir)  +  Ib*math.cos(phib)  +  Iy*math.cos(phiy)\n",
+      "Inv  =  IR*math.sin(phir)  +  Ib*math.sin(phib)  +  Iy*math.sin(phiy)\n",
+      "In  =  (Inh**2  +  Inv**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)cuurnt  in  R  line  is  \",round(IR,2),\"  A,  cuurnt  in  Y  line  is  \",round(Iy,2),\"  A  \"\n",
+      "print   \"and  cuurnt  in  B  line  is  \",round(Ib,2),\"  A\"\n",
+      "print  \"\\n  (b)cuurnt  in  neutral  line  is  \",round(In,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)cuurnt  in  R  line  is   100.0   A,  cuurnt  in  Y  line  is   75.0   A  and  cuurnt  in  B  line  is   50.0   A\n",
+        "\n",
+        "  (b)cuurnt  in  neutral  line  is   43.3   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 302</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the phase current, and (b) the line current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  0.1273;#  in  Henry\n",
+      "VL  =  440;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      " #For  a  delta  connection,\n",
+      "IL  =  Ip*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  phase  current  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (b)line  current  \",round(IL,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  phase  current   8.8   A\n",
+        "\n",
+        "  (b)line  current   15.24   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 302</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitance of each of the capacitors.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "IL  =  15;#  in  Amperes\n",
+      "VL  =  415;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  delta  connection\n",
+      "Ip  =  IL/(3**0.5)#phase  current\n",
+      "Vp  =  VL\n",
+      " #Capacitive  reactance  per  phase\n",
+      "Xc  =  Vp/Ip\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   66.43 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 303</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  3;#  in  ohms\n",
+      "XL  =  4;#  in  ohms\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star  connection:\n",
+      " #IL  =  Ip\n",
+      " #VL  =  Vp*(3**0.5)\n",
+      "VLs  =  VL\n",
+      "Vps  =  VLs/(3**0.5)\n",
+      " #Impedance  per  phase,\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Ips  =  Vps/Zp\n",
+      "ILs  =  Ips\n",
+      " #For  a  delta  connection:\n",
+      " #VL  =  Vp\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "VLd  =  VL\n",
+      "Vpd  =  VLd\n",
+      "Ipd  =  Vpd/Zp\n",
+      "ILd  =  Ipd*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  line  voltage  for  star  connection  is  \",round(VLs,2),\"  V  \"\n",
+      "print   \"and the  phase  voltage  for  star  connection  is  \",round(Vps,2),\"  V  \"\n",
+      "print   \"and the  line  voltage  for  delta  connection  is  \",round(VLd,2),\"  V  \"\n",
+      "print   \"and the  phase  voltage  for  delta  connection  is  \",round(Vpd,2),\"  V\"\n",
+      "print  \"\\n  (b)the  line  current  for  star  connection  is  \",round(ILs,2),\"  A  \"\n",
+      "print   \"and  the  phase  current  for  star  connection  is  \",round(Ips,2),\"  A  \"\n",
+      "print   \"and the  line  current  for  delta  connection  is  \",round(ILd,2),\"  A  \"\n",
+      "print   \"and the  phase  current  for  delta  connection  is  \",round(Ipd,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  line  voltage  for  star  connection  is   415.0   V  \n",
+        "and the  phase  voltage  for  star  connection  is   239.6   V  \n",
+        "and the  line  voltage  for  delta  connection  is   415.0   V  \n",
+        "and the  phase  voltage  for  delta  connection  is   415.0   V\n",
+        "\n",
+        "  (b)the  line  current  for  star  connection  is   47.92   A  \n",
+        "and  the  phase  current  for  star  connection  is   47.92   A  \n",
+        "and the  line  current  for  delta  connection  is   143.76   A  \n",
+        "and the  phase  current  for  delta  connection  is   83.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the total power dissipated by the resistors.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rp  =  12;#  in  ohms\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "Vp  =  VL/(3**0.5)#  since  the  resistors  are  star-connected\n",
+      " #Phase  current,  Ip\n",
+      "Zp  =  Rp\n",
+      "Ip  =  Vp/Zp\n",
+      " #For  a  star  connection\n",
+      "IL  =  Ip\n",
+      " #  For  a  purely  resistive  load,  the  power  factor  cos(phi)  =  1\n",
+      "pf  =  1\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)total  power  dissipated  by  the  resistors  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)total  power  dissipated  by  the  resistors  is   14352.08   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power factor of the system.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "P  =  5000;#  in  Watts\n",
+      "IL  =  8.6;#  in  amperes\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pf  =  P/(VL*IL*(3**0.5))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  factor  is  \",round(pf,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  factor  is   0.839"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the total power dissipated in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "L  =  0.042;#  in  Henry\n",
+      "VL  =  415;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star  connection:\n",
+      " #IL  =  Ip\n",
+      " #VL  =  Vp*(3**0.5)\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "VLs  =  VL\n",
+      "Vps  =  VLs/(3**0.5)\n",
+      " #Impedance  per  phase,\n",
+      "Ips  =  Vps/Zp\n",
+      "ILs  =  Ips\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pfs  =  R/Zp\n",
+      "Ps  =  VLs*ILs*(3**0.5)*pfs\n",
+      "\n",
+      " #For  a  delta  connection:\n",
+      " #VL  =  Vp\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "VLd  =  VL\n",
+      "Vpd  =  VLd\n",
+      "Ipd  =  Vpd/Zp\n",
+      "ILd  =  Ipd*(3**0.5)\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pfd  =  R/Zp\n",
+      "Pd  =  VLd*ILd*(3**0.5)*pfd\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  total  power  dissipated  in  star  is  \",round(Ps,2),\" W  and  in  delta  is  \",round(Pd,2),\" W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  total  power  dissipated  in  star  is   6283.29  W  and  in  delta  is   18849.88  W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 305</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the power input, (b) the line current and (c) the phase current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Po  =  12750;#  in  Watts\n",
+      "pf  =  0.77;#  power  factor\n",
+      "eff  =  0.85;\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #eff  =  power_out/power_in\n",
+      "Pi  =  Po/eff\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "IL  =  Pi/(VL*(3**0.5)*pf)#  line  current\n",
+      " #For  a  delta  connection:\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "Ip  =  IL/(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\" W\"\n",
+      "print  \"\\n  (b)line  current  is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (c)phase  current  is  \",round(Ip,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   15000.0  W\n",
+        "\n",
+        "  (b)line  current  is   27.1   A\n",
+        "\n",
+        "  (c)phase  current  is   15.65   A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 308</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "XL  =  40;#  in  ohms\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      " #a  delta-connected  load\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      "IL  =  Ip*(3**0.5)\n",
+      " #Alternator  output  power  is  equal  to  the  power  dissipated  by  the  load.\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pf  =  R/Zp\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      " #Alternator  output  kVA,\n",
+      "S  =  VL*IL*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  supplied  by  the  alternator  is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (b)output  power  is  \",round(P/1000,2),\"KW  and  kVA  of  the  alternator  is  \",round(S/1000,2),\"kVA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  supplied  by  the  alternator  is   13.86   A\n",
+        "\n",
+        "  (b)output  power  is   5.76 KW  and  kVA  of  the  alternator  is   9.6 kVA"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 308</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the phase current, (b) the line current, (c) the total power dissipated and \n",
+      "#(d) the kVA rating of the load. Draw the complete phasor diagram for the load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "C  =  80E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Capacitive  reactance\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Zp  =  (R*R  +  Xc*Xc)**0.5\n",
+      "pf  =  R/Zp\n",
+      " #a  delta-connected  load\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      "IL  =  Ip*(3**0.5)\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      " #Alternator  output  kVA,\n",
+      "S  =  VL*IL*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  phase  current    is  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (b)the  line  current    is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (c) power  is  \",round(P/1000,2),\"kW\"\n",
+      "print  \"\\n  (d)kVA  of  the  alternator  is  \", round(S/1000,2),\"kVA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  phase  current    is   8.03   A\n",
+        "\n",
+        "  (b)the  line  current    is   13.9   A\n",
+        "\n",
+        "  (c) power  is   5.8 kW\n",
+        "\n",
+        "  (d)kVA  of  the  alternator  is   9.63 kVA"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 309</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the total power input and (b) the load power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi1  =  8000;#  in  Watts\n",
+      "Pi2  =  4000;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Total  input  power\n",
+      "Pi  =  Pi1  +  Pi2\n",
+      "phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))\n",
+      " #Power  factor\n",
+      "pf  =  math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\"W\"\n",
+      "print  \"\\n  (b)power  factor  is  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   12000.0 W\n",
+        "\n",
+        "  (b)power  factor  is   0.87"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 310</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the readings of each wattmeter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi  =  12000;#  in  Watts\n",
+      "pf  =  0.6;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #If  the  two  wattmeters  indicate  Pi1  and  Pi2  respectively\n",
+      " #  Pit  =  Pi1  +  Pi2\n",
+      "Pit  =  Pi\n",
+      " #  Pid  =  Pi1  -  Pi2\n",
+      " #power  factor  =  0.6  =  cos(phi)\n",
+      "phi  =  math.acos(pf)\n",
+      "Pid  =  Pit*math.tan(phi)/(3**0.5)\n",
+      " #Hence  wattmeter  1  reads\n",
+      "Pi1  =  (Pid  +  Pit)/2\n",
+      " #wattmeter  2  reads\n",
+      "Pi2  =  Pit  -  Pi1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reading  in  each  wattameter  are  \",round(Pi1,2),\"W  and  \",round(Pi2,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reading  in  each  wattameter  are   10618.8 W  and   1381.2 W"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 310</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the input power and (b) the load power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi1  =  10000;#  in  Watts\n",
+      "Pi2  =  -3000;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Total  input  power\n",
+      "Pi  =  Pi1  +  Pi2\n",
+      "phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))\n",
+      " #Power  factor\n",
+      "pf  =  math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\"W\"\n",
+      "print  \"\\n  (b)power  factor  is  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   7000.0 W\n",
+        "\n",
+        "  (b)power  factor  is   0.3"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 311</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing the variables:\n",
+      "R = 8; # in ohms\n",
+      "XL = 8; # in ohms\n",
+      "VL = 415; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#For a star connection:\n",
+      "#IL = Ip\n",
+      "#VL = Vp*(3**0.5)\n",
+      "VLs = VL\n",
+      "Vps = VLs/(3**0.5)\n",
+      "#Impedance per phase,\n",
+      "Zp = (R*R + XL*XL)**0.5\n",
+      "Ips = Vps/Zp\n",
+      "ILs = Ips\n",
+      "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)\n",
+      "pf = R/Zp\n",
+      "Ps = VLs*ILs*(3**0.5)*pf\n",
+      "#If wattmeter readings are P1 and P2 then P1 + P2 = Pst\n",
+      "Pst = Ps\n",
+      "# Pid = Pi1 - Pi2\n",
+      "phi = math.acos(pf)\n",
+      "Psd = Pst*math.tan(phi)/(3**0.5)\n",
+      "#Hence wattmeter 1 reads\n",
+      "Ps1 = (Psd + Pst)/2\n",
+      "#wattmeter 2 reads\n",
+      "Ps2 = Pst - Ps1\n",
+      "\n",
+      "#For a delta connection:\n",
+      "#VL = Vp\n",
+      "#IL = Ip*(3**0.5)\n",
+      "VLd = VL\n",
+      "Vpd = VLd\n",
+      "Ipd = Vpd/Zp\n",
+      "ILd = Ipd*(3**0.5)\n",
+      "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)\n",
+      "Pd = VLd*ILd*(3**0.5)*pf\n",
+      "#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt\n",
+      "Pdt = Pd\n",
+      "# Pid = Pi1 - Pi2\n",
+      "Pdd = Pdt*math.tan(phi)/(3**0.5)\n",
+      "#Hence wattmeter 1 reads\n",
+      "Pd1 = (Pdd + Pdt)/2\n",
+      "#wattmeter 2 reads\n",
+      "Pd2 = Pdt - Pd1\n",
+      "\n",
+      "#results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"(a)When the coils are star-connected the wattmeter readings are\", round(Ps1,2),\"W and \",round(Ps2,2),\"W\"\n",
+      "print \"(b)When the coils are delta-connected the wattmeter readings are are\", round(Pd1,2),\"W and\", round(Pd2,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "(a)When the coils are star-connected the wattmeter readings are 8489.35 W and  2274.71 W\n",
+        "(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_3.ipynb
new file mode 100755
index 00000000..a3d9112a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_3.ipynb
@@ -0,0 +1,974 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ff0801e05c7bf0796ff3208e43c2911eddcab1546881143bf5f93af7c58ebca6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 19: Three phase systems</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 299</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Vl  =  415;#  in  Volts\n",
+      "Rp  =  30;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "Vp  =  Vl/(3**0.5)\n",
+      "Ip  =  Vp/Rp\n",
+      "Il  =  Ip\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  system  phase  voltage  is  \",round(Vp,2),\"  V\"\n",
+      "print  \"\\n  (b)phase  current  is  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (c)line  current  is  \",round(Il,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  system  phase  voltage  is   239.6   V\n",
+        "\n",
+        "  (b)phase  current  is   7.99   A\n",
+        "\n",
+        "  (c)line  current  is   7.99   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 299</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  0.1273;#  in  Henry\n",
+      "Ip  =  5.08;#  in  Amperes\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Il  =  Ip\n",
+      "Vp  =  Ip*Zp\n",
+      "Vl  =  Vp*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)line  voltage  is  \",round(Vl,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)line  voltage  is   439.89   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 301</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "PR  =  24000;#  in  Watt\n",
+      "Py  =  18000;#  in  Watt\n",
+      "Pb  =  12000;#  in  Watt\n",
+      "VR  =  240;#  in  Volts\n",
+      "Vy  =  240;#  in  Volts\n",
+      "Vb  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star-connected  system  VL  =  Vp*(3**0.5)\n",
+      "Vp  =  V/(3**0.5)\n",
+      "phir  =  90*math.pi/180\n",
+      "phiy  =  330*math.pi/180\n",
+      "phib  =  210*math.pi/180\n",
+      " #  I  =  P/V    for  a  resistive  load\n",
+      "IR  =  PR/VR\n",
+      "Iy  =  Py/Vy\n",
+      "Ib  =  Pb/Vb\n",
+      "Inh  =  IR*math.cos(phir)  +  Ib*math.cos(phib)  +  Iy*math.cos(phiy)\n",
+      "Inv  =  IR*math.sin(phir)  +  Ib*math.sin(phib)  +  Iy*math.sin(phiy)\n",
+      "In  =  (Inh**2  +  Inv**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)cuurnt  in  R  line  is  \",round(IR,2),\"  A,  cuurnt  in  Y  line  is  \",round(Iy,2),\"  A  \"\n",
+      "print   \"and  cuurnt  in  B  line  is  \",round(Ib,2),\"  A\"\n",
+      "print  \"\\n  (b)cuurnt  in  neutral  line  is  \",round(In,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)cuurnt  in  R  line  is   100.0   A,  cuurnt  in  Y  line  is   75.0   A  and  cuurnt  in  B  line  is   50.0   A\n",
+        "\n",
+        "  (b)cuurnt  in  neutral  line  is   43.3   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 302</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "L  =  0.1273;#  in  Henry\n",
+      "VL  =  440;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      " #For  a  delta  connection,\n",
+      "IL  =  Ip*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  phase  current  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (b)line  current  \",round(IL,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  phase  current   8.8   A\n",
+        "\n",
+        "  (b)line  current   15.24   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 302</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "IL  =  15;#  in  Amperes\n",
+      "VL  =  415;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  delta  connection\n",
+      "Ip  =  IL/(3**0.5)#phase  current\n",
+      "Vp  =  VL\n",
+      " #Capacitive  reactance  per  phase\n",
+      "Xc  =  Vp/Ip\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   66.43 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 303</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  3;#  in  ohms\n",
+      "XL  =  4;#  in  ohms\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star  connection:\n",
+      " #IL  =  Ip\n",
+      " #VL  =  Vp*(3**0.5)\n",
+      "VLs  =  VL\n",
+      "Vps  =  VLs/(3**0.5)\n",
+      " #Impedance  per  phase,\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "Ips  =  Vps/Zp\n",
+      "ILs  =  Ips\n",
+      " #For  a  delta  connection:\n",
+      " #VL  =  Vp\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "VLd  =  VL\n",
+      "Vpd  =  VLd\n",
+      "Ipd  =  Vpd/Zp\n",
+      "ILd  =  Ipd*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  line  voltage  for  star  connection  is  \",round(VLs,2),\"  V  \"\n",
+      "print   \"and the  phase  voltage  for  star  connection  is  \",round(Vps,2),\"  V  \"\n",
+      "print   \"and the  line  voltage  for  delta  connection  is  \",round(VLd,2),\"  V  \"\n",
+      "print   \"and the  phase  voltage  for  delta  connection  is  \",round(Vpd,2),\"  V\"\n",
+      "print  \"\\n  (b)the  line  current  for  star  connection  is  \",round(ILs,2),\"  A  \"\n",
+      "print   \"and  the  phase  current  for  star  connection  is  \",round(Ips,2),\"  A  \"\n",
+      "print   \"and the  line  current  for  delta  connection  is  \",round(ILd,2),\"  A  \"\n",
+      "print   \"and the  phase  current  for  delta  connection  is  \",round(Ipd,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  line  voltage  for  star  connection  is   415.0   V  \n",
+        "and the  phase  voltage  for  star  connection  is   239.6   V  \n",
+        "and the  line  voltage  for  delta  connection  is   415.0   V  \n",
+        "and the  phase  voltage  for  delta  connection  is   415.0   V\n",
+        "\n",
+        "  (b)the  line  current  for  star  connection  is   47.92   A  \n",
+        "and  the  phase  current  for  star  connection  is   47.92   A  \n",
+        "and the  line  current  for  delta  connection  is   143.76   A  \n",
+        "and the  phase  current  for  delta  connection  is   83.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rp  =  12;#  in  ohms\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "Vp  =  VL/(3**0.5)#  since  the  resistors  are  star-connected\n",
+      " #Phase  current,  Ip\n",
+      "Zp  =  Rp\n",
+      "Ip  =  Vp/Zp\n",
+      " #For  a  star  connection\n",
+      "IL  =  Ip\n",
+      " #  For  a  purely  resistive  load,  the  power  factor  cos(phi)  =  1\n",
+      "pf  =  1\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)total  power  dissipated  by  the  resistors  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)total  power  dissipated  by  the  resistors  is   14352.08   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "P  =  5000;#  in  Watts\n",
+      "IL  =  8.6;#  in  amperes\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pf  =  P/(VL*IL*(3**0.5))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  factor  is  \",round(pf,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  factor  is   0.839"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 304</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "L  =  0.042;#  in  Henry\n",
+      "VL  =  415;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star  connection:\n",
+      " #IL  =  Ip\n",
+      " #VL  =  Vp*(3**0.5)\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      "VLs  =  VL\n",
+      "Vps  =  VLs/(3**0.5)\n",
+      " #Impedance  per  phase,\n",
+      "Ips  =  Vps/Zp\n",
+      "ILs  =  Ips\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pfs  =  R/Zp\n",
+      "Ps  =  VLs*ILs*(3**0.5)*pfs\n",
+      "\n",
+      " #For  a  delta  connection:\n",
+      " #VL  =  Vp\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "VLd  =  VL\n",
+      "Vpd  =  VLd\n",
+      "Ipd  =  Vpd/Zp\n",
+      "ILd  =  Ipd*(3**0.5)\n",
+      " #Power  dissipated,  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pfd  =  R/Zp\n",
+      "Pd  =  VLd*ILd*(3**0.5)*pfd\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  total  power  dissipated  in  star  is  \",round(Ps,2),\" W  and  in  delta  is  \",round(Pd,2),\" W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  total  power  dissipated  in  star  is   6283.29  W  and  in  delta  is   18849.88  W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 305</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Po  =  12750;#  in  Watts\n",
+      "pf  =  0.77;#  power  factor\n",
+      "eff  =  0.85;\n",
+      "VL  =  415;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #eff  =  power_out/power_in\n",
+      "Pi  =  Po/eff\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "IL  =  Pi/(VL*(3**0.5)*pf)#  line  current\n",
+      " #For  a  delta  connection:\n",
+      " #IL  =  Ip*(3**0.5)\n",
+      "Ip  =  IL/(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\" W\"\n",
+      "print  \"\\n  (b)line  current  is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (c)phase  current  is  \",round(Ip,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   15000.0  W\n",
+        "\n",
+        "  (b)line  current  is   27.1   A\n",
+        "\n",
+        "  (c)phase  current  is   15.65   A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 308</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "XL  =  40;#  in  ohms\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "Zp  =  (R*R  +  XL*XL)**0.5\n",
+      " #a  delta-connected  load\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      "IL  =  Ip*(3**0.5)\n",
+      " #Alternator  output  power  is  equal  to  the  power  dissipated  by  the  load.\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "pf  =  R/Zp\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      " #Alternator  output  kVA,\n",
+      "S  =  VL*IL*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  supplied  by  the  alternator  is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (b)output  power  is  \",round(P/1000,2),\"KW  and  kVA  of  the  alternator  is  \",round(S/1000,2),\"kVA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  supplied  by  the  alternator  is   13.86   A\n",
+        "\n",
+        "  (b)output  power  is   5.76 KW  and  kVA  of  the  alternator  is   9.6 kVA"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 308</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  30;#  in  ohms\n",
+      "C  =  80E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "VL  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Capacitive  reactance\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      "Zp  =  (R*R  +  Xc*Xc)**0.5\n",
+      "pf  =  R/Zp\n",
+      " #a  delta-connected  load\n",
+      "Vp  =  VL\n",
+      " #Phase  current\n",
+      "Ip  =  Vp/Zp\n",
+      "IL  =  Ip*(3**0.5)\n",
+      " #Power  P  =  VL*IL*(3**0.5)*cos(phi)    or    P  =  3*Ip*Ip*Rp)\n",
+      "P  =  VL*IL*(3**0.5)*pf\n",
+      " #Alternator  output  kVA,\n",
+      "S  =  VL*IL*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  phase  current    is  \",round(Ip,2),\"  A\"\n",
+      "print  \"\\n  (b)the  line  current    is  \",round(IL,2),\"  A\"\n",
+      "print  \"\\n  (c) power  is  \",round(P/1000,2),\"kW\"\n",
+      "print  \"\\n  (d)kVA  of  the  alternator  is  \", round(S/1000,2),\"kVA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  phase  current    is   8.03   A\n",
+        "\n",
+        "  (b)the  line  current    is   13.9   A\n",
+        "\n",
+        "  (c) power  is   5.8 kW\n",
+        "\n",
+        "  (d)kVA  of  the  alternator  is   9.63 kVA"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 309</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi1  =  8000;#  in  Watts\n",
+      "Pi2  =  4000;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Total  input  power\n",
+      "Pi  =  Pi1  +  Pi2\n",
+      "phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))\n",
+      " #Power  factor\n",
+      "pf  =  math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\"W\"\n",
+      "print  \"\\n  (b)power  factor  is  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   12000.0 W\n",
+        "\n",
+        "  (b)power  factor  is   0.87"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 310</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi  =  12000;#  in  Watts\n",
+      "pf  =  0.6;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #If  the  two  wattmeters  indicate  Pi1  and  Pi2  respectively\n",
+      " #  Pit  =  Pi1  +  Pi2\n",
+      "Pit  =  Pi\n",
+      " #  Pid  =  Pi1  -  Pi2\n",
+      " #power  factor  =  0.6  =  cos(phi)\n",
+      "phi  =  math.acos(pf)\n",
+      "Pid  =  Pit*math.tan(phi)/(3**0.5)\n",
+      " #Hence  wattmeter  1  reads\n",
+      "Pi1  =  (Pid  +  Pit)/2\n",
+      " #wattmeter  2  reads\n",
+      "Pi2  =  Pit  -  Pi1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reading  in  each  wattameter  are  \",round(Pi1,2),\"W  and  \",round(Pi2,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reading  in  each  wattameter  are   10618.8 W  and   1381.2 W"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 310</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Pi1  =  10000;#  in  Watts\n",
+      "Pi2  =  -3000;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Total  input  power\n",
+      "Pi  =  Pi1  +  Pi2\n",
+      "phi  =  math.atan((Pi1  -  Pi2)*(3**0.5)/(Pi1  +  Pi2))\n",
+      " #Power  factor\n",
+      "pf  =  math.cos(phi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)power  input  is  \",round(Pi,2),\"W\"\n",
+      "print  \"\\n  (b)power  factor  is  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)power  input  is   7000.0 W\n",
+        "\n",
+        "  (b)power  factor  is   0.3"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 311</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing the variables:\n",
+      "R = 8; # in ohms\n",
+      "XL = 8; # in ohms\n",
+      "VL = 415; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "#For a star connection:\n",
+      "#IL = Ip\n",
+      "#VL = Vp*(3**0.5)\n",
+      "VLs = VL\n",
+      "Vps = VLs/(3**0.5)\n",
+      "#Impedance per phase,\n",
+      "Zp = (R*R + XL*XL)**0.5\n",
+      "Ips = Vps/Zp\n",
+      "ILs = Ips\n",
+      "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)\n",
+      "pf = R/Zp\n",
+      "Ps = VLs*ILs*(3**0.5)*pf\n",
+      "#If wattmeter readings are P1 and P2 then P1 + P2 = Pst\n",
+      "Pst = Ps\n",
+      "# Pid = Pi1 - Pi2\n",
+      "phi = math.acos(pf)\n",
+      "Psd = Pst*math.tan(phi)/(3**0.5)\n",
+      "#Hence wattmeter 1 reads\n",
+      "Ps1 = (Psd + Pst)/2\n",
+      "#wattmeter 2 reads\n",
+      "Ps2 = Pst - Ps1\n",
+      "\n",
+      "#For a delta connection:\n",
+      "#VL = Vp\n",
+      "#IL = Ip*(3**0.5)\n",
+      "VLd = VL\n",
+      "Vpd = VLd\n",
+      "Ipd = Vpd/Zp\n",
+      "ILd = Ipd*(3**0.5)\n",
+      "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi)  or  P = 3*Ip*Ip*Rp)\n",
+      "Pd = VLd*ILd*(3**0.5)*pf\n",
+      "#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt\n",
+      "Pdt = Pd\n",
+      "# Pid = Pi1 - Pi2\n",
+      "Pdd = Pdt*math.tan(phi)/(3**0.5)\n",
+      "#Hence wattmeter 1 reads\n",
+      "Pd1 = (Pdd + Pdt)/2\n",
+      "#wattmeter 2 reads\n",
+      "Pd2 = Pdt - Pd1\n",
+      "\n",
+      "#results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"(a)When the coils are star-connected the wattmeter readings are\", round(Ps1,2),\"W and \",round(Ps2,2),\"W\"\n",
+      "print \"(b)When the coils are delta-connected the wattmeter readings are are\", round(Pd1,2),\"W and\", round(Pd2,2),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "(a)When the coils are star-connected the wattmeter readings are 8489.35 W and  2274.71 W\n",
+        "(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb
new file mode 100755
index 00000000..5b0974af
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint.ipynb
@@ -0,0 +1,1627 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 20: Transformers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the secondary voltage, assuming an ideal transformer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  500;#  primary  turns\n",
+      "N2  =  3000;#  secondary  turns\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  an  ideal  transformer,  voltage  ratio  =  turns  ratio\n",
+      "V2  =  V1*N2/N1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  voltage  \",round(V2,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  voltage   1440.0 V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its output voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  2/7;#  turns  ratio\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  2:7  means  that  the  transformer  has  2  turns  on  the  primary  \n",
+      "    #for  every  7  turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  \",round(V2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage   840.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine secondary voltage and current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  8/1;#  turns  ratio\n",
+      "I1  =  3;#  in  Amperes\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  \n",
+      "    #primary  for  every  1turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      " #secondary  current\n",
+      "I2  =  I1*tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  voltage  is  \",round(V2,2),\"  V  and  secondary  current  is  \", round(I2,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  voltage  is   30.0   V  and  secondary  current  is   24.0   A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 318</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the transformer turns ratio and the current taken from the supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  240;#  in  Volts\n",
+      "V2  =  12;#  in  Volts\n",
+      "P  =  150;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      "I2  =  P/V2\n",
+      " #A  turns  ratio  =  Vp/Vs\n",
+      "tr  =  V1/V2#  turn  ratio\n",
+      " #    V1/V2  =  I2/I1\n",
+      " #current  taken  from  the  supply\n",
+      "I1  =  I2*V2/V1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  turn  ratio  is  \",round(tr,2),\"  and  current  taken  from  the  supply  is  \",round(I1,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  turn  ratio  is   20.0   and  current  taken  from  the  supply  is   0.63   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 318</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the full-load secondary current,\n",
+      "#(b) the minimum load resistance which can be connected across the secondary winding to give full load kVA, \n",
+      "#(c) the primary current at full load kVA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  5000;#  in  VA\n",
+      "tr  =  10;#  turn  ratio\n",
+      "V1  =  2500;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  primary  for  every  1turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      " #transformer  rating  in  volt-amperes  =  Vs*Is\n",
+      "I2  =  S/V2\n",
+      " #Minimum  value  of  load  resistance\n",
+      "RL  =  V2/I2\n",
+      " #    tr  =  I2/I1\n",
+      "I1  =  I2/tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)full-load  secondary  current  is  \",round(I2,2),\"  A\"\n",
+      "print  \"\\n  (b)minimum  load  resistance  is  \",round(RL,2),\"  ohm\"\n",
+      "print  \"\\n  (c)  primary  current  is  \",round(I1,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)full-load  secondary  current  is   20.0   A\n",
+        "\n",
+        "  (b)minimum  load  resistance  is   12.5   ohm\n",
+        "\n",
+        "  (c)  primary  current  is   2.0   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 319</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of the magnetizing and core loss components of the noload current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  2400;#  in  Volts\n",
+      "V2  =  400;#  in  Volts\n",
+      "I0  =  0.5;#  in  Amperes\n",
+      "Pc  =  400;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Core  loss  (i.e.  iron  loss)  P  =  V1*I0*cos(phi0)\n",
+      "pf  =  Pc/(V1*I0)\n",
+      "phi0  =  math.acos(pf)\n",
+      " #Magnetizing  component\n",
+      "Im  =  I0*math.sin(phi0)\n",
+      " #Core  loss  component\n",
+      "Ic  =  I0*math.cos(phi0)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)magnetizing  component  is  \",round(Im,3),\"  A  and  Core  loss  component  is  \",round(Ic,3),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)magnetizing  component  is   0.471   A  and  Core  loss  component  is   0.167   A"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 320</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the iron loss current, \n",
+      "#(b) the power factor on no-load, and (c) the magnetizing current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "I0  =  0.8;#  in  Amperes\n",
+      "P  =  72;#  in  Watts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  absorbed  =  total  core  loss,  P  =  V*I0*cos(phi0)\n",
+      " #Ic  =  I0*cos(phi0)\n",
+      "Ic  =  P/V\n",
+      "pf  =  Ic/I0\n",
+      " #From  the  right-angled  triangle  in  Figure  20.2(b)  and  using\n",
+      " #Pythagoras\u2019  theorem,  \n",
+      "Im  =  (I0*I0  -  Ic*Ic)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Core  loss  component  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)  power  factor  is  \",round(  pf,2),\"\"\n",
+      "print  \"\\n  (c)magnetizing  component  is  \",round(Im,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Core  loss  component  is   0.3   A\n",
+        "\n",
+        "  (b)  power  factor  is   0.37 \n",
+        "\n",
+        "  (c)magnetizing  component  is   0.74   A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 321</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  100000;#  in  VA\n",
+      "V1  =  4000;#  in  Volts\n",
+      "V2  =  200;#  in  Volts\n",
+      "N2  =  100;#  sec  turns\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Transformer  rating  =  V1*I1  =  V2*I2\n",
+      " #primary  current\n",
+      "I1  =  S/V1\n",
+      " #secondary  current\n",
+      "I2  =  S/V2\n",
+      " #primary  turns\n",
+      "N1  =  N2*V1/V2\n",
+      " #maximum  flux\n",
+      " #assuming  E2  =  V2\n",
+      "Phim  =  V2/(4.44*f*N2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)primary  current  is  \",round(  I1,2),\"  A  and  secondary  current  is  \",round(  I2,2),\"  A\"\n",
+      "print  \"\\n  (b)number  of  primary  turns  is  \",round(  N1,2),\"\"\n",
+      "print  \"\\n  (c)maximum  value  of  the  flux  is  \",round(Phim*1000,2),\"mWb\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)primary  current  is   25.0   A  and  secondary  current  is   500.0   A\n",
+        "\n",
+        "  (b)number  of  primary  turns  is   2000.0 \n",
+        "\n",
+        "  (c)maximum  value  of  the  flux  is   9.01 mWb"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 322</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the maximum value of the flux density in the core, and\n",
+      "#(b) the voltage induced in the secondary winding.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  250;#  in  Volts\n",
+      "A  =  0.03;#  in  m2\n",
+      "N2  =  300;#  sec  turns\n",
+      "N1  =  25;#  prim  turns\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #maximum  flux  density,\n",
+      "Phim  =  V1/(4.44*f*N1)\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #maximum  core  flux  density\n",
+      "Bm  =  Phim/A\n",
+      " #voltage  induced  in  the  secondary  winding,\n",
+      "V2  =  V1*N2/N1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  core  flux  density  \",round(  Bm,2),\"  T\"\n",
+      "print  \"\\n  (b)voltage  induced  in  the  secondary  winding  is  \",round(  V2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  core  flux  density   1.5   T\n",
+        "\n",
+        "  (b)voltage  induced  in  the  secondary  winding  is   3000.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 323</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the number of primary and secondary turns\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  500;#  in  Volts\n",
+      "V2  =  100;#  in  Volts\n",
+      "Bm  =  1.5;#  in  Tesla\n",
+      "A  =  0.005;#  in  m2\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #maximum  core  flux  density\n",
+      "Phim  =  Bm*A\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #primary  turns,\n",
+      "N1  =  V1/(4.44*f*Phim)\n",
+      " #secondary  turns,\n",
+      "N2  =  V2*N1/V1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  no.  of  primary  and  secondary  turns  are  \",round(N1,2),\"  turns,  and  \",round(N2,2),\"  turns  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  no.  of  primary  and  secondary  turns  are   300.3   turns,  and   60.06   turns  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 323</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the number of primary and secondary turns and \n",
+      "#(b) the cross-sectional area of the core.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "emfpt  =  15;#  in  Volts\n",
+      "V1  =  4500;#  in  Volts\n",
+      "V2  =  225;#  in  Volts\n",
+      "Bm  =  1.4;#  in  Tesla\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #E.m.f.  per  turn,  V1/N1  =  V2/N2  =  emfpt\n",
+      " #primary  turns,\n",
+      "N1  =  V1/emfpt\n",
+      " #secondary  turns,\n",
+      "N2  =  V2/emfpt\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #maximum  flux  density,\n",
+      "Phim  =  V1/(4.44*f*N1)\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #cross-sectional  area\n",
+      "A  =  Phim/Bm\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)no.  of  primary  and  secondary  turns  are  \",  N1,\"  turns,  and  \",  N2,\"  turns  respectively\"\n",
+      "print  \"\\n  (b)cross-sectional  area  is  \",  round(A,4),\"m2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)no.  of  primary  and  secondary  turns  are   300.0   turns,  and   15.0   turns  respectively\n",
+        "\n",
+        "  (b)cross-sectional  area  is   0.0483 m2"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 324</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the primary current and power factor when the secondary current is 100 A at a power factor of 0.85 lagging\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  2000;#  prim  turns\n",
+      "N2  =  800;#  sec  turns\n",
+      "I0  =  5;#  in  Amperes\n",
+      "pf0  =  0.20;#  power  factor\n",
+      "I2  =  100;#  in  Amperes\n",
+      "pf2  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Let  I01  be  the  component  of  the  primary  current  which  provides  the  restoring  mmf.  Then  I01*N1  =  I2*N2\n",
+      "I01  =  I2*N2/N1\n",
+      " #If  the  power  factor  of  the  secondary  is  0.85\n",
+      "phi2  =  math.acos(pf2)\n",
+      " #If  the  power  factor  on  no-load  is  0.20,\n",
+      "phi0  =  math.acos(pf0)\n",
+      "I1h  =  I0*math.cos(phi0)  +  I01*math.cos(phi2)\n",
+      "I1v  =  I0*math.sin(phi0)  +  I01*math.sin(phi2)\n",
+      " #Hence  the  magnitude  of  I1\n",
+      "I1  =  (I1h*I1h  +  I1v*I1v)**0.5\n",
+      "pf1  =  math.cos(math.atan(I1v/I1h))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Primary  current  is  \", round(I1,2),\"  A,  and  Power  factor  is  \",round(pf1,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Primary  current  is   43.58   A,  and  Power  factor  is   0.8"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 328</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the equivalent resistance referred to the primary winding, \n",
+      "#(b) the equivalent reactance referred to the primary winding,\n",
+      "#(c) the equivalent impedance referred to the primary winding, and \n",
+      "#(d) the phase angle of the impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  600;#  prim  turns\n",
+      "N2  =  150;#  sec  turns\n",
+      "R1  =  0.25;#  in  ohms\n",
+      "R2  =  0.01;#  in  ohms\n",
+      "X1  =  1.0;#  in  ohms\n",
+      "X2  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "tr  =  N1/N2#  turn  ratio\n",
+      "vr  =  tr#  voltage  ratio  =  turn  raio,  vr  =  V1/V2\n",
+      " #equivalent  resistance  Re\n",
+      "Re  =  R1  +  R2*(vr**2)\n",
+      " #equivalent  reactance,  Xe\n",
+      "Xe  =  X1  +  X2*(vr**2)\n",
+      " #equivalent  impedance,  Ze\n",
+      "Ze  =  (Re*Re  +  Xe*Xe)**0.5\n",
+      " #cos(phie)  =  Re/Ze\n",
+      "pfe  =  Re/Ze\n",
+      "phie  =  math.acos(pfe)\n",
+      "phied  =  phie*180/math.pi#  in  \u00b0(deg)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  resistance  referred  to  the  primary  winding  is  \",round(  Re,2),\"  ohm\"\n",
+      "print  \"\\n  (b)the  equivalent  reactance  referred  to  the  primary  winding  is  \",round(  Xe,2),\"  ohm\"\n",
+      "print  \"\\n  (c)the  equivalent  impedance  referred  to  the  primary  winding  is  \",round(  Ze,2),\"  ohm\"\n",
+      "print  \"\\n  (d)phase  angle  is  \",round(  phied,2),\"deg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  resistance  referred  to  the  primary  winding  is   0.41   ohm\n",
+        "\n",
+        "  (b)the  equivalent  reactance  referred  to  the  primary  winding  is   1.64   ohm\n",
+        "\n",
+        "  (c)the  equivalent  impedance  referred  to  the  primary  winding  is   1.69   ohm\n",
+        "\n",
+        "  (d)phase  angle  is   75.96 deg"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 329</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the regulation of the transformer.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  200;#  in  Volts\n",
+      "V2  =  400;#  in  Volts\n",
+      "V2L  =  387.6;#  in  Volts\n",
+      "S  =  5000;#  in  VA\n",
+      "\n",
+      "#calculation:\n",
+      " #regulation  =(No-load  secondary  voltage  -\u0003  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %\n",
+      "reg  =  (V2  -  V2L)*100/V2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  regulation  of  the  transformer  is  \",round(reg,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  regulation  of  the  transformer  is   3.1   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 329</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the load voltage at which the mechanism operates.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VnL  =  240;#  in  Volts\n",
+      "reg  =  2.5;#  in  percent\n",
+      "\n",
+      "#calculation:\n",
+      " #regulation  =(No-load  secondary  voltage  -\u0003  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %\n",
+      "VL  =  VnL  -  reg*VnL/100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  load  voltage  at  which  the  mechanism  operates  is  \",round(VL,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  load  voltage  at  which  the  mechanism  operates  is   234.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 331</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the transformer efficiency at full load and 0.85 power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  200000;#  in  VA\n",
+      "Pc  =  1500;#  in  Watt\n",
+      "Pi  =  1000;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Full-load  output  power  =  V*I*pf\n",
+      "Po  =  S*pf\n",
+      " #Total  losses\n",
+      "Pl  =  Pc  +  Pi\n",
+      " #Input  power  =  output  power  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #efficiency\n",
+      "eff  =  1-(Pl/PI)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  efficiency  at  full  load  is  \",round(eff,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  efficiency  at  full  load  is   0.99"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 331</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the efficiency of the transformer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  200000;#  in  VA\n",
+      "Pc  =  1500;#  in  Watt\n",
+      "Pi  =  1000;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Half  full-load  power  output  =  V*I*pf/2\n",
+      "Po  =  S*pf/2\n",
+      " #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared\n",
+      " #Hence  the  copper  loss  at  half  full-load  is\n",
+      "Pch  =  Pc/(2*2)\n",
+      " #Iron  loss  =  1000  W  (constant)\n",
+      " #Total  losses\n",
+      "Pl  =  Pch  +  Pi\n",
+      " #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #efficiency\n",
+      "eff  =  (1-(Pl/PI))*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  efficiency  at  half  full  load  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  efficiency  at  half  full  load  is   98.41   percent"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 332</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the efficiency of the transformer (a) on full load, and (b) on half load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  400000;#  in  VA\n",
+      "R1  =  0.5;#  in  Ohm\n",
+      "R2  =  0.001;#  in  Ohm\n",
+      "V1  =  5000;#  in  Volts\n",
+      "V2  =  320;#  in  Volts\n",
+      "Pi  =  2500;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Rating  =  400  kVA  =  V1*I1  =  V2*I2\n",
+      " #Hence  primary  current\n",
+      "I1  =  S/V1\n",
+      " #secondary  current\n",
+      "I2  =  S/V2\n",
+      " #Total  copper  loss  =  I1*I1*R1  +  I2*I2*R2,\n",
+      "Pcf  =  I1*I1*R1  +  I2*I2*R2\n",
+      " #On  full  load,  total  loss  =  copper  loss  +  iron  loss\n",
+      "Plf  =  Pcf  +  Pi\n",
+      " #  full-load  power  output  =  V2*I2*pf\n",
+      "Pof  =  S*pf\n",
+      " #Input  power  at  full-load  =  output  power  at  full-load  +  losses\n",
+      "PIf  =  Pof  +  Plf\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      "efff  =  (1-(Plf/PIf))*100\n",
+      "\n",
+      " #Half  full-load  power  output  =  V*I*pf/2\n",
+      "Poh  =  S*pf/2\n",
+      " #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared\n",
+      " #Hence  the  copper  loss  at  half  full-load  is\n",
+      "Pch  =  Pcf/(2*2)\n",
+      " #Iron  loss  =  2500  W  (constant)\n",
+      " #Total  losses\n",
+      "Plh  =  Pch  +  Pi\n",
+      " #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses\n",
+      "PIh  =  Poh  +  Plh\n",
+      " #efficiency\n",
+      "effh  =  (1-(Plh/PIh))*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  transformer  efficiency  at  full  load  is  \", round(efff,2),\"  percent\"\n",
+      "print  \"\\n  (b)the  transformer  efficiency  at  half  full  load  is  \",  round(effh,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  transformer  efficiency  at  full  load  is   97.91   percent\n",
+        "\n",
+        "  (b)the  transformer  efficiency  at  half  full  load  is   97.88   percent"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 333</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the maximum efficiency,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  500000;#  in  VA\n",
+      "Pcf  =  4000;#  in  Watt\n",
+      "Pi  =  2500;#  in  Watt\n",
+      "pf  =  0.75;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Let  x  be  the  fraction  of  full  load  kVA  at  which  the  efficiency  is  a  maximum.\n",
+      " #The  corresponding  total  copper  loss  =  (4  kW)*(x**2)\n",
+      " #At  maximum  efficiency,  copper  loss  =  iron  loss  Hence\n",
+      "x  =  (Pi/Pcf)**0.5\n",
+      " #Hence  the  output  kVA  at  maximum  efficiency\n",
+      "So  =  x*S\n",
+      " #Total  loss  at  maximum  efficiency\n",
+      "Pl  =  2*Pi\n",
+      " #Output  power\n",
+      "Po  =  So*pf\n",
+      " #Input  power  =  output  power  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u00e2\u20ac\u201dlosses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Maximum  efficiency\n",
+      "effm  =  (1  -  Pl/PI)*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the output kVA at maximum efficiency is  \",round(So/1000,2),\"kVA\"\n",
+      "print  \"\\n  max.  efficiency  is  \",round(effm,2),\"  pecent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the output kVA at maximum efficiency is   395.28 kVA\n",
+        "\n",
+        "  max.  efficiency  is   98.34   pecent"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the equivalent input resistance of the transformer.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  4;#  turn  ratio\n",
+      "RL  =  100;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #the  equivalent  input  resistance,\n",
+      "Ri  =  RL*(tr**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  equivalent  input  resistance  is  \",round(Ri,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  equivalent  input  resistance  is   1600.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the optimum turns ratio of a transformer which would match a \n",
+      "#load resistance of 7 ohmto the output resistance of the amplifier.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  112;#  in  Ohms\n",
+      "RL  =  7;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  112  ohm  for  maximum  power  transfer.\n",
+      " #R1  =  RL*(tr**2)\n",
+      " #  tr  =  N1/N2  turn  ratio\n",
+      "tr  =  (R1/RL)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  turns  ratio  is  \",tr,\": 1.0\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  turns  ratio  is   4.0 : 1.0"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the optimum value of load resistance for maximum power transfer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  5;#  turn  ratio\n",
+      "R1  =  150;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  150  ohm  for  maximum  power  transfer.\n",
+      " #R1  =  RL*(tr**2)\n",
+      "RL  =  R1/(tr**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  value  of  load  resistance  is  \",round(RL,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  value  of  load  resistance  is   6.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the primary current flowing and (b) the power dissipated in the load resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  220;#  in  Volts\n",
+      "V2  =  1760;#  in  Volts\n",
+      "V  =  220;#  in  Volts\n",
+      "RL  =  1280;#  in  Ohms\n",
+      "R  =  2;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Turns  ratio,  tr  =  N1/N2  =  V1/V2\n",
+      "tr  =  V1/V2\n",
+      " #Equivalent  input  resistance  of  the  transformer,\n",
+      " #R1  =  RL*(tr**2)\n",
+      "R1  =  RL*(tr**2)\n",
+      " #Total  input  resistance\n",
+      "Rin  =  R  +  R1\n",
+      " #  Primary  current\n",
+      "I1  =  V1/Rin\n",
+      " #For  an  ideal  transformer  V1/V2  =  I2/I1,\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistor  RL\n",
+      "P  =  I2*I2*RL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  primary  current  flowing  is  \",round(I1,2),\"  A\"\n",
+      "print  \"\\n  (b)    power  dissipated  in  the  load  resistor  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  primary  current  flowing  is   10.0   A\n",
+        "\n",
+        "  (b)    power  dissipated  in  the  load  resistor  is   2000.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 336</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of the load resistance and (b) the power dissipated in the load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  25;#  teurn  ratio\n",
+      "V  =  24;#  in  Volts\n",
+      "R1  =  15000;#  in  Ohms\n",
+      "Rin  =  15000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Turns  ratio,  tr  =  N1/N2  =  V1/V2\n",
+      " #For  maximum  power  transfer  R1  needs  to  be  equal  to  15  kohm\n",
+      "RL  =  R1/(tr**2)\n",
+      " #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is\n",
+      "Rt  =  Rin  +  R1\n",
+      " #Primary  current,\n",
+      "I1  =  V/Rt\n",
+      " #N1/N2  =  I2/I1\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistor  RL\n",
+      "P  =  I2*I2*RL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  load  resistance  is  \",round(RL,2),\"ohm\"\n",
+      "print  \"\\n  (b)    power  dissipated  in  the  load  resistor  is  \",round(P*1000,2),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  load  resistance  is   24.0 ohm\n",
+        "\n",
+        "  (b)    power  dissipated  in  the  load  resistor  is   9.6 mW"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 337</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in each section of the winding.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  320;#  in  Volts\n",
+      "V2  =  250;#  in  Volts\n",
+      "S  =  20000;#  in  VA\n",
+      "\n",
+      "#calculation:\n",
+      " #Rating  =  20  kVA  =  V1*I1  =  V2*I2\n",
+      " #Hence  primary  current,  I1\n",
+      "I1  =  S/V1\n",
+      " #secondary  current,  I2\n",
+      "I2  =  S/V2\n",
+      " #Hence  current  in  common  part  of  the  winding\n",
+      "I  =  I2  -  I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  in  common  part  of  the  winding  is  \", round(I,2),\"  A\"\n",
+      "print  \"\\n  primary  current  and  secondary  current  are  \",round(I1,2),\"  A  and  \",round(I2,2),\"  A  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  in  common  part  of  the  winding  is   17.5   A\n",
+        "\n",
+        "  primary  current  and  secondary  current  are   62.5   A  and   80.0   A  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 339</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the saving in the volume of copper used in an auto transformer compared with a double-wound transformer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1a  =  200;#  in  Volts\n",
+      "V2a  =  150;#  in  Volts\n",
+      "V1b  =  500;#  in  Volts\n",
+      "V2b  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  200  V:150  V  transformer,  xa\n",
+      "xa  =  V2a/V1a\n",
+      " #volume  of  copper  in  auto  transformer\n",
+      "vca  =  (1  -  xa)*100#  of  copper  in  a  double-wound  transformer\n",
+      " #the  saving  is\n",
+      "vsa  =  100  -  vca\n",
+      " #For  a  500  V:100  V  transformer,  xb\n",
+      "xb  =  V2b/V1b\n",
+      " #volume  of  copper  in  auto  transformer\n",
+      "vcb  =  (1  -  xb)*100#  of  copper  in  a  double-wound  transformer\n",
+      " #the  saving  is\n",
+      "vsb  =  100  -  vcb\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)For  a  200  V:150  V  transformer,  the  saving  is  \", round(vsa,2),\"  percent\"\n",
+      "print  \"\\n  (b)For  a  500  V:100  V  transformer,  the  saving  is  \", round(vsb,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For  a  200  V:150  V  transformer,  the  saving  is   75.0   percent\n",
+        "\n",
+        "  (b)For  a  500  V:100  V  transformer,  the  saving  is   20.0   percent"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 340</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the secondary line voltage on no-load when the windings are connected (a) star-delta, (b) delta-star.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  500;#  prim  turns\n",
+      "N2  =  50;#  sec  turns\n",
+      "VL  =  2400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star-connection,  VL  =  Vp*(3**0.5)\n",
+      "VL1s  =  VL\n",
+      " #Primary  phase  voltage\n",
+      "Vp1s  =  VL1s/(3**0.5)\n",
+      " #For  a  delta-connection,  VL  =  Vp\n",
+      " #N1/N2  =  V1/V2,  from  which,\n",
+      " #secondary  phase  voltage,  Vp2s\n",
+      "Vp2s  =  Vp1s*N2/N1\n",
+      "VL2d  =  Vp2s\n",
+      "\n",
+      " #For  a  delta-connection,  VL  =  Vp\n",
+      "VL1d  =  VL\n",
+      " #primary  phase  voltage  Vp1d\n",
+      "Vp1d  =  VL1d\n",
+      " #Secondary  phase  voltage,  Vp2d\n",
+      "Vp2d  =  Vp1d*N2/N1\n",
+      " #For  a  star-connection,  VL  =  Vp*(3**0.5)\n",
+      "VL2s  =  Vp2d*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  secondary  line  voltage  for  star  and  delta  connection  are  \",round(Vp2s,1),\"  V \"\n",
+      "print   \" and  \",round(VL2s,0),\"  V  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  secondary  line  voltage  for  star  and  delta  connection  are   138.6   V \n",
+        " and   416.0   V  respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 343</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the reading on the ammeter, \n",
+      "#(b) the potential difference across the ammeter and\n",
+      "#(c) the total load (in VA) on the secondary.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  1;#  prim  turns\n",
+      "N2  =  60;#  sec  turns\n",
+      "I1  =  300;#  in  amperes\n",
+      "Ra  =  0.15;#  in  ohms\n",
+      "R2  =  0.25;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Reading  on  the  ammeter,\n",
+      "I2  =  I1*(N1/N2)\n",
+      " #P.d.  across  the  ammeter  =  I2*RA,  where  RA  is  the  ammeter  resistance\n",
+      "pd  =  I2*Ra\n",
+      " #Total  resistance  of  secondary  circuit\n",
+      "Rt  =  Ra  +  R2\n",
+      " #Induced  e.m.f.  in  secondary\n",
+      "V2  =  I2*Rt\n",
+      " #Total  load  on  secondary\n",
+      "S  =  V2*I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  reading  on  the  ammeter  is  \",round(I2,2),\"  A  \"\n",
+      "print  \"\\n  (b)potential  difference  across  the  ammeter  is  \",round(pd,2),\"  V  \"\n",
+      "print  \"\\n  (c)total  load  (in  VA)  on  the  secondary  is  \",round(S,2),\"  VA  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  reading  on  the  ammeter  is   5.0   A  \n",
+        "\n",
+        "  (b)potential  difference  across  the  ammeter  is   0.75   V  \n",
+        "\n",
+        "  (c)total  load  (in  VA)  on  the  secondary  is   10.0   VA  "
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_1.ipynb
new file mode 100755
index 00000000..5b0974af
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_1.ipynb
@@ -0,0 +1,1627 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 20: Transformers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the secondary voltage, assuming an ideal transformer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  500;#  primary  turns\n",
+      "N2  =  3000;#  secondary  turns\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  an  ideal  transformer,  voltage  ratio  =  turns  ratio\n",
+      "V2  =  V1*N2/N1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  voltage  \",round(V2,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  voltage   1440.0 V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its output voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  2/7;#  turns  ratio\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  2:7  means  that  the  transformer  has  2  turns  on  the  primary  \n",
+      "    #for  every  7  turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  \",round(V2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage   840.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine secondary voltage and current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  8/1;#  turns  ratio\n",
+      "I1  =  3;#  in  Amperes\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  \n",
+      "    #primary  for  every  1turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      " #secondary  current\n",
+      "I2  =  I1*tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  voltage  is  \",round(V2,2),\"  V  and  secondary  current  is  \", round(I2,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  voltage  is   30.0   V  and  secondary  current  is   24.0   A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 318</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the transformer turns ratio and the current taken from the supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  240;#  in  Volts\n",
+      "V2  =  12;#  in  Volts\n",
+      "P  =  150;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      "I2  =  P/V2\n",
+      " #A  turns  ratio  =  Vp/Vs\n",
+      "tr  =  V1/V2#  turn  ratio\n",
+      " #    V1/V2  =  I2/I1\n",
+      " #current  taken  from  the  supply\n",
+      "I1  =  I2*V2/V1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  turn  ratio  is  \",round(tr,2),\"  and  current  taken  from  the  supply  is  \",round(I1,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  turn  ratio  is   20.0   and  current  taken  from  the  supply  is   0.63   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 318</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the full-load secondary current,\n",
+      "#(b) the minimum load resistance which can be connected across the secondary winding to give full load kVA, \n",
+      "#(c) the primary current at full load kVA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  5000;#  in  VA\n",
+      "tr  =  10;#  turn  ratio\n",
+      "V1  =  2500;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  primary  for  every  1turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      " #transformer  rating  in  volt-amperes  =  Vs*Is\n",
+      "I2  =  S/V2\n",
+      " #Minimum  value  of  load  resistance\n",
+      "RL  =  V2/I2\n",
+      " #    tr  =  I2/I1\n",
+      "I1  =  I2/tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)full-load  secondary  current  is  \",round(I2,2),\"  A\"\n",
+      "print  \"\\n  (b)minimum  load  resistance  is  \",round(RL,2),\"  ohm\"\n",
+      "print  \"\\n  (c)  primary  current  is  \",round(I1,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)full-load  secondary  current  is   20.0   A\n",
+        "\n",
+        "  (b)minimum  load  resistance  is   12.5   ohm\n",
+        "\n",
+        "  (c)  primary  current  is   2.0   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 319</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of the magnetizing and core loss components of the noload current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  2400;#  in  Volts\n",
+      "V2  =  400;#  in  Volts\n",
+      "I0  =  0.5;#  in  Amperes\n",
+      "Pc  =  400;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Core  loss  (i.e.  iron  loss)  P  =  V1*I0*cos(phi0)\n",
+      "pf  =  Pc/(V1*I0)\n",
+      "phi0  =  math.acos(pf)\n",
+      " #Magnetizing  component\n",
+      "Im  =  I0*math.sin(phi0)\n",
+      " #Core  loss  component\n",
+      "Ic  =  I0*math.cos(phi0)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)magnetizing  component  is  \",round(Im,3),\"  A  and  Core  loss  component  is  \",round(Ic,3),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)magnetizing  component  is   0.471   A  and  Core  loss  component  is   0.167   A"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 320</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the iron loss current, \n",
+      "#(b) the power factor on no-load, and (c) the magnetizing current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "I0  =  0.8;#  in  Amperes\n",
+      "P  =  72;#  in  Watts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  absorbed  =  total  core  loss,  P  =  V*I0*cos(phi0)\n",
+      " #Ic  =  I0*cos(phi0)\n",
+      "Ic  =  P/V\n",
+      "pf  =  Ic/I0\n",
+      " #From  the  right-angled  triangle  in  Figure  20.2(b)  and  using\n",
+      " #Pythagoras\u2019  theorem,  \n",
+      "Im  =  (I0*I0  -  Ic*Ic)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Core  loss  component  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)  power  factor  is  \",round(  pf,2),\"\"\n",
+      "print  \"\\n  (c)magnetizing  component  is  \",round(Im,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Core  loss  component  is   0.3   A\n",
+        "\n",
+        "  (b)  power  factor  is   0.37 \n",
+        "\n",
+        "  (c)magnetizing  component  is   0.74   A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 321</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  100000;#  in  VA\n",
+      "V1  =  4000;#  in  Volts\n",
+      "V2  =  200;#  in  Volts\n",
+      "N2  =  100;#  sec  turns\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Transformer  rating  =  V1*I1  =  V2*I2\n",
+      " #primary  current\n",
+      "I1  =  S/V1\n",
+      " #secondary  current\n",
+      "I2  =  S/V2\n",
+      " #primary  turns\n",
+      "N1  =  N2*V1/V2\n",
+      " #maximum  flux\n",
+      " #assuming  E2  =  V2\n",
+      "Phim  =  V2/(4.44*f*N2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)primary  current  is  \",round(  I1,2),\"  A  and  secondary  current  is  \",round(  I2,2),\"  A\"\n",
+      "print  \"\\n  (b)number  of  primary  turns  is  \",round(  N1,2),\"\"\n",
+      "print  \"\\n  (c)maximum  value  of  the  flux  is  \",round(Phim*1000,2),\"mWb\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)primary  current  is   25.0   A  and  secondary  current  is   500.0   A\n",
+        "\n",
+        "  (b)number  of  primary  turns  is   2000.0 \n",
+        "\n",
+        "  (c)maximum  value  of  the  flux  is   9.01 mWb"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 322</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the maximum value of the flux density in the core, and\n",
+      "#(b) the voltage induced in the secondary winding.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  250;#  in  Volts\n",
+      "A  =  0.03;#  in  m2\n",
+      "N2  =  300;#  sec  turns\n",
+      "N1  =  25;#  prim  turns\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #maximum  flux  density,\n",
+      "Phim  =  V1/(4.44*f*N1)\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #maximum  core  flux  density\n",
+      "Bm  =  Phim/A\n",
+      " #voltage  induced  in  the  secondary  winding,\n",
+      "V2  =  V1*N2/N1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  core  flux  density  \",round(  Bm,2),\"  T\"\n",
+      "print  \"\\n  (b)voltage  induced  in  the  secondary  winding  is  \",round(  V2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  core  flux  density   1.5   T\n",
+        "\n",
+        "  (b)voltage  induced  in  the  secondary  winding  is   3000.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 323</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the number of primary and secondary turns\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  500;#  in  Volts\n",
+      "V2  =  100;#  in  Volts\n",
+      "Bm  =  1.5;#  in  Tesla\n",
+      "A  =  0.005;#  in  m2\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #maximum  core  flux  density\n",
+      "Phim  =  Bm*A\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #primary  turns,\n",
+      "N1  =  V1/(4.44*f*Phim)\n",
+      " #secondary  turns,\n",
+      "N2  =  V2*N1/V1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  no.  of  primary  and  secondary  turns  are  \",round(N1,2),\"  turns,  and  \",round(N2,2),\"  turns  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  no.  of  primary  and  secondary  turns  are   300.3   turns,  and   60.06   turns  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 323</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the number of primary and secondary turns and \n",
+      "#(b) the cross-sectional area of the core.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "emfpt  =  15;#  in  Volts\n",
+      "V1  =  4500;#  in  Volts\n",
+      "V2  =  225;#  in  Volts\n",
+      "Bm  =  1.4;#  in  Tesla\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #E.m.f.  per  turn,  V1/N1  =  V2/N2  =  emfpt\n",
+      " #primary  turns,\n",
+      "N1  =  V1/emfpt\n",
+      " #secondary  turns,\n",
+      "N2  =  V2/emfpt\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #maximum  flux  density,\n",
+      "Phim  =  V1/(4.44*f*N1)\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #cross-sectional  area\n",
+      "A  =  Phim/Bm\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)no.  of  primary  and  secondary  turns  are  \",  N1,\"  turns,  and  \",  N2,\"  turns  respectively\"\n",
+      "print  \"\\n  (b)cross-sectional  area  is  \",  round(A,4),\"m2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)no.  of  primary  and  secondary  turns  are   300.0   turns,  and   15.0   turns  respectively\n",
+        "\n",
+        "  (b)cross-sectional  area  is   0.0483 m2"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 324</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the primary current and power factor when the secondary current is 100 A at a power factor of 0.85 lagging\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  2000;#  prim  turns\n",
+      "N2  =  800;#  sec  turns\n",
+      "I0  =  5;#  in  Amperes\n",
+      "pf0  =  0.20;#  power  factor\n",
+      "I2  =  100;#  in  Amperes\n",
+      "pf2  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Let  I01  be  the  component  of  the  primary  current  which  provides  the  restoring  mmf.  Then  I01*N1  =  I2*N2\n",
+      "I01  =  I2*N2/N1\n",
+      " #If  the  power  factor  of  the  secondary  is  0.85\n",
+      "phi2  =  math.acos(pf2)\n",
+      " #If  the  power  factor  on  no-load  is  0.20,\n",
+      "phi0  =  math.acos(pf0)\n",
+      "I1h  =  I0*math.cos(phi0)  +  I01*math.cos(phi2)\n",
+      "I1v  =  I0*math.sin(phi0)  +  I01*math.sin(phi2)\n",
+      " #Hence  the  magnitude  of  I1\n",
+      "I1  =  (I1h*I1h  +  I1v*I1v)**0.5\n",
+      "pf1  =  math.cos(math.atan(I1v/I1h))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Primary  current  is  \", round(I1,2),\"  A,  and  Power  factor  is  \",round(pf1,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Primary  current  is   43.58   A,  and  Power  factor  is   0.8"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 328</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the equivalent resistance referred to the primary winding, \n",
+      "#(b) the equivalent reactance referred to the primary winding,\n",
+      "#(c) the equivalent impedance referred to the primary winding, and \n",
+      "#(d) the phase angle of the impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  600;#  prim  turns\n",
+      "N2  =  150;#  sec  turns\n",
+      "R1  =  0.25;#  in  ohms\n",
+      "R2  =  0.01;#  in  ohms\n",
+      "X1  =  1.0;#  in  ohms\n",
+      "X2  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "tr  =  N1/N2#  turn  ratio\n",
+      "vr  =  tr#  voltage  ratio  =  turn  raio,  vr  =  V1/V2\n",
+      " #equivalent  resistance  Re\n",
+      "Re  =  R1  +  R2*(vr**2)\n",
+      " #equivalent  reactance,  Xe\n",
+      "Xe  =  X1  +  X2*(vr**2)\n",
+      " #equivalent  impedance,  Ze\n",
+      "Ze  =  (Re*Re  +  Xe*Xe)**0.5\n",
+      " #cos(phie)  =  Re/Ze\n",
+      "pfe  =  Re/Ze\n",
+      "phie  =  math.acos(pfe)\n",
+      "phied  =  phie*180/math.pi#  in  \u00b0(deg)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  resistance  referred  to  the  primary  winding  is  \",round(  Re,2),\"  ohm\"\n",
+      "print  \"\\n  (b)the  equivalent  reactance  referred  to  the  primary  winding  is  \",round(  Xe,2),\"  ohm\"\n",
+      "print  \"\\n  (c)the  equivalent  impedance  referred  to  the  primary  winding  is  \",round(  Ze,2),\"  ohm\"\n",
+      "print  \"\\n  (d)phase  angle  is  \",round(  phied,2),\"deg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  resistance  referred  to  the  primary  winding  is   0.41   ohm\n",
+        "\n",
+        "  (b)the  equivalent  reactance  referred  to  the  primary  winding  is   1.64   ohm\n",
+        "\n",
+        "  (c)the  equivalent  impedance  referred  to  the  primary  winding  is   1.69   ohm\n",
+        "\n",
+        "  (d)phase  angle  is   75.96 deg"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 329</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the regulation of the transformer.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  200;#  in  Volts\n",
+      "V2  =  400;#  in  Volts\n",
+      "V2L  =  387.6;#  in  Volts\n",
+      "S  =  5000;#  in  VA\n",
+      "\n",
+      "#calculation:\n",
+      " #regulation  =(No-load  secondary  voltage  -\u0003  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %\n",
+      "reg  =  (V2  -  V2L)*100/V2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  regulation  of  the  transformer  is  \",round(reg,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  regulation  of  the  transformer  is   3.1   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 329</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the load voltage at which the mechanism operates.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VnL  =  240;#  in  Volts\n",
+      "reg  =  2.5;#  in  percent\n",
+      "\n",
+      "#calculation:\n",
+      " #regulation  =(No-load  secondary  voltage  -\u0003  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %\n",
+      "VL  =  VnL  -  reg*VnL/100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  load  voltage  at  which  the  mechanism  operates  is  \",round(VL,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  load  voltage  at  which  the  mechanism  operates  is   234.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 331</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the transformer efficiency at full load and 0.85 power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  200000;#  in  VA\n",
+      "Pc  =  1500;#  in  Watt\n",
+      "Pi  =  1000;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Full-load  output  power  =  V*I*pf\n",
+      "Po  =  S*pf\n",
+      " #Total  losses\n",
+      "Pl  =  Pc  +  Pi\n",
+      " #Input  power  =  output  power  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #efficiency\n",
+      "eff  =  1-(Pl/PI)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  efficiency  at  full  load  is  \",round(eff,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  efficiency  at  full  load  is   0.99"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 331</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the efficiency of the transformer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  200000;#  in  VA\n",
+      "Pc  =  1500;#  in  Watt\n",
+      "Pi  =  1000;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Half  full-load  power  output  =  V*I*pf/2\n",
+      "Po  =  S*pf/2\n",
+      " #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared\n",
+      " #Hence  the  copper  loss  at  half  full-load  is\n",
+      "Pch  =  Pc/(2*2)\n",
+      " #Iron  loss  =  1000  W  (constant)\n",
+      " #Total  losses\n",
+      "Pl  =  Pch  +  Pi\n",
+      " #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #efficiency\n",
+      "eff  =  (1-(Pl/PI))*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  efficiency  at  half  full  load  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  efficiency  at  half  full  load  is   98.41   percent"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 332</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the efficiency of the transformer (a) on full load, and (b) on half load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  400000;#  in  VA\n",
+      "R1  =  0.5;#  in  Ohm\n",
+      "R2  =  0.001;#  in  Ohm\n",
+      "V1  =  5000;#  in  Volts\n",
+      "V2  =  320;#  in  Volts\n",
+      "Pi  =  2500;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Rating  =  400  kVA  =  V1*I1  =  V2*I2\n",
+      " #Hence  primary  current\n",
+      "I1  =  S/V1\n",
+      " #secondary  current\n",
+      "I2  =  S/V2\n",
+      " #Total  copper  loss  =  I1*I1*R1  +  I2*I2*R2,\n",
+      "Pcf  =  I1*I1*R1  +  I2*I2*R2\n",
+      " #On  full  load,  total  loss  =  copper  loss  +  iron  loss\n",
+      "Plf  =  Pcf  +  Pi\n",
+      " #  full-load  power  output  =  V2*I2*pf\n",
+      "Pof  =  S*pf\n",
+      " #Input  power  at  full-load  =  output  power  at  full-load  +  losses\n",
+      "PIf  =  Pof  +  Plf\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      "efff  =  (1-(Plf/PIf))*100\n",
+      "\n",
+      " #Half  full-load  power  output  =  V*I*pf/2\n",
+      "Poh  =  S*pf/2\n",
+      " #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared\n",
+      " #Hence  the  copper  loss  at  half  full-load  is\n",
+      "Pch  =  Pcf/(2*2)\n",
+      " #Iron  loss  =  2500  W  (constant)\n",
+      " #Total  losses\n",
+      "Plh  =  Pch  +  Pi\n",
+      " #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses\n",
+      "PIh  =  Poh  +  Plh\n",
+      " #efficiency\n",
+      "effh  =  (1-(Plh/PIh))*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  transformer  efficiency  at  full  load  is  \", round(efff,2),\"  percent\"\n",
+      "print  \"\\n  (b)the  transformer  efficiency  at  half  full  load  is  \",  round(effh,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  transformer  efficiency  at  full  load  is   97.91   percent\n",
+        "\n",
+        "  (b)the  transformer  efficiency  at  half  full  load  is   97.88   percent"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 333</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the maximum efficiency,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  500000;#  in  VA\n",
+      "Pcf  =  4000;#  in  Watt\n",
+      "Pi  =  2500;#  in  Watt\n",
+      "pf  =  0.75;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Let  x  be  the  fraction  of  full  load  kVA  at  which  the  efficiency  is  a  maximum.\n",
+      " #The  corresponding  total  copper  loss  =  (4  kW)*(x**2)\n",
+      " #At  maximum  efficiency,  copper  loss  =  iron  loss  Hence\n",
+      "x  =  (Pi/Pcf)**0.5\n",
+      " #Hence  the  output  kVA  at  maximum  efficiency\n",
+      "So  =  x*S\n",
+      " #Total  loss  at  maximum  efficiency\n",
+      "Pl  =  2*Pi\n",
+      " #Output  power\n",
+      "Po  =  So*pf\n",
+      " #Input  power  =  output  power  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u00e2\u20ac\u201dlosses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Maximum  efficiency\n",
+      "effm  =  (1  -  Pl/PI)*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the output kVA at maximum efficiency is  \",round(So/1000,2),\"kVA\"\n",
+      "print  \"\\n  max.  efficiency  is  \",round(effm,2),\"  pecent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the output kVA at maximum efficiency is   395.28 kVA\n",
+        "\n",
+        "  max.  efficiency  is   98.34   pecent"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the equivalent input resistance of the transformer.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  4;#  turn  ratio\n",
+      "RL  =  100;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #the  equivalent  input  resistance,\n",
+      "Ri  =  RL*(tr**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  equivalent  input  resistance  is  \",round(Ri,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  equivalent  input  resistance  is   1600.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the optimum turns ratio of a transformer which would match a \n",
+      "#load resistance of 7 ohmto the output resistance of the amplifier.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  112;#  in  Ohms\n",
+      "RL  =  7;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  112  ohm  for  maximum  power  transfer.\n",
+      " #R1  =  RL*(tr**2)\n",
+      " #  tr  =  N1/N2  turn  ratio\n",
+      "tr  =  (R1/RL)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  turns  ratio  is  \",tr,\": 1.0\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  turns  ratio  is   4.0 : 1.0"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the optimum value of load resistance for maximum power transfer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  5;#  turn  ratio\n",
+      "R1  =  150;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  150  ohm  for  maximum  power  transfer.\n",
+      " #R1  =  RL*(tr**2)\n",
+      "RL  =  R1/(tr**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  value  of  load  resistance  is  \",round(RL,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  value  of  load  resistance  is   6.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the primary current flowing and (b) the power dissipated in the load resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  220;#  in  Volts\n",
+      "V2  =  1760;#  in  Volts\n",
+      "V  =  220;#  in  Volts\n",
+      "RL  =  1280;#  in  Ohms\n",
+      "R  =  2;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Turns  ratio,  tr  =  N1/N2  =  V1/V2\n",
+      "tr  =  V1/V2\n",
+      " #Equivalent  input  resistance  of  the  transformer,\n",
+      " #R1  =  RL*(tr**2)\n",
+      "R1  =  RL*(tr**2)\n",
+      " #Total  input  resistance\n",
+      "Rin  =  R  +  R1\n",
+      " #  Primary  current\n",
+      "I1  =  V1/Rin\n",
+      " #For  an  ideal  transformer  V1/V2  =  I2/I1,\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistor  RL\n",
+      "P  =  I2*I2*RL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  primary  current  flowing  is  \",round(I1,2),\"  A\"\n",
+      "print  \"\\n  (b)    power  dissipated  in  the  load  resistor  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  primary  current  flowing  is   10.0   A\n",
+        "\n",
+        "  (b)    power  dissipated  in  the  load  resistor  is   2000.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 336</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of the load resistance and (b) the power dissipated in the load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  25;#  teurn  ratio\n",
+      "V  =  24;#  in  Volts\n",
+      "R1  =  15000;#  in  Ohms\n",
+      "Rin  =  15000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Turns  ratio,  tr  =  N1/N2  =  V1/V2\n",
+      " #For  maximum  power  transfer  R1  needs  to  be  equal  to  15  kohm\n",
+      "RL  =  R1/(tr**2)\n",
+      " #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is\n",
+      "Rt  =  Rin  +  R1\n",
+      " #Primary  current,\n",
+      "I1  =  V/Rt\n",
+      " #N1/N2  =  I2/I1\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistor  RL\n",
+      "P  =  I2*I2*RL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  load  resistance  is  \",round(RL,2),\"ohm\"\n",
+      "print  \"\\n  (b)    power  dissipated  in  the  load  resistor  is  \",round(P*1000,2),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  load  resistance  is   24.0 ohm\n",
+        "\n",
+        "  (b)    power  dissipated  in  the  load  resistor  is   9.6 mW"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 337</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in each section of the winding.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  320;#  in  Volts\n",
+      "V2  =  250;#  in  Volts\n",
+      "S  =  20000;#  in  VA\n",
+      "\n",
+      "#calculation:\n",
+      " #Rating  =  20  kVA  =  V1*I1  =  V2*I2\n",
+      " #Hence  primary  current,  I1\n",
+      "I1  =  S/V1\n",
+      " #secondary  current,  I2\n",
+      "I2  =  S/V2\n",
+      " #Hence  current  in  common  part  of  the  winding\n",
+      "I  =  I2  -  I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  in  common  part  of  the  winding  is  \", round(I,2),\"  A\"\n",
+      "print  \"\\n  primary  current  and  secondary  current  are  \",round(I1,2),\"  A  and  \",round(I2,2),\"  A  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  in  common  part  of  the  winding  is   17.5   A\n",
+        "\n",
+        "  primary  current  and  secondary  current  are   62.5   A  and   80.0   A  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 339</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the saving in the volume of copper used in an auto transformer compared with a double-wound transformer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1a  =  200;#  in  Volts\n",
+      "V2a  =  150;#  in  Volts\n",
+      "V1b  =  500;#  in  Volts\n",
+      "V2b  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  200  V:150  V  transformer,  xa\n",
+      "xa  =  V2a/V1a\n",
+      " #volume  of  copper  in  auto  transformer\n",
+      "vca  =  (1  -  xa)*100#  of  copper  in  a  double-wound  transformer\n",
+      " #the  saving  is\n",
+      "vsa  =  100  -  vca\n",
+      " #For  a  500  V:100  V  transformer,  xb\n",
+      "xb  =  V2b/V1b\n",
+      " #volume  of  copper  in  auto  transformer\n",
+      "vcb  =  (1  -  xb)*100#  of  copper  in  a  double-wound  transformer\n",
+      " #the  saving  is\n",
+      "vsb  =  100  -  vcb\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)For  a  200  V:150  V  transformer,  the  saving  is  \", round(vsa,2),\"  percent\"\n",
+      "print  \"\\n  (b)For  a  500  V:100  V  transformer,  the  saving  is  \", round(vsb,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For  a  200  V:150  V  transformer,  the  saving  is   75.0   percent\n",
+        "\n",
+        "  (b)For  a  500  V:100  V  transformer,  the  saving  is   20.0   percent"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 340</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the secondary line voltage on no-load when the windings are connected (a) star-delta, (b) delta-star.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  500;#  prim  turns\n",
+      "N2  =  50;#  sec  turns\n",
+      "VL  =  2400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star-connection,  VL  =  Vp*(3**0.5)\n",
+      "VL1s  =  VL\n",
+      " #Primary  phase  voltage\n",
+      "Vp1s  =  VL1s/(3**0.5)\n",
+      " #For  a  delta-connection,  VL  =  Vp\n",
+      " #N1/N2  =  V1/V2,  from  which,\n",
+      " #secondary  phase  voltage,  Vp2s\n",
+      "Vp2s  =  Vp1s*N2/N1\n",
+      "VL2d  =  Vp2s\n",
+      "\n",
+      " #For  a  delta-connection,  VL  =  Vp\n",
+      "VL1d  =  VL\n",
+      " #primary  phase  voltage  Vp1d\n",
+      "Vp1d  =  VL1d\n",
+      " #Secondary  phase  voltage,  Vp2d\n",
+      "Vp2d  =  Vp1d*N2/N1\n",
+      " #For  a  star-connection,  VL  =  Vp*(3**0.5)\n",
+      "VL2s  =  Vp2d*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  secondary  line  voltage  for  star  and  delta  connection  are  \",round(Vp2s,1),\"  V \"\n",
+      "print   \" and  \",round(VL2s,0),\"  V  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  secondary  line  voltage  for  star  and  delta  connection  are   138.6   V \n",
+        " and   416.0   V  respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 343</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the reading on the ammeter, \n",
+      "#(b) the potential difference across the ammeter and\n",
+      "#(c) the total load (in VA) on the secondary.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  1;#  prim  turns\n",
+      "N2  =  60;#  sec  turns\n",
+      "I1  =  300;#  in  amperes\n",
+      "Ra  =  0.15;#  in  ohms\n",
+      "R2  =  0.25;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Reading  on  the  ammeter,\n",
+      "I2  =  I1*(N1/N2)\n",
+      " #P.d.  across  the  ammeter  =  I2*RA,  where  RA  is  the  ammeter  resistance\n",
+      "pd  =  I2*Ra\n",
+      " #Total  resistance  of  secondary  circuit\n",
+      "Rt  =  Ra  +  R2\n",
+      " #Induced  e.m.f.  in  secondary\n",
+      "V2  =  I2*Rt\n",
+      " #Total  load  on  secondary\n",
+      "S  =  V2*I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  reading  on  the  ammeter  is  \",round(I2,2),\"  A  \"\n",
+      "print  \"\\n  (b)potential  difference  across  the  ammeter  is  \",round(pd,2),\"  V  \"\n",
+      "print  \"\\n  (c)total  load  (in  VA)  on  the  secondary  is  \",round(S,2),\"  VA  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  reading  on  the  ammeter  is   5.0   A  \n",
+        "\n",
+        "  (b)potential  difference  across  the  ammeter  is   0.75   V  \n",
+        "\n",
+        "  (c)total  load  (in  VA)  on  the  secondary  is   10.0   VA  "
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_2.ipynb
new file mode 100755
index 00000000..5b0974af
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_2.ipynb
@@ -0,0 +1,1627 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 20: Transformers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the secondary voltage, assuming an ideal transformer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  500;#  primary  turns\n",
+      "N2  =  3000;#  secondary  turns\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  an  ideal  transformer,  voltage  ratio  =  turns  ratio\n",
+      "V2  =  V1*N2/N1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  voltage  \",round(V2,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  voltage   1440.0 V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine its output voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  2/7;#  turns  ratio\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  2:7  means  that  the  transformer  has  2  turns  on  the  primary  \n",
+      "    #for  every  7  turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  \",round(V2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage   840.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine secondary voltage and current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  8/1;#  turns  ratio\n",
+      "I1  =  3;#  in  Amperes\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  \n",
+      "    #primary  for  every  1turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      " #secondary  current\n",
+      "I2  =  I1*tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  voltage  is  \",round(V2,2),\"  V  and  secondary  current  is  \", round(I2,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  voltage  is   30.0   V  and  secondary  current  is   24.0   A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 318</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the transformer turns ratio and the current taken from the supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  240;#  in  Volts\n",
+      "V2  =  12;#  in  Volts\n",
+      "P  =  150;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      "I2  =  P/V2\n",
+      " #A  turns  ratio  =  Vp/Vs\n",
+      "tr  =  V1/V2#  turn  ratio\n",
+      " #    V1/V2  =  I2/I1\n",
+      " #current  taken  from  the  supply\n",
+      "I1  =  I2*V2/V1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  turn  ratio  is  \",round(tr,2),\"  and  current  taken  from  the  supply  is  \",round(I1,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  turn  ratio  is   20.0   and  current  taken  from  the  supply  is   0.63   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 318</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the full-load secondary current,\n",
+      "#(b) the minimum load resistance which can be connected across the secondary winding to give full load kVA, \n",
+      "#(c) the primary current at full load kVA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  5000;#  in  VA\n",
+      "tr  =  10;#  turn  ratio\n",
+      "V1  =  2500;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  primary  for  every  1turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      " #transformer  rating  in  volt-amperes  =  Vs*Is\n",
+      "I2  =  S/V2\n",
+      " #Minimum  value  of  load  resistance\n",
+      "RL  =  V2/I2\n",
+      " #    tr  =  I2/I1\n",
+      "I1  =  I2/tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)full-load  secondary  current  is  \",round(I2,2),\"  A\"\n",
+      "print  \"\\n  (b)minimum  load  resistance  is  \",round(RL,2),\"  ohm\"\n",
+      "print  \"\\n  (c)  primary  current  is  \",round(I1,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)full-load  secondary  current  is   20.0   A\n",
+        "\n",
+        "  (b)minimum  load  resistance  is   12.5   ohm\n",
+        "\n",
+        "  (c)  primary  current  is   2.0   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 319</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of the magnetizing and core loss components of the noload current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  2400;#  in  Volts\n",
+      "V2  =  400;#  in  Volts\n",
+      "I0  =  0.5;#  in  Amperes\n",
+      "Pc  =  400;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Core  loss  (i.e.  iron  loss)  P  =  V1*I0*cos(phi0)\n",
+      "pf  =  Pc/(V1*I0)\n",
+      "phi0  =  math.acos(pf)\n",
+      " #Magnetizing  component\n",
+      "Im  =  I0*math.sin(phi0)\n",
+      " #Core  loss  component\n",
+      "Ic  =  I0*math.cos(phi0)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)magnetizing  component  is  \",round(Im,3),\"  A  and  Core  loss  component  is  \",round(Ic,3),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)magnetizing  component  is   0.471   A  and  Core  loss  component  is   0.167   A"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 320</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the iron loss current, \n",
+      "#(b) the power factor on no-load, and (c) the magnetizing current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "I0  =  0.8;#  in  Amperes\n",
+      "P  =  72;#  in  Watts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  absorbed  =  total  core  loss,  P  =  V*I0*cos(phi0)\n",
+      " #Ic  =  I0*cos(phi0)\n",
+      "Ic  =  P/V\n",
+      "pf  =  Ic/I0\n",
+      " #From  the  right-angled  triangle  in  Figure  20.2(b)  and  using\n",
+      " #Pythagoras\u2019  theorem,  \n",
+      "Im  =  (I0*I0  -  Ic*Ic)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Core  loss  component  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)  power  factor  is  \",round(  pf,2),\"\"\n",
+      "print  \"\\n  (c)magnetizing  component  is  \",round(Im,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Core  loss  component  is   0.3   A\n",
+        "\n",
+        "  (b)  power  factor  is   0.37 \n",
+        "\n",
+        "  (c)magnetizing  component  is   0.74   A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 321</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  100000;#  in  VA\n",
+      "V1  =  4000;#  in  Volts\n",
+      "V2  =  200;#  in  Volts\n",
+      "N2  =  100;#  sec  turns\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Transformer  rating  =  V1*I1  =  V2*I2\n",
+      " #primary  current\n",
+      "I1  =  S/V1\n",
+      " #secondary  current\n",
+      "I2  =  S/V2\n",
+      " #primary  turns\n",
+      "N1  =  N2*V1/V2\n",
+      " #maximum  flux\n",
+      " #assuming  E2  =  V2\n",
+      "Phim  =  V2/(4.44*f*N2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)primary  current  is  \",round(  I1,2),\"  A  and  secondary  current  is  \",round(  I2,2),\"  A\"\n",
+      "print  \"\\n  (b)number  of  primary  turns  is  \",round(  N1,2),\"\"\n",
+      "print  \"\\n  (c)maximum  value  of  the  flux  is  \",round(Phim*1000,2),\"mWb\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)primary  current  is   25.0   A  and  secondary  current  is   500.0   A\n",
+        "\n",
+        "  (b)number  of  primary  turns  is   2000.0 \n",
+        "\n",
+        "  (c)maximum  value  of  the  flux  is   9.01 mWb"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 322</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the maximum value of the flux density in the core, and\n",
+      "#(b) the voltage induced in the secondary winding.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  250;#  in  Volts\n",
+      "A  =  0.03;#  in  m2\n",
+      "N2  =  300;#  sec  turns\n",
+      "N1  =  25;#  prim  turns\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #maximum  flux  density,\n",
+      "Phim  =  V1/(4.44*f*N1)\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #maximum  core  flux  density\n",
+      "Bm  =  Phim/A\n",
+      " #voltage  induced  in  the  secondary  winding,\n",
+      "V2  =  V1*N2/N1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  core  flux  density  \",round(  Bm,2),\"  T\"\n",
+      "print  \"\\n  (b)voltage  induced  in  the  secondary  winding  is  \",round(  V2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  core  flux  density   1.5   T\n",
+        "\n",
+        "  (b)voltage  induced  in  the  secondary  winding  is   3000.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 323</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the number of primary and secondary turns\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  500;#  in  Volts\n",
+      "V2  =  100;#  in  Volts\n",
+      "Bm  =  1.5;#  in  Tesla\n",
+      "A  =  0.005;#  in  m2\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #maximum  core  flux  density\n",
+      "Phim  =  Bm*A\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #primary  turns,\n",
+      "N1  =  V1/(4.44*f*Phim)\n",
+      " #secondary  turns,\n",
+      "N2  =  V2*N1/V1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  no.  of  primary  and  secondary  turns  are  \",round(N1,2),\"  turns,  and  \",round(N2,2),\"  turns  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  no.  of  primary  and  secondary  turns  are   300.3   turns,  and   60.06   turns  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 323</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the number of primary and secondary turns and \n",
+      "#(b) the cross-sectional area of the core.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "emfpt  =  15;#  in  Volts\n",
+      "V1  =  4500;#  in  Volts\n",
+      "V2  =  225;#  in  Volts\n",
+      "Bm  =  1.4;#  in  Tesla\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #E.m.f.  per  turn,  V1/N1  =  V2/N2  =  emfpt\n",
+      " #primary  turns,\n",
+      "N1  =  V1/emfpt\n",
+      " #secondary  turns,\n",
+      "N2  =  V2/emfpt\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #maximum  flux  density,\n",
+      "Phim  =  V1/(4.44*f*N1)\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #cross-sectional  area\n",
+      "A  =  Phim/Bm\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)no.  of  primary  and  secondary  turns  are  \",  N1,\"  turns,  and  \",  N2,\"  turns  respectively\"\n",
+      "print  \"\\n  (b)cross-sectional  area  is  \",  round(A,4),\"m2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)no.  of  primary  and  secondary  turns  are   300.0   turns,  and   15.0   turns  respectively\n",
+        "\n",
+        "  (b)cross-sectional  area  is   0.0483 m2"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 324</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the primary current and power factor when the secondary current is 100 A at a power factor of 0.85 lagging\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  2000;#  prim  turns\n",
+      "N2  =  800;#  sec  turns\n",
+      "I0  =  5;#  in  Amperes\n",
+      "pf0  =  0.20;#  power  factor\n",
+      "I2  =  100;#  in  Amperes\n",
+      "pf2  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Let  I01  be  the  component  of  the  primary  current  which  provides  the  restoring  mmf.  Then  I01*N1  =  I2*N2\n",
+      "I01  =  I2*N2/N1\n",
+      " #If  the  power  factor  of  the  secondary  is  0.85\n",
+      "phi2  =  math.acos(pf2)\n",
+      " #If  the  power  factor  on  no-load  is  0.20,\n",
+      "phi0  =  math.acos(pf0)\n",
+      "I1h  =  I0*math.cos(phi0)  +  I01*math.cos(phi2)\n",
+      "I1v  =  I0*math.sin(phi0)  +  I01*math.sin(phi2)\n",
+      " #Hence  the  magnitude  of  I1\n",
+      "I1  =  (I1h*I1h  +  I1v*I1v)**0.5\n",
+      "pf1  =  math.cos(math.atan(I1v/I1h))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Primary  current  is  \", round(I1,2),\"  A,  and  Power  factor  is  \",round(pf1,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Primary  current  is   43.58   A,  and  Power  factor  is   0.8"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 328</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the equivalent resistance referred to the primary winding, \n",
+      "#(b) the equivalent reactance referred to the primary winding,\n",
+      "#(c) the equivalent impedance referred to the primary winding, and \n",
+      "#(d) the phase angle of the impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  600;#  prim  turns\n",
+      "N2  =  150;#  sec  turns\n",
+      "R1  =  0.25;#  in  ohms\n",
+      "R2  =  0.01;#  in  ohms\n",
+      "X1  =  1.0;#  in  ohms\n",
+      "X2  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "tr  =  N1/N2#  turn  ratio\n",
+      "vr  =  tr#  voltage  ratio  =  turn  raio,  vr  =  V1/V2\n",
+      " #equivalent  resistance  Re\n",
+      "Re  =  R1  +  R2*(vr**2)\n",
+      " #equivalent  reactance,  Xe\n",
+      "Xe  =  X1  +  X2*(vr**2)\n",
+      " #equivalent  impedance,  Ze\n",
+      "Ze  =  (Re*Re  +  Xe*Xe)**0.5\n",
+      " #cos(phie)  =  Re/Ze\n",
+      "pfe  =  Re/Ze\n",
+      "phie  =  math.acos(pfe)\n",
+      "phied  =  phie*180/math.pi#  in  \u00b0(deg)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  resistance  referred  to  the  primary  winding  is  \",round(  Re,2),\"  ohm\"\n",
+      "print  \"\\n  (b)the  equivalent  reactance  referred  to  the  primary  winding  is  \",round(  Xe,2),\"  ohm\"\n",
+      "print  \"\\n  (c)the  equivalent  impedance  referred  to  the  primary  winding  is  \",round(  Ze,2),\"  ohm\"\n",
+      "print  \"\\n  (d)phase  angle  is  \",round(  phied,2),\"deg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  resistance  referred  to  the  primary  winding  is   0.41   ohm\n",
+        "\n",
+        "  (b)the  equivalent  reactance  referred  to  the  primary  winding  is   1.64   ohm\n",
+        "\n",
+        "  (c)the  equivalent  impedance  referred  to  the  primary  winding  is   1.69   ohm\n",
+        "\n",
+        "  (d)phase  angle  is   75.96 deg"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 329</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the regulation of the transformer.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  200;#  in  Volts\n",
+      "V2  =  400;#  in  Volts\n",
+      "V2L  =  387.6;#  in  Volts\n",
+      "S  =  5000;#  in  VA\n",
+      "\n",
+      "#calculation:\n",
+      " #regulation  =(No-load  secondary  voltage  -\u0003  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %\n",
+      "reg  =  (V2  -  V2L)*100/V2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  regulation  of  the  transformer  is  \",round(reg,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  regulation  of  the  transformer  is   3.1   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 329</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the load voltage at which the mechanism operates.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VnL  =  240;#  in  Volts\n",
+      "reg  =  2.5;#  in  percent\n",
+      "\n",
+      "#calculation:\n",
+      " #regulation  =(No-load  secondary  voltage  -\u0003  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %\n",
+      "VL  =  VnL  -  reg*VnL/100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  load  voltage  at  which  the  mechanism  operates  is  \",round(VL,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  load  voltage  at  which  the  mechanism  operates  is   234.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 331</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the transformer efficiency at full load and 0.85 power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  200000;#  in  VA\n",
+      "Pc  =  1500;#  in  Watt\n",
+      "Pi  =  1000;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Full-load  output  power  =  V*I*pf\n",
+      "Po  =  S*pf\n",
+      " #Total  losses\n",
+      "Pl  =  Pc  +  Pi\n",
+      " #Input  power  =  output  power  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #efficiency\n",
+      "eff  =  1-(Pl/PI)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  efficiency  at  full  load  is  \",round(eff,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  efficiency  at  full  load  is   0.99"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 331</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the efficiency of the transformer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  200000;#  in  VA\n",
+      "Pc  =  1500;#  in  Watt\n",
+      "Pi  =  1000;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Half  full-load  power  output  =  V*I*pf/2\n",
+      "Po  =  S*pf/2\n",
+      " #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared\n",
+      " #Hence  the  copper  loss  at  half  full-load  is\n",
+      "Pch  =  Pc/(2*2)\n",
+      " #Iron  loss  =  1000  W  (constant)\n",
+      " #Total  losses\n",
+      "Pl  =  Pch  +  Pi\n",
+      " #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #efficiency\n",
+      "eff  =  (1-(Pl/PI))*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  efficiency  at  half  full  load  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  efficiency  at  half  full  load  is   98.41   percent"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 332</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the efficiency of the transformer (a) on full load, and (b) on half load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  400000;#  in  VA\n",
+      "R1  =  0.5;#  in  Ohm\n",
+      "R2  =  0.001;#  in  Ohm\n",
+      "V1  =  5000;#  in  Volts\n",
+      "V2  =  320;#  in  Volts\n",
+      "Pi  =  2500;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Rating  =  400  kVA  =  V1*I1  =  V2*I2\n",
+      " #Hence  primary  current\n",
+      "I1  =  S/V1\n",
+      " #secondary  current\n",
+      "I2  =  S/V2\n",
+      " #Total  copper  loss  =  I1*I1*R1  +  I2*I2*R2,\n",
+      "Pcf  =  I1*I1*R1  +  I2*I2*R2\n",
+      " #On  full  load,  total  loss  =  copper  loss  +  iron  loss\n",
+      "Plf  =  Pcf  +  Pi\n",
+      " #  full-load  power  output  =  V2*I2*pf\n",
+      "Pof  =  S*pf\n",
+      " #Input  power  at  full-load  =  output  power  at  full-load  +  losses\n",
+      "PIf  =  Pof  +  Plf\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      "efff  =  (1-(Plf/PIf))*100\n",
+      "\n",
+      " #Half  full-load  power  output  =  V*I*pf/2\n",
+      "Poh  =  S*pf/2\n",
+      " #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared\n",
+      " #Hence  the  copper  loss  at  half  full-load  is\n",
+      "Pch  =  Pcf/(2*2)\n",
+      " #Iron  loss  =  2500  W  (constant)\n",
+      " #Total  losses\n",
+      "Plh  =  Pch  +  Pi\n",
+      " #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses\n",
+      "PIh  =  Poh  +  Plh\n",
+      " #efficiency\n",
+      "effh  =  (1-(Plh/PIh))*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  transformer  efficiency  at  full  load  is  \", round(efff,2),\"  percent\"\n",
+      "print  \"\\n  (b)the  transformer  efficiency  at  half  full  load  is  \",  round(effh,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  transformer  efficiency  at  full  load  is   97.91   percent\n",
+        "\n",
+        "  (b)the  transformer  efficiency  at  half  full  load  is   97.88   percent"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 333</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the maximum efficiency,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  500000;#  in  VA\n",
+      "Pcf  =  4000;#  in  Watt\n",
+      "Pi  =  2500;#  in  Watt\n",
+      "pf  =  0.75;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Let  x  be  the  fraction  of  full  load  kVA  at  which  the  efficiency  is  a  maximum.\n",
+      " #The  corresponding  total  copper  loss  =  (4  kW)*(x**2)\n",
+      " #At  maximum  efficiency,  copper  loss  =  iron  loss  Hence\n",
+      "x  =  (Pi/Pcf)**0.5\n",
+      " #Hence  the  output  kVA  at  maximum  efficiency\n",
+      "So  =  x*S\n",
+      " #Total  loss  at  maximum  efficiency\n",
+      "Pl  =  2*Pi\n",
+      " #Output  power\n",
+      "Po  =  So*pf\n",
+      " #Input  power  =  output  power  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u00e2\u20ac\u201dlosses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Maximum  efficiency\n",
+      "effm  =  (1  -  Pl/PI)*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the output kVA at maximum efficiency is  \",round(So/1000,2),\"kVA\"\n",
+      "print  \"\\n  max.  efficiency  is  \",round(effm,2),\"  pecent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the output kVA at maximum efficiency is   395.28 kVA\n",
+        "\n",
+        "  max.  efficiency  is   98.34   pecent"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the equivalent input resistance of the transformer.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  4;#  turn  ratio\n",
+      "RL  =  100;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #the  equivalent  input  resistance,\n",
+      "Ri  =  RL*(tr**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  equivalent  input  resistance  is  \",round(Ri,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  equivalent  input  resistance  is   1600.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the optimum turns ratio of a transformer which would match a \n",
+      "#load resistance of 7 ohmto the output resistance of the amplifier.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  112;#  in  Ohms\n",
+      "RL  =  7;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  112  ohm  for  maximum  power  transfer.\n",
+      " #R1  =  RL*(tr**2)\n",
+      " #  tr  =  N1/N2  turn  ratio\n",
+      "tr  =  (R1/RL)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  turns  ratio  is  \",tr,\": 1.0\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  turns  ratio  is   4.0 : 1.0"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the optimum value of load resistance for maximum power transfer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  5;#  turn  ratio\n",
+      "R1  =  150;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  150  ohm  for  maximum  power  transfer.\n",
+      " #R1  =  RL*(tr**2)\n",
+      "RL  =  R1/(tr**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  value  of  load  resistance  is  \",round(RL,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  value  of  load  resistance  is   6.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the primary current flowing and (b) the power dissipated in the load resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  220;#  in  Volts\n",
+      "V2  =  1760;#  in  Volts\n",
+      "V  =  220;#  in  Volts\n",
+      "RL  =  1280;#  in  Ohms\n",
+      "R  =  2;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Turns  ratio,  tr  =  N1/N2  =  V1/V2\n",
+      "tr  =  V1/V2\n",
+      " #Equivalent  input  resistance  of  the  transformer,\n",
+      " #R1  =  RL*(tr**2)\n",
+      "R1  =  RL*(tr**2)\n",
+      " #Total  input  resistance\n",
+      "Rin  =  R  +  R1\n",
+      " #  Primary  current\n",
+      "I1  =  V1/Rin\n",
+      " #For  an  ideal  transformer  V1/V2  =  I2/I1,\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistor  RL\n",
+      "P  =  I2*I2*RL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  primary  current  flowing  is  \",round(I1,2),\"  A\"\n",
+      "print  \"\\n  (b)    power  dissipated  in  the  load  resistor  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  primary  current  flowing  is   10.0   A\n",
+        "\n",
+        "  (b)    power  dissipated  in  the  load  resistor  is   2000.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 336</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of the load resistance and (b) the power dissipated in the load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  25;#  teurn  ratio\n",
+      "V  =  24;#  in  Volts\n",
+      "R1  =  15000;#  in  Ohms\n",
+      "Rin  =  15000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Turns  ratio,  tr  =  N1/N2  =  V1/V2\n",
+      " #For  maximum  power  transfer  R1  needs  to  be  equal  to  15  kohm\n",
+      "RL  =  R1/(tr**2)\n",
+      " #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is\n",
+      "Rt  =  Rin  +  R1\n",
+      " #Primary  current,\n",
+      "I1  =  V/Rt\n",
+      " #N1/N2  =  I2/I1\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistor  RL\n",
+      "P  =  I2*I2*RL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  load  resistance  is  \",round(RL,2),\"ohm\"\n",
+      "print  \"\\n  (b)    power  dissipated  in  the  load  resistor  is  \",round(P*1000,2),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  load  resistance  is   24.0 ohm\n",
+        "\n",
+        "  (b)    power  dissipated  in  the  load  resistor  is   9.6 mW"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 337</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current in each section of the winding.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  320;#  in  Volts\n",
+      "V2  =  250;#  in  Volts\n",
+      "S  =  20000;#  in  VA\n",
+      "\n",
+      "#calculation:\n",
+      " #Rating  =  20  kVA  =  V1*I1  =  V2*I2\n",
+      " #Hence  primary  current,  I1\n",
+      "I1  =  S/V1\n",
+      " #secondary  current,  I2\n",
+      "I2  =  S/V2\n",
+      " #Hence  current  in  common  part  of  the  winding\n",
+      "I  =  I2  -  I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  in  common  part  of  the  winding  is  \", round(I,2),\"  A\"\n",
+      "print  \"\\n  primary  current  and  secondary  current  are  \",round(I1,2),\"  A  and  \",round(I2,2),\"  A  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  in  common  part  of  the  winding  is   17.5   A\n",
+        "\n",
+        "  primary  current  and  secondary  current  are   62.5   A  and   80.0   A  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 339</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the saving in the volume of copper used in an auto transformer compared with a double-wound transformer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1a  =  200;#  in  Volts\n",
+      "V2a  =  150;#  in  Volts\n",
+      "V1b  =  500;#  in  Volts\n",
+      "V2b  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  200  V:150  V  transformer,  xa\n",
+      "xa  =  V2a/V1a\n",
+      " #volume  of  copper  in  auto  transformer\n",
+      "vca  =  (1  -  xa)*100#  of  copper  in  a  double-wound  transformer\n",
+      " #the  saving  is\n",
+      "vsa  =  100  -  vca\n",
+      " #For  a  500  V:100  V  transformer,  xb\n",
+      "xb  =  V2b/V1b\n",
+      " #volume  of  copper  in  auto  transformer\n",
+      "vcb  =  (1  -  xb)*100#  of  copper  in  a  double-wound  transformer\n",
+      " #the  saving  is\n",
+      "vsb  =  100  -  vcb\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)For  a  200  V:150  V  transformer,  the  saving  is  \", round(vsa,2),\"  percent\"\n",
+      "print  \"\\n  (b)For  a  500  V:100  V  transformer,  the  saving  is  \", round(vsb,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For  a  200  V:150  V  transformer,  the  saving  is   75.0   percent\n",
+        "\n",
+        "  (b)For  a  500  V:100  V  transformer,  the  saving  is   20.0   percent"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 340</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the secondary line voltage on no-load when the windings are connected (a) star-delta, (b) delta-star.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  500;#  prim  turns\n",
+      "N2  =  50;#  sec  turns\n",
+      "VL  =  2400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star-connection,  VL  =  Vp*(3**0.5)\n",
+      "VL1s  =  VL\n",
+      " #Primary  phase  voltage\n",
+      "Vp1s  =  VL1s/(3**0.5)\n",
+      " #For  a  delta-connection,  VL  =  Vp\n",
+      " #N1/N2  =  V1/V2,  from  which,\n",
+      " #secondary  phase  voltage,  Vp2s\n",
+      "Vp2s  =  Vp1s*N2/N1\n",
+      "VL2d  =  Vp2s\n",
+      "\n",
+      " #For  a  delta-connection,  VL  =  Vp\n",
+      "VL1d  =  VL\n",
+      " #primary  phase  voltage  Vp1d\n",
+      "Vp1d  =  VL1d\n",
+      " #Secondary  phase  voltage,  Vp2d\n",
+      "Vp2d  =  Vp1d*N2/N1\n",
+      " #For  a  star-connection,  VL  =  Vp*(3**0.5)\n",
+      "VL2s  =  Vp2d*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  secondary  line  voltage  for  star  and  delta  connection  are  \",round(Vp2s,1),\"  V \"\n",
+      "print   \" and  \",round(VL2s,0),\"  V  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  secondary  line  voltage  for  star  and  delta  connection  are   138.6   V \n",
+        " and   416.0   V  respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 343</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the reading on the ammeter, \n",
+      "#(b) the potential difference across the ammeter and\n",
+      "#(c) the total load (in VA) on the secondary.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  1;#  prim  turns\n",
+      "N2  =  60;#  sec  turns\n",
+      "I1  =  300;#  in  amperes\n",
+      "Ra  =  0.15;#  in  ohms\n",
+      "R2  =  0.25;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Reading  on  the  ammeter,\n",
+      "I2  =  I1*(N1/N2)\n",
+      " #P.d.  across  the  ammeter  =  I2*RA,  where  RA  is  the  ammeter  resistance\n",
+      "pd  =  I2*Ra\n",
+      " #Total  resistance  of  secondary  circuit\n",
+      "Rt  =  Ra  +  R2\n",
+      " #Induced  e.m.f.  in  secondary\n",
+      "V2  =  I2*Rt\n",
+      " #Total  load  on  secondary\n",
+      "S  =  V2*I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  reading  on  the  ammeter  is  \",round(I2,2),\"  A  \"\n",
+      "print  \"\\n  (b)potential  difference  across  the  ammeter  is  \",round(pd,2),\"  V  \"\n",
+      "print  \"\\n  (c)total  load  (in  VA)  on  the  secondary  is  \",round(S,2),\"  VA  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  reading  on  the  ammeter  is   5.0   A  \n",
+        "\n",
+        "  (b)potential  difference  across  the  ammeter  is   0.75   V  \n",
+        "\n",
+        "  (c)total  load  (in  VA)  on  the  secondary  is   10.0   VA  "
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_3.ipynb
new file mode 100755
index 00000000..538ace20
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_20-checkpoint_3.ipynb
@@ -0,0 +1,1617 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e3b742f593657a54a0584c9ec4333c4c34458354568e4b8a1e97b9d029596f5b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 20: Transformers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  500;#  primary  turns\n",
+      "N2  =  3000;#  secondary  turns\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  an  ideal  transformer,  voltage  ratio  =  turns  ratio\n",
+      "V2  =  V1*N2/N1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  voltage  \",round(V2,2),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  voltage   1440.0 V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  2/7;#  turns  ratio\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  2:7  means  that  the  transformer  has  2  turns  on  the  primary  \n",
+      "    #for  every  7  turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  output  voltage  \",round(V2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  output  voltage   840.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 317</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  8/1;#  turns  ratio\n",
+      "I1  =  3;#  in  Amperes\n",
+      "V1  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  \n",
+      "    #primary  for  every  1turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      " #secondary  current\n",
+      "I2  =  I1*tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  voltage  is  \",round(V2,2),\"  V  and  secondary  current  is  \", round(I2,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  voltage  is   30.0   V  and  secondary  current  is   24.0   A"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 318</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  240;#  in  Volts\n",
+      "V2  =  12;#  in  Volts\n",
+      "P  =  150;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      "I2  =  P/V2\n",
+      " #A  turns  ratio  =  Vp/Vs\n",
+      "tr  =  V1/V2#  turn  ratio\n",
+      " #    V1/V2  =  I2/I1\n",
+      " #current  taken  from  the  supply\n",
+      "I1  =  I2*V2/V1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  turn  ratio  is  \",round(tr,2),\"  and  current  taken  from  the  supply  is  \",round(I1,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  turn  ratio  is   20.0   and  current  taken  from  the  supply  is   0.63   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 318</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  5000;#  in  VA\n",
+      "tr  =  10;#  turn  ratio\n",
+      "V1  =  2500;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #A  turns  ratio  of  8:1  means  that  the  transformer  has  28  turns  on  the  primary  for  every  1turns  on  the  secondary\n",
+      "V2  =  V1/tr\n",
+      " #transformer  rating  in  volt-amperes  =  Vs*Is\n",
+      "I2  =  S/V2\n",
+      " #Minimum  value  of  load  resistance\n",
+      "RL  =  V2/I2\n",
+      " #    tr  =  I2/I1\n",
+      "I1  =  I2/tr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)full-load  secondary  current  is  \",round(I2,2),\"  A\"\n",
+      "print  \"\\n  (b)minimum  load  resistance  is  \",round(RL,2),\"  ohm\"\n",
+      "print  \"\\n  (c)  primary  current  is  \",round(I1,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)full-load  secondary  current  is   20.0   A\n",
+        "\n",
+        "  (b)minimum  load  resistance  is   12.5   ohm\n",
+        "\n",
+        "  (c)  primary  current  is   2.0   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 319</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  2400;#  in  Volts\n",
+      "V2  =  400;#  in  Volts\n",
+      "I0  =  0.5;#  in  Amperes\n",
+      "Pc  =  400;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Core  loss  (i.e.  iron  loss)  P  =  V1*I0*cos(phi0)\n",
+      "pf  =  Pc/(V1*I0)\n",
+      "phi0  =  math.acos(pf)\n",
+      " #Magnetizing  component\n",
+      "Im  =  I0*math.sin(phi0)\n",
+      " #Core  loss  component\n",
+      "Ic  =  I0*math.cos(phi0)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)magnetizing  component  is  \",round(Im,3),\"  A  and  Core  loss  component  is  \",round(Ic,3),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)magnetizing  component  is   0.471   A  and  Core  loss  component  is   0.167   A"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 320</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "I0  =  0.8;#  in  Amperes\n",
+      "P  =  72;#  in  Watts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Power  absorbed  =  total  core  loss,  P  =  V*I0*cos(phi0)\n",
+      " #Ic  =  I0*cos(phi0)\n",
+      "Ic  =  P/V\n",
+      "pf  =  Ic/I0\n",
+      " #From  the  right-angled  triangle  in  Figure  20.2(b)  and  using\n",
+      " #Pythagoras\u2019  theorem,  \n",
+      "Im  =  (I0*I0  -  Ic*Ic)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  Core  loss  component  is  \",round(  Ic,2),\"  A\"\n",
+      "print  \"\\n  (b)  power  factor  is  \",round(  pf,2),\"\"\n",
+      "print  \"\\n  (c)magnetizing  component  is  \",round(Im,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  Core  loss  component  is   0.3   A\n",
+        "\n",
+        "  (b)  power  factor  is   0.37 \n",
+        "\n",
+        "  (c)magnetizing  component  is   0.74   A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 321</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  100000;#  in  VA\n",
+      "V1  =  4000;#  in  Volts\n",
+      "V2  =  200;#  in  Volts\n",
+      "N2  =  100;#  sec  turns\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Transformer  rating  =  V1*I1  =  V2*I2\n",
+      " #primary  current\n",
+      "I1  =  S/V1\n",
+      " #secondary  current\n",
+      "I2  =  S/V2\n",
+      " #primary  turns\n",
+      "N1  =  N2*V1/V2\n",
+      " #maximum  flux\n",
+      " #assuming  E2  =  V2\n",
+      "Phim  =  V2/(4.44*f*N2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)primary  current  is  \",round(  I1,2),\"  A  and  secondary  current  is  \",round(  I2,2),\"  A\"\n",
+      "print  \"\\n  (b)number  of  primary  turns  is  \",round(  N1,2),\"\"\n",
+      "print  \"\\n  (c)maximum  value  of  the  flux  is  \",round(Phim*1000,2),\"mWb\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)primary  current  is   25.0   A  and  secondary  current  is   500.0   A\n",
+        "\n",
+        "  (b)number  of  primary  turns  is   2000.0 \n",
+        "\n",
+        "  (c)maximum  value  of  the  flux  is   9.01 mWb"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 322</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  250;#  in  Volts\n",
+      "A  =  0.03;#  in  m2\n",
+      "N2  =  300;#  sec  turns\n",
+      "N1  =  25;#  prim  turns\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #maximum  flux  density,\n",
+      "Phim  =  V1/(4.44*f*N1)\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #maximum  core  flux  density\n",
+      "Bm  =  Phim/A\n",
+      " #voltage  induced  in  the  secondary  winding,\n",
+      "V2  =  V1*N2/N1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  core  flux  density  \",round(  Bm,2),\"  T\"\n",
+      "print  \"\\n  (b)voltage  induced  in  the  secondary  winding  is  \",round(  V2,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  core  flux  density   1.5   T\n",
+        "\n",
+        "  (b)voltage  induced  in  the  secondary  winding  is   3000.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 323</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  500;#  in  Volts\n",
+      "V2  =  100;#  in  Volts\n",
+      "Bm  =  1.5;#  in  Tesla\n",
+      "A  =  0.005;#  in  m2\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #maximum  core  flux  density\n",
+      "Phim  =  Bm*A\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #primary  turns,\n",
+      "N1  =  V1/(4.44*f*Phim)\n",
+      " #secondary  turns,\n",
+      "N2  =  V2*N1/V1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  no.  of  primary  and  secondary  turns  are  \",round(N1,2),\"  turns,  and  \",round(N2,2),\"  turns  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  no.  of  primary  and  secondary  turns  are   300.3   turns,  and   60.06   turns  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 323</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "emfpt  =  15;#  in  Volts\n",
+      "V1  =  4500;#  in  Volts\n",
+      "V2  =  225;#  in  Volts\n",
+      "Bm  =  1.4;#  in  Tesla\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #E.m.f.  per  turn,  V1/N1  =  V2/N2  =  emfpt\n",
+      " #primary  turns,\n",
+      "N1  =  V1/emfpt\n",
+      " #secondary  turns,\n",
+      "N2  =  V2/emfpt\n",
+      " #e.m.f.  E1  =  4.44*f*Phim*N1\n",
+      " #maximum  flux  density,\n",
+      "Phim  =  V1/(4.44*f*N1)\n",
+      " #Phim  =  Bm*A,  where  Bm  =  maximum  core  flux  density  and  A  =  cross-sectional  area  of  the  core\n",
+      " #cross-sectional  area\n",
+      "A  =  Phim/Bm\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)no.  of  primary  and  secondary  turns  are  \",  N1,\"  turns,  and  \",  N2,\"  turns  respectively\"\n",
+      "print  \"\\n  (b)cross-sectional  area  is  \",  round(A,4),\"m2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)no.  of  primary  and  secondary  turns  are   300.0   turns,  and   15.0   turns  respectively\n",
+        "\n",
+        "  (b)cross-sectional  area  is   0.0483 m2"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 324</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  2000;#  prim  turns\n",
+      "N2  =  800;#  sec  turns\n",
+      "I0  =  5;#  in  Amperes\n",
+      "pf0  =  0.20;#  power  factor\n",
+      "I2  =  100;#  in  Amperes\n",
+      "pf2  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Let  I01  be  the  component  of  the  primary  current  which  provides  the  restoring  mmf.  Then  I01*N1  =  I2*N2\n",
+      "I01  =  I2*N2/N1\n",
+      " #If  the  power  factor  of  the  secondary  is  0.85\n",
+      "phi2  =  math.acos(pf2)\n",
+      " #If  the  power  factor  on  no-load  is  0.20,\n",
+      "phi0  =  math.acos(pf0)\n",
+      "I1h  =  I0*math.cos(phi0)  +  I01*math.cos(phi2)\n",
+      "I1v  =  I0*math.sin(phi0)  +  I01*math.sin(phi2)\n",
+      " #Hence  the  magnitude  of  I1\n",
+      "I1  =  (I1h*I1h  +  I1v*I1v)**0.5\n",
+      "pf1  =  math.cos(math.atan(I1v/I1h))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Primary  current  is  \", round(I1,2),\"  A,  and  Power  factor  is  \",round(pf1,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Primary  current  is   43.58   A,  and  Power  factor  is   0.8"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 328</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  600;#  prim  turns\n",
+      "N2  =  150;#  sec  turns\n",
+      "R1  =  0.25;#  in  ohms\n",
+      "R2  =  0.01;#  in  ohms\n",
+      "X1  =  1.0;#  in  ohms\n",
+      "X2  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "tr  =  N1/N2#  turn  ratio\n",
+      "vr  =  tr#  voltage  ratio  =  turn  raio,  vr  =  V1/V2\n",
+      " #equivalent  resistance  Re\n",
+      "Re  =  R1  +  R2*(vr**2)\n",
+      " #equivalent  reactance,  Xe\n",
+      "Xe  =  X1  +  X2*(vr**2)\n",
+      " #equivalent  impedance,  Ze\n",
+      "Ze  =  (Re*Re  +  Xe*Xe)**0.5\n",
+      " #cos(phie)  =  Re/Ze\n",
+      "pfe  =  Re/Ze\n",
+      "phie  =  math.acos(pfe)\n",
+      "phied  =  phie*180/math.pi#  in  \u00b0(deg)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  resistance  referred  to  the  primary  winding  is  \",round(  Re,2),\"  ohm\"\n",
+      "print  \"\\n  (b)the  equivalent  reactance  referred  to  the  primary  winding  is  \",round(  Xe,2),\"  ohm\"\n",
+      "print  \"\\n  (c)the  equivalent  impedance  referred  to  the  primary  winding  is  \",round(  Ze,2),\"  ohm\"\n",
+      "print  \"\\n  (d)phase  angle  is  \",round(  phied,2),\"deg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  resistance  referred  to  the  primary  winding  is   0.41   ohm\n",
+        "\n",
+        "  (b)the  equivalent  reactance  referred  to  the  primary  winding  is   1.64   ohm\n",
+        "\n",
+        "  (c)the  equivalent  impedance  referred  to  the  primary  winding  is   1.69   ohm\n",
+        "\n",
+        "  (d)phase  angle  is   75.96 deg"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 329</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  200;#  in  Volts\n",
+      "V2  =  400;#  in  Volts\n",
+      "V2L  =  387.6;#  in  Volts\n",
+      "S  =  5000;#  in  VA\n",
+      "\n",
+      "#calculation:\n",
+      " #regulation  =(No-load  secondary  voltage  -\u0003  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %\n",
+      "reg  =  (V2  -  V2L)*100/V2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  regulation  of  the  transformer  is  \",round(reg,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  regulation  of  the  transformer  is   3.1   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 329</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "VnL  =  240;#  in  Volts\n",
+      "reg  =  2.5;#  in  percent\n",
+      "\n",
+      "#calculation:\n",
+      " #regulation  =(No-load  secondary  voltage  -\u0003  terminal  voltage  on  load)*100/no-load  secondary  voltage    in  %\n",
+      "VL  =  VnL  -  reg*VnL/100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  load  voltage  at  which  the  mechanism  operates  is  \",round(VL,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  load  voltage  at  which  the  mechanism  operates  is   234.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 331</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  200000;#  in  VA\n",
+      "Pc  =  1500;#  in  Watt\n",
+      "Pi  =  1000;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Full-load  output  power  =  V*I*pf\n",
+      "Po  =  S*pf\n",
+      " #Total  losses\n",
+      "Pl  =  Pc  +  Pi\n",
+      " #Input  power  =  output  power  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #efficiency\n",
+      "eff  =  1-(Pl/PI)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  efficiency  at  full  load  is  \",round(eff,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  efficiency  at  full  load  is   0.99"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 331</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  200000;#  in  VA\n",
+      "Pc  =  1500;#  in  Watt\n",
+      "Pi  =  1000;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Half  full-load  power  output  =  V*I*pf/2\n",
+      "Po  =  S*pf/2\n",
+      " #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared\n",
+      " #Hence  the  copper  loss  at  half  full-load  is\n",
+      "Pch  =  Pc/(2*2)\n",
+      " #Iron  loss  =  1000  W  (constant)\n",
+      " #Total  losses\n",
+      "Pl  =  Pch  +  Pi\n",
+      " #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #efficiency\n",
+      "eff  =  (1-(Pl/PI))*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  efficiency  at  half  full  load  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  efficiency  at  half  full  load  is   98.41   percent"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 332</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  400000;#  in  VA\n",
+      "R1  =  0.5;#  in  Ohm\n",
+      "R2  =  0.001;#  in  Ohm\n",
+      "V1  =  5000;#  in  Volts\n",
+      "V2  =  320;#  in  Volts\n",
+      "Pi  =  2500;#  in  Watt\n",
+      "pf  =  0.85;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Rating  =  400  kVA  =  V1*I1  =  V2*I2\n",
+      " #Hence  primary  current\n",
+      "I1  =  S/V1\n",
+      " #secondary  current\n",
+      "I2  =  S/V2\n",
+      " #Total  copper  loss  =  I1*I1*R1  +  I2*I2*R2,\n",
+      "Pcf  =  I1*I1*R1  +  I2*I2*R2\n",
+      " #On  full  load,  total  loss  =  copper  loss  +  iron  loss\n",
+      "Plf  =  Pcf  +  Pi\n",
+      " #  full-load  power  output  =  V2*I2*pf\n",
+      "Pof  =  S*pf\n",
+      " #Input  power  at  full-load  =  output  power  at  full-load  +  losses\n",
+      "PIf  =  Pof  +  Plf\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u2014losses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      "efff  =  (1-(Plf/PIf))*100\n",
+      "\n",
+      " #Half  full-load  power  output  =  V*I*pf/2\n",
+      "Poh  =  S*pf/2\n",
+      " #Copper  loss  (or  I*I*R  loss)  is  proportional  to  current  squared\n",
+      " #Hence  the  copper  loss  at  half  full-load  is\n",
+      "Pch  =  Pcf/(2*2)\n",
+      " #Iron  loss  =  2500  W  (constant)\n",
+      " #Total  losses\n",
+      "Plh  =  Pch  +  Pi\n",
+      " #Input  power  at  half  full-load  =  output  power  at  half  full-load  +  losses\n",
+      "PIh  =  Poh  +  Plh\n",
+      " #efficiency\n",
+      "effh  =  (1-(Plh/PIh))*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  transformer  efficiency  at  full  load  is  \", round(efff,2),\"  percent\"\n",
+      "print  \"\\n  (b)the  transformer  efficiency  at  half  full  load  is  \",  round(effh,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  transformer  efficiency  at  full  load  is   97.91   percent\n",
+        "\n",
+        "  (b)the  transformer  efficiency  at  half  full  load  is   97.88   percent"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 333</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "S  =  500000;#  in  VA\n",
+      "Pcf  =  4000;#  in  Watt\n",
+      "Pi  =  2500;#  in  Watt\n",
+      "pf  =  0.75;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #Let  x  be  the  fraction  of  full  load  kVA  at  which  the  efficiency  is  a  maximum.\n",
+      " #The  corresponding  total  copper  loss  =  (4  kW)*(x**2)\n",
+      " #At  maximum  efficiency,  copper  loss  =  iron  loss  Hence\n",
+      "x  =  (Pi/Pcf)**0.5\n",
+      " #Hence  the  output  kVA  at  maximum  efficiency\n",
+      "So  =  x*S\n",
+      " #Total  loss  at  maximum  efficiency\n",
+      "Pl  =  2*Pi\n",
+      " #Output  power\n",
+      "Po  =  So*pf\n",
+      " #Input  power  =  output  power  +  losses\n",
+      "PI  =  Po  +  Pl\n",
+      " #Efficiency  =  output  power/input  power  =  (input  power\u00e2\u20ac\u201dlosses)/input  power\n",
+      " #Efficiency  =  1  -  losses/input  power\n",
+      " #Maximum  efficiency\n",
+      "effm  =  (1  -  Pl/PI)*100\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the output kVA at maximum efficiency is  \",round(So/1000,2),\"kVA\"\n",
+      "print  \"\\n  max.  efficiency  is  \",round(effm,2),\"  pecent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the output kVA at maximum efficiency is   395.28 kVA\n",
+        "\n",
+        "  max.  efficiency  is   98.34   pecent"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  4;#  turn  ratio\n",
+      "RL  =  100;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #the  equivalent  input  resistance,\n",
+      "Ri  =  RL*(tr**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  equivalent  input  resistance  is  \",round(Ri,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  equivalent  input  resistance  is   1600.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  112;#  in  Ohms\n",
+      "RL  =  7;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  112  ohm  for  maximum  power  transfer.\n",
+      " #R1  =  RL*(tr**2)\n",
+      " #  tr  =  N1/N2  turn  ratio\n",
+      "tr  =  (R1/RL)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  turns  ratio  is  \",tr,\": 1.0\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  turns  ratio  is   4.0 : 1.0"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  5;#  turn  ratio\n",
+      "R1  =  150;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #The  equivalent  input  resistance,  R1  of  the  transformer  needs  to  be  150  ohm  for  maximum  power  transfer.\n",
+      " #R1  =  RL*(tr**2)\n",
+      "RL  =  R1/(tr**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  value  of  load  resistance  is  \",round(RL,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  value  of  load  resistance  is   6.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 335</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  220;#  in  Volts\n",
+      "V2  =  1760;#  in  Volts\n",
+      "V  =  220;#  in  Volts\n",
+      "RL  =  1280;#  in  Ohms\n",
+      "R  =  2;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Turns  ratio,  tr  =  N1/N2  =  V1/V2\n",
+      "tr  =  V1/V2\n",
+      " #Equivalent  input  resistance  of  the  transformer,\n",
+      " #R1  =  RL*(tr**2)\n",
+      "R1  =  RL*(tr**2)\n",
+      " #Total  input  resistance\n",
+      "Rin  =  R  +  R1\n",
+      " #  Primary  current\n",
+      "I1  =  V1/Rin\n",
+      " #For  an  ideal  transformer  V1/V2  =  I2/I1,\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistor  RL\n",
+      "P  =  I2*I2*RL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  primary  current  flowing  is  \",round(I1,2),\"  A\"\n",
+      "print  \"\\n  (b)    power  dissipated  in  the  load  resistor  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  primary  current  flowing  is   10.0   A\n",
+        "\n",
+        "  (b)    power  dissipated  in  the  load  resistor  is   2000.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 336</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "tr  =  25;#  teurn  ratio\n",
+      "V  =  24;#  in  Volts\n",
+      "R1  =  15000;#  in  Ohms\n",
+      "Rin  =  15000;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Turns  ratio,  tr  =  N1/N2  =  V1/V2\n",
+      " #For  maximum  power  transfer  R1  needs  to  be  equal  to  15  kohm\n",
+      "RL  =  R1/(tr**2)\n",
+      " #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is\n",
+      "Rt  =  Rin  +  R1\n",
+      " #Primary  current,\n",
+      "I1  =  V/Rt\n",
+      " #N1/N2  =  I2/I1\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistor  RL\n",
+      "P  =  I2*I2*RL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  load  resistance  is  \",round(RL,2),\"ohm\"\n",
+      "print  \"\\n  (b)    power  dissipated  in  the  load  resistor  is  \",round(P*1000,2),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  load  resistance  is   24.0 ohm\n",
+        "\n",
+        "  (b)    power  dissipated  in  the  load  resistor  is   9.6 mW"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 337</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1  =  320;#  in  Volts\n",
+      "V2  =  250;#  in  Volts\n",
+      "S  =  20000;#  in  VA\n",
+      "\n",
+      "#calculation:\n",
+      " #Rating  =  20  kVA  =  V1*I1  =  V2*I2\n",
+      " #Hence  primary  current,  I1\n",
+      "I1  =  S/V1\n",
+      " #secondary  current,  I2\n",
+      "I2  =  S/V2\n",
+      " #Hence  current  in  common  part  of  the  winding\n",
+      "I  =  I2  -  I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  in  common  part  of  the  winding  is  \", round(I,2),\"  A\"\n",
+      "print  \"\\n  primary  current  and  secondary  current  are  \",round(I1,2),\"  A  and  \",round(I2,2),\"  A  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  in  common  part  of  the  winding  is   17.5   A\n",
+        "\n",
+        "  primary  current  and  secondary  current  are   62.5   A  and   80.0   A  respectively"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 339</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V1a  =  200;#  in  Volts\n",
+      "V2a  =  150;#  in  Volts\n",
+      "V1b  =  500;#  in  Volts\n",
+      "V2b  =  100;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  200  V:150  V  transformer,  xa\n",
+      "xa  =  V2a/V1a\n",
+      " #volume  of  copper  in  auto  transformer\n",
+      "vca  =  (1  -  xa)*100#  of  copper  in  a  double-wound  transformer\n",
+      " #the  saving  is\n",
+      "vsa  =  100  -  vca\n",
+      " #For  a  500  V:100  V  transformer,  xb\n",
+      "xb  =  V2b/V1b\n",
+      " #volume  of  copper  in  auto  transformer\n",
+      "vcb  =  (1  -  xb)*100#  of  copper  in  a  double-wound  transformer\n",
+      " #the  saving  is\n",
+      "vsb  =  100  -  vcb\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)For  a  200  V:150  V  transformer,  the  saving  is  \", round(vsa,2),\"  percent\"\n",
+      "print  \"\\n  (b)For  a  500  V:100  V  transformer,  the  saving  is  \", round(vsb,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For  a  200  V:150  V  transformer,  the  saving  is   75.0   percent\n",
+        "\n",
+        "  (b)For  a  500  V:100  V  transformer,  the  saving  is   20.0   percent"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 340</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  500;#  prim  turns\n",
+      "N2  =  50;#  sec  turns\n",
+      "VL  =  2400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  star-connection,  VL  =  Vp*(3**0.5)\n",
+      "VL1s  =  VL\n",
+      " #Primary  phase  voltage\n",
+      "Vp1s  =  VL1s/(3**0.5)\n",
+      " #For  a  delta-connection,  VL  =  Vp\n",
+      " #N1/N2  =  V1/V2,  from  which,\n",
+      " #secondary  phase  voltage,  Vp2s\n",
+      "Vp2s  =  Vp1s*N2/N1\n",
+      "VL2d  =  Vp2s\n",
+      "\n",
+      " #For  a  delta-connection,  VL  =  Vp\n",
+      "VL1d  =  VL\n",
+      " #primary  phase  voltage  Vp1d\n",
+      "Vp1d  =  VL1d\n",
+      " #Secondary  phase  voltage,  Vp2d\n",
+      "Vp2d  =  Vp1d*N2/N1\n",
+      " #For  a  star-connection,  VL  =  Vp*(3**0.5)\n",
+      "VL2s  =  Vp2d*(3**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  secondary  line  voltage  for  star  and  delta  connection  are  \",round(Vp2s,1),\"  V \"\n",
+      "print   \" and  \",round(VL2s,0),\"  V  respectively\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  secondary  line  voltage  for  star  and  delta  connection  are   138.6   V \n",
+        " and   416.0   V  respectively\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 343</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "N1  =  1;#  prim  turns\n",
+      "N2  =  60;#  sec  turns\n",
+      "I1  =  300;#  in  amperes\n",
+      "Ra  =  0.15;#  in  ohms\n",
+      "R2  =  0.25;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Reading  on  the  ammeter,\n",
+      "I2  =  I1*(N1/N2)\n",
+      " #P.d.  across  the  ammeter  =  I2*RA,  where  RA  is  the  ammeter  resistance\n",
+      "pd  =  I2*Ra\n",
+      " #Total  resistance  of  secondary  circuit\n",
+      "Rt  =  Ra  +  R2\n",
+      " #Induced  e.m.f.  in  secondary\n",
+      "V2  =  I2*Rt\n",
+      " #Total  load  on  secondary\n",
+      "S  =  V2*I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  reading  on  the  ammeter  is  \",round(I2,2),\"  A  \"\n",
+      "print  \"\\n  (b)potential  difference  across  the  ammeter  is  \",round(pd,2),\"  V  \"\n",
+      "print  \"\\n  (c)total  load  (in  VA)  on  the  secondary  is  \",round(S,2),\"  VA  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  reading  on  the  ammeter  is   5.0   A  \n",
+        "\n",
+        "  (b)potential  difference  across  the  ammeter  is   0.75   V  \n",
+        "\n",
+        "  (c)total  load  (in  VA)  on  the  secondary  is   10.0   VA  "
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint.ipynb
new file mode 100755
index 00000000..28433f9f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint.ipynb
@@ -0,0 +1,1641 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 21: D.c. machines</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the generated e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  600;#  no.  of  conductors\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "p  =  4;#  no.  of  pairs\n",
+      "n  =  625/60;#  in  rev/sec\n",
+      "Phi  =  20E-3;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  generated  e.m.f  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  generated  e.m.f  is   500.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the speed at which the machine must be driven to generate an e.m.f. of 240 V\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  50*16;#  no.  of  conductors\n",
+      "p  =  1;#  let  no.  of  pairs\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "E  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      " #Rearranging  gives,  speed\n",
+      "n  =  E*c/(2*p*Phi*Z)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  speed  at  which  the  machine  must  be  driven  is  \",round(n,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  speed  at  which  the  machine  must  be  driven  is   10.0   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. generated when running at 500 rev/min.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  1200;#  no.  of  conductors\n",
+      "p  =  1;#  let,  no.  of  pairs\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "n  =  500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   300.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 355</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the generated e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  1200;#  no.  of  conductors\n",
+      "p  =  4;#  let,  no.  of  pairs\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "n  =  500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   1200.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 355</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the change in the generated voltage when the field current is reduced by 20%, \n",
+      "#assuming the flux is proportional to the field current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1 = 150; # in Volts\n",
+      "x = 0.2;\n",
+      "\n",
+      "#calculation:\n",
+      "E2  =  E1*(1- x)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E2,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   120.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 356</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "n1  =  30;#  in  rev/sec\n",
+      "E1  =  200;#  in  Volts\n",
+      "n2  =  20;#  in  rev/sec\n",
+      "E2  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E  proportional  to  phi*w  and  since  w  =  2*pi*n,  then\n",
+      " #  E  proportional  to  phi*n\n",
+      " #  E1/E2  =  Phi1*n1/(Phi2*n2)\n",
+      " #  let  Phi2/Phi1  =  Phi\n",
+      "Phi  =  E2*n1/(E1*n2)\n",
+      "Phi_inc  =  (Phi  -  1)*100#/in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  percentage  increase  in  the  flux  per  pole  is  \",round(Phi_inc,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  percentage  increase  in  the  flux  per  pole  is   87.5   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the terminal voltage of a generator\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.30;#  in  ohms\n",
+      "Ia  =  30;#  in  Amperes\n",
+      "E  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #terminal  voltage,\n",
+      " #V  =  E  -  Ia*Ra\n",
+      "V  =  E  -  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  terminal  voltage  of  a  generator  is  \",round(V,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  terminal  voltage  of  a  generator  is   191.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the terminal voltage, and (b) the generated e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "RL  =  60;#  in  ohms\n",
+      "Ia  =  8;#  in  Amperes\n",
+      "Ra  =  1;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #terminal  voltage,\n",
+      " #V  =  Ia*RL\n",
+      "V  =  Ia*RL\n",
+      " #Generated  e.m.f.,  E\n",
+      "E  =  V  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)terminal  voltage  is  \",round(V,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)terminal  voltage  is   480.0   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   488.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the generated e.m.f. when \n",
+      "#(a) the speed increases to 25 rev/s and the pole flux remains unchanged, \n",
+      "#(b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08 Wb,\n",
+      "#and (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07 Wb.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  150;#  in  Volts\n",
+      "n1  =  20;#  in  rev/sec\n",
+      "Phi1  =  0.10;#  in  Wb\n",
+      "n2  =  25;#  in  rev/sec\n",
+      "Phi2  =  0.10;#  in  Wb\n",
+      "n3  =  20;#  in  rev/sec\n",
+      "Phi3  =  0.08;#  in  Wb\n",
+      "n4  =  24;#  in  rev/sec\n",
+      "Phi4  =  0.07;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E  proportional  to  phi*w  and  since  w  =  2*pi*n,  then\n",
+      " #  E  proportional  to  phi*n\n",
+      " #  E1/E2  =  Phi1*n1/(Phi2*n2)\n",
+      "E2  =  E1*Phi2*n2/(Phi1*n1)\n",
+      "E3  =  E1*Phi3*n3/(Phi1*n1)\n",
+      "E4  =  E1*Phi4*n4/(Phi1*n1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  generated  e.m.f  is  \",round(E2,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E3,2),\"  V  \"\n",
+      "print  \"\\n  (c)generated  e.m.f.  is  \",round(E4,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  generated  e.m.f  is   187.5   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   120.0   V  \n",
+        "\n",
+        "  (c)generated  e.m.f.  is   126.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 359</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ps  =  20000;#  in  Watts\n",
+      "Vs  =  200;#  in  Volts\n",
+      "Rs  =  0.1;#  in  ohms\n",
+      "Rf  =  50;#  in  ohms\n",
+      "Ra  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Load  current,  I\n",
+      "Is  =  Ps/Vs\n",
+      " #Volt  drop  in  the  cables  to  the  load\n",
+      "Vd  =  Is*Rs\n",
+      " #Hence  terminal  voltage,\n",
+      "V  =  Vs  +  Vd\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  Is\n",
+      " #Generated  e.m.f.  E\n",
+      "E  =  V  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)terminal  voltage  is  \",round(V,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)terminal  voltage  is   210.0   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   214.17   V  "
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 361</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the e.m.f. generated.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Is  =  80;#  in  amperes\n",
+      "Vs  =  200;#  in  Volts\n",
+      "Rf  =  40;#  in  ohms\n",
+      "Rse  =  0.02;#  in  ohms\n",
+      "Ra  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Volt  drop  in  series  winding\n",
+      "Vse  =  Is*Rse\n",
+      " #P.d.  across  the  field  winding  =  p.d.  across  armature\n",
+      "V1  =  Vs  +  Vse\n",
+      " #Field  current,  If\n",
+      "If  =  V1/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  Is\n",
+      " #Generated  e.m.f.  E\n",
+      "E  =  V1  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  generated  e.m.f.  is   205.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 363</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the efficiency of the generator at full load\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ps  =  10000;#  in  Watt\n",
+      "Pl  =  600;#  in  Watt\n",
+      "Ra  =  0.75;#  in  ohms\n",
+      "Rf  =  125;#  in  ohms\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Output  power  Ps  =  V*I\n",
+      " #from  which,  load  current  I\n",
+      "I  =  Ps/V\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  I\n",
+      " #Efficiency,\n",
+      "eff  =  Ps*100/((V*I)  +  (Ia*Ia*Ra)  +  (If*V)  +  (Pl))#  in  Percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Efficiency  is  \",round(eff,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Efficiency  is   80.5   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 364</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the back e.m.f. when the armature current is 50 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.2;#  in  ohms\n",
+      "V  =  240;#  in  Volts\n",
+      "Ia  =  50;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  motor,  V  =  E  +  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  back  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  back  e.m.f.  is   230.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 365</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A, and\n",
+      "#(b) as a motor taking 80 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.25;#  in  ohms\n",
+      "V  =  300;#  in  Volts\n",
+      "Ig  =  100;#  in  Amperes\n",
+      "Im  =  80;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #As  a  generator,  generated  e.m.f.,\n",
+      " #  E  =  V  +  Ia*Ra\n",
+      "Eg  =  V  +  Ig*Ra\n",
+      " #For  a  motor,  generated  e.m.f.  (or  back  e.m.f.),\n",
+      " #  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Im*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)As  a  generator,  generated  e.m.f.  is  \",round(Eg,2),\"  V  \"\n",
+      "print  \"\\n  (b)back  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)As  a  generator,  generated  e.m.f.  is   325.0   V  \n",
+        "\n",
+        "  (b)back  e.m.f.  is   280.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the torque exerted when a current of 30 A flows in each armature conductor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  4;\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "Phi  =  25E-3;#  Wb\n",
+      "Z  =  900;\n",
+      "Ia  =  30;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #torque  T  =  p*Phi*Z*Ia/(pi*c)\n",
+      "T  =  p*Phi*Z*Ia/(1*math.pi*c)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  torque  exerted  is   429.72   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the torque\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  350;#  in  Volts\n",
+      "Ra  =  0.5;#  in  ohms\n",
+      "n  =  15;#  in  rev/sec\n",
+      "Ia  =  60;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Back  e.m.f.  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      " #torque  T  =  E*Ia/(2*n*pi)\n",
+      "T  =  E*Ia/(2*n*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  torque  exerted  is   203.72   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the speed and (b) the torque developed when the armature current is 40 A\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  1;#  let\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  20E-3;#  Wb\n",
+      "Z  =  500;\n",
+      "V  =  250;#  in  Volts\n",
+      "Ra  =  1;#  in  ohms\n",
+      "Ia  =  40;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Back  e.m.f.  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      " #E.m.f.  E  =  2*p*Phi*n*Z/c\n",
+      " #  rearrange,\n",
+      "n  =  E*c/(2*p*Phi*Z)\n",
+      " #torque  T  =  E*Ia/(2*n*pi)\n",
+      "T  =  E*Ia/(2*n*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)speed  n  is  \",round(n,2),\"  rev/sec  \"\n",
+      "print  \"\\n  (b)the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)speed  n  is   21.0   rev/sec  \n",
+        "\n",
+        "  (b)the  torque  exerted  is   63.66   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 367</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the armature current at this new value of torque.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T1  =  25;#  in  Nm\n",
+      "T2  =  35;#  in  Nm\n",
+      "Ia1  =  16;#  in  Amperes\n",
+      "V  =  100;#  in  Volts\n",
+      "x  =  0.15;\n",
+      "\n",
+      " #calculation:\n",
+      " #the  shaft  torque  T  of  a  generator  is  proportional  to  (phi*Ia),\n",
+      "    #where  Phi  is  the  flux  and  Ia  is  the  armature  current.  Thus,  T  =  k*Phi*Ia,  where  k  is  a  constant.\n",
+      " #The  torque  at  flux  phi1  and  armature  current  Ia1  is  T1  =  k*Phi1*Ia1.\n",
+      " #similarly  T2  =  k*Phi2*Ia2\n",
+      "\n",
+      "Ia2  =  T2*Ia1/(0.85*T1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  armature  current  at  the  new  value  of  torque  is  \",round(Ia2,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  armature  current  at  the  new  value  of  torque  is   26.35   A  "
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 367</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the efficiency of the generator and (b) the power loss in the generator.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  12;#  in  Nm\n",
+      "I  =  15;#  in  Amperes\n",
+      "V  =  100;#  in  Volts\n",
+      "n  =  1500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #the  efficiency  of  a  generator  =  (output  power/input  power)*100  %\n",
+      " #The  output  power  is  the  electrical  output,  i.e.  VI  watts.  \n",
+      "    #The  input  power  to  a  generator  is  the  mechanical  power  in  the  shaft  driving  the  generator, \n",
+      "    #i.e.  T*w  or  T(2*pi*n)  watts,  where  T  is  the  torque  in  Nm  and  n  is  speed  of  rotation  in  rev/s.  Hence,  for  a  generator  \n",
+      " #efficiency  =  V*I*100/(T*2*pi*n) %\n",
+      "eff  =  V*I*100/(T*2*math.pi*n)#  in    Percent\n",
+      " #The  input  power  =  output  power  +  losses\n",
+      " #  hence,  T*2*math.pi*n  =  V*I  +  losses\n",
+      "Pl  =  T*2*math.pi*n  -  V*I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  efficiency  is  \",round(eff,2),\" % \"\n",
+      "print  \"\\n  (b)  power  loss  is  \",round(Pl,2),\"  W  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  efficiency  is   79.58  % \n",
+        "\n",
+        "  (b)  power  loss  is   384.96   W  "
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 368</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current in the armature, and (b) the back e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rf  =  150;#  in  Ohms\n",
+      "Ra  =  0.4;#  in  Ohms\n",
+      "I  =  30;#  in  Amperes\n",
+      "V  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Field  current  If\n",
+      "If  =  V/Rf\n",
+      " #Supply  current  I  =  Ia  +  If\n",
+      " #Hence  armature  current,  Ia\n",
+      "Ia  =  I  -  If\n",
+      " #Back  e.m.f.  E  =  V  -\u0004  Ia*Ra\n",
+      "E  =  V  -  (Ia*Ra)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)    current  in  the  armature  is  \",round(Ia,2),\"  A  \"\n",
+      "print  \"\\n  (b)  Back  e.m.f.  E  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)    current  in  the  armature  is   28.4   A  \n",
+        "\n",
+        "  (b)  Back  e.m.f.  E  is   228.64   V  "
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 370</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the speed of the motor, assuming the flux remains constant.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  30;#  in  Amperes\n",
+      "Ia2  =  45;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "n1  =  1350/60;#  in  Rev/sec\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #The  relationship  E  proportional  to  (Phi*n)  applies  to  both  generators  and  motors.  For  a  motor,\n",
+      " #E  =  V  -  (Ia*Ra)\n",
+      "E1  =    V  -  (Ia1*Ra)\n",
+      "E2  =    V  -  (Ia2*Ra)\n",
+      " #The  relationship,  E1/E2  =  Phi1*n1/Phi2*n2,    applies  to  both  generators  and  motors.\n",
+      "    #Since  the  flux  is  constant,  Phi1  =  Phi2\n",
+      "n2  =  E2*n1/(E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  speed  of  the  motor  is  \",round(n2,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  speed  of  the  motor  is   21.78   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 370</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and \n",
+      "#(b) the steadystate value of the armature current at the new value of flux,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  30;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "n  =  800/60;#  in  Rev/sec\n",
+      "V  =  220;#  in  Volts\n",
+      "x=  0.1;\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  d.c.  shunt-wound  motor,  E  =  V  -  (Ia*Ra),Hence  initial  generated  e.m.f.,\n",
+      "E1  =    V  -  (Ia1*Ra)\n",
+      " #The  generated  e.m.f.  is  also  such  that  E  proportional  to  (Phi*n)  \n",
+      "    #so  at  the  instant  the  flux  is  reduced,  the  speed  has  not  had  time  to  change,  and\n",
+      "E  =  E1*(1-x)\n",
+      " #Hence,  the  voltage  drop  due  to  the  armature  resistance  is\n",
+      "Vd  =  V  -  E\n",
+      " #The  instantaneous  value  of  the  current  is\n",
+      "Ia  =  Vd/Ra\n",
+      " #T  proportional  to  (Phi*Ia),  since  the  torque  is  constant,\n",
+      " #Phi1*Ia1  =  Phi2*Ia2,    The  flux  8  is  reduced  by  10%,  hence\n",
+      "Ia2  =  Ia1/0.9\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)instantaneous  value  of  the  current  \",round(Ia,2),\"  A  \"\n",
+      "print  \"\\n  (b)steady  state  value  of  armature  current,  \",round(Ia2,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)instantaneous  value  of  the  current   82.0   A  \n",
+        "\n",
+        "  (b)steady  state  value  of  armature  current,   33.33   A  "
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 372</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the generated e.m.f. at this load. \n",
+      "#(b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A.\n",
+      "#Assume that this causes a doubling of the flux.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  15;#  in  Amperes\n",
+      "Ia2  =  30;#  in  Amperes\n",
+      "Rf  =  0.3;#  in  ohms\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n1  =  24;#  in  Rev/sec\n",
+      "V  =  240;#  in  Volts\n",
+      "x=  2;\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E,  at  initial  load,  is  given  by\n",
+      "E1  =    V  -  Ia1*(Ra  +  Rf)\n",
+      " #When  the  current  is  increased  to  30  A,  the  generated  e.m.f.  is  given  by:\n",
+      "E2  =    V  -  Ia2*(Ra  +  Rf)\n",
+      " #E  proportional  to  (Phi*n)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  =  E2*n1/(2*E1)  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)generated  e.m.f.,  E  is  \",round(E1,2),\"  V  \"\n",
+      "print  \"\\n  (b)speed  of  motor,  n2,  \",round(n2,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)generated  e.m.f.,  E  is   232.5   V  \n",
+        "\n",
+        "  (b)speed  of  motor,  n2,   11.61   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 374</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the overall efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  80;#  in  Amperes\n",
+      "C  =  1500;#  in  Watt\n",
+      "Rf  =  40;#  in  ohms\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n  =  1000/60;#  in  Rev/sec\n",
+      "V  =  320;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current  Ia\n",
+      "Ia  =  I  -  If\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (Ia*Ia*Ra) - (If*V) - C)/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 80.09 %"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 374</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the maximum efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  40;#  in  Amperes\n",
+      "Rf  =  0.05;#  in  ohms\n",
+      "Ra  =  0.15;#  in  ohm\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #However  for  a  series  motor,  If  =  0  and  the  Ia*Ia*Ra  loss  needs  to  be  I*I*(\u0011Ra  +  Rf)\n",
+      " #For  maximum  efficiency  I*I*\u0011(Ra  +  Rf)  =  C\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 93.6"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 375</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current supplied to the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  15;#  in  Nm\n",
+      "n  =  1200/60;#  in  rev/sec\n",
+      "eff  =  0.8;\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "I = T*2*math.pi*n/(V*eff)\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n current supplied, I is \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " current supplied, I is  11.78 A"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 376</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  2;#  in  ohm\n",
+      "n  =  30;#  in  rev/sec\n",
+      "I  =  10;#  in  A\n",
+      "C  =  300;#  in  Watt\n",
+      "V  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (I*I*R) - C)/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 87.5 %"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 378</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the speed when the current is 60 A and a resistance of 0.5 ohmis connected in series with the armature, \n",
+      "#the shunt field remaining constant. \n",
+      "#(b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value \n",
+      "#by increasing resistance in the field circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  120;#  in  A\n",
+      "Ia2  =  60;#  in  A\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n1  =  10;#  in  rev/sec\n",
+      "R  =  0.5;#  in  ohm\n",
+      "x  =  0.8;\n",
+      "V  =  500;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #back  e.m.f.  at  Ia1\n",
+      "E1  =  V  -  Ia1*Ra\n",
+      " #at  Ia2\n",
+      "E2  =  V  -  Ia2*(Ra  +  R)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  = n1*E2/E1\n",
+      " #Back  e.m.f.  when  Ia2\n",
+      "E3  =  V  -  Ia2*Ra\n",
+      "n3  =  n1*E3/(x*E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)speed  n2  is  \",round(n2,2),\"  rev/sec\"\n",
+      "print  \"\\n  (b)speed  n3  is  \",round(n3,2),\"  rev/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)speed  n2  is   9.62   rev/sec\n",
+        "\n",
+        "  (b)speed  n3  is   12.82   rev/sec"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 29, page no. 379</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the speed when developing full load torque but with a 0.2 ohm diverter in parallel with the field winding.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  90;#  in  Amperes\n",
+      "Ra  =  0.1;#  in  ohm\n",
+      "Rse  =  0.05;#  in  ohm\n",
+      "Rd  =  0.2;#  in  Ohm\n",
+      "n1  =  15;#  in  rev/sec\n",
+      "V  =  300;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1\n",
+      "E1  =  V  -  Ia1*(Ra  +  Rse)\n",
+      " #With  the  Rd  diverter  in  parallel  with  Rse\n",
+      " #equivalent  resistance,  Re\n",
+      "Re  =  Rd*Rse/(Rd+Rse)\n",
+      " #Torque,  T  proprtional  to  Ia*Phi  and  for  full  load  torque,  Ia1*Phi1  =  Ia2*Phi2\n",
+      " #Since  flux  is  proportional  to  field  current  Phi1  proportional  to  Ta1  and  Phi2  Proportional  to  I1\n",
+      "I1  =    (Ia1*Ia1*0.8)**0.5\n",
+      " #By  current  division,  current  I1\n",
+      "Ia2  =  I1/(Rd/(Rd  +  Rse))\n",
+      " #Hence  e.m.f.  E2\n",
+      "E2  =  V  -  Ia2*(Ra  +  Re)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  =  E2*Ia1*n1/(I1*E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  speed  n2  is  \",round(n2,2),\"  rev/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  speed  n2  is   16.74   rev/sec"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 30, page no. 380</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  25;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "Rse  =  0.2;#  in  ohm\n",
+      "n1  =  800/60;#  in  rev/sec\n",
+      "n2  =  600/60;#  in  rev/sec\n",
+      "V  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1\n",
+      "E1  =  V  -  Ia1*(Ra  +  Rse)\n",
+      " #At  n2,  since  the  current  is  unchanged,  the  flux  is  unchanged.\n",
+      " #E1/E2  =  n1/n2\n",
+      "E2  =  E1*n2/n1\n",
+      " #and  E2  =  V  -  Ia1(\u0011Ra  +  Rse  +  R)\n",
+      "R  =  (V  -  E2)/Ia1  -  Ra  -  Rse\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  is  \",round(R,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  is   3.85   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_1.ipynb
new file mode 100755
index 00000000..28433f9f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_1.ipynb
@@ -0,0 +1,1641 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 21: D.c. machines</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the generated e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  600;#  no.  of  conductors\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "p  =  4;#  no.  of  pairs\n",
+      "n  =  625/60;#  in  rev/sec\n",
+      "Phi  =  20E-3;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  generated  e.m.f  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  generated  e.m.f  is   500.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the speed at which the machine must be driven to generate an e.m.f. of 240 V\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  50*16;#  no.  of  conductors\n",
+      "p  =  1;#  let  no.  of  pairs\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "E  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      " #Rearranging  gives,  speed\n",
+      "n  =  E*c/(2*p*Phi*Z)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  speed  at  which  the  machine  must  be  driven  is  \",round(n,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  speed  at  which  the  machine  must  be  driven  is   10.0   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. generated when running at 500 rev/min.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  1200;#  no.  of  conductors\n",
+      "p  =  1;#  let,  no.  of  pairs\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "n  =  500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   300.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 355</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the generated e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  1200;#  no.  of  conductors\n",
+      "p  =  4;#  let,  no.  of  pairs\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "n  =  500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   1200.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 355</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the change in the generated voltage when the field current is reduced by 20%, \n",
+      "#assuming the flux is proportional to the field current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1 = 150; # in Volts\n",
+      "x = 0.2;\n",
+      "\n",
+      "#calculation:\n",
+      "E2  =  E1*(1- x)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E2,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   120.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 356</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "n1  =  30;#  in  rev/sec\n",
+      "E1  =  200;#  in  Volts\n",
+      "n2  =  20;#  in  rev/sec\n",
+      "E2  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E  proportional  to  phi*w  and  since  w  =  2*pi*n,  then\n",
+      " #  E  proportional  to  phi*n\n",
+      " #  E1/E2  =  Phi1*n1/(Phi2*n2)\n",
+      " #  let  Phi2/Phi1  =  Phi\n",
+      "Phi  =  E2*n1/(E1*n2)\n",
+      "Phi_inc  =  (Phi  -  1)*100#/in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  percentage  increase  in  the  flux  per  pole  is  \",round(Phi_inc,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  percentage  increase  in  the  flux  per  pole  is   87.5   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the terminal voltage of a generator\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.30;#  in  ohms\n",
+      "Ia  =  30;#  in  Amperes\n",
+      "E  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #terminal  voltage,\n",
+      " #V  =  E  -  Ia*Ra\n",
+      "V  =  E  -  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  terminal  voltage  of  a  generator  is  \",round(V,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  terminal  voltage  of  a  generator  is   191.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the terminal voltage, and (b) the generated e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "RL  =  60;#  in  ohms\n",
+      "Ia  =  8;#  in  Amperes\n",
+      "Ra  =  1;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #terminal  voltage,\n",
+      " #V  =  Ia*RL\n",
+      "V  =  Ia*RL\n",
+      " #Generated  e.m.f.,  E\n",
+      "E  =  V  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)terminal  voltage  is  \",round(V,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)terminal  voltage  is   480.0   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   488.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the generated e.m.f. when \n",
+      "#(a) the speed increases to 25 rev/s and the pole flux remains unchanged, \n",
+      "#(b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08 Wb,\n",
+      "#and (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07 Wb.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  150;#  in  Volts\n",
+      "n1  =  20;#  in  rev/sec\n",
+      "Phi1  =  0.10;#  in  Wb\n",
+      "n2  =  25;#  in  rev/sec\n",
+      "Phi2  =  0.10;#  in  Wb\n",
+      "n3  =  20;#  in  rev/sec\n",
+      "Phi3  =  0.08;#  in  Wb\n",
+      "n4  =  24;#  in  rev/sec\n",
+      "Phi4  =  0.07;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E  proportional  to  phi*w  and  since  w  =  2*pi*n,  then\n",
+      " #  E  proportional  to  phi*n\n",
+      " #  E1/E2  =  Phi1*n1/(Phi2*n2)\n",
+      "E2  =  E1*Phi2*n2/(Phi1*n1)\n",
+      "E3  =  E1*Phi3*n3/(Phi1*n1)\n",
+      "E4  =  E1*Phi4*n4/(Phi1*n1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  generated  e.m.f  is  \",round(E2,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E3,2),\"  V  \"\n",
+      "print  \"\\n  (c)generated  e.m.f.  is  \",round(E4,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  generated  e.m.f  is   187.5   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   120.0   V  \n",
+        "\n",
+        "  (c)generated  e.m.f.  is   126.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 359</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ps  =  20000;#  in  Watts\n",
+      "Vs  =  200;#  in  Volts\n",
+      "Rs  =  0.1;#  in  ohms\n",
+      "Rf  =  50;#  in  ohms\n",
+      "Ra  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Load  current,  I\n",
+      "Is  =  Ps/Vs\n",
+      " #Volt  drop  in  the  cables  to  the  load\n",
+      "Vd  =  Is*Rs\n",
+      " #Hence  terminal  voltage,\n",
+      "V  =  Vs  +  Vd\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  Is\n",
+      " #Generated  e.m.f.  E\n",
+      "E  =  V  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)terminal  voltage  is  \",round(V,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)terminal  voltage  is   210.0   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   214.17   V  "
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 361</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the e.m.f. generated.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Is  =  80;#  in  amperes\n",
+      "Vs  =  200;#  in  Volts\n",
+      "Rf  =  40;#  in  ohms\n",
+      "Rse  =  0.02;#  in  ohms\n",
+      "Ra  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Volt  drop  in  series  winding\n",
+      "Vse  =  Is*Rse\n",
+      " #P.d.  across  the  field  winding  =  p.d.  across  armature\n",
+      "V1  =  Vs  +  Vse\n",
+      " #Field  current,  If\n",
+      "If  =  V1/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  Is\n",
+      " #Generated  e.m.f.  E\n",
+      "E  =  V1  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  generated  e.m.f.  is   205.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 363</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the efficiency of the generator at full load\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ps  =  10000;#  in  Watt\n",
+      "Pl  =  600;#  in  Watt\n",
+      "Ra  =  0.75;#  in  ohms\n",
+      "Rf  =  125;#  in  ohms\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Output  power  Ps  =  V*I\n",
+      " #from  which,  load  current  I\n",
+      "I  =  Ps/V\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  I\n",
+      " #Efficiency,\n",
+      "eff  =  Ps*100/((V*I)  +  (Ia*Ia*Ra)  +  (If*V)  +  (Pl))#  in  Percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Efficiency  is  \",round(eff,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Efficiency  is   80.5   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 364</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the back e.m.f. when the armature current is 50 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.2;#  in  ohms\n",
+      "V  =  240;#  in  Volts\n",
+      "Ia  =  50;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  motor,  V  =  E  +  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  back  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  back  e.m.f.  is   230.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 365</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A, and\n",
+      "#(b) as a motor taking 80 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.25;#  in  ohms\n",
+      "V  =  300;#  in  Volts\n",
+      "Ig  =  100;#  in  Amperes\n",
+      "Im  =  80;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #As  a  generator,  generated  e.m.f.,\n",
+      " #  E  =  V  +  Ia*Ra\n",
+      "Eg  =  V  +  Ig*Ra\n",
+      " #For  a  motor,  generated  e.m.f.  (or  back  e.m.f.),\n",
+      " #  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Im*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)As  a  generator,  generated  e.m.f.  is  \",round(Eg,2),\"  V  \"\n",
+      "print  \"\\n  (b)back  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)As  a  generator,  generated  e.m.f.  is   325.0   V  \n",
+        "\n",
+        "  (b)back  e.m.f.  is   280.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the torque exerted when a current of 30 A flows in each armature conductor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  4;\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "Phi  =  25E-3;#  Wb\n",
+      "Z  =  900;\n",
+      "Ia  =  30;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #torque  T  =  p*Phi*Z*Ia/(pi*c)\n",
+      "T  =  p*Phi*Z*Ia/(1*math.pi*c)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  torque  exerted  is   429.72   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the torque\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  350;#  in  Volts\n",
+      "Ra  =  0.5;#  in  ohms\n",
+      "n  =  15;#  in  rev/sec\n",
+      "Ia  =  60;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Back  e.m.f.  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      " #torque  T  =  E*Ia/(2*n*pi)\n",
+      "T  =  E*Ia/(2*n*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  torque  exerted  is   203.72   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the speed and (b) the torque developed when the armature current is 40 A\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  1;#  let\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  20E-3;#  Wb\n",
+      "Z  =  500;\n",
+      "V  =  250;#  in  Volts\n",
+      "Ra  =  1;#  in  ohms\n",
+      "Ia  =  40;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Back  e.m.f.  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      " #E.m.f.  E  =  2*p*Phi*n*Z/c\n",
+      " #  rearrange,\n",
+      "n  =  E*c/(2*p*Phi*Z)\n",
+      " #torque  T  =  E*Ia/(2*n*pi)\n",
+      "T  =  E*Ia/(2*n*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)speed  n  is  \",round(n,2),\"  rev/sec  \"\n",
+      "print  \"\\n  (b)the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)speed  n  is   21.0   rev/sec  \n",
+        "\n",
+        "  (b)the  torque  exerted  is   63.66   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 367</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the armature current at this new value of torque.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T1  =  25;#  in  Nm\n",
+      "T2  =  35;#  in  Nm\n",
+      "Ia1  =  16;#  in  Amperes\n",
+      "V  =  100;#  in  Volts\n",
+      "x  =  0.15;\n",
+      "\n",
+      " #calculation:\n",
+      " #the  shaft  torque  T  of  a  generator  is  proportional  to  (phi*Ia),\n",
+      "    #where  Phi  is  the  flux  and  Ia  is  the  armature  current.  Thus,  T  =  k*Phi*Ia,  where  k  is  a  constant.\n",
+      " #The  torque  at  flux  phi1  and  armature  current  Ia1  is  T1  =  k*Phi1*Ia1.\n",
+      " #similarly  T2  =  k*Phi2*Ia2\n",
+      "\n",
+      "Ia2  =  T2*Ia1/(0.85*T1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  armature  current  at  the  new  value  of  torque  is  \",round(Ia2,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  armature  current  at  the  new  value  of  torque  is   26.35   A  "
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 367</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the efficiency of the generator and (b) the power loss in the generator.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  12;#  in  Nm\n",
+      "I  =  15;#  in  Amperes\n",
+      "V  =  100;#  in  Volts\n",
+      "n  =  1500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #the  efficiency  of  a  generator  =  (output  power/input  power)*100  %\n",
+      " #The  output  power  is  the  electrical  output,  i.e.  VI  watts.  \n",
+      "    #The  input  power  to  a  generator  is  the  mechanical  power  in  the  shaft  driving  the  generator, \n",
+      "    #i.e.  T*w  or  T(2*pi*n)  watts,  where  T  is  the  torque  in  Nm  and  n  is  speed  of  rotation  in  rev/s.  Hence,  for  a  generator  \n",
+      " #efficiency  =  V*I*100/(T*2*pi*n) %\n",
+      "eff  =  V*I*100/(T*2*math.pi*n)#  in    Percent\n",
+      " #The  input  power  =  output  power  +  losses\n",
+      " #  hence,  T*2*math.pi*n  =  V*I  +  losses\n",
+      "Pl  =  T*2*math.pi*n  -  V*I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  efficiency  is  \",round(eff,2),\" % \"\n",
+      "print  \"\\n  (b)  power  loss  is  \",round(Pl,2),\"  W  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  efficiency  is   79.58  % \n",
+        "\n",
+        "  (b)  power  loss  is   384.96   W  "
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 368</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current in the armature, and (b) the back e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rf  =  150;#  in  Ohms\n",
+      "Ra  =  0.4;#  in  Ohms\n",
+      "I  =  30;#  in  Amperes\n",
+      "V  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Field  current  If\n",
+      "If  =  V/Rf\n",
+      " #Supply  current  I  =  Ia  +  If\n",
+      " #Hence  armature  current,  Ia\n",
+      "Ia  =  I  -  If\n",
+      " #Back  e.m.f.  E  =  V  -\u0004  Ia*Ra\n",
+      "E  =  V  -  (Ia*Ra)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)    current  in  the  armature  is  \",round(Ia,2),\"  A  \"\n",
+      "print  \"\\n  (b)  Back  e.m.f.  E  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)    current  in  the  armature  is   28.4   A  \n",
+        "\n",
+        "  (b)  Back  e.m.f.  E  is   228.64   V  "
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 370</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the speed of the motor, assuming the flux remains constant.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  30;#  in  Amperes\n",
+      "Ia2  =  45;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "n1  =  1350/60;#  in  Rev/sec\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #The  relationship  E  proportional  to  (Phi*n)  applies  to  both  generators  and  motors.  For  a  motor,\n",
+      " #E  =  V  -  (Ia*Ra)\n",
+      "E1  =    V  -  (Ia1*Ra)\n",
+      "E2  =    V  -  (Ia2*Ra)\n",
+      " #The  relationship,  E1/E2  =  Phi1*n1/Phi2*n2,    applies  to  both  generators  and  motors.\n",
+      "    #Since  the  flux  is  constant,  Phi1  =  Phi2\n",
+      "n2  =  E2*n1/(E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  speed  of  the  motor  is  \",round(n2,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  speed  of  the  motor  is   21.78   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 370</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and \n",
+      "#(b) the steadystate value of the armature current at the new value of flux,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  30;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "n  =  800/60;#  in  Rev/sec\n",
+      "V  =  220;#  in  Volts\n",
+      "x=  0.1;\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  d.c.  shunt-wound  motor,  E  =  V  -  (Ia*Ra),Hence  initial  generated  e.m.f.,\n",
+      "E1  =    V  -  (Ia1*Ra)\n",
+      " #The  generated  e.m.f.  is  also  such  that  E  proportional  to  (Phi*n)  \n",
+      "    #so  at  the  instant  the  flux  is  reduced,  the  speed  has  not  had  time  to  change,  and\n",
+      "E  =  E1*(1-x)\n",
+      " #Hence,  the  voltage  drop  due  to  the  armature  resistance  is\n",
+      "Vd  =  V  -  E\n",
+      " #The  instantaneous  value  of  the  current  is\n",
+      "Ia  =  Vd/Ra\n",
+      " #T  proportional  to  (Phi*Ia),  since  the  torque  is  constant,\n",
+      " #Phi1*Ia1  =  Phi2*Ia2,    The  flux  8  is  reduced  by  10%,  hence\n",
+      "Ia2  =  Ia1/0.9\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)instantaneous  value  of  the  current  \",round(Ia,2),\"  A  \"\n",
+      "print  \"\\n  (b)steady  state  value  of  armature  current,  \",round(Ia2,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)instantaneous  value  of  the  current   82.0   A  \n",
+        "\n",
+        "  (b)steady  state  value  of  armature  current,   33.33   A  "
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 372</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the generated e.m.f. at this load. \n",
+      "#(b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A.\n",
+      "#Assume that this causes a doubling of the flux.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  15;#  in  Amperes\n",
+      "Ia2  =  30;#  in  Amperes\n",
+      "Rf  =  0.3;#  in  ohms\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n1  =  24;#  in  Rev/sec\n",
+      "V  =  240;#  in  Volts\n",
+      "x=  2;\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E,  at  initial  load,  is  given  by\n",
+      "E1  =    V  -  Ia1*(Ra  +  Rf)\n",
+      " #When  the  current  is  increased  to  30  A,  the  generated  e.m.f.  is  given  by:\n",
+      "E2  =    V  -  Ia2*(Ra  +  Rf)\n",
+      " #E  proportional  to  (Phi*n)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  =  E2*n1/(2*E1)  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)generated  e.m.f.,  E  is  \",round(E1,2),\"  V  \"\n",
+      "print  \"\\n  (b)speed  of  motor,  n2,  \",round(n2,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)generated  e.m.f.,  E  is   232.5   V  \n",
+        "\n",
+        "  (b)speed  of  motor,  n2,   11.61   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 374</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the overall efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  80;#  in  Amperes\n",
+      "C  =  1500;#  in  Watt\n",
+      "Rf  =  40;#  in  ohms\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n  =  1000/60;#  in  Rev/sec\n",
+      "V  =  320;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current  Ia\n",
+      "Ia  =  I  -  If\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (Ia*Ia*Ra) - (If*V) - C)/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 80.09 %"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 374</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the maximum efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  40;#  in  Amperes\n",
+      "Rf  =  0.05;#  in  ohms\n",
+      "Ra  =  0.15;#  in  ohm\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #However  for  a  series  motor,  If  =  0  and  the  Ia*Ia*Ra  loss  needs  to  be  I*I*(\u0011Ra  +  Rf)\n",
+      " #For  maximum  efficiency  I*I*\u0011(Ra  +  Rf)  =  C\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 93.6"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 375</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current supplied to the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  15;#  in  Nm\n",
+      "n  =  1200/60;#  in  rev/sec\n",
+      "eff  =  0.8;\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "I = T*2*math.pi*n/(V*eff)\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n current supplied, I is \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " current supplied, I is  11.78 A"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 376</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  2;#  in  ohm\n",
+      "n  =  30;#  in  rev/sec\n",
+      "I  =  10;#  in  A\n",
+      "C  =  300;#  in  Watt\n",
+      "V  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (I*I*R) - C)/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 87.5 %"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 378</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the speed when the current is 60 A and a resistance of 0.5 ohmis connected in series with the armature, \n",
+      "#the shunt field remaining constant. \n",
+      "#(b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value \n",
+      "#by increasing resistance in the field circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  120;#  in  A\n",
+      "Ia2  =  60;#  in  A\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n1  =  10;#  in  rev/sec\n",
+      "R  =  0.5;#  in  ohm\n",
+      "x  =  0.8;\n",
+      "V  =  500;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #back  e.m.f.  at  Ia1\n",
+      "E1  =  V  -  Ia1*Ra\n",
+      " #at  Ia2\n",
+      "E2  =  V  -  Ia2*(Ra  +  R)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  = n1*E2/E1\n",
+      " #Back  e.m.f.  when  Ia2\n",
+      "E3  =  V  -  Ia2*Ra\n",
+      "n3  =  n1*E3/(x*E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)speed  n2  is  \",round(n2,2),\"  rev/sec\"\n",
+      "print  \"\\n  (b)speed  n3  is  \",round(n3,2),\"  rev/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)speed  n2  is   9.62   rev/sec\n",
+        "\n",
+        "  (b)speed  n3  is   12.82   rev/sec"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 29, page no. 379</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the speed when developing full load torque but with a 0.2 ohm diverter in parallel with the field winding.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  90;#  in  Amperes\n",
+      "Ra  =  0.1;#  in  ohm\n",
+      "Rse  =  0.05;#  in  ohm\n",
+      "Rd  =  0.2;#  in  Ohm\n",
+      "n1  =  15;#  in  rev/sec\n",
+      "V  =  300;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1\n",
+      "E1  =  V  -  Ia1*(Ra  +  Rse)\n",
+      " #With  the  Rd  diverter  in  parallel  with  Rse\n",
+      " #equivalent  resistance,  Re\n",
+      "Re  =  Rd*Rse/(Rd+Rse)\n",
+      " #Torque,  T  proprtional  to  Ia*Phi  and  for  full  load  torque,  Ia1*Phi1  =  Ia2*Phi2\n",
+      " #Since  flux  is  proportional  to  field  current  Phi1  proportional  to  Ta1  and  Phi2  Proportional  to  I1\n",
+      "I1  =    (Ia1*Ia1*0.8)**0.5\n",
+      " #By  current  division,  current  I1\n",
+      "Ia2  =  I1/(Rd/(Rd  +  Rse))\n",
+      " #Hence  e.m.f.  E2\n",
+      "E2  =  V  -  Ia2*(Ra  +  Re)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  =  E2*Ia1*n1/(I1*E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  speed  n2  is  \",round(n2,2),\"  rev/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  speed  n2  is   16.74   rev/sec"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 30, page no. 380</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  25;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "Rse  =  0.2;#  in  ohm\n",
+      "n1  =  800/60;#  in  rev/sec\n",
+      "n2  =  600/60;#  in  rev/sec\n",
+      "V  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1\n",
+      "E1  =  V  -  Ia1*(Ra  +  Rse)\n",
+      " #At  n2,  since  the  current  is  unchanged,  the  flux  is  unchanged.\n",
+      " #E1/E2  =  n1/n2\n",
+      "E2  =  E1*n2/n1\n",
+      " #and  E2  =  V  -  Ia1(\u0011Ra  +  Rse  +  R)\n",
+      "R  =  (V  -  E2)/Ia1  -  Ra  -  Rse\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  is  \",round(R,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  is   3.85   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_2.ipynb
new file mode 100755
index 00000000..28433f9f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_2.ipynb
@@ -0,0 +1,1641 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 21: D.c. machines</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the generated e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  600;#  no.  of  conductors\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "p  =  4;#  no.  of  pairs\n",
+      "n  =  625/60;#  in  rev/sec\n",
+      "Phi  =  20E-3;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  generated  e.m.f  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  generated  e.m.f  is   500.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the speed at which the machine must be driven to generate an e.m.f. of 240 V\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  50*16;#  no.  of  conductors\n",
+      "p  =  1;#  let  no.  of  pairs\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "E  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      " #Rearranging  gives,  speed\n",
+      "n  =  E*c/(2*p*Phi*Z)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  speed  at  which  the  machine  must  be  driven  is  \",round(n,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  speed  at  which  the  machine  must  be  driven  is   10.0   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. generated when running at 500 rev/min.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  1200;#  no.  of  conductors\n",
+      "p  =  1;#  let,  no.  of  pairs\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "n  =  500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   300.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 355</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the generated e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  1200;#  no.  of  conductors\n",
+      "p  =  4;#  let,  no.  of  pairs\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "n  =  500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   1200.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 355</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the change in the generated voltage when the field current is reduced by 20%, \n",
+      "#assuming the flux is proportional to the field current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1 = 150; # in Volts\n",
+      "x = 0.2;\n",
+      "\n",
+      "#calculation:\n",
+      "E2  =  E1*(1- x)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E2,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   120.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 356</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "n1  =  30;#  in  rev/sec\n",
+      "E1  =  200;#  in  Volts\n",
+      "n2  =  20;#  in  rev/sec\n",
+      "E2  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E  proportional  to  phi*w  and  since  w  =  2*pi*n,  then\n",
+      " #  E  proportional  to  phi*n\n",
+      " #  E1/E2  =  Phi1*n1/(Phi2*n2)\n",
+      " #  let  Phi2/Phi1  =  Phi\n",
+      "Phi  =  E2*n1/(E1*n2)\n",
+      "Phi_inc  =  (Phi  -  1)*100#/in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  percentage  increase  in  the  flux  per  pole  is  \",round(Phi_inc,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  percentage  increase  in  the  flux  per  pole  is   87.5   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the terminal voltage of a generator\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.30;#  in  ohms\n",
+      "Ia  =  30;#  in  Amperes\n",
+      "E  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #terminal  voltage,\n",
+      " #V  =  E  -  Ia*Ra\n",
+      "V  =  E  -  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  terminal  voltage  of  a  generator  is  \",round(V,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  terminal  voltage  of  a  generator  is   191.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the terminal voltage, and (b) the generated e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "RL  =  60;#  in  ohms\n",
+      "Ia  =  8;#  in  Amperes\n",
+      "Ra  =  1;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #terminal  voltage,\n",
+      " #V  =  Ia*RL\n",
+      "V  =  Ia*RL\n",
+      " #Generated  e.m.f.,  E\n",
+      "E  =  V  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)terminal  voltage  is  \",round(V,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)terminal  voltage  is   480.0   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   488.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the generated e.m.f. when \n",
+      "#(a) the speed increases to 25 rev/s and the pole flux remains unchanged, \n",
+      "#(b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08 Wb,\n",
+      "#and (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07 Wb.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  150;#  in  Volts\n",
+      "n1  =  20;#  in  rev/sec\n",
+      "Phi1  =  0.10;#  in  Wb\n",
+      "n2  =  25;#  in  rev/sec\n",
+      "Phi2  =  0.10;#  in  Wb\n",
+      "n3  =  20;#  in  rev/sec\n",
+      "Phi3  =  0.08;#  in  Wb\n",
+      "n4  =  24;#  in  rev/sec\n",
+      "Phi4  =  0.07;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E  proportional  to  phi*w  and  since  w  =  2*pi*n,  then\n",
+      " #  E  proportional  to  phi*n\n",
+      " #  E1/E2  =  Phi1*n1/(Phi2*n2)\n",
+      "E2  =  E1*Phi2*n2/(Phi1*n1)\n",
+      "E3  =  E1*Phi3*n3/(Phi1*n1)\n",
+      "E4  =  E1*Phi4*n4/(Phi1*n1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  generated  e.m.f  is  \",round(E2,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E3,2),\"  V  \"\n",
+      "print  \"\\n  (c)generated  e.m.f.  is  \",round(E4,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  generated  e.m.f  is   187.5   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   120.0   V  \n",
+        "\n",
+        "  (c)generated  e.m.f.  is   126.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 359</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ps  =  20000;#  in  Watts\n",
+      "Vs  =  200;#  in  Volts\n",
+      "Rs  =  0.1;#  in  ohms\n",
+      "Rf  =  50;#  in  ohms\n",
+      "Ra  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Load  current,  I\n",
+      "Is  =  Ps/Vs\n",
+      " #Volt  drop  in  the  cables  to  the  load\n",
+      "Vd  =  Is*Rs\n",
+      " #Hence  terminal  voltage,\n",
+      "V  =  Vs  +  Vd\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  Is\n",
+      " #Generated  e.m.f.  E\n",
+      "E  =  V  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)terminal  voltage  is  \",round(V,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)terminal  voltage  is   210.0   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   214.17   V  "
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 361</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the e.m.f. generated.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Is  =  80;#  in  amperes\n",
+      "Vs  =  200;#  in  Volts\n",
+      "Rf  =  40;#  in  ohms\n",
+      "Rse  =  0.02;#  in  ohms\n",
+      "Ra  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Volt  drop  in  series  winding\n",
+      "Vse  =  Is*Rse\n",
+      " #P.d.  across  the  field  winding  =  p.d.  across  armature\n",
+      "V1  =  Vs  +  Vse\n",
+      " #Field  current,  If\n",
+      "If  =  V1/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  Is\n",
+      " #Generated  e.m.f.  E\n",
+      "E  =  V1  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  generated  e.m.f.  is   205.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 363</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the efficiency of the generator at full load\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ps  =  10000;#  in  Watt\n",
+      "Pl  =  600;#  in  Watt\n",
+      "Ra  =  0.75;#  in  ohms\n",
+      "Rf  =  125;#  in  ohms\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Output  power  Ps  =  V*I\n",
+      " #from  which,  load  current  I\n",
+      "I  =  Ps/V\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  I\n",
+      " #Efficiency,\n",
+      "eff  =  Ps*100/((V*I)  +  (Ia*Ia*Ra)  +  (If*V)  +  (Pl))#  in  Percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Efficiency  is  \",round(eff,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Efficiency  is   80.5   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 364</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the back e.m.f. when the armature current is 50 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.2;#  in  ohms\n",
+      "V  =  240;#  in  Volts\n",
+      "Ia  =  50;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  motor,  V  =  E  +  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  back  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  back  e.m.f.  is   230.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 365</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A, and\n",
+      "#(b) as a motor taking 80 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.25;#  in  ohms\n",
+      "V  =  300;#  in  Volts\n",
+      "Ig  =  100;#  in  Amperes\n",
+      "Im  =  80;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #As  a  generator,  generated  e.m.f.,\n",
+      " #  E  =  V  +  Ia*Ra\n",
+      "Eg  =  V  +  Ig*Ra\n",
+      " #For  a  motor,  generated  e.m.f.  (or  back  e.m.f.),\n",
+      " #  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Im*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)As  a  generator,  generated  e.m.f.  is  \",round(Eg,2),\"  V  \"\n",
+      "print  \"\\n  (b)back  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)As  a  generator,  generated  e.m.f.  is   325.0   V  \n",
+        "\n",
+        "  (b)back  e.m.f.  is   280.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the torque exerted when a current of 30 A flows in each armature conductor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  4;\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "Phi  =  25E-3;#  Wb\n",
+      "Z  =  900;\n",
+      "Ia  =  30;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #torque  T  =  p*Phi*Z*Ia/(pi*c)\n",
+      "T  =  p*Phi*Z*Ia/(1*math.pi*c)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  torque  exerted  is   429.72   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the torque\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  350;#  in  Volts\n",
+      "Ra  =  0.5;#  in  ohms\n",
+      "n  =  15;#  in  rev/sec\n",
+      "Ia  =  60;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Back  e.m.f.  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      " #torque  T  =  E*Ia/(2*n*pi)\n",
+      "T  =  E*Ia/(2*n*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  torque  exerted  is   203.72   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the speed and (b) the torque developed when the armature current is 40 A\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  1;#  let\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  20E-3;#  Wb\n",
+      "Z  =  500;\n",
+      "V  =  250;#  in  Volts\n",
+      "Ra  =  1;#  in  ohms\n",
+      "Ia  =  40;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Back  e.m.f.  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      " #E.m.f.  E  =  2*p*Phi*n*Z/c\n",
+      " #  rearrange,\n",
+      "n  =  E*c/(2*p*Phi*Z)\n",
+      " #torque  T  =  E*Ia/(2*n*pi)\n",
+      "T  =  E*Ia/(2*n*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)speed  n  is  \",round(n,2),\"  rev/sec  \"\n",
+      "print  \"\\n  (b)the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)speed  n  is   21.0   rev/sec  \n",
+        "\n",
+        "  (b)the  torque  exerted  is   63.66   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 367</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the armature current at this new value of torque.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T1  =  25;#  in  Nm\n",
+      "T2  =  35;#  in  Nm\n",
+      "Ia1  =  16;#  in  Amperes\n",
+      "V  =  100;#  in  Volts\n",
+      "x  =  0.15;\n",
+      "\n",
+      " #calculation:\n",
+      " #the  shaft  torque  T  of  a  generator  is  proportional  to  (phi*Ia),\n",
+      "    #where  Phi  is  the  flux  and  Ia  is  the  armature  current.  Thus,  T  =  k*Phi*Ia,  where  k  is  a  constant.\n",
+      " #The  torque  at  flux  phi1  and  armature  current  Ia1  is  T1  =  k*Phi1*Ia1.\n",
+      " #similarly  T2  =  k*Phi2*Ia2\n",
+      "\n",
+      "Ia2  =  T2*Ia1/(0.85*T1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  armature  current  at  the  new  value  of  torque  is  \",round(Ia2,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  armature  current  at  the  new  value  of  torque  is   26.35   A  "
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 367</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the efficiency of the generator and (b) the power loss in the generator.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  12;#  in  Nm\n",
+      "I  =  15;#  in  Amperes\n",
+      "V  =  100;#  in  Volts\n",
+      "n  =  1500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #the  efficiency  of  a  generator  =  (output  power/input  power)*100  %\n",
+      " #The  output  power  is  the  electrical  output,  i.e.  VI  watts.  \n",
+      "    #The  input  power  to  a  generator  is  the  mechanical  power  in  the  shaft  driving  the  generator, \n",
+      "    #i.e.  T*w  or  T(2*pi*n)  watts,  where  T  is  the  torque  in  Nm  and  n  is  speed  of  rotation  in  rev/s.  Hence,  for  a  generator  \n",
+      " #efficiency  =  V*I*100/(T*2*pi*n) %\n",
+      "eff  =  V*I*100/(T*2*math.pi*n)#  in    Percent\n",
+      " #The  input  power  =  output  power  +  losses\n",
+      " #  hence,  T*2*math.pi*n  =  V*I  +  losses\n",
+      "Pl  =  T*2*math.pi*n  -  V*I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  efficiency  is  \",round(eff,2),\" % \"\n",
+      "print  \"\\n  (b)  power  loss  is  \",round(Pl,2),\"  W  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  efficiency  is   79.58  % \n",
+        "\n",
+        "  (b)  power  loss  is   384.96   W  "
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 368</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current in the armature, and (b) the back e.m.f.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rf  =  150;#  in  Ohms\n",
+      "Ra  =  0.4;#  in  Ohms\n",
+      "I  =  30;#  in  Amperes\n",
+      "V  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Field  current  If\n",
+      "If  =  V/Rf\n",
+      " #Supply  current  I  =  Ia  +  If\n",
+      " #Hence  armature  current,  Ia\n",
+      "Ia  =  I  -  If\n",
+      " #Back  e.m.f.  E  =  V  -\u0004  Ia*Ra\n",
+      "E  =  V  -  (Ia*Ra)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)    current  in  the  armature  is  \",round(Ia,2),\"  A  \"\n",
+      "print  \"\\n  (b)  Back  e.m.f.  E  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)    current  in  the  armature  is   28.4   A  \n",
+        "\n",
+        "  (b)  Back  e.m.f.  E  is   228.64   V  "
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 370</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the speed of the motor, assuming the flux remains constant.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  30;#  in  Amperes\n",
+      "Ia2  =  45;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "n1  =  1350/60;#  in  Rev/sec\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #The  relationship  E  proportional  to  (Phi*n)  applies  to  both  generators  and  motors.  For  a  motor,\n",
+      " #E  =  V  -  (Ia*Ra)\n",
+      "E1  =    V  -  (Ia1*Ra)\n",
+      "E2  =    V  -  (Ia2*Ra)\n",
+      " #The  relationship,  E1/E2  =  Phi1*n1/Phi2*n2,    applies  to  both  generators  and  motors.\n",
+      "    #Since  the  flux  is  constant,  Phi1  =  Phi2\n",
+      "n2  =  E2*n1/(E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  speed  of  the  motor  is  \",round(n2,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  speed  of  the  motor  is   21.78   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 370</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and \n",
+      "#(b) the steadystate value of the armature current at the new value of flux,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  30;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "n  =  800/60;#  in  Rev/sec\n",
+      "V  =  220;#  in  Volts\n",
+      "x=  0.1;\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  d.c.  shunt-wound  motor,  E  =  V  -  (Ia*Ra),Hence  initial  generated  e.m.f.,\n",
+      "E1  =    V  -  (Ia1*Ra)\n",
+      " #The  generated  e.m.f.  is  also  such  that  E  proportional  to  (Phi*n)  \n",
+      "    #so  at  the  instant  the  flux  is  reduced,  the  speed  has  not  had  time  to  change,  and\n",
+      "E  =  E1*(1-x)\n",
+      " #Hence,  the  voltage  drop  due  to  the  armature  resistance  is\n",
+      "Vd  =  V  -  E\n",
+      " #The  instantaneous  value  of  the  current  is\n",
+      "Ia  =  Vd/Ra\n",
+      " #T  proportional  to  (Phi*Ia),  since  the  torque  is  constant,\n",
+      " #Phi1*Ia1  =  Phi2*Ia2,    The  flux  8  is  reduced  by  10%,  hence\n",
+      "Ia2  =  Ia1/0.9\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)instantaneous  value  of  the  current  \",round(Ia,2),\"  A  \"\n",
+      "print  \"\\n  (b)steady  state  value  of  armature  current,  \",round(Ia2,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)instantaneous  value  of  the  current   82.0   A  \n",
+        "\n",
+        "  (b)steady  state  value  of  armature  current,   33.33   A  "
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 372</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the generated e.m.f. at this load. \n",
+      "#(b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A.\n",
+      "#Assume that this causes a doubling of the flux.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  15;#  in  Amperes\n",
+      "Ia2  =  30;#  in  Amperes\n",
+      "Rf  =  0.3;#  in  ohms\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n1  =  24;#  in  Rev/sec\n",
+      "V  =  240;#  in  Volts\n",
+      "x=  2;\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E,  at  initial  load,  is  given  by\n",
+      "E1  =    V  -  Ia1*(Ra  +  Rf)\n",
+      " #When  the  current  is  increased  to  30  A,  the  generated  e.m.f.  is  given  by:\n",
+      "E2  =    V  -  Ia2*(Ra  +  Rf)\n",
+      " #E  proportional  to  (Phi*n)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  =  E2*n1/(2*E1)  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)generated  e.m.f.,  E  is  \",round(E1,2),\"  V  \"\n",
+      "print  \"\\n  (b)speed  of  motor,  n2,  \",round(n2,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)generated  e.m.f.,  E  is   232.5   V  \n",
+        "\n",
+        "  (b)speed  of  motor,  n2,   11.61   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 374</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the overall efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  80;#  in  Amperes\n",
+      "C  =  1500;#  in  Watt\n",
+      "Rf  =  40;#  in  ohms\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n  =  1000/60;#  in  Rev/sec\n",
+      "V  =  320;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current  Ia\n",
+      "Ia  =  I  -  If\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (Ia*Ia*Ra) - (If*V) - C)/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 80.09 %"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 374</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the maximum efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  40;#  in  Amperes\n",
+      "Rf  =  0.05;#  in  ohms\n",
+      "Ra  =  0.15;#  in  ohm\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #However  for  a  series  motor,  If  =  0  and  the  Ia*Ia*Ra  loss  needs  to  be  I*I*(\u0011Ra  +  Rf)\n",
+      " #For  maximum  efficiency  I*I*\u0011(Ra  +  Rf)  =  C\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 93.6"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 375</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current supplied to the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  15;#  in  Nm\n",
+      "n  =  1200/60;#  in  rev/sec\n",
+      "eff  =  0.8;\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "I = T*2*math.pi*n/(V*eff)\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n current supplied, I is \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " current supplied, I is  11.78 A"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 376</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  2;#  in  ohm\n",
+      "n  =  30;#  in  rev/sec\n",
+      "I  =  10;#  in  A\n",
+      "C  =  300;#  in  Watt\n",
+      "V  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (I*I*R) - C)/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 87.5 %"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 378</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the speed when the current is 60 A and a resistance of 0.5 ohmis connected in series with the armature, \n",
+      "#the shunt field remaining constant. \n",
+      "#(b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value \n",
+      "#by increasing resistance in the field circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  120;#  in  A\n",
+      "Ia2  =  60;#  in  A\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n1  =  10;#  in  rev/sec\n",
+      "R  =  0.5;#  in  ohm\n",
+      "x  =  0.8;\n",
+      "V  =  500;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #back  e.m.f.  at  Ia1\n",
+      "E1  =  V  -  Ia1*Ra\n",
+      " #at  Ia2\n",
+      "E2  =  V  -  Ia2*(Ra  +  R)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  = n1*E2/E1\n",
+      " #Back  e.m.f.  when  Ia2\n",
+      "E3  =  V  -  Ia2*Ra\n",
+      "n3  =  n1*E3/(x*E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)speed  n2  is  \",round(n2,2),\"  rev/sec\"\n",
+      "print  \"\\n  (b)speed  n3  is  \",round(n3,2),\"  rev/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)speed  n2  is   9.62   rev/sec\n",
+        "\n",
+        "  (b)speed  n3  is   12.82   rev/sec"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 29, page no. 379</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the speed when developing full load torque but with a 0.2 ohm diverter in parallel with the field winding.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  90;#  in  Amperes\n",
+      "Ra  =  0.1;#  in  ohm\n",
+      "Rse  =  0.05;#  in  ohm\n",
+      "Rd  =  0.2;#  in  Ohm\n",
+      "n1  =  15;#  in  rev/sec\n",
+      "V  =  300;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1\n",
+      "E1  =  V  -  Ia1*(Ra  +  Rse)\n",
+      " #With  the  Rd  diverter  in  parallel  with  Rse\n",
+      " #equivalent  resistance,  Re\n",
+      "Re  =  Rd*Rse/(Rd+Rse)\n",
+      " #Torque,  T  proprtional  to  Ia*Phi  and  for  full  load  torque,  Ia1*Phi1  =  Ia2*Phi2\n",
+      " #Since  flux  is  proportional  to  field  current  Phi1  proportional  to  Ta1  and  Phi2  Proportional  to  I1\n",
+      "I1  =    (Ia1*Ia1*0.8)**0.5\n",
+      " #By  current  division,  current  I1\n",
+      "Ia2  =  I1/(Rd/(Rd  +  Rse))\n",
+      " #Hence  e.m.f.  E2\n",
+      "E2  =  V  -  Ia2*(Ra  +  Re)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  =  E2*Ia1*n1/(I1*E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  speed  n2  is  \",round(n2,2),\"  rev/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  speed  n2  is   16.74   rev/sec"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 30, page no. 380</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  25;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "Rse  =  0.2;#  in  ohm\n",
+      "n1  =  800/60;#  in  rev/sec\n",
+      "n2  =  600/60;#  in  rev/sec\n",
+      "V  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1\n",
+      "E1  =  V  -  Ia1*(Ra  +  Rse)\n",
+      " #At  n2,  since  the  current  is  unchanged,  the  flux  is  unchanged.\n",
+      " #E1/E2  =  n1/n2\n",
+      "E2  =  E1*n2/n1\n",
+      " #and  E2  =  V  -  Ia1(\u0011Ra  +  Rse  +  R)\n",
+      "R  =  (V  -  E2)/Ia1  -  Ra  -  Rse\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  is  \",round(R,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  is   3.85   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_3.ipynb
new file mode 100755
index 00000000..373f5943
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_3.ipynb
@@ -0,0 +1,1631 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:18b2ae9034d59ad1d2a71a370085420c43adb19c1da344e7757efdb47969ddb0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 21: D.c. machines</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  600;#  no.  of  conductors\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "p  =  4;#  no.  of  pairs\n",
+      "n  =  625/60;#  in  rev/sec\n",
+      "Phi  =  20E-3;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  generated  e.m.f  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  generated  e.m.f  is   500.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  50*16;#  no.  of  conductors\n",
+      "p  =  1;#  let  no.  of  pairs\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "E  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      " #Rearranging  gives,  speed\n",
+      "n  =  E*c/(2*p*Phi*Z)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  speed  at  which  the  machine  must  be  driven  is  \",round(n,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  speed  at  which  the  machine  must  be  driven  is   10.0   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 354</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  1200;#  no.  of  conductors\n",
+      "p  =  1;#  let,  no.  of  pairs\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "n  =  500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   300.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 355</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Z  =  1200;#  no.  of  conductors\n",
+      "p  =  4;#  let,  no.  of  pairs\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "Phi  =  30E-3;#  in  Wb\n",
+      "n  =  500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #Generated  e.m.f.,  E  =  2*p*Phi*n*Z/c\n",
+      "E  =  2*p*Phi*n*Z/c\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   1200.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 355</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1 = 150; # in Volts\n",
+      "x = 0.2;\n",
+      "\n",
+      "#calculation:\n",
+      "E2  =  E1*(1- x)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Generated  e.m.f.  is  \",round(E2,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Generated  e.m.f.  is   120.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 356</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "n1  =  30;#  in  rev/sec\n",
+      "E1  =  200;#  in  Volts\n",
+      "n2  =  20;#  in  rev/sec\n",
+      "E2  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E  proportional  to  phi*w  and  since  w  =  2*pi*n,  then\n",
+      " #  E  proportional  to  phi*n\n",
+      " #  E1/E2  =  Phi1*n1/(Phi2*n2)\n",
+      " #  let  Phi2/Phi1  =  Phi\n",
+      "Phi  =  E2*n1/(E1*n2)\n",
+      "Phi_inc  =  (Phi  -  1)*100#/in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  percentage  increase  in  the  flux  per  pole  is  \",round(Phi_inc,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  percentage  increase  in  the  flux  per  pole  is   87.5   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.30;#  in  ohms\n",
+      "Ia  =  30;#  in  Amperes\n",
+      "E  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #terminal  voltage,\n",
+      " #V  =  E  -  Ia*Ra\n",
+      "V  =  E  -  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  terminal  voltage  of  a  generator  is  \",round(V,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  terminal  voltage  of  a  generator  is   191.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "RL  =  60;#  in  ohms\n",
+      "Ia  =  8;#  in  Amperes\n",
+      "Ra  =  1;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #terminal  voltage,\n",
+      " #V  =  Ia*RL\n",
+      "V  =  Ia*RL\n",
+      " #Generated  e.m.f.,  E\n",
+      "E  =  V  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)terminal  voltage  is  \",round(V,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)terminal  voltage  is   480.0   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   488.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 357</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "E1  =  150;#  in  Volts\n",
+      "n1  =  20;#  in  rev/sec\n",
+      "Phi1  =  0.10;#  in  Wb\n",
+      "n2  =  25;#  in  rev/sec\n",
+      "Phi2  =  0.10;#  in  Wb\n",
+      "n3  =  20;#  in  rev/sec\n",
+      "Phi3  =  0.08;#  in  Wb\n",
+      "n4  =  24;#  in  rev/sec\n",
+      "Phi4  =  0.07;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E  proportional  to  phi*w  and  since  w  =  2*pi*n,  then\n",
+      " #  E  proportional  to  phi*n\n",
+      " #  E1/E2  =  Phi1*n1/(Phi2*n2)\n",
+      "E2  =  E1*Phi2*n2/(Phi1*n1)\n",
+      "E3  =  E1*Phi3*n3/(Phi1*n1)\n",
+      "E4  =  E1*Phi4*n4/(Phi1*n1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  generated  e.m.f  is  \",round(E2,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E3,2),\"  V  \"\n",
+      "print  \"\\n  (c)generated  e.m.f.  is  \",round(E4,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  generated  e.m.f  is   187.5   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   120.0   V  \n",
+        "\n",
+        "  (c)generated  e.m.f.  is   126.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 359</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ps  =  20000;#  in  Watts\n",
+      "Vs  =  200;#  in  Volts\n",
+      "Rs  =  0.1;#  in  ohms\n",
+      "Rf  =  50;#  in  ohms\n",
+      "Ra  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Load  current,  I\n",
+      "Is  =  Ps/Vs\n",
+      " #Volt  drop  in  the  cables  to  the  load\n",
+      "Vd  =  Is*Rs\n",
+      " #Hence  terminal  voltage,\n",
+      "V  =  Vs  +  Vd\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  Is\n",
+      " #Generated  e.m.f.  E\n",
+      "E  =  V  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)terminal  voltage  is  \",round(V,2),\"  V  \"\n",
+      "print  \"\\n  (b)generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)terminal  voltage  is   210.0   V  \n",
+        "\n",
+        "  (b)generated  e.m.f.  is   214.17   V  "
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 361</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Is  =  80;#  in  amperes\n",
+      "Vs  =  200;#  in  Volts\n",
+      "Rf  =  40;#  in  ohms\n",
+      "Rse  =  0.02;#  in  ohms\n",
+      "Ra  =  0.04;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Volt  drop  in  series  winding\n",
+      "Vse  =  Is*Rse\n",
+      " #P.d.  across  the  field  winding  =  p.d.  across  armature\n",
+      "V1  =  Vs  +  Vse\n",
+      " #Field  current,  If\n",
+      "If  =  V1/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  Is\n",
+      " #Generated  e.m.f.  E\n",
+      "E  =  V1  +  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  generated  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  generated  e.m.f.  is   205.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 363</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ps  =  10000;#  in  Watt\n",
+      "Pl  =  600;#  in  Watt\n",
+      "Ra  =  0.75;#  in  ohms\n",
+      "Rf  =  125;#  in  ohms\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Output  power  Ps  =  V*I\n",
+      " #from  which,  load  current  I\n",
+      "I  =  Ps/V\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current\n",
+      "Ia  =  If  +  I\n",
+      " #Efficiency,\n",
+      "eff  =  Ps*100/((V*I)  +  (Ia*Ia*Ra)  +  (If*V)  +  (Pl))#  in  Percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Efficiency  is  \",round(eff,2),\"  percent  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Efficiency  is   80.5   percent  "
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 364</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.2;#  in  ohms\n",
+      "V  =  240;#  in  Volts\n",
+      "Ia  =  50;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  motor,  V  =  E  +  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  back  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  back  e.m.f.  is   230.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 365</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ra  =  0.25;#  in  ohms\n",
+      "V  =  300;#  in  Volts\n",
+      "Ig  =  100;#  in  Amperes\n",
+      "Im  =  80;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #As  a  generator,  generated  e.m.f.,\n",
+      " #  E  =  V  +  Ia*Ra\n",
+      "Eg  =  V  +  Ig*Ra\n",
+      " #For  a  motor,  generated  e.m.f.  (or  back  e.m.f.),\n",
+      " #  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Im*Ra\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)As  a  generator,  generated  e.m.f.  is  \",round(Eg,2),\"  V  \"\n",
+      "print  \"\\n  (b)back  e.m.f.  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)As  a  generator,  generated  e.m.f.  is   325.0   V  \n",
+        "\n",
+        "  (b)back  e.m.f.  is   280.0   V  "
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  4;\n",
+      "c  =  2;#  for  a  wave  winding\n",
+      "Phi  =  25E-3;#  Wb\n",
+      "Z  =  900;\n",
+      "Ia  =  30;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #torque  T  =  p*Phi*Z*Ia/(pi*c)\n",
+      "T  =  p*Phi*Z*Ia/(1*math.pi*c)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  torque  exerted  is   429.72   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  350;#  in  Volts\n",
+      "Ra  =  0.5;#  in  ohms\n",
+      "n  =  15;#  in  rev/sec\n",
+      "Ia  =  60;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Back  e.m.f.  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      " #torque  T  =  E*Ia/(2*n*pi)\n",
+      "T  =  E*Ia/(2*n*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  torque  exerted  is   203.72   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 366</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  1;#  let\n",
+      "c  =  2*p;#  for  a  lap  winding\n",
+      "Phi  =  20E-3;#  Wb\n",
+      "Z  =  500;\n",
+      "V  =  250;#  in  Volts\n",
+      "Ra  =  1;#  in  ohms\n",
+      "Ia  =  40;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Back  e.m.f.  E  =  V  -  Ia*Ra\n",
+      "E  =  V  -  Ia*Ra\n",
+      " #E.m.f.  E  =  2*p*Phi*n*Z/c\n",
+      " #  rearrange,\n",
+      "n  =  E*c/(2*p*Phi*Z)\n",
+      " #torque  T  =  E*Ia/(2*n*pi)\n",
+      "T  =  E*Ia/(2*n*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)speed  n  is  \",round(n,2),\"  rev/sec  \"\n",
+      "print  \"\\n  (b)the  torque  exerted  is  \",round(T,2),\"  Nm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)speed  n  is   21.0   rev/sec  \n",
+        "\n",
+        "  (b)the  torque  exerted  is   63.66   Nm  "
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 367</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T1  =  25;#  in  Nm\n",
+      "T2  =  35;#  in  Nm\n",
+      "Ia1  =  16;#  in  Amperes\n",
+      "V  =  100;#  in  Volts\n",
+      "x  =  0.15;\n",
+      "\n",
+      " #calculation:\n",
+      " #the  shaft  torque  T  of  a  generator  is  proportional  to  (phi*Ia),\n",
+      "    #where  Phi  is  the  flux  and  Ia  is  the  armature  current.  Thus,  T  =  k*Phi*Ia,  where  k  is  a  constant.\n",
+      " #The  torque  at  flux  phi1  and  armature  current  Ia1  is  T1  =  k*Phi1*Ia1.\n",
+      " #similarly  T2  =  k*Phi2*Ia2\n",
+      "\n",
+      "Ia2  =  T2*Ia1/(0.85*T1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  armature  current  at  the  new  value  of  torque  is  \",round(Ia2,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  armature  current  at  the  new  value  of  torque  is   26.35   A  "
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 367</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  12;#  in  Nm\n",
+      "I  =  15;#  in  Amperes\n",
+      "V  =  100;#  in  Volts\n",
+      "n  =  1500/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #the  efficiency  of  a  generator  =  (output  power/input  power)*100  %\n",
+      " #The  output  power  is  the  electrical  output,  i.e.  VI  watts.  \n",
+      "    #The  input  power  to  a  generator  is  the  mechanical  power  in  the  shaft  driving  the  generator, \n",
+      "    #i.e.  T*w  or  T(2*pi*n)  watts,  where  T  is  the  torque  in  Nm  and  n  is  speed  of  rotation  in  rev/s.  Hence,  for  a  generator  \n",
+      " #efficiency  =  V*I*100/(T*2*pi*n) %\n",
+      "eff  =  V*I*100/(T*2*math.pi*n)#  in    Percent\n",
+      " #The  input  power  =  output  power  +  losses\n",
+      " #  hence,  T*2*math.pi*n  =  V*I  +  losses\n",
+      "Pl  =  T*2*math.pi*n  -  V*I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  efficiency  is  \",round(eff,2),\" % \"\n",
+      "print  \"\\n  (b)  power  loss  is  \",round(Pl,2),\"  W  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  efficiency  is   79.58  % \n",
+        "\n",
+        "  (b)  power  loss  is   384.96   W  "
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 368</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rf  =  150;#  in  Ohms\n",
+      "Ra  =  0.4;#  in  Ohms\n",
+      "I  =  30;#  in  Amperes\n",
+      "V  =  240;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Field  current  If\n",
+      "If  =  V/Rf\n",
+      " #Supply  current  I  =  Ia  +  If\n",
+      " #Hence  armature  current,  Ia\n",
+      "Ia  =  I  -  If\n",
+      " #Back  e.m.f.  E  =  V  -\u0004  Ia*Ra\n",
+      "E  =  V  -  (Ia*Ra)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)    current  in  the  armature  is  \",round(Ia,2),\"  A  \"\n",
+      "print  \"\\n  (b)  Back  e.m.f.  E  is  \",round(E,2),\"  V  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)    current  in  the  armature  is   28.4   A  \n",
+        "\n",
+        "  (b)  Back  e.m.f.  E  is   228.64   V  "
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 370</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  30;#  in  Amperes\n",
+      "Ia2  =  45;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "n1  =  1350/60;#  in  Rev/sec\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #The  relationship  E  proportional  to  (Phi*n)  applies  to  both  generators  and  motors.  For  a  motor,\n",
+      " #E  =  V  -  (Ia*Ra)\n",
+      "E1  =    V  -  (Ia1*Ra)\n",
+      "E2  =    V  -  (Ia2*Ra)\n",
+      " #The  relationship,  E1/E2  =  Phi1*n1/Phi2*n2,    applies  to  both  generators  and  motors.\n",
+      "    #Since  the  flux  is  constant,  Phi1  =  Phi2\n",
+      "n2  =  E2*n1/(E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  speed  of  the  motor  is  \",round(n2,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  speed  of  the  motor  is   21.78   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 370</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  30;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "n  =  800/60;#  in  Rev/sec\n",
+      "V  =  220;#  in  Volts\n",
+      "x=  0.1;\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  d.c.  shunt-wound  motor,  E  =  V  -  (Ia*Ra),Hence  initial  generated  e.m.f.,\n",
+      "E1  =    V  -  (Ia1*Ra)\n",
+      " #The  generated  e.m.f.  is  also  such  that  E  proportional  to  (Phi*n)  \n",
+      "    #so  at  the  instant  the  flux  is  reduced,  the  speed  has  not  had  time  to  change,  and\n",
+      "E  =  E1*(1-x)\n",
+      " #Hence,  the  voltage  drop  due  to  the  armature  resistance  is\n",
+      "Vd  =  V  -  E\n",
+      " #The  instantaneous  value  of  the  current  is\n",
+      "Ia  =  Vd/Ra\n",
+      " #T  proportional  to  (Phi*Ia),  since  the  torque  is  constant,\n",
+      " #Phi1*Ia1  =  Phi2*Ia2,    The  flux  8  is  reduced  by  10%,  hence\n",
+      "Ia2  =  Ia1/0.9\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)instantaneous  value  of  the  current  \",round(Ia,2),\"  A  \"\n",
+      "print  \"\\n  (b)steady  state  value  of  armature  current,  \",round(Ia2,2),\"  A  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)instantaneous  value  of  the  current   82.0   A  \n",
+        "\n",
+        "  (b)steady  state  value  of  armature  current,   33.33   A  "
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 372</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  15;#  in  Amperes\n",
+      "Ia2  =  30;#  in  Amperes\n",
+      "Rf  =  0.3;#  in  ohms\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n1  =  24;#  in  Rev/sec\n",
+      "V  =  240;#  in  Volts\n",
+      "x=  2;\n",
+      "\n",
+      "#calculation:\n",
+      " #generated  e.m.f.,  E,  at  initial  load,  is  given  by\n",
+      "E1  =    V  -  Ia1*(Ra  +  Rf)\n",
+      " #When  the  current  is  increased  to  30  A,  the  generated  e.m.f.  is  given  by:\n",
+      "E2  =    V  -  Ia2*(Ra  +  Rf)\n",
+      " #E  proportional  to  (Phi*n)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  =  E2*n1/(2*E1)  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)generated  e.m.f.,  E  is  \",round(E1,2),\"  V  \"\n",
+      "print  \"\\n  (b)speed  of  motor,  n2,  \",round(n2,2),\"  rev/sec  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)generated  e.m.f.,  E  is   232.5   V  \n",
+        "\n",
+        "  (b)speed  of  motor,  n2,   11.61   rev/sec  "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 374</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  80;#  in  Amperes\n",
+      "C  =  1500;#  in  Watt\n",
+      "Rf  =  40;#  in  ohms\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n  =  1000/60;#  in  Rev/sec\n",
+      "V  =  320;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Field  current,  If\n",
+      "If  =  V/Rf\n",
+      " #Armature  current  Ia\n",
+      "Ia  =  I  -  If\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (Ia*Ia*Ra) - (If*V) - C)/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 80.09 %"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 25, page no. 374</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "I  =  40;#  in  Amperes\n",
+      "Rf  =  0.05;#  in  ohms\n",
+      "Ra  =  0.15;#  in  ohm\n",
+      "V  =  250;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #However  for  a  series  motor,  If  =  0  and  the  Ia*Ia*Ra  loss  needs  to  be  I*I*(\u0011Ra  +  Rf)\n",
+      " #For  maximum  efficiency  I*I*\u0011(Ra  +  Rf)  =  C\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 93.6"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 26, page no. 375</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "T  =  15;#  in  Nm\n",
+      "n  =  1200/60;#  in  rev/sec\n",
+      "eff  =  0.8;\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      "I = T*2*math.pi*n/(V*eff)\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n current supplied, I is \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " current supplied, I is  11.78 A"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 27, page no. 376</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R  =  2;#  in  ohm\n",
+      "n  =  30;#  in  rev/sec\n",
+      "I  =  10;#  in  A\n",
+      "C  =  300;#  in  Watt\n",
+      "V  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Efficiency\n",
+      "eff = ((V*I - (I*I*R) - C)/(V*I))*100 # in percent\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n Result \\n\\n\"\n",
+      "print \"\\n efficiency is\",round(eff,2),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Result \n",
+        "\n",
+        "\n",
+        "\n",
+        " efficiency is 87.5 %"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 28, page no. 378</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  120;#  in  A\n",
+      "Ia2  =  60;#  in  A\n",
+      "Ra  =  0.2;#  in  ohm\n",
+      "n1  =  10;#  in  rev/sec\n",
+      "R  =  0.5;#  in  ohm\n",
+      "x  =  0.8;\n",
+      "V  =  500;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #back  e.m.f.  at  Ia1\n",
+      "E1  =  V  -  Ia1*Ra\n",
+      " #at  Ia2\n",
+      "E2  =  V  -  Ia2*(Ra  +  R)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  = n1*E2/E1\n",
+      " #Back  e.m.f.  when  Ia2\n",
+      "E3  =  V  -  Ia2*Ra\n",
+      "n3  =  n1*E3/(x*E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)speed  n2  is  \",round(n2,2),\"  rev/sec\"\n",
+      "print  \"\\n  (b)speed  n3  is  \",round(n3,2),\"  rev/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)speed  n2  is   9.62   rev/sec\n",
+        "\n",
+        "  (b)speed  n3  is   12.82   rev/sec"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 29, page no. 379</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  90;#  in  Amperes\n",
+      "Ra  =  0.1;#  in  ohm\n",
+      "Rse  =  0.05;#  in  ohm\n",
+      "Rd  =  0.2;#  in  Ohm\n",
+      "n1  =  15;#  in  rev/sec\n",
+      "V  =  300;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1\n",
+      "E1  =  V  -  Ia1*(Ra  +  Rse)\n",
+      " #With  the  Rd  diverter  in  parallel  with  Rse\n",
+      " #equivalent  resistance,  Re\n",
+      "Re  =  Rd*Rse/(Rd+Rse)\n",
+      " #Torque,  T  proprtional  to  Ia*Phi  and  for  full  load  torque,  Ia1*Phi1  =  Ia2*Phi2\n",
+      " #Since  flux  is  proportional  to  field  current  Phi1  proportional  to  Ta1  and  Phi2  Proportional  to  I1\n",
+      "I1  =    (Ia1*Ia1*0.8)**0.5\n",
+      " #By  current  division,  current  I1\n",
+      "Ia2  =  I1/(Rd/(Rd  +  Rse))\n",
+      " #Hence  e.m.f.  E2\n",
+      "E2  =  V  -  Ia2*(Ra  +  Re)\n",
+      " #E1/E2  =  Phi1*n1/Phi2*n2\n",
+      "n2  =  E2*Ia1*n1/(I1*E1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  speed  n2  is  \",round(n2,2),\"  rev/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  speed  n2  is   16.74   rev/sec"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 30, page no. 380</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  25;#  in  Amperes\n",
+      "Ra  =  0.4;#  in  ohm\n",
+      "Rse  =  0.2;#  in  ohm\n",
+      "n1  =  800/60;#  in  rev/sec\n",
+      "n2  =  600/60;#  in  rev/sec\n",
+      "V  =  400;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #e.m.f.  E1\n",
+      "E1  =  V  -  Ia1*(Ra  +  Rse)\n",
+      " #At  n2,  since  the  current  is  unchanged,  the  flux  is  unchanged.\n",
+      " #E1/E2  =  n1/n2\n",
+      "E2  =  E1*n2/n1\n",
+      " #and  E2  =  V  -  Ia1(\u0011Ra  +  Rse  +  R)\n",
+      "R  =  (V  -  E2)/Ia1  -  Ra  -  Rse\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Resistance  is  \",round(R,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Resistance  is   3.85   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb
new file mode 100755
index 00000000..4bad2773
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb
@@ -0,0 +1,791 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 22: Three-phase induction motors</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 389</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the synchronous speed of the motor in rev/min.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  Hz\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  \n",
+      "    #f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      "nsrpm  =  ns*60\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nsynchronous  speed  of  the  motor  is  \",nsrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "synchronous  speed  of  the  motor  is   3000.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 389</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the number of poles.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  60;#  in  Hz\n",
+      "ns  =  900/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "p  =  f/ns\n",
+      "np  =  p*2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nnumber  of  poles  is  \", round(np,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "number  of  poles  is   8.0"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 390</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the frequency of the supply voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "ns  =  6000/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"frequency  is  \",f,\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "frequency  is   100.0   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 391</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the synchronous speed and (b) the slip at full load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  4/2;#  number  of  pairs  of  poles\n",
+      "f  =  50;#  in  Hz\n",
+      "nr  =  1455/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  synchronous  speed  is  \",ns,\"  rev/sec\"\n",
+      "print  \"\\n(b)  slip  is  \",s,\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  synchronous  speed  is   25.0   rev/sec\n",
+        "\n",
+        "(b)  slip  is   3.0   percent"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 392</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the synchronous speed,\n",
+      "#(b) the speed of the rotor and (c) the frequency of the induced e.m.f.\u2019s in the rotor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "f  =  60;#  in  Hz\n",
+      "s  =  0.02;#  slip\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  the  rotor  runs  at\n",
+      "nr  =  ns*(1  -  s)\n",
+      " #frequency  of  the  e.m.f.  induced  in  the  rotor  bars  is\n",
+      "fr  =  ns  -  nr\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  synchronous  speed  is  \",ns,\"  rev/sec\"\n",
+      "print  \"\\n(b)  rotor  speed  is  \",nr,\"  rev/sec\"\n",
+      "print  \"\\n(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is  \",fr,\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  synchronous  speed  is   60.0   rev/sec\n",
+        "\n",
+        "(b)  rotor  speed  is   58.8   rev/sec\n",
+        "\n",
+        "(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is   1.2   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 392</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the synchronous speed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  Hz\n",
+      "nr  =  1200/60;#  in  rev/min\n",
+      "s  =  0.04;#  slip\n",
+      "\n",
+      "#calculation:\n",
+      " #the  synchronous  speed.\n",
+      "ns  =  nr/(1  -  s)\n",
+      "nsrpm  =  ns*60\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  synchronous  speed  is  \",nsrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  synchronous  speed  is   1250.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 394</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the slip, and (b) the rotor speed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  8/2;#  number  of  pairs  of  poles\n",
+      "f  =  50;#  in  Hz\n",
+      "fr  =  3;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #fr  =  s*f\n",
+      "s  =  (fr/f)\n",
+      " #the  rotor  speed.\n",
+      "nr  =  ns*(1  -  s)\n",
+      "nrrpm  =  nr*60\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  slip  is  \",s*100,\"  percent\"\n",
+      "print  \"\\n  (b)  rotor  speed  is  \",nrrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  slip  is   6.0   percent\n",
+        "\n",
+        "  (b)  rotor  speed  is   705.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 396</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the rotor copper loss, \n",
+      "#(b) the total mechanical power developed by the rotor,\n",
+      "#(c) the output power of the motor if friction and windage losses are 750 W, and \n",
+      "#(d) the efficiency of the motor, neglecting rotor iron loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Psi  =  32000;#  in  Watts\n",
+      "Psl  =  1200;#  in  Watts\n",
+      "s  =  0.05;#  slip\n",
+      "Pfl  =  750;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Input  power  to  rotor  =  stator  input  power  -\u0006  stator  losses\n",
+      "Pi  =    Psi  -  Psl\n",
+      " #slip  =  rotor  copper  loss/rotor  input\n",
+      "Pl  =  s*Pi\n",
+      " #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -\u0006  rotor  losses\n",
+      "Pr  =  Pi  -  Pl\n",
+      " #Output  power  of  motor  =  power  developed  by  the  rotor  -\u0006  friction  and  windage  losses\n",
+      "Po  =  Pr  -  Pfl\n",
+      " #Efficiency  of  induction  motor  =  (output  power/input  power)*100\n",
+      "eff  =  (Po/Psi)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  rotor  copper  loss  is  \",Pl,\"  Watt\"\n",
+      "print  \"\\n(b)  Total  mechanical  power  developed  by  the  rotor  is  \",Pr,\"  W\"\n",
+      "print  \"\\n(c)  Output  power  of  motor  is  \",Po,\"  Watt\"\n",
+      "print  \"\\n(d)  efficiency  of  induction  motor  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  rotor  copper  loss  is   1540.0   Watt\n",
+        "\n",
+        "(b)  Total  mechanical  power  developed  by  the  rotor  is   29260.0   W\n",
+        "\n",
+        "(c)  Output  power  of  motor  is   28510.0   Watt\n",
+        "\n",
+        "(d)  efficiency  of  induction  motor  is   89.09   percent"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 397</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the rotor copper loss, and (b) the efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Psi  =  32000;#  in  Watts\n",
+      "Psl  =  1200;#  in  Watts\n",
+      "Pfl  =  750;#  in  Watts\n",
+      "x  =  0.35;\n",
+      "\n",
+      "#calculation:\n",
+      " #The  slip,  s\n",
+      "s  =  1-x\n",
+      " #Input  power  to  rotor  =  stator  input  power  -  stator  losses\n",
+      "Pi  =    Psi  -  Psl\n",
+      " #slip  =  rotor  copper  loss/rotor  input\n",
+      "Pl  =  s*Pi\n",
+      " #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -\u0006  rotor  losses\n",
+      "Pr  =  Pi  -  Pl\n",
+      " #Output  power  of  motor  =  power  developed  by  the  rotor  -\u0006  friction  and  windage  losses\n",
+      "Po  =  Pr  -  Pfl\n",
+      " #Efficiency  of  induction  motor  =  (output  power/input  power)*100\n",
+      "eff  =  (Po/Psi)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  rotor  copper  loss  is  \",Pl,\"  Watt\"\n",
+      "print  \"\\n(b)  efficiency  of  induction  motor  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  rotor  copper  loss  is   20020.0   Watt\n",
+        "\n",
+        "(b)  efficiency  of  induction  motor  is   31.34   percent"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 398</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the synchronous speed, (b) the slip, \n",
+      "#(c) the full load torque, (d) the power output if mechanical losses amount to 770 W, \n",
+      "#(e) the maximum torque, (f) the speed at which maximum torque occurs,\n",
+      "#and (g) the starting torque.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "tr  =  0.85;#  turn  ratio  N2/N1\n",
+      "Pl  =  770;#  in  Watt\n",
+      "m  =  3;#  no.  of  phases\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      " #Phase  voltage,  E1  =  V/(3**0.5)\n",
+      "E1  =  V/(3**0.5)\n",
+      " #Full  load  torque\n",
+      "T  =  (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2  +  (X2*(s/100))**2))\n",
+      " #Output  power,  including  friction  losses\n",
+      "Pm  =  2*math.pi*nr*T\n",
+      " #power  output\n",
+      "Po  =  Pm  -  Pl\n",
+      " #Maximum  torque  occurs  when  R2  =  Xr  =  0.35  ohm\n",
+      " #Slip  \n",
+      "sm  =  R2/X2\n",
+      " #maximum  torque,  Tm  \n",
+      "Tm  =  (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2  +  (X2*sm)**2))\n",
+      " #speed  at  which  maximum  torque  occurs\n",
+      "nrm  =  ns*(1  -  sm)\n",
+      "nrmrpm  =  nrm*60\n",
+      " #At  the  start,  i.e.,  at  standstill,  slip,  s=1\n",
+      "ss  =  1\n",
+      " #starting  torque\n",
+      "Ts  =  (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2  +  (X2*ss)**2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)Synchronous  speed  is  \",round(ns,2),\"  rev/sec\"\n",
+      "print  \"\\n(b)Slip  is  \",round(s,2),\"  percent\"\n",
+      "print  \"\\n(c)Full  load  torque  is  \",round(T,2),\"  Nm\"\n",
+      "print  \"\\n(d)power  output  is  \",round(Po,2),\"W\"\n",
+      "print  \"\\n(e)maximum  torque  is  \",round(Tm,2),\"  Nm\"\n",
+      "print  \"\\n(f)speed  at  which  maximum  torque  occurs  is  \",round(nrmrpm,2),\"rev/min\"\n",
+      "print  \"\\n(g)starting  torque  is  \",round(Ts,2),\"  Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)Synchronous  speed  is   25.0   rev/sec\n",
+        "\n",
+        "(b)Slip  is   4.0   percent\n",
+        "\n",
+        "(c)Full  load  torque  is   78.05   Nm\n",
+        "\n",
+        "(d)power  output  is   10998.99 W\n",
+        "\n",
+        "(e)maximum  torque  is   113.17   Nm\n",
+        "\n",
+        "(f)speed  at  which  maximum  torque  occurs  is   1350.0 rev/min\n",
+        "\n",
+        "(g)starting  torque  is   22.41   Nm"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 400</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the induction motor in problem 10 at full load, \n",
+      "#(a) the rotor current, (b) the rotor copper loss, and (c) the starting current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "tr  =  0.85;#  turn  ratio  N2/N1\n",
+      "m  =  3;#  no.  of  phases\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      " #Phase  voltage,  E1  =  V/(3**0.5)\n",
+      "E1  =  V/(3**0.5)\n",
+      " #rotor  current,\n",
+      "Ir  =  (s/100)*E1*tr/((R2**2  +  (X2*(s/100))**2)**0.5)\n",
+      " #Rotor  copper  loss  \n",
+      "Pcl  =  m*R2*(Ir**2)\n",
+      " #starting  current,\n",
+      "ss  =1\n",
+      "I2  =  ss*tr*E1/((R2**2  +  (X2*ss)**2)**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)rotor  current  is  \",round(Ir,2),\"  A\"\n",
+      "print  \"\\n(b)Total  copper  loss  is  \",round(Pcl,2),\"  W\"\n",
+      "print  \"\\n(c)starting  current  is  \",round(I2,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)rotor  current  is   21.61   A\n",
+        "\n",
+        "(b)Total  copper  loss  is   490.37   W\n",
+        "\n",
+        "(c)starting  current  is   57.9   A"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 401</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the power input at full load, \n",
+      "#(b) the efficiency of the motor at full load and \n",
+      "#(c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "Psl  =  650;#  in  Watt\n",
+      "pf  =  0.87;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "Pm  =  11770;#  watts  from  part  (d),  Problem  22.10\n",
+      "Pcl  =  490.35;#  watts,  Rotor  copper  loss,  from  part  (b),  Problem  22.11\n",
+      " #Stator  input  power\n",
+      "P1  =  Pm  +  Pcl  +  Psl\n",
+      "Po  =  11000#  watts,  Net  power  output,  from  part  (d),  Problem  22.10\n",
+      " #efficiency  =  (output/input)  *100\n",
+      "eff  =  (Po/P1)*100#  in  percent\n",
+      " #Power  input,  P1  =  (3**0.5)*VL*IL*cos(phi)\u000e\n",
+      " #  pf  =  cos(phi)\n",
+      " #supply  current,  IL\n",
+      "I  =  P1/((3**0.5)*V*pf)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(aStator  input  power  is  \",round(P1,2),\" W\"\n",
+      "print  \"\\n(b)efficiency  is  \",round(eff,2),\"  percent\"\n",
+      "print  \"\\n(c)supply  current  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(aStator  input  power  is   12910.35  W\n",
+        "\n",
+        "(b)efficiency  is   85.2   percent\n",
+        "\n",
+        "(c)supply  current  is   20.64   A"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 401</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the resistance of the rotor winding required for maximum starting torque.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #At  the  moment  of  starting,  slip,  \n",
+      "s  =  1\n",
+      " #Maximum  torque  occurs  when  rotor  reactance  equals  rotor  resistance\n",
+      " #for  maximum  torque\n",
+      "R2  =  s*X2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nresistance  of  the  rotor  is  \",R2,\"  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "resistance  of  the  rotor  is   3.5   Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_1.ipynb
new file mode 100755
index 00000000..4bad2773
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_1.ipynb
@@ -0,0 +1,791 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 22: Three-phase induction motors</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 389</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the synchronous speed of the motor in rev/min.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  Hz\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  \n",
+      "    #f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      "nsrpm  =  ns*60\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nsynchronous  speed  of  the  motor  is  \",nsrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "synchronous  speed  of  the  motor  is   3000.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 389</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the number of poles.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  60;#  in  Hz\n",
+      "ns  =  900/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "p  =  f/ns\n",
+      "np  =  p*2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nnumber  of  poles  is  \", round(np,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "number  of  poles  is   8.0"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 390</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the frequency of the supply voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "ns  =  6000/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"frequency  is  \",f,\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "frequency  is   100.0   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 391</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the synchronous speed and (b) the slip at full load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  4/2;#  number  of  pairs  of  poles\n",
+      "f  =  50;#  in  Hz\n",
+      "nr  =  1455/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  synchronous  speed  is  \",ns,\"  rev/sec\"\n",
+      "print  \"\\n(b)  slip  is  \",s,\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  synchronous  speed  is   25.0   rev/sec\n",
+        "\n",
+        "(b)  slip  is   3.0   percent"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 392</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the synchronous speed,\n",
+      "#(b) the speed of the rotor and (c) the frequency of the induced e.m.f.\u2019s in the rotor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "f  =  60;#  in  Hz\n",
+      "s  =  0.02;#  slip\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  the  rotor  runs  at\n",
+      "nr  =  ns*(1  -  s)\n",
+      " #frequency  of  the  e.m.f.  induced  in  the  rotor  bars  is\n",
+      "fr  =  ns  -  nr\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  synchronous  speed  is  \",ns,\"  rev/sec\"\n",
+      "print  \"\\n(b)  rotor  speed  is  \",nr,\"  rev/sec\"\n",
+      "print  \"\\n(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is  \",fr,\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  synchronous  speed  is   60.0   rev/sec\n",
+        "\n",
+        "(b)  rotor  speed  is   58.8   rev/sec\n",
+        "\n",
+        "(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is   1.2   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 392</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the synchronous speed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  Hz\n",
+      "nr  =  1200/60;#  in  rev/min\n",
+      "s  =  0.04;#  slip\n",
+      "\n",
+      "#calculation:\n",
+      " #the  synchronous  speed.\n",
+      "ns  =  nr/(1  -  s)\n",
+      "nsrpm  =  ns*60\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  synchronous  speed  is  \",nsrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  synchronous  speed  is   1250.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 394</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the slip, and (b) the rotor speed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  8/2;#  number  of  pairs  of  poles\n",
+      "f  =  50;#  in  Hz\n",
+      "fr  =  3;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #fr  =  s*f\n",
+      "s  =  (fr/f)\n",
+      " #the  rotor  speed.\n",
+      "nr  =  ns*(1  -  s)\n",
+      "nrrpm  =  nr*60\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  slip  is  \",s*100,\"  percent\"\n",
+      "print  \"\\n  (b)  rotor  speed  is  \",nrrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  slip  is   6.0   percent\n",
+        "\n",
+        "  (b)  rotor  speed  is   705.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 396</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the rotor copper loss, \n",
+      "#(b) the total mechanical power developed by the rotor,\n",
+      "#(c) the output power of the motor if friction and windage losses are 750 W, and \n",
+      "#(d) the efficiency of the motor, neglecting rotor iron loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Psi  =  32000;#  in  Watts\n",
+      "Psl  =  1200;#  in  Watts\n",
+      "s  =  0.05;#  slip\n",
+      "Pfl  =  750;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Input  power  to  rotor  =  stator  input  power  -\u0006  stator  losses\n",
+      "Pi  =    Psi  -  Psl\n",
+      " #slip  =  rotor  copper  loss/rotor  input\n",
+      "Pl  =  s*Pi\n",
+      " #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -\u0006  rotor  losses\n",
+      "Pr  =  Pi  -  Pl\n",
+      " #Output  power  of  motor  =  power  developed  by  the  rotor  -\u0006  friction  and  windage  losses\n",
+      "Po  =  Pr  -  Pfl\n",
+      " #Efficiency  of  induction  motor  =  (output  power/input  power)*100\n",
+      "eff  =  (Po/Psi)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  rotor  copper  loss  is  \",Pl,\"  Watt\"\n",
+      "print  \"\\n(b)  Total  mechanical  power  developed  by  the  rotor  is  \",Pr,\"  W\"\n",
+      "print  \"\\n(c)  Output  power  of  motor  is  \",Po,\"  Watt\"\n",
+      "print  \"\\n(d)  efficiency  of  induction  motor  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  rotor  copper  loss  is   1540.0   Watt\n",
+        "\n",
+        "(b)  Total  mechanical  power  developed  by  the  rotor  is   29260.0   W\n",
+        "\n",
+        "(c)  Output  power  of  motor  is   28510.0   Watt\n",
+        "\n",
+        "(d)  efficiency  of  induction  motor  is   89.09   percent"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 397</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the rotor copper loss, and (b) the efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Psi  =  32000;#  in  Watts\n",
+      "Psl  =  1200;#  in  Watts\n",
+      "Pfl  =  750;#  in  Watts\n",
+      "x  =  0.35;\n",
+      "\n",
+      "#calculation:\n",
+      " #The  slip,  s\n",
+      "s  =  1-x\n",
+      " #Input  power  to  rotor  =  stator  input  power  -  stator  losses\n",
+      "Pi  =    Psi  -  Psl\n",
+      " #slip  =  rotor  copper  loss/rotor  input\n",
+      "Pl  =  s*Pi\n",
+      " #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -\u0006  rotor  losses\n",
+      "Pr  =  Pi  -  Pl\n",
+      " #Output  power  of  motor  =  power  developed  by  the  rotor  -\u0006  friction  and  windage  losses\n",
+      "Po  =  Pr  -  Pfl\n",
+      " #Efficiency  of  induction  motor  =  (output  power/input  power)*100\n",
+      "eff  =  (Po/Psi)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  rotor  copper  loss  is  \",Pl,\"  Watt\"\n",
+      "print  \"\\n(b)  efficiency  of  induction  motor  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  rotor  copper  loss  is   20020.0   Watt\n",
+        "\n",
+        "(b)  efficiency  of  induction  motor  is   31.34   percent"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 398</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the synchronous speed, (b) the slip, \n",
+      "#(c) the full load torque, (d) the power output if mechanical losses amount to 770 W, \n",
+      "#(e) the maximum torque, (f) the speed at which maximum torque occurs,\n",
+      "#and (g) the starting torque.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "tr  =  0.85;#  turn  ratio  N2/N1\n",
+      "Pl  =  770;#  in  Watt\n",
+      "m  =  3;#  no.  of  phases\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      " #Phase  voltage,  E1  =  V/(3**0.5)\n",
+      "E1  =  V/(3**0.5)\n",
+      " #Full  load  torque\n",
+      "T  =  (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2  +  (X2*(s/100))**2))\n",
+      " #Output  power,  including  friction  losses\n",
+      "Pm  =  2*math.pi*nr*T\n",
+      " #power  output\n",
+      "Po  =  Pm  -  Pl\n",
+      " #Maximum  torque  occurs  when  R2  =  Xr  =  0.35  ohm\n",
+      " #Slip  \n",
+      "sm  =  R2/X2\n",
+      " #maximum  torque,  Tm  \n",
+      "Tm  =  (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2  +  (X2*sm)**2))\n",
+      " #speed  at  which  maximum  torque  occurs\n",
+      "nrm  =  ns*(1  -  sm)\n",
+      "nrmrpm  =  nrm*60\n",
+      " #At  the  start,  i.e.,  at  standstill,  slip,  s=1\n",
+      "ss  =  1\n",
+      " #starting  torque\n",
+      "Ts  =  (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2  +  (X2*ss)**2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)Synchronous  speed  is  \",round(ns,2),\"  rev/sec\"\n",
+      "print  \"\\n(b)Slip  is  \",round(s,2),\"  percent\"\n",
+      "print  \"\\n(c)Full  load  torque  is  \",round(T,2),\"  Nm\"\n",
+      "print  \"\\n(d)power  output  is  \",round(Po,2),\"W\"\n",
+      "print  \"\\n(e)maximum  torque  is  \",round(Tm,2),\"  Nm\"\n",
+      "print  \"\\n(f)speed  at  which  maximum  torque  occurs  is  \",round(nrmrpm,2),\"rev/min\"\n",
+      "print  \"\\n(g)starting  torque  is  \",round(Ts,2),\"  Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)Synchronous  speed  is   25.0   rev/sec\n",
+        "\n",
+        "(b)Slip  is   4.0   percent\n",
+        "\n",
+        "(c)Full  load  torque  is   78.05   Nm\n",
+        "\n",
+        "(d)power  output  is   10998.99 W\n",
+        "\n",
+        "(e)maximum  torque  is   113.17   Nm\n",
+        "\n",
+        "(f)speed  at  which  maximum  torque  occurs  is   1350.0 rev/min\n",
+        "\n",
+        "(g)starting  torque  is   22.41   Nm"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 400</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the induction motor in problem 10 at full load, \n",
+      "#(a) the rotor current, (b) the rotor copper loss, and (c) the starting current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "tr  =  0.85;#  turn  ratio  N2/N1\n",
+      "m  =  3;#  no.  of  phases\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      " #Phase  voltage,  E1  =  V/(3**0.5)\n",
+      "E1  =  V/(3**0.5)\n",
+      " #rotor  current,\n",
+      "Ir  =  (s/100)*E1*tr/((R2**2  +  (X2*(s/100))**2)**0.5)\n",
+      " #Rotor  copper  loss  \n",
+      "Pcl  =  m*R2*(Ir**2)\n",
+      " #starting  current,\n",
+      "ss  =1\n",
+      "I2  =  ss*tr*E1/((R2**2  +  (X2*ss)**2)**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)rotor  current  is  \",round(Ir,2),\"  A\"\n",
+      "print  \"\\n(b)Total  copper  loss  is  \",round(Pcl,2),\"  W\"\n",
+      "print  \"\\n(c)starting  current  is  \",round(I2,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)rotor  current  is   21.61   A\n",
+        "\n",
+        "(b)Total  copper  loss  is   490.37   W\n",
+        "\n",
+        "(c)starting  current  is   57.9   A"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 401</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the power input at full load, \n",
+      "#(b) the efficiency of the motor at full load and \n",
+      "#(c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "Psl  =  650;#  in  Watt\n",
+      "pf  =  0.87;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "Pm  =  11770;#  watts  from  part  (d),  Problem  22.10\n",
+      "Pcl  =  490.35;#  watts,  Rotor  copper  loss,  from  part  (b),  Problem  22.11\n",
+      " #Stator  input  power\n",
+      "P1  =  Pm  +  Pcl  +  Psl\n",
+      "Po  =  11000#  watts,  Net  power  output,  from  part  (d),  Problem  22.10\n",
+      " #efficiency  =  (output/input)  *100\n",
+      "eff  =  (Po/P1)*100#  in  percent\n",
+      " #Power  input,  P1  =  (3**0.5)*VL*IL*cos(phi)\u000e\n",
+      " #  pf  =  cos(phi)\n",
+      " #supply  current,  IL\n",
+      "I  =  P1/((3**0.5)*V*pf)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(aStator  input  power  is  \",round(P1,2),\" W\"\n",
+      "print  \"\\n(b)efficiency  is  \",round(eff,2),\"  percent\"\n",
+      "print  \"\\n(c)supply  current  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(aStator  input  power  is   12910.35  W\n",
+        "\n",
+        "(b)efficiency  is   85.2   percent\n",
+        "\n",
+        "(c)supply  current  is   20.64   A"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 401</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the resistance of the rotor winding required for maximum starting torque.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #At  the  moment  of  starting,  slip,  \n",
+      "s  =  1\n",
+      " #Maximum  torque  occurs  when  rotor  reactance  equals  rotor  resistance\n",
+      " #for  maximum  torque\n",
+      "R2  =  s*X2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nresistance  of  the  rotor  is  \",R2,\"  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "resistance  of  the  rotor  is   3.5   Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_2.ipynb
new file mode 100755
index 00000000..4bad2773
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_2.ipynb
@@ -0,0 +1,791 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 22: Three-phase induction motors</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 389</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the synchronous speed of the motor in rev/min.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  Hz\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  \n",
+      "    #f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      "nsrpm  =  ns*60\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nsynchronous  speed  of  the  motor  is  \",nsrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "synchronous  speed  of  the  motor  is   3000.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 389</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the number of poles.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  60;#  in  Hz\n",
+      "ns  =  900/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "p  =  f/ns\n",
+      "np  =  p*2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nnumber  of  poles  is  \", round(np,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "number  of  poles  is   8.0"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 390</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the frequency of the supply voltage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "ns  =  6000/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"frequency  is  \",f,\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "frequency  is   100.0   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 391</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the synchronous speed and (b) the slip at full load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  4/2;#  number  of  pairs  of  poles\n",
+      "f  =  50;#  in  Hz\n",
+      "nr  =  1455/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  synchronous  speed  is  \",ns,\"  rev/sec\"\n",
+      "print  \"\\n(b)  slip  is  \",s,\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  synchronous  speed  is   25.0   rev/sec\n",
+        "\n",
+        "(b)  slip  is   3.0   percent"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 392</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the synchronous speed,\n",
+      "#(b) the speed of the rotor and (c) the frequency of the induced e.m.f.\u2019s in the rotor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "f  =  60;#  in  Hz\n",
+      "s  =  0.02;#  slip\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  the  rotor  runs  at\n",
+      "nr  =  ns*(1  -  s)\n",
+      " #frequency  of  the  e.m.f.  induced  in  the  rotor  bars  is\n",
+      "fr  =  ns  -  nr\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  synchronous  speed  is  \",ns,\"  rev/sec\"\n",
+      "print  \"\\n(b)  rotor  speed  is  \",nr,\"  rev/sec\"\n",
+      "print  \"\\n(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is  \",fr,\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  synchronous  speed  is   60.0   rev/sec\n",
+        "\n",
+        "(b)  rotor  speed  is   58.8   rev/sec\n",
+        "\n",
+        "(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is   1.2   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 392</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the synchronous speed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  Hz\n",
+      "nr  =  1200/60;#  in  rev/min\n",
+      "s  =  0.04;#  slip\n",
+      "\n",
+      "#calculation:\n",
+      " #the  synchronous  speed.\n",
+      "ns  =  nr/(1  -  s)\n",
+      "nsrpm  =  ns*60\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  synchronous  speed  is  \",nsrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  synchronous  speed  is   1250.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 394</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the slip, and (b) the rotor speed.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  8/2;#  number  of  pairs  of  poles\n",
+      "f  =  50;#  in  Hz\n",
+      "fr  =  3;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #fr  =  s*f\n",
+      "s  =  (fr/f)\n",
+      " #the  rotor  speed.\n",
+      "nr  =  ns*(1  -  s)\n",
+      "nrrpm  =  nr*60\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  slip  is  \",s*100,\"  percent\"\n",
+      "print  \"\\n  (b)  rotor  speed  is  \",nrrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  slip  is   6.0   percent\n",
+        "\n",
+        "  (b)  rotor  speed  is   705.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 396</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the rotor copper loss, \n",
+      "#(b) the total mechanical power developed by the rotor,\n",
+      "#(c) the output power of the motor if friction and windage losses are 750 W, and \n",
+      "#(d) the efficiency of the motor, neglecting rotor iron loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Psi  =  32000;#  in  Watts\n",
+      "Psl  =  1200;#  in  Watts\n",
+      "s  =  0.05;#  slip\n",
+      "Pfl  =  750;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Input  power  to  rotor  =  stator  input  power  -\u0006  stator  losses\n",
+      "Pi  =    Psi  -  Psl\n",
+      " #slip  =  rotor  copper  loss/rotor  input\n",
+      "Pl  =  s*Pi\n",
+      " #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -\u0006  rotor  losses\n",
+      "Pr  =  Pi  -  Pl\n",
+      " #Output  power  of  motor  =  power  developed  by  the  rotor  -\u0006  friction  and  windage  losses\n",
+      "Po  =  Pr  -  Pfl\n",
+      " #Efficiency  of  induction  motor  =  (output  power/input  power)*100\n",
+      "eff  =  (Po/Psi)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  rotor  copper  loss  is  \",Pl,\"  Watt\"\n",
+      "print  \"\\n(b)  Total  mechanical  power  developed  by  the  rotor  is  \",Pr,\"  W\"\n",
+      "print  \"\\n(c)  Output  power  of  motor  is  \",Po,\"  Watt\"\n",
+      "print  \"\\n(d)  efficiency  of  induction  motor  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  rotor  copper  loss  is   1540.0   Watt\n",
+        "\n",
+        "(b)  Total  mechanical  power  developed  by  the  rotor  is   29260.0   W\n",
+        "\n",
+        "(c)  Output  power  of  motor  is   28510.0   Watt\n",
+        "\n",
+        "(d)  efficiency  of  induction  motor  is   89.09   percent"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 397</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the rotor copper loss, and (b) the efficiency of the motor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Psi  =  32000;#  in  Watts\n",
+      "Psl  =  1200;#  in  Watts\n",
+      "Pfl  =  750;#  in  Watts\n",
+      "x  =  0.35;\n",
+      "\n",
+      "#calculation:\n",
+      " #The  slip,  s\n",
+      "s  =  1-x\n",
+      " #Input  power  to  rotor  =  stator  input  power  -  stator  losses\n",
+      "Pi  =    Psi  -  Psl\n",
+      " #slip  =  rotor  copper  loss/rotor  input\n",
+      "Pl  =  s*Pi\n",
+      " #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -\u0006  rotor  losses\n",
+      "Pr  =  Pi  -  Pl\n",
+      " #Output  power  of  motor  =  power  developed  by  the  rotor  -\u0006  friction  and  windage  losses\n",
+      "Po  =  Pr  -  Pfl\n",
+      " #Efficiency  of  induction  motor  =  (output  power/input  power)*100\n",
+      "eff  =  (Po/Psi)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  rotor  copper  loss  is  \",Pl,\"  Watt\"\n",
+      "print  \"\\n(b)  efficiency  of  induction  motor  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  rotor  copper  loss  is   20020.0   Watt\n",
+        "\n",
+        "(b)  efficiency  of  induction  motor  is   31.34   percent"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 398</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the synchronous speed, (b) the slip, \n",
+      "#(c) the full load torque, (d) the power output if mechanical losses amount to 770 W, \n",
+      "#(e) the maximum torque, (f) the speed at which maximum torque occurs,\n",
+      "#and (g) the starting torque.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "tr  =  0.85;#  turn  ratio  N2/N1\n",
+      "Pl  =  770;#  in  Watt\n",
+      "m  =  3;#  no.  of  phases\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      " #Phase  voltage,  E1  =  V/(3**0.5)\n",
+      "E1  =  V/(3**0.5)\n",
+      " #Full  load  torque\n",
+      "T  =  (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2  +  (X2*(s/100))**2))\n",
+      " #Output  power,  including  friction  losses\n",
+      "Pm  =  2*math.pi*nr*T\n",
+      " #power  output\n",
+      "Po  =  Pm  -  Pl\n",
+      " #Maximum  torque  occurs  when  R2  =  Xr  =  0.35  ohm\n",
+      " #Slip  \n",
+      "sm  =  R2/X2\n",
+      " #maximum  torque,  Tm  \n",
+      "Tm  =  (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2  +  (X2*sm)**2))\n",
+      " #speed  at  which  maximum  torque  occurs\n",
+      "nrm  =  ns*(1  -  sm)\n",
+      "nrmrpm  =  nrm*60\n",
+      " #At  the  start,  i.e.,  at  standstill,  slip,  s=1\n",
+      "ss  =  1\n",
+      " #starting  torque\n",
+      "Ts  =  (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2  +  (X2*ss)**2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)Synchronous  speed  is  \",round(ns,2),\"  rev/sec\"\n",
+      "print  \"\\n(b)Slip  is  \",round(s,2),\"  percent\"\n",
+      "print  \"\\n(c)Full  load  torque  is  \",round(T,2),\"  Nm\"\n",
+      "print  \"\\n(d)power  output  is  \",round(Po,2),\"W\"\n",
+      "print  \"\\n(e)maximum  torque  is  \",round(Tm,2),\"  Nm\"\n",
+      "print  \"\\n(f)speed  at  which  maximum  torque  occurs  is  \",round(nrmrpm,2),\"rev/min\"\n",
+      "print  \"\\n(g)starting  torque  is  \",round(Ts,2),\"  Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)Synchronous  speed  is   25.0   rev/sec\n",
+        "\n",
+        "(b)Slip  is   4.0   percent\n",
+        "\n",
+        "(c)Full  load  torque  is   78.05   Nm\n",
+        "\n",
+        "(d)power  output  is   10998.99 W\n",
+        "\n",
+        "(e)maximum  torque  is   113.17   Nm\n",
+        "\n",
+        "(f)speed  at  which  maximum  torque  occurs  is   1350.0 rev/min\n",
+        "\n",
+        "(g)starting  torque  is   22.41   Nm"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 400</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the induction motor in problem 10 at full load, \n",
+      "#(a) the rotor current, (b) the rotor copper loss, and (c) the starting current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "tr  =  0.85;#  turn  ratio  N2/N1\n",
+      "m  =  3;#  no.  of  phases\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      " #Phase  voltage,  E1  =  V/(3**0.5)\n",
+      "E1  =  V/(3**0.5)\n",
+      " #rotor  current,\n",
+      "Ir  =  (s/100)*E1*tr/((R2**2  +  (X2*(s/100))**2)**0.5)\n",
+      " #Rotor  copper  loss  \n",
+      "Pcl  =  m*R2*(Ir**2)\n",
+      " #starting  current,\n",
+      "ss  =1\n",
+      "I2  =  ss*tr*E1/((R2**2  +  (X2*ss)**2)**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)rotor  current  is  \",round(Ir,2),\"  A\"\n",
+      "print  \"\\n(b)Total  copper  loss  is  \",round(Pcl,2),\"  W\"\n",
+      "print  \"\\n(c)starting  current  is  \",round(I2,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)rotor  current  is   21.61   A\n",
+        "\n",
+        "(b)Total  copper  loss  is   490.37   W\n",
+        "\n",
+        "(c)starting  current  is   57.9   A"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 401</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the power input at full load, \n",
+      "#(b) the efficiency of the motor at full load and \n",
+      "#(c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "Psl  =  650;#  in  Watt\n",
+      "pf  =  0.87;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "Pm  =  11770;#  watts  from  part  (d),  Problem  22.10\n",
+      "Pcl  =  490.35;#  watts,  Rotor  copper  loss,  from  part  (b),  Problem  22.11\n",
+      " #Stator  input  power\n",
+      "P1  =  Pm  +  Pcl  +  Psl\n",
+      "Po  =  11000#  watts,  Net  power  output,  from  part  (d),  Problem  22.10\n",
+      " #efficiency  =  (output/input)  *100\n",
+      "eff  =  (Po/P1)*100#  in  percent\n",
+      " #Power  input,  P1  =  (3**0.5)*VL*IL*cos(phi)\u000e\n",
+      " #  pf  =  cos(phi)\n",
+      " #supply  current,  IL\n",
+      "I  =  P1/((3**0.5)*V*pf)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(aStator  input  power  is  \",round(P1,2),\" W\"\n",
+      "print  \"\\n(b)efficiency  is  \",round(eff,2),\"  percent\"\n",
+      "print  \"\\n(c)supply  current  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(aStator  input  power  is   12910.35  W\n",
+        "\n",
+        "(b)efficiency  is   85.2   percent\n",
+        "\n",
+        "(c)supply  current  is   20.64   A"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 401</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the resistance of the rotor winding required for maximum starting torque.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #At  the  moment  of  starting,  slip,  \n",
+      "s  =  1\n",
+      " #Maximum  torque  occurs  when  rotor  reactance  equals  rotor  resistance\n",
+      " #for  maximum  torque\n",
+      "R2  =  s*X2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nresistance  of  the  rotor  is  \",R2,\"  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "resistance  of  the  rotor  is   3.5   Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_3.ipynb
new file mode 100755
index 00000000..531badfb
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint_3.ipynb
@@ -0,0 +1,782 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d3dead09e75deb08529dd3e4c9a6e20d157d996c5db2df7bb50efc38f7ff7a89"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 22: Three-phase induction motors</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 389</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  Hz\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  \n",
+      "    #f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      "nsrpm  =  ns*60\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nsynchronous  speed  of  the  motor  is  \",nsrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "synchronous  speed  of  the  motor  is   3000.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 389</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  60;#  in  Hz\n",
+      "ns  =  900/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "p  =  f/ns\n",
+      "np  =  p*2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nnumber  of  poles  is  \", round(np,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "number  of  poles  is   8.0"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 390</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "ns  =  6000/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"frequency  is  \",f,\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "frequency  is   100.0   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 391</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  4/2;#  number  of  pairs  of  poles\n",
+      "f  =  50;#  in  Hz\n",
+      "nr  =  1455/60;#  in  rev/sec\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  synchronous  speed  is  \",ns,\"  rev/sec\"\n",
+      "print  \"\\n(b)  slip  is  \",s,\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  synchronous  speed  is   25.0   rev/sec\n",
+        "\n",
+        "(b)  slip  is   3.0   percent"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 392</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  2/2;#  number  of  pairs  of  poles\n",
+      "f  =  60;#  in  Hz\n",
+      "s  =  0.02;#  slip\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  the  rotor  runs  at\n",
+      "nr  =  ns*(1  -  s)\n",
+      " #frequency  of  the  e.m.f.  induced  in  the  rotor  bars  is\n",
+      "fr  =  ns  -  nr\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  synchronous  speed  is  \",ns,\"  rev/sec\"\n",
+      "print  \"\\n(b)  rotor  speed  is  \",nr,\"  rev/sec\"\n",
+      "print  \"\\n(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is  \",fr,\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  synchronous  speed  is   60.0   rev/sec\n",
+        "\n",
+        "(b)  rotor  speed  is   58.8   rev/sec\n",
+        "\n",
+        "(c)  frequency  of  the  e.m.f. induced  in  the  rotor  bars  is  is   1.2   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 392</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "f  =  50;#  in  Hz\n",
+      "nr  =  1200/60;#  in  rev/min\n",
+      "s  =  0.04;#  slip\n",
+      "\n",
+      "#calculation:\n",
+      " #the  synchronous  speed.\n",
+      "ns  =  nr/(1  -  s)\n",
+      "nsrpm  =  ns*60\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  synchronous  speed  is  \",nsrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  synchronous  speed  is   1250.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 394</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "p  =  8/2;#  number  of  pairs  of  poles\n",
+      "f  =  50;#  in  Hz\n",
+      "fr  =  3;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #fr  =  s*f\n",
+      "s  =  (fr/f)\n",
+      " #the  rotor  speed.\n",
+      "nr  =  ns*(1  -  s)\n",
+      "nrrpm  =  nr*60\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  slip  is  \",s*100,\"  percent\"\n",
+      "print  \"\\n  (b)  rotor  speed  is  \",nrrpm,\"  rev/min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  slip  is   6.0   percent\n",
+        "\n",
+        "  (b)  rotor  speed  is   705.0   rev/min"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 396</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Psi  =  32000;#  in  Watts\n",
+      "Psl  =  1200;#  in  Watts\n",
+      "s  =  0.05;#  slip\n",
+      "Pfl  =  750;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #Input  power  to  rotor  =  stator  input  power  -\u0006  stator  losses\n",
+      "Pi  =    Psi  -  Psl\n",
+      " #slip  =  rotor  copper  loss/rotor  input\n",
+      "Pl  =  s*Pi\n",
+      " #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -\u0006  rotor  losses\n",
+      "Pr  =  Pi  -  Pl\n",
+      " #Output  power  of  motor  =  power  developed  by  the  rotor  -\u0006  friction  and  windage  losses\n",
+      "Po  =  Pr  -  Pfl\n",
+      " #Efficiency  of  induction  motor  =  (output  power/input  power)*100\n",
+      "eff  =  (Po/Psi)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  rotor  copper  loss  is  \",Pl,\"  Watt\"\n",
+      "print  \"\\n(b)  Total  mechanical  power  developed  by  the  rotor  is  \",Pr,\"  W\"\n",
+      "print  \"\\n(c)  Output  power  of  motor  is  \",Po,\"  Watt\"\n",
+      "print  \"\\n(d)  efficiency  of  induction  motor  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  rotor  copper  loss  is   1540.0   Watt\n",
+        "\n",
+        "(b)  Total  mechanical  power  developed  by  the  rotor  is   29260.0   W\n",
+        "\n",
+        "(c)  Output  power  of  motor  is   28510.0   Watt\n",
+        "\n",
+        "(d)  efficiency  of  induction  motor  is   89.09   percent"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 397</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Psi  =  32000;#  in  Watts\n",
+      "Psl  =  1200;#  in  Watts\n",
+      "Pfl  =  750;#  in  Watts\n",
+      "x  =  0.35;\n",
+      "\n",
+      "#calculation:\n",
+      " #The  slip,  s\n",
+      "s  =  1-x\n",
+      " #Input  power  to  rotor  =  stator  input  power  -  stator  losses\n",
+      "Pi  =    Psi  -  Psl\n",
+      " #slip  =  rotor  copper  loss/rotor  input\n",
+      "Pl  =  s*Pi\n",
+      " #Total  mechanical  power  developed  by  the  rotor  =  rotor  input  power  -\u0006  rotor  losses\n",
+      "Pr  =  Pi  -  Pl\n",
+      " #Output  power  of  motor  =  power  developed  by  the  rotor  -\u0006  friction  and  windage  losses\n",
+      "Po  =  Pr  -  Pfl\n",
+      " #Efficiency  of  induction  motor  =  (output  power/input  power)*100\n",
+      "eff  =  (Po/Psi)*100#  in  percent\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)  rotor  copper  loss  is  \",Pl,\"  Watt\"\n",
+      "print  \"\\n(b)  efficiency  of  induction  motor  is  \",round(eff,2),\"  percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)  rotor  copper  loss  is   20020.0   Watt\n",
+        "\n",
+        "(b)  efficiency  of  induction  motor  is   31.34   percent"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 398</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "tr  =  0.85;#  turn  ratio  N2/N1\n",
+      "Pl  =  770;#  in  Watt\n",
+      "m  =  3;#  no.  of  phases\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and  \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      " #Phase  voltage,  E1  =  V/(3**0.5)\n",
+      "E1  =  V/(3**0.5)\n",
+      " #Full  load  torque\n",
+      "T  =  (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2  +  (X2*(s/100))**2))\n",
+      " #Output  power,  including  friction  losses\n",
+      "Pm  =  2*math.pi*nr*T\n",
+      " #power  output\n",
+      "Po  =  Pm  -  Pl\n",
+      " #Maximum  torque  occurs  when  R2  =  Xr  =  0.35  ohm\n",
+      " #Slip  \n",
+      "sm  =  R2/X2\n",
+      " #maximum  torque,  Tm  \n",
+      "Tm  =  (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2  +  (X2*sm)**2))\n",
+      " #speed  at  which  maximum  torque  occurs\n",
+      "nrm  =  ns*(1  -  sm)\n",
+      "nrmrpm  =  nrm*60\n",
+      " #At  the  start,  i.e.,  at  standstill,  slip,  s=1\n",
+      "ss  =  1\n",
+      " #starting  torque\n",
+      "Ts  =  (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2  +  (X2*ss)**2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)Synchronous  speed  is  \",round(ns,2),\"  rev/sec\"\n",
+      "print  \"\\n(b)Slip  is  \",round(s,2),\"  percent\"\n",
+      "print  \"\\n(c)Full  load  torque  is  \",round(T,2),\"  Nm\"\n",
+      "print  \"\\n(d)power  output  is  \",round(Po,2),\"W\"\n",
+      "print  \"\\n(e)maximum  torque  is  \",round(Tm,2),\"  Nm\"\n",
+      "print  \"\\n(f)speed  at  which  maximum  torque  occurs  is  \",round(nrmrpm,2),\"rev/min\"\n",
+      "print  \"\\n(g)starting  torque  is  \",round(Ts,2),\"  Nm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)Synchronous  speed  is   25.0   rev/sec\n",
+        "\n",
+        "(b)Slip  is   4.0   percent\n",
+        "\n",
+        "(c)Full  load  torque  is   78.05   Nm\n",
+        "\n",
+        "(d)power  output  is   10998.99 W\n",
+        "\n",
+        "(e)maximum  torque  is   113.17   Nm\n",
+        "\n",
+        "(f)speed  at  which  maximum  torque  occurs  is   1350.0 rev/min\n",
+        "\n",
+        "(g)starting  torque  is   22.41   Nm"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 400</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "tr  =  0.85;#  turn  ratio  N2/N1\n",
+      "m  =  3;#  no.  of  phases\n",
+      "\n",
+      "#calculation:\n",
+      " #ns  is  the  synchronous  speed,  f  is  the  frequency  in  hertz  of  the  supply  to  the  stator  and \n",
+      "    #p  is  the  number  of  pairs  of  poles.\n",
+      "ns  =  f/p\n",
+      " #The  slip,  s\n",
+      "s  =  ((ns  -  nr)/ns)*100#  in  percent\n",
+      " #Phase  voltage,  E1  =  V/(3**0.5)\n",
+      "E1  =  V/(3**0.5)\n",
+      " #rotor  current,\n",
+      "Ir  =  (s/100)*E1*tr/((R2**2  +  (X2*(s/100))**2)**0.5)\n",
+      " #Rotor  copper  loss  \n",
+      "Pcl  =  m*R2*(Ir**2)\n",
+      " #starting  current,\n",
+      "ss  =1\n",
+      "I2  =  ss*tr*E1/((R2**2  +  (X2*ss)**2)**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)rotor  current  is  \",round(Ir,2),\"  A\"\n",
+      "print  \"\\n(b)Total  copper  loss  is  \",round(Pcl,2),\"  W\"\n",
+      "print  \"\\n(c)starting  current  is  \",round(I2,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)rotor  current  is   21.61   A\n",
+        "\n",
+        "(b)Total  copper  loss  is   490.37   W\n",
+        "\n",
+        "(c)starting  current  is   57.9   A"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 401</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "Psl  =  650;#  in  Watt\n",
+      "pf  =  0.87;#  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "Pm  =  11770;#  watts  from  part  (d),  Problem  22.10\n",
+      "Pcl  =  490.35;#  watts,  Rotor  copper  loss,  from  part  (b),  Problem  22.11\n",
+      " #Stator  input  power\n",
+      "P1  =  Pm  +  Pcl  +  Psl\n",
+      "Po  =  11000#  watts,  Net  power  output,  from  part  (d),  Problem  22.10\n",
+      " #efficiency  =  (output/input)  *100\n",
+      "eff  =  (Po/P1)*100#  in  percent\n",
+      " #Power  input,  P1  =  (3**0.5)*VL*IL*cos(phi)\u000e\n",
+      " #  pf  =  cos(phi)\n",
+      " #supply  current,  IL\n",
+      "I  =  P1/((3**0.5)*V*pf)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(aStator  input  power  is  \",round(P1,2),\" W\"\n",
+      "print  \"\\n(b)efficiency  is  \",round(eff,2),\"  percent\"\n",
+      "print  \"\\n(c)supply  current  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(aStator  input  power  is   12910.35  W\n",
+        "\n",
+        "(b)efficiency  is   85.2   percent\n",
+        "\n",
+        "(c)supply  current  is   20.64   A"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 401</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  415;#  in  Volts\n",
+      "f  =  50  ;#  in  Hz\n",
+      "nr  =  24;#  in  rev/sec\n",
+      "p  =  4/2;#  no.  of  pole  pairs\n",
+      "R2  =  0.35;#  in  Ohms\n",
+      "X2  =  3.5;#  in  Ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #At  the  moment  of  starting,  slip,  \n",
+      "s  =  1\n",
+      " #Maximum  torque  occurs  when  rotor  reactance  equals  rotor  resistance\n",
+      " #for  maximum  torque\n",
+      "R2  =  s*X2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nresistance  of  the  rotor  is  \",R2,\"  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "resistance  of  the  rotor  is   3.5   Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint.ipynb
new file mode 100755
index 00000000..043d0bdd
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint.ipynb
@@ -0,0 +1,434 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e4e0b2e74a0b30b96b3a586de7b740bbefabf9c41230a1638409bfb7b309d066"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 23: Revision of complex numbers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 418</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine ZT\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  5  -  3j;\n",
+      "Z2  =  4  +  7j;\n",
+      "Z3  =  3.9  -  6.7j;\n",
+      "\n",
+      " #calculation:\n",
+      "ZT  =  (Z1*Z2/(Z1  +  Z2))+  Z3\n",
+      "y  =  ZT.imag\n",
+      "x  =  ZT.real\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  ZT  is  \",round(x,2),\"  +  (\",round(y,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  ZT  is   8.65   +  ( -6.26 )i"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 418</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine in cartesian form correct to three decimal places:\n",
+      "#(a)1/Z1 (b)1/Z2 (c) 1/Z1 * 1/Z2 (d) 1/(1/Z1 + 1/Z2)\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  3  +  4j;\n",
+      "Z2  =  2  -  5j;\n",
+      "\n",
+      "#calculation:\n",
+      "za  =  1/Z1\n",
+      "zb  =  1/Z2\n",
+      "zc  =  za  +  zb\n",
+      "zd  =  1/zc\n",
+      "zax  =  za.real\n",
+      "zay  =  za.imag\n",
+      "zbx  =  zb.real\n",
+      "zby  =  zb.imag\n",
+      "zcx  =  zc.real\n",
+      "zcy  =  zc.imag\n",
+      "zdx  =  zd.real\n",
+      "zdy  =  zd.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)1/Z1  is  \",round(  zax,2),\"  +  (\",round(zay,2),\")i\"\n",
+      "print  \"\\n  (b)1/Z2  is  \",round(  zbx,2),\"  +  (\",round(zby,2),\")i\"\n",
+      "print  \"\\n  (c)1/Z1  +  1/Z2  is  \",round(  zcx,2),\"  +  (\",round(zcy,2),\")i\"\n",
+      "print  \"\\n  (d)1/(1/Z1  +  1/Z2)  is  \",round(  zdx,2),\"  +  (\",round(zdy,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)1/Z1  is   0.12   +  ( -0.16 )i\n",
+        "\n",
+        "  (b)1/Z2  is   0.07   +  ( 0.17 )i\n",
+        "\n",
+        "  (c)1/Z1  +  1/Z2  is   0.19   +  ( 0.01 )i\n",
+        "\n",
+        "  (d)1/(1/Z1  +  1/Z2)  is   5.27   +  ( -0.35 )i"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 419</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find a, b, x and y?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  9  -  2j;\n",
+      "Z2  =  2  +  1j;\n",
+      "Z3  =  -2  +  1j;\n",
+      "Z4  =  5  +  2j;\n",
+      "\n",
+      "#calculation:\n",
+      "za  =  Z1/3\n",
+      "zb  =  Z2*Z3\n",
+      "zca  =  (2*Z4.real  +  Z4.imag)/-1\n",
+      "zcb  =  Z4.real  -  zca\n",
+      "zaa  =  za.real\n",
+      "zab  =  za.imag\n",
+      "zbx  =  zb.real\n",
+      "zby  =  zb.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)a  and  b  are  \",  zaa,\"  and  \",round(zab,2),\"  resp.\"\n",
+      "print  \"\\n  (b)x  and  y  are  \",  zbx,\"  and  \",zby,\"  resp.\"\n",
+      "print  \"\\n  (c)a  and  b  are  \",  zca,\"  and  \",zcb,\"  resp.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)a  and  b  are   3.0   and   -0.67   resp.\n",
+        "\n",
+        "  (b)x  and  y  are   -5.0   and   0.0   resp.\n",
+        "\n",
+        "  (c)a  and  b  are   -12.0   and   17.0   resp."
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 422</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert 5/_132\u00b0 into a + jb form correct to four significant figures.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  5;#  magnitude\n",
+      "theta  =  -132;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      "x  =  r*math.sin(theta*math.pi/180)\n",
+      "y  =  r*math.cos(theta*math.pi/180)\n",
+      "z  =  x + y*1j\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Z  is  \",round(x,2),\"  +  (\",round(y,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Z  is   -3.72   +  ( -3.35 )i"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 422</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine in polar form the total impedance ZT given that ZT = Z1Z2/\u0007(Z1 + Z2\b)\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r1  =  4.76;#  magnitude\n",
+      "theta1  =  35;#  in  degree\n",
+      "r2  =  7.36;#  magnitude\n",
+      "theta2  =  -48;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      "x1  =  r1*cmath.cos(theta1*math.pi/180)\n",
+      "y1  =  r1*cmath.sin(theta1*math.pi/180)\n",
+      "z1  =  x1 + y1*1j\n",
+      "x2  =  r2*cmath.cos(theta2*math.pi/180)\n",
+      "y2  =  r2*cmath.sin(theta2*math.pi/180)\n",
+      "z2  =  x2 + y2*1j\n",
+      "z3  =  z1*z2/(z1  +  z2)\n",
+      "x3  =  z3.real\n",
+      "y3  =  z3.imag\n",
+      "r3  =  (x3**2  +  y3**2)**0.5\n",
+      "theta3  =  cmath.phase(complex(x3,y3))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  ZT  is  (\",round( r3,2),\",round(/_\",round(theta3,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  ZT  is  ( 3.79 ,round(/_ 4.25 deg)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 423</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine \u0007\u0003(2 +j3\b)^5 in polar and in cartesian form.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z  =  -2  +  3j;\n",
+      "\n",
+      "#calculation:\n",
+      "zc  =  z**5\n",
+      "x  =  zc.real\n",
+      "y  =  zc.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Z  is  \",round(  x,2),\"  +  (\",round(y,2),\")i\"\n",
+      "print  \"\\n  ZT  is  (\",round(  r,2),\"round/_\",round(theta,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Z  is   -122.0   +  ( -597.0 )i\n",
+        "\n",
+        "  ZT  is  ( 609.34 round/_ -101.55 deg)"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 423</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the two square roots of the complex number \u0007(12 + j5)\b in cartesian and polar form\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z  =  12  +  5j;\n",
+      "\n",
+      "#calculation:\n",
+      "x  =  z.real\n",
+      "y  =  z.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta1  =  cmath.atan(y/x)*180/math.pi\n",
+      "'''\n",
+      "if  ((x<0)&(y<0))\n",
+      "         theta1  =  theta1  -180;\n",
+      "elif  ((x<0)&(y>0))\n",
+      "         theta1  =  theta1  +180;\n",
+      "'''\n",
+      "theta2  =  theta1  +  360\n",
+      "rtheta1  =  theta1/2\n",
+      "rtheta2  =  theta2/2\n",
+      "'''\n",
+      "if  (rtheta2  >  180)\n",
+      "         rtheta2  =  rtheta2  -360;\n",
+      "elif  ((x<0)&(y>0))\n",
+      "         rtheta2  =  rtheta2  +360;\n",
+      "'''\n",
+      "rr  =  r**0.5\n",
+      "x1  =  rr*cmath.cos(rtheta1*math.pi/180)\n",
+      "y1  =  rr*cmath.sin(rtheta1*math.pi/180)\n",
+      "z1  =  x1  +  y1*1j\n",
+      "\n",
+      "x2  =  rr*cmath.cos(rtheta2*math.pi/180)\n",
+      "y2  =  rr*cmath.sin(rtheta2*math.pi/180)\n",
+      "z2  =  x2  +  y2*1j\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  two  roots  are  (\",round(z1.real,2),\"  +  (\",round(z1.imag,2),\")i) \"\n",
+      "print   \" and  (\",round(z2.real,2),\"  +  (\",round(z2.imag,2),\")i)\"\n",
+      "print  \"\\n  two  roots  are  (\",round(  rr,2),\"/_\",round((cmath.phase(complex(z1.real,z1.imag)))*180/math.pi,2),\"deg) \"\n",
+      "print   \" and  (\",round(  rr,2),\"/_\",round((cmath.phase(complex(z2.real,z2.imag)))*180/math.pi,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  two  roots  are  ( 3.54   +  ( 0.71 )i) \n",
+        " and  ( -3.54   +  ( -0.71 )i)\n",
+        "\n",
+        "  two  roots  are  ( 3.61 /_ 11.31 deg) \n",
+        " and  ( 3.61 /_ -168.69 deg)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_1.ipynb
new file mode 100755
index 00000000..043d0bdd
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_1.ipynb
@@ -0,0 +1,434 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e4e0b2e74a0b30b96b3a586de7b740bbefabf9c41230a1638409bfb7b309d066"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 23: Revision of complex numbers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 418</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine ZT\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  5  -  3j;\n",
+      "Z2  =  4  +  7j;\n",
+      "Z3  =  3.9  -  6.7j;\n",
+      "\n",
+      " #calculation:\n",
+      "ZT  =  (Z1*Z2/(Z1  +  Z2))+  Z3\n",
+      "y  =  ZT.imag\n",
+      "x  =  ZT.real\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  ZT  is  \",round(x,2),\"  +  (\",round(y,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  ZT  is   8.65   +  ( -6.26 )i"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 418</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine in cartesian form correct to three decimal places:\n",
+      "#(a)1/Z1 (b)1/Z2 (c) 1/Z1 * 1/Z2 (d) 1/(1/Z1 + 1/Z2)\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  3  +  4j;\n",
+      "Z2  =  2  -  5j;\n",
+      "\n",
+      "#calculation:\n",
+      "za  =  1/Z1\n",
+      "zb  =  1/Z2\n",
+      "zc  =  za  +  zb\n",
+      "zd  =  1/zc\n",
+      "zax  =  za.real\n",
+      "zay  =  za.imag\n",
+      "zbx  =  zb.real\n",
+      "zby  =  zb.imag\n",
+      "zcx  =  zc.real\n",
+      "zcy  =  zc.imag\n",
+      "zdx  =  zd.real\n",
+      "zdy  =  zd.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)1/Z1  is  \",round(  zax,2),\"  +  (\",round(zay,2),\")i\"\n",
+      "print  \"\\n  (b)1/Z2  is  \",round(  zbx,2),\"  +  (\",round(zby,2),\")i\"\n",
+      "print  \"\\n  (c)1/Z1  +  1/Z2  is  \",round(  zcx,2),\"  +  (\",round(zcy,2),\")i\"\n",
+      "print  \"\\n  (d)1/(1/Z1  +  1/Z2)  is  \",round(  zdx,2),\"  +  (\",round(zdy,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)1/Z1  is   0.12   +  ( -0.16 )i\n",
+        "\n",
+        "  (b)1/Z2  is   0.07   +  ( 0.17 )i\n",
+        "\n",
+        "  (c)1/Z1  +  1/Z2  is   0.19   +  ( 0.01 )i\n",
+        "\n",
+        "  (d)1/(1/Z1  +  1/Z2)  is   5.27   +  ( -0.35 )i"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 419</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find a, b, x and y?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  9  -  2j;\n",
+      "Z2  =  2  +  1j;\n",
+      "Z3  =  -2  +  1j;\n",
+      "Z4  =  5  +  2j;\n",
+      "\n",
+      "#calculation:\n",
+      "za  =  Z1/3\n",
+      "zb  =  Z2*Z3\n",
+      "zca  =  (2*Z4.real  +  Z4.imag)/-1\n",
+      "zcb  =  Z4.real  -  zca\n",
+      "zaa  =  za.real\n",
+      "zab  =  za.imag\n",
+      "zbx  =  zb.real\n",
+      "zby  =  zb.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)a  and  b  are  \",  zaa,\"  and  \",round(zab,2),\"  resp.\"\n",
+      "print  \"\\n  (b)x  and  y  are  \",  zbx,\"  and  \",zby,\"  resp.\"\n",
+      "print  \"\\n  (c)a  and  b  are  \",  zca,\"  and  \",zcb,\"  resp.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)a  and  b  are   3.0   and   -0.67   resp.\n",
+        "\n",
+        "  (b)x  and  y  are   -5.0   and   0.0   resp.\n",
+        "\n",
+        "  (c)a  and  b  are   -12.0   and   17.0   resp."
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 422</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert 5/_132\u00b0 into a + jb form correct to four significant figures.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  5;#  magnitude\n",
+      "theta  =  -132;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      "x  =  r*math.sin(theta*math.pi/180)\n",
+      "y  =  r*math.cos(theta*math.pi/180)\n",
+      "z  =  x + y*1j\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Z  is  \",round(x,2),\"  +  (\",round(y,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Z  is   -3.72   +  ( -3.35 )i"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 422</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine in polar form the total impedance ZT given that ZT = Z1Z2/\u0007(Z1 + Z2\b)\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r1  =  4.76;#  magnitude\n",
+      "theta1  =  35;#  in  degree\n",
+      "r2  =  7.36;#  magnitude\n",
+      "theta2  =  -48;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      "x1  =  r1*cmath.cos(theta1*math.pi/180)\n",
+      "y1  =  r1*cmath.sin(theta1*math.pi/180)\n",
+      "z1  =  x1 + y1*1j\n",
+      "x2  =  r2*cmath.cos(theta2*math.pi/180)\n",
+      "y2  =  r2*cmath.sin(theta2*math.pi/180)\n",
+      "z2  =  x2 + y2*1j\n",
+      "z3  =  z1*z2/(z1  +  z2)\n",
+      "x3  =  z3.real\n",
+      "y3  =  z3.imag\n",
+      "r3  =  (x3**2  +  y3**2)**0.5\n",
+      "theta3  =  cmath.phase(complex(x3,y3))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  ZT  is  (\",round( r3,2),\",round(/_\",round(theta3,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  ZT  is  ( 3.79 ,round(/_ 4.25 deg)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 423</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine \u0007\u0003(2 +j3\b)^5 in polar and in cartesian form.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z  =  -2  +  3j;\n",
+      "\n",
+      "#calculation:\n",
+      "zc  =  z**5\n",
+      "x  =  zc.real\n",
+      "y  =  zc.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Z  is  \",round(  x,2),\"  +  (\",round(y,2),\")i\"\n",
+      "print  \"\\n  ZT  is  (\",round(  r,2),\"round/_\",round(theta,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Z  is   -122.0   +  ( -597.0 )i\n",
+        "\n",
+        "  ZT  is  ( 609.34 round/_ -101.55 deg)"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 423</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the two square roots of the complex number \u0007(12 + j5)\b in cartesian and polar form\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z  =  12  +  5j;\n",
+      "\n",
+      "#calculation:\n",
+      "x  =  z.real\n",
+      "y  =  z.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta1  =  cmath.atan(y/x)*180/math.pi\n",
+      "'''\n",
+      "if  ((x<0)&(y<0))\n",
+      "         theta1  =  theta1  -180;\n",
+      "elif  ((x<0)&(y>0))\n",
+      "         theta1  =  theta1  +180;\n",
+      "'''\n",
+      "theta2  =  theta1  +  360\n",
+      "rtheta1  =  theta1/2\n",
+      "rtheta2  =  theta2/2\n",
+      "'''\n",
+      "if  (rtheta2  >  180)\n",
+      "         rtheta2  =  rtheta2  -360;\n",
+      "elif  ((x<0)&(y>0))\n",
+      "         rtheta2  =  rtheta2  +360;\n",
+      "'''\n",
+      "rr  =  r**0.5\n",
+      "x1  =  rr*cmath.cos(rtheta1*math.pi/180)\n",
+      "y1  =  rr*cmath.sin(rtheta1*math.pi/180)\n",
+      "z1  =  x1  +  y1*1j\n",
+      "\n",
+      "x2  =  rr*cmath.cos(rtheta2*math.pi/180)\n",
+      "y2  =  rr*cmath.sin(rtheta2*math.pi/180)\n",
+      "z2  =  x2  +  y2*1j\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  two  roots  are  (\",round(z1.real,2),\"  +  (\",round(z1.imag,2),\")i) \"\n",
+      "print   \" and  (\",round(z2.real,2),\"  +  (\",round(z2.imag,2),\")i)\"\n",
+      "print  \"\\n  two  roots  are  (\",round(  rr,2),\"/_\",round((cmath.phase(complex(z1.real,z1.imag)))*180/math.pi,2),\"deg) \"\n",
+      "print   \" and  (\",round(  rr,2),\"/_\",round((cmath.phase(complex(z2.real,z2.imag)))*180/math.pi,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  two  roots  are  ( 3.54   +  ( 0.71 )i) \n",
+        " and  ( -3.54   +  ( -0.71 )i)\n",
+        "\n",
+        "  two  roots  are  ( 3.61 /_ 11.31 deg) \n",
+        " and  ( 3.61 /_ -168.69 deg)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_2.ipynb
new file mode 100755
index 00000000..043d0bdd
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_2.ipynb
@@ -0,0 +1,434 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e4e0b2e74a0b30b96b3a586de7b740bbefabf9c41230a1638409bfb7b309d066"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 23: Revision of complex numbers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 418</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine ZT\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  5  -  3j;\n",
+      "Z2  =  4  +  7j;\n",
+      "Z3  =  3.9  -  6.7j;\n",
+      "\n",
+      " #calculation:\n",
+      "ZT  =  (Z1*Z2/(Z1  +  Z2))+  Z3\n",
+      "y  =  ZT.imag\n",
+      "x  =  ZT.real\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  ZT  is  \",round(x,2),\"  +  (\",round(y,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  ZT  is   8.65   +  ( -6.26 )i"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 418</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine in cartesian form correct to three decimal places:\n",
+      "#(a)1/Z1 (b)1/Z2 (c) 1/Z1 * 1/Z2 (d) 1/(1/Z1 + 1/Z2)\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  3  +  4j;\n",
+      "Z2  =  2  -  5j;\n",
+      "\n",
+      "#calculation:\n",
+      "za  =  1/Z1\n",
+      "zb  =  1/Z2\n",
+      "zc  =  za  +  zb\n",
+      "zd  =  1/zc\n",
+      "zax  =  za.real\n",
+      "zay  =  za.imag\n",
+      "zbx  =  zb.real\n",
+      "zby  =  zb.imag\n",
+      "zcx  =  zc.real\n",
+      "zcy  =  zc.imag\n",
+      "zdx  =  zd.real\n",
+      "zdy  =  zd.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)1/Z1  is  \",round(  zax,2),\"  +  (\",round(zay,2),\")i\"\n",
+      "print  \"\\n  (b)1/Z2  is  \",round(  zbx,2),\"  +  (\",round(zby,2),\")i\"\n",
+      "print  \"\\n  (c)1/Z1  +  1/Z2  is  \",round(  zcx,2),\"  +  (\",round(zcy,2),\")i\"\n",
+      "print  \"\\n  (d)1/(1/Z1  +  1/Z2)  is  \",round(  zdx,2),\"  +  (\",round(zdy,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)1/Z1  is   0.12   +  ( -0.16 )i\n",
+        "\n",
+        "  (b)1/Z2  is   0.07   +  ( 0.17 )i\n",
+        "\n",
+        "  (c)1/Z1  +  1/Z2  is   0.19   +  ( 0.01 )i\n",
+        "\n",
+        "  (d)1/(1/Z1  +  1/Z2)  is   5.27   +  ( -0.35 )i"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 419</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find a, b, x and y?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  9  -  2j;\n",
+      "Z2  =  2  +  1j;\n",
+      "Z3  =  -2  +  1j;\n",
+      "Z4  =  5  +  2j;\n",
+      "\n",
+      "#calculation:\n",
+      "za  =  Z1/3\n",
+      "zb  =  Z2*Z3\n",
+      "zca  =  (2*Z4.real  +  Z4.imag)/-1\n",
+      "zcb  =  Z4.real  -  zca\n",
+      "zaa  =  za.real\n",
+      "zab  =  za.imag\n",
+      "zbx  =  zb.real\n",
+      "zby  =  zb.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)a  and  b  are  \",  zaa,\"  and  \",round(zab,2),\"  resp.\"\n",
+      "print  \"\\n  (b)x  and  y  are  \",  zbx,\"  and  \",zby,\"  resp.\"\n",
+      "print  \"\\n  (c)a  and  b  are  \",  zca,\"  and  \",zcb,\"  resp.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)a  and  b  are   3.0   and   -0.67   resp.\n",
+        "\n",
+        "  (b)x  and  y  are   -5.0   and   0.0   resp.\n",
+        "\n",
+        "  (c)a  and  b  are   -12.0   and   17.0   resp."
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 422</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert 5/_132\u00b0 into a + jb form correct to four significant figures.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  5;#  magnitude\n",
+      "theta  =  -132;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      "x  =  r*math.sin(theta*math.pi/180)\n",
+      "y  =  r*math.cos(theta*math.pi/180)\n",
+      "z  =  x + y*1j\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Z  is  \",round(x,2),\"  +  (\",round(y,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Z  is   -3.72   +  ( -3.35 )i"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 422</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine in polar form the total impedance ZT given that ZT = Z1Z2/\u0007(Z1 + Z2\b)\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r1  =  4.76;#  magnitude\n",
+      "theta1  =  35;#  in  degree\n",
+      "r2  =  7.36;#  magnitude\n",
+      "theta2  =  -48;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      "x1  =  r1*cmath.cos(theta1*math.pi/180)\n",
+      "y1  =  r1*cmath.sin(theta1*math.pi/180)\n",
+      "z1  =  x1 + y1*1j\n",
+      "x2  =  r2*cmath.cos(theta2*math.pi/180)\n",
+      "y2  =  r2*cmath.sin(theta2*math.pi/180)\n",
+      "z2  =  x2 + y2*1j\n",
+      "z3  =  z1*z2/(z1  +  z2)\n",
+      "x3  =  z3.real\n",
+      "y3  =  z3.imag\n",
+      "r3  =  (x3**2  +  y3**2)**0.5\n",
+      "theta3  =  cmath.phase(complex(x3,y3))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  ZT  is  (\",round( r3,2),\",round(/_\",round(theta3,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  ZT  is  ( 3.79 ,round(/_ 4.25 deg)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 423</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine \u0007\u0003(2 +j3\b)^5 in polar and in cartesian form.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z  =  -2  +  3j;\n",
+      "\n",
+      "#calculation:\n",
+      "zc  =  z**5\n",
+      "x  =  zc.real\n",
+      "y  =  zc.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Z  is  \",round(  x,2),\"  +  (\",round(y,2),\")i\"\n",
+      "print  \"\\n  ZT  is  (\",round(  r,2),\"round/_\",round(theta,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Z  is   -122.0   +  ( -597.0 )i\n",
+        "\n",
+        "  ZT  is  ( 609.34 round/_ -101.55 deg)"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 423</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the two square roots of the complex number \u0007(12 + j5)\b in cartesian and polar form\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z  =  12  +  5j;\n",
+      "\n",
+      "#calculation:\n",
+      "x  =  z.real\n",
+      "y  =  z.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta1  =  cmath.atan(y/x)*180/math.pi\n",
+      "'''\n",
+      "if  ((x<0)&(y<0))\n",
+      "         theta1  =  theta1  -180;\n",
+      "elif  ((x<0)&(y>0))\n",
+      "         theta1  =  theta1  +180;\n",
+      "'''\n",
+      "theta2  =  theta1  +  360\n",
+      "rtheta1  =  theta1/2\n",
+      "rtheta2  =  theta2/2\n",
+      "'''\n",
+      "if  (rtheta2  >  180)\n",
+      "         rtheta2  =  rtheta2  -360;\n",
+      "elif  ((x<0)&(y>0))\n",
+      "         rtheta2  =  rtheta2  +360;\n",
+      "'''\n",
+      "rr  =  r**0.5\n",
+      "x1  =  rr*cmath.cos(rtheta1*math.pi/180)\n",
+      "y1  =  rr*cmath.sin(rtheta1*math.pi/180)\n",
+      "z1  =  x1  +  y1*1j\n",
+      "\n",
+      "x2  =  rr*cmath.cos(rtheta2*math.pi/180)\n",
+      "y2  =  rr*cmath.sin(rtheta2*math.pi/180)\n",
+      "z2  =  x2  +  y2*1j\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  two  roots  are  (\",round(z1.real,2),\"  +  (\",round(z1.imag,2),\")i) \"\n",
+      "print   \" and  (\",round(z2.real,2),\"  +  (\",round(z2.imag,2),\")i)\"\n",
+      "print  \"\\n  two  roots  are  (\",round(  rr,2),\"/_\",round((cmath.phase(complex(z1.real,z1.imag)))*180/math.pi,2),\"deg) \"\n",
+      "print   \" and  (\",round(  rr,2),\"/_\",round((cmath.phase(complex(z2.real,z2.imag)))*180/math.pi,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  two  roots  are  ( 3.54   +  ( 0.71 )i) \n",
+        " and  ( -3.54   +  ( -0.71 )i)\n",
+        "\n",
+        "  two  roots  are  ( 3.61 /_ 11.31 deg) \n",
+        " and  ( 3.61 /_ -168.69 deg)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_3.ipynb
new file mode 100755
index 00000000..7d18b803
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_23-checkpoint_3.ipynb
@@ -0,0 +1,433 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f74164c35ff607aa01d36f8c7437d6d39765d83d37afb55e71acad01e24d293a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 23: Revision of complex numbers</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 418</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  5  -  3j;\n",
+      "Z2  =  4  +  7j;\n",
+      "Z3  =  3.9  -  6.7j;\n",
+      "\n",
+      " #calculation:\n",
+      "ZT  =  (Z1*Z2/(Z1  +  Z2))+  Z3\n",
+      "y  =  ZT.imag\n",
+      "x  =  ZT.real\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  ZT  is  \",round(x,2),\"  +  (\",round(y,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  ZT  is   8.65   +  ( -6.26 )i"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 418</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  3  +  4j;\n",
+      "Z2  =  2  -  5j;\n",
+      "\n",
+      "#calculation:\n",
+      "za  =  1/Z1\n",
+      "zb  =  1/Z2\n",
+      "zc  =  za  +  zb\n",
+      "zd  =  1/zc\n",
+      "zax  =  za.real\n",
+      "zay  =  za.imag\n",
+      "zbx  =  zb.real\n",
+      "zby  =  zb.imag\n",
+      "zcx  =  zc.real\n",
+      "zcy  =  zc.imag\n",
+      "zdx  =  zd.real\n",
+      "zdy  =  zd.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)1/Z1  is  \",round(  zax,2),\"  +  (\",round(zay,2),\")i\"\n",
+      "print  \"\\n  (b)1/Z2  is  \",round(  zbx,2),\"  +  (\",round(zby,2),\")i\"\n",
+      "print  \"\\n  (c)1/Z1  +  1/Z2  is  \",round(  zcx,2),\"  +  (\",round(zcy,2),\")i\"\n",
+      "print  \"\\n  (d)1/(1/Z1  +  1/Z2)  is  \",round(  zdx,2),\"  +  (\",round(zdy,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)1/Z1  is   0.12   +  ( -0.16 )i\n",
+        "\n",
+        "  (b)1/Z2  is   0.07   +  ( 0.17 )i\n",
+        "\n",
+        "  (c)1/Z1  +  1/Z2  is   0.19   +  ( 0.01 )i\n",
+        "\n",
+        "  (d)1/(1/Z1  +  1/Z2)  is   5.27   +  ( -0.35 )i"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 419</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  9  -  2j;\n",
+      "Z2  =  2  +  1j;\n",
+      "Z3  =  -2  +  1j;\n",
+      "Z4  =  5  +  2j;\n",
+      "\n",
+      "#calculation:\n",
+      "za  =  Z1/3\n",
+      "zb  =  Z2*Z3\n",
+      "zca  =  (2*Z4.real  +  Z4.imag)/-1\n",
+      "zcb  =  Z4.real  -  zca\n",
+      "zaa  =  za.real\n",
+      "zab  =  za.imag\n",
+      "zbx  =  zb.real\n",
+      "zby  =  zb.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)a  and  b  are  \",  zaa,\"  and  \",round(zab,2),\"  resp.\"\n",
+      "print  \"\\n  (b)x  and  y  are  \",  zbx,\"  and  \",zby,\"  resp.\"\n",
+      "print  \"\\n  (c)a  and  b  are  \",  zca,\"  and  \",zcb,\"  resp.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)a  and  b  are   3.0   and   -0.67   resp.\n",
+        "\n",
+        "  (b)x  and  y  are   -5.0   and   0.0   resp.\n",
+        "\n",
+        "  (c)a  and  b  are   -12.0   and   17.0   resp."
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 422</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  5;#  magnitude\n",
+      "theta  =  -132;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      "x  =  r*math.sin(theta*math.pi/180)\n",
+      "y  =  r*math.cos(theta*math.pi/180)\n",
+      "z  =  x + y*1j\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Z  is  \",round(x,2),\"  +  (\",round(y,2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Z  is   -3.72   +  ( -3.35 )i"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 422</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r1  =  4.76;#  magnitude\n",
+      "theta1  =  35;#  in  degree\n",
+      "r2  =  7.36;#  magnitude\n",
+      "theta2  =  -48;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      "x1  =  r1*cmath.cos(theta1*math.pi/180)\n",
+      "y1  =  r1*cmath.sin(theta1*math.pi/180)\n",
+      "z1  =  x1 + y1*1j\n",
+      "x2  =  r2*cmath.cos(theta2*math.pi/180)\n",
+      "y2  =  r2*cmath.sin(theta2*math.pi/180)\n",
+      "z2  =  x2 + y2*1j\n",
+      "z3  =  z1*z2/(z1  +  z2)\n",
+      "x3  =  z3.real\n",
+      "y3  =  z3.imag\n",
+      "r3  =  (x3**2  +  y3**2)**0.5\n",
+      "theta3  =  cmath.phase(complex(x3,y3))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  ZT  is  (\",round( r3,2),\",round(/_\",round(theta3,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  ZT  is  ( 3.79 ,round(/_ 4.25 deg)"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 423</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z  =  -2  +  3j;\n",
+      "\n",
+      "#calculation:\n",
+      "zc  =  z**5\n",
+      "x  =  zc.real\n",
+      "y  =  zc.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Z  is  \",round(  x,2),\"  +  (\",round(y,2),\")i\"\n",
+      "print  \"\\n  ZT  is  (\",round(  r,2),\"round/_\",round(theta,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Z  is   -122.0   +  ( -597.0 )i\n",
+        "\n",
+        "  ZT  is  ( 609.34 round/_ -101.55 deg)"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 423</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z  =  12  +  5j;\n",
+      "\n",
+      "#calculation:\n",
+      "x  =  z.real\n",
+      "y  =  z.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta1  =  cmath.atan(y/x)*180/math.pi\n",
+      "'''\n",
+      "if  ((x<0)&(y<0))\n",
+      "         theta1  =  theta1  -180;\n",
+      "elif  ((x<0)&(y>0))\n",
+      "         theta1  =  theta1  +180;\n",
+      "'''\n",
+      "theta2  =  theta1  +  360\n",
+      "rtheta1  =  theta1/2\n",
+      "rtheta2  =  theta2/2\n",
+      "'''\n",
+      "if  (rtheta2  >  180)\n",
+      "         rtheta2  =  rtheta2  -360;\n",
+      "elif  ((x<0)&(y>0))\n",
+      "         rtheta2  =  rtheta2  +360;\n",
+      "'''\n",
+      "rr  =  r**0.5\n",
+      "x1  =  rr*cmath.cos(rtheta1*math.pi/180)\n",
+      "y1  =  rr*cmath.sin(rtheta1*math.pi/180)\n",
+      "z1  =  x1  +  y1*1j\n",
+      "\n",
+      "x2  =  rr*cmath.cos(rtheta2*math.pi/180)\n",
+      "y2  =  rr*cmath.sin(rtheta2*math.pi/180)\n",
+      "z2  =  x2  +  y2*1j\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  two  roots  are  (\",round(z1.real,2),\"  +  (\",round(z1.imag,2),\")i) \"\n",
+      "print   \" and  (\",round(z2.real,2),\"  +  (\",round(z2.imag,2),\")i)\"\n",
+      "print  \"\\n  two  roots  are  (\",round(  rr,2),\"/_\",round((cmath.phase(complex(z1.real,z1.imag)))*180/math.pi,2),\"deg) \"\n",
+      "print   \" and  (\",round(  rr,2),\"/_\",round((cmath.phase(complex(z2.real,z2.imag)))*180/math.pi,2),\"deg)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  two  roots  are  ( 3.54   +  ( 0.71 )i) \n",
+        " and  ( -3.54   +  ( -0.71 )i)\n",
+        "\n",
+        "  two  roots  are  ( 3.61 /_ 11.31 deg) \n",
+        " and  ( 3.61 /_ -168.69 deg)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint.ipynb
new file mode 100755
index 00000000..e4305467
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint.ipynb
@@ -0,0 +1,843 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 24: Application of complex numbers to series a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 433</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances:\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z1  =  12  +  5j;\n",
+      "z2  =  -40j;\n",
+      "r3  =  30;\n",
+      "theta3  =  60;#  in  degrees\n",
+      "r4  =  2.20E6;  \n",
+      "theta4  =  -30;#  in  degrees\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #for  an  R-L  series  circuit,  impedance\n",
+      " #  Z  =  R  +  iXL\n",
+      "Ra  =  z1.real\n",
+      "XLa  =  z1.imag\n",
+      "La  =  XLa/(2*math.pi*f)\n",
+      " #for  a  purely  capacitive  circuit,  impedance  Z  =  -iXc\n",
+      "Xcb  =  abs(z2.imag)\n",
+      "Cb  =  1/(2*math.pi*f*Xcb)\n",
+      "z3  =  r3*cmath.cos(theta3*math.pi/180)  +  (r3*cmath.sin(theta3*math.pi/180))*1j\n",
+      "Rc  =  z3.real\n",
+      "XLc  =  z3.imag\n",
+      "Lc  =  XLc/(2*math.pi*f)\n",
+      "z4  =  r4*cmath.cos(theta4*math.pi/180)  +  (r4*cmath.sin(theta4*math.pi/180))*1j\n",
+      "Rd  =  z4.real\n",
+      "Xcd  =  abs(z4.imag)\n",
+      "Cd  =  1/(2*math.pi*f*Xcd)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of  \",round(  Ra,2),\"  ohm  \"\n",
+      "print   \"in  series  with  an  inductance  of  \",round(La*1000,2),\"mH\"\n",
+      "print  \"\\n  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance  \",round(Cb*1E6,2),\"uF\"\n",
+      "print  \"\\n  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of  \",round(Rc,2),\"  ohm  \"\n",
+      "print   \"in  series  with  an  inductance  of  \",round(Lc*1000,2),\"mH\"\n",
+      "print  \"\\n  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of  \",round(Rd/1000,2),\"kohm \"\n",
+      "print   \" in  series  with  a  capacitor  of  capacitance  \",round(Cd*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of   12.0   ohm  \n",
+        "in  series  with  an  inductance  of   15.92 mH\n",
+        "\n",
+        "  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance   79.58 uF\n",
+        "\n",
+        "  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of   15.0   ohm  \n",
+        "in  series  with  an  inductance  of   82.7 mH\n",
+        "\n",
+        "  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of   1905.26 kohm \n",
+        " in  series  with  a  capacitor  of  capacitance   2.89 nF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 434</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.1592  ;#  in  Henry\n",
+      "V  =  250;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  0;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #for  an  R\u00e2\u20ac\u201cL  series  circuit,  impedance\n",
+      " #  Z  =  R  +  iXL\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  R  +  1j*XL\n",
+      "I  =  V/Z\n",
+      "x  =  I.real\n",
+      "y  =  I.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "if  ((x==0)&(y<0)):\n",
+      "    theta  =  -90\n",
+      "elif  ((x==0)&(y>0)):\n",
+      "    theta  =  +90\n",
+      "else:\n",
+      "    theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  is  (\",round(r,2),\"/_\",theta,\"deg)  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  is  ( 5.0 /_ -90 deg)  A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 435</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the supply p.d.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  3E-6  ;#  in  farad\n",
+      "f  =  1000;#  in  Hz\n",
+      "ri  =  2.83;\n",
+      "thetai  =  90;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #Capacitive  reactance  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #  circuit  impedance  Z\n",
+      "Z  =  -1*1j*Xc\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      "V  =  I*Z\n",
+      "x  =  V.real\n",
+      "y  =  V.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  p.d.  is \",round(abs(V),0),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  p.d.  is  150.0 V"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 435</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resistance, (b) the capacitance, \n",
+      "#(c) the modulus of the impedance, and (d) the current flowing and its phase angle,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "Z  =  30  -  50j;\n",
+      "\n",
+      "#calculation:\n",
+      " #Since  impedance  Z  =  30  -  i50,\n",
+      " #resistance\n",
+      "R  =  Z.real\n",
+      " #capacitive  reactance\n",
+      "Xc  =  abs(Z.imag)\n",
+      " #capacitance\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      " #modulus  of  impedance\n",
+      "modZ  =  (R**2  +  Xc**2)**0.5\n",
+      "I  =  V/Z\n",
+      "x  =  I.real\n",
+      "y  =  I.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  is  \",round(  R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)capacitance  is  \",round(C*1E6,2),\"uFarad\"\n",
+      "print  \"\\n  (c)modulus  of  impedance  is  \",round(modZ,2),\"  ohm\"\n",
+      "print  \"\\n  (d)current  flowing  and  its  phase  angle  is  (\",round(  r,2),\"/_\",round(  theta,2),\"deg)  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  is   30.0   ohm\n",
+        "\n",
+        "  (b)capacitance  is   63.66 uFarad\n",
+        "\n",
+        "  (c)modulus  of  impedance  is   58.31   ohm\n",
+        "\n",
+        "  (d)current  flowing  and  its  phase  angle  is  ( 4.12 /_ 59.04 deg)  A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 436</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the impedance of the circuit,\n",
+      "#(b) the current and circuit phase angle, \n",
+      "#(c) the p.d. across the 32 ohm resistor, and (d) the p.d. across the coil\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  200;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  32;#  in  ohms\n",
+      "L  =  0.15;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #impedance,  Z\n",
+      "Z  =  R  +  1j*XL\n",
+      " #Current  I\n",
+      "I  =  V/Z\n",
+      "xi  =  I.real\n",
+      "yi  =  I.imag\n",
+      "ri  =  (xi**2  +  yi**2)**0.5\n",
+      "if  ((xi==0)&(yi<0)):\n",
+      "         thetai  =  -90\n",
+      "elif  ((xi==0)&(yi>0)):\n",
+      "         thetai  =  +90\n",
+      "else:\n",
+      "         thetai  =  cmath.phase(complex(xi,yi))*180/math.pi\n",
+      "\n",
+      " #P.d.  across  the  resistor\n",
+      "VR  =  I*R\n",
+      "xr  =  VR.real\n",
+      "yr  =  VR.imag\n",
+      "rr  =  (xr**2  +  yr**2)**0.5\n",
+      "thetar  =  cmath.phase(complex(xr,yr))*180/math.pi\n",
+      " #P.d.  across  the  coil,  VL\n",
+      "VL  =  I*1j*XL\n",
+      "xl  =  VL.real\n",
+      "yl  =  VL.imag\n",
+      "rl  =  (xl**2  +  yl**2)**0.5\n",
+      "thetal  =  cmath.phase(complex(xl,yl))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)impedance  is  \",round(Z.real,2),\"  + \",round(  Z.imag,2),\")i  ohm\"\n",
+      "print  \"\\n  (b)current  flowing  and  its  phase  angle  is lagging the voltage = (\",round(  ri,2),\"/_\",round(  thetai,2),\"deg)  A\"\n",
+      "print  \"\\n  (c)P.d.  across  the  resistor  is  (\",round(rr,2),\"/_\",round(thetar,2),\"deg)  V\"\n",
+      "print  \"\\n  (d)P.d.  across  the  coil,  VL  is  (\",round(rl,2),\"/_\",round(thetal,2),\"deg)  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)impedance  is   32.0   +  47.12 )i  ohm\n",
+        "\n",
+        "  (b)current  flowing  and  its  phase  angle  is lagging the voltage = ( 3.51 /_ -55.82 deg)  A\n",
+        "\n",
+        "  (c)P.d.  across  the  resistor  is  ( 112.36 /_ -55.82 deg)  V\n",
+        "\n",
+        "  (d)P.d.  across  the  coil,  VL  is  ( 165.46 /_ 34.18 deg)  V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 436</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of impedance\n",
+      "#determine the value of the components forming the series circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  120  +  200j;#  in  Volts\n",
+      "f  =  5E6;#  in  Hz\n",
+      "I  =  7  +  16j;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z\n",
+      "Z  =  V/I\n",
+      "R  =  Z.real\n",
+      "X  =  Z.imag  \n",
+      "if  ((R>0)&(X<0)):\n",
+      "    C  =  -1/(2*math.pi*f*X)\n",
+      "#Results\n",
+      "    print  \"\\n\\n  Result  \\n\\n\"\n",
+      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm  \"\n",
+      "    print   \"and  a  capacitor  of  capacitive reactance\", round(X*-1,3),\"ohm and capacitance is\",round(C*1E9,2),\" nFarad\\n\"\n",
+      "elif  ((R>0)&(X>0)):\n",
+      "    L  =  2*math.pi*f*X\n",
+      "#Results\n",
+      "    print  \"\\n\\n  Result  \\n\\n\"\n",
+      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm \"\n",
+      "    print   \" and  a  inductor  of  insuctance  \",round(L*100,2),\" mHenry\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  The  series  circuit  thus  consists  of  a  resistor  of  resistance   13.25   ohm  \n",
+        "and  a  capacitor  of  capacitive reactance 1.705 ohm and capacitance is 18.67  nFarad\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 437</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of impedance Z2.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  70;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "ri  =  3.5;#  in  amperes\n",
+      "thetai  =  -20;#  in  degrees\n",
+      " #z1  consist  of  two  resistance\n",
+      "R1  =  4.36;#  in  ohms\n",
+      "R2  =  -2.1j;#  in  ohms\n",
+      "\n",
+      " #calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #impedance,  Z\n",
+      "Z  =  V/I\n",
+      " #Total  impedance  Z  =  z1  +  z2\n",
+      "Z1  =  R1  +  R2\n",
+      "Z2  =  Z  -  Z1\n",
+      "x  =  Z2.real\n",
+      "y  =  Z2.imag  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  impedance  Z2  is  \",round(x,2),\"  +  (\",round(y,2),\")i  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  impedance  Z2  is   8.5   +  ( 17.42 )i  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 437</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the supply voltage, (b) the voltage across the 90 \u000e resistance, \n",
+      "#(c) the voltage across the inductance, and (d) the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  90;#  in  ohms\n",
+      "XL  =  150;#  in  ohms\n",
+      "ri  =  1.35;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  1j*XL\n",
+      " #Supply  voltage,  V\n",
+      "V  =  I*Z\n",
+      " #Voltage  across  90  ohm\u000e  resistor\n",
+      "VR  =  V.real\n",
+      "#Voltage  across  inductance,  VL\n",
+      "VL  =  V.imag\n",
+      "xv  =  V.real\n",
+      "yv  =  V.imag\n",
+      "rv  =  (xv**2  +  yv**2)**0.5\n",
+      "thetav  = cmath.phase(complex(xv,yv))*180/math.pi\n",
+      "phi  =  thetav  -  thetai\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Supply  voltage,  V  is  \",xv,\"  +  (\",yv,\")i  V\\n\"\n",
+      "print  \"\\n  (b)Voltage  across  90  ohm resistor,  VR  is  \",VR,\"  V\\n\"\n",
+      "print  \"\\n  (c)Voltage  across  inductance,  VL  is  \",VL,\"  V\\n\"\n",
+      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Supply  voltage,  V  is   121.5   +  ( 202.5 )i  V\n",
+        "\n",
+        "\n",
+        "  (b)Voltage  across  90  ohm resistor,  VR  is   121.5   V\n",
+        "\n",
+        "\n",
+        "  (c)Voltage  across  inductance,  VL  is   202.5   V\n",
+        "\n",
+        "\n",
+        "  (d)Circuit  phase  angle  is   59.04 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 438</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the rms value of voltage (in polar form),\n",
+      "#(b) the circuit impedance, (c) the rms current flowing, and\n",
+      "#(d) the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohms\n",
+      "L  =  0.02;#  in  henry\n",
+      "Vm  =  282.8;#  in  volts\n",
+      "w  =  628.4;#  in  rad/sec\n",
+      "phiv  =  math.pi/3;#  phase  angle\n",
+      "\n",
+      "#calculation:\n",
+      " #rms  voltage\n",
+      "Vrms  =  0.707*Vm*math.cos(phiv)  +  0.707*Vm*math.sin(phiv)*1j\n",
+      " #frequency\n",
+      "f  =  w/(2*math.pi)\n",
+      " #Inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  XL*1j\n",
+      " #Rms  current\n",
+      "Irms  =  Vrms/Z\n",
+      "phii  =  cmath.phase(complex(Irms.real, Irms.imag))*180/math.pi\n",
+      "phi  =  phiv*180/math.pi  -  phii\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms.real,2),\"  +  (\",round(  Vrms.imag,2),\")i  V\\n\"\n",
+      "print  \"\\n  (b)the  circuit  impedance  is  \",round(R,2),\"  +  (\",round(  XL,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (c)the  rms  current  flowing  is  \",round(Irms.real,2),\"  +  (\",round(  Irms.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  rms  value  of  voltage  is   99.97   +  ( 173.15 )i  V\n",
+        "\n",
+        "\n",
+        "  (b)the  circuit  impedance  is   25.0   +  ( 12.57 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (c)the  rms  current  flowing  is   5.97   +  ( 3.92 )i  A\n",
+        "\n",
+        "\n",
+        "  (d)Circuit  phase  angle  is   26.69 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 438</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  12;#  in  ohms\n",
+      "L  =  0.10;#  in  henry\n",
+      "C  =  120E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance,  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  1j*(XL  -  Xc)\n",
+      "I  =  V/Z\n",
+      "phii  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phiv  =  0#  in  degrees\n",
+      "phi  =  phiv  -  phii\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  is  \",round(abs(I),1),\"/_\",round(cmath.phase(complex(I.real,I.imag))*180/math.pi,1),\"deg  A\\n\"\n",
+      "print  \"and  Circuit  phase  angle  is  \",round(phi,1),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  is   18.5 /_ -22.2 deg  A\n",
+        "\n",
+        "and  Circuit  phase  angle  is   22.2 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 439</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of R and L, Determine also the voltage across the coil and the voltage across the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  50E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  225;#  in  volts\n",
+      "ri  =  1.5;#  in  Amperes\n",
+      "thetai  =  -30;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  V/I\n",
+      "R  =  Z.real\n",
+      "XL  =  Z.imag  +  Xc\n",
+      " #inductance  L\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      " #Voltage  across  coil\n",
+      "Zcoil  =  R  +  1j*XL\n",
+      "Vcoil  =  I*Zcoil\n",
+      " #Voltage  across  capacitor,\n",
+      "Vc  =  I*(-1j*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  is  \",round(R,2),\"  ohm  and  inductance  is  \",round(  L,3),\" H\\n\"\n",
+      "print  \"\\n  (b)voltage  across  the  coil  is  \",round(Vcoil.real,2),\"  +  (\",round(  Vcoil.imag,2),\")i  V\\n\"\n",
+      "print  \"\\n  (c)voltage  across  the  capacitor  is  \",round(Vc.real,2),\"  +  (\",round(  Vc.imag,2),\")i  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  is   129.9   ohm  and  inductance  is   0.441  H\n",
+        "\n",
+        "\n",
+        "  (b)voltage  across  the  coil  is   272.75   +  ( 82.7 )i  V\n",
+        "\n",
+        "\n",
+        "  (c)voltage  across  the  capacitor  is   -47.75   +  ( -82.7 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 440</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine also the value of the supply voltage V and the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  2.653E-6;#  in  Farads\n",
+      "R1  =  8;#  in  ohms\n",
+      "R2  =  5;#  in  ohms\n",
+      "L  =  0.477E-3;#  in  Henry\n",
+      "f  =  4000;#  in  Hz\n",
+      "ri  =  6;#  in  Amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #impedance  Z1\n",
+      "Z1  =  R1  -  1j*Xc\n",
+      " #inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #impedance  Z2,\n",
+      "Z2  =  R2  +  1j*XL\n",
+      " #voltage  V1\n",
+      "V1  =  I*Z1\n",
+      " #voltage  V2\n",
+      "V2  =  I*Z2\n",
+      " #Supply  voltage,  V\n",
+      "V  =  V1  +  V2\n",
+      "phiv  =  cmath.phase(complex(V.real, V.imag))*180/math.pi\n",
+      "phi  =  phiv  -  thetai\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  voltage  is  \",round(V.real,2),\"  +  (\",round(  V.imag,2),\")i  V\\n\"\n",
+      "print  \"and  Circuit  phase  angle  is  \",round(phi,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  voltage  is   78.0   +  ( -18.06 )i  V\n",
+        "\n",
+        "and  Circuit  phase  angle  is   -13.03 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_1.ipynb
new file mode 100755
index 00000000..e4305467
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_1.ipynb
@@ -0,0 +1,843 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 24: Application of complex numbers to series a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 433</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances:\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z1  =  12  +  5j;\n",
+      "z2  =  -40j;\n",
+      "r3  =  30;\n",
+      "theta3  =  60;#  in  degrees\n",
+      "r4  =  2.20E6;  \n",
+      "theta4  =  -30;#  in  degrees\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #for  an  R-L  series  circuit,  impedance\n",
+      " #  Z  =  R  +  iXL\n",
+      "Ra  =  z1.real\n",
+      "XLa  =  z1.imag\n",
+      "La  =  XLa/(2*math.pi*f)\n",
+      " #for  a  purely  capacitive  circuit,  impedance  Z  =  -iXc\n",
+      "Xcb  =  abs(z2.imag)\n",
+      "Cb  =  1/(2*math.pi*f*Xcb)\n",
+      "z3  =  r3*cmath.cos(theta3*math.pi/180)  +  (r3*cmath.sin(theta3*math.pi/180))*1j\n",
+      "Rc  =  z3.real\n",
+      "XLc  =  z3.imag\n",
+      "Lc  =  XLc/(2*math.pi*f)\n",
+      "z4  =  r4*cmath.cos(theta4*math.pi/180)  +  (r4*cmath.sin(theta4*math.pi/180))*1j\n",
+      "Rd  =  z4.real\n",
+      "Xcd  =  abs(z4.imag)\n",
+      "Cd  =  1/(2*math.pi*f*Xcd)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of  \",round(  Ra,2),\"  ohm  \"\n",
+      "print   \"in  series  with  an  inductance  of  \",round(La*1000,2),\"mH\"\n",
+      "print  \"\\n  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance  \",round(Cb*1E6,2),\"uF\"\n",
+      "print  \"\\n  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of  \",round(Rc,2),\"  ohm  \"\n",
+      "print   \"in  series  with  an  inductance  of  \",round(Lc*1000,2),\"mH\"\n",
+      "print  \"\\n  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of  \",round(Rd/1000,2),\"kohm \"\n",
+      "print   \" in  series  with  a  capacitor  of  capacitance  \",round(Cd*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of   12.0   ohm  \n",
+        "in  series  with  an  inductance  of   15.92 mH\n",
+        "\n",
+        "  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance   79.58 uF\n",
+        "\n",
+        "  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of   15.0   ohm  \n",
+        "in  series  with  an  inductance  of   82.7 mH\n",
+        "\n",
+        "  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of   1905.26 kohm \n",
+        " in  series  with  a  capacitor  of  capacitance   2.89 nF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 434</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.1592  ;#  in  Henry\n",
+      "V  =  250;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  0;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #for  an  R\u00e2\u20ac\u201cL  series  circuit,  impedance\n",
+      " #  Z  =  R  +  iXL\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  R  +  1j*XL\n",
+      "I  =  V/Z\n",
+      "x  =  I.real\n",
+      "y  =  I.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "if  ((x==0)&(y<0)):\n",
+      "    theta  =  -90\n",
+      "elif  ((x==0)&(y>0)):\n",
+      "    theta  =  +90\n",
+      "else:\n",
+      "    theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  is  (\",round(r,2),\"/_\",theta,\"deg)  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  is  ( 5.0 /_ -90 deg)  A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 435</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the supply p.d.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  3E-6  ;#  in  farad\n",
+      "f  =  1000;#  in  Hz\n",
+      "ri  =  2.83;\n",
+      "thetai  =  90;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #Capacitive  reactance  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #  circuit  impedance  Z\n",
+      "Z  =  -1*1j*Xc\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      "V  =  I*Z\n",
+      "x  =  V.real\n",
+      "y  =  V.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  p.d.  is \",round(abs(V),0),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  p.d.  is  150.0 V"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 435</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resistance, (b) the capacitance, \n",
+      "#(c) the modulus of the impedance, and (d) the current flowing and its phase angle,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "Z  =  30  -  50j;\n",
+      "\n",
+      "#calculation:\n",
+      " #Since  impedance  Z  =  30  -  i50,\n",
+      " #resistance\n",
+      "R  =  Z.real\n",
+      " #capacitive  reactance\n",
+      "Xc  =  abs(Z.imag)\n",
+      " #capacitance\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      " #modulus  of  impedance\n",
+      "modZ  =  (R**2  +  Xc**2)**0.5\n",
+      "I  =  V/Z\n",
+      "x  =  I.real\n",
+      "y  =  I.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  is  \",round(  R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)capacitance  is  \",round(C*1E6,2),\"uFarad\"\n",
+      "print  \"\\n  (c)modulus  of  impedance  is  \",round(modZ,2),\"  ohm\"\n",
+      "print  \"\\n  (d)current  flowing  and  its  phase  angle  is  (\",round(  r,2),\"/_\",round(  theta,2),\"deg)  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  is   30.0   ohm\n",
+        "\n",
+        "  (b)capacitance  is   63.66 uFarad\n",
+        "\n",
+        "  (c)modulus  of  impedance  is   58.31   ohm\n",
+        "\n",
+        "  (d)current  flowing  and  its  phase  angle  is  ( 4.12 /_ 59.04 deg)  A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 436</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the impedance of the circuit,\n",
+      "#(b) the current and circuit phase angle, \n",
+      "#(c) the p.d. across the 32 ohm resistor, and (d) the p.d. across the coil\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  200;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  32;#  in  ohms\n",
+      "L  =  0.15;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #impedance,  Z\n",
+      "Z  =  R  +  1j*XL\n",
+      " #Current  I\n",
+      "I  =  V/Z\n",
+      "xi  =  I.real\n",
+      "yi  =  I.imag\n",
+      "ri  =  (xi**2  +  yi**2)**0.5\n",
+      "if  ((xi==0)&(yi<0)):\n",
+      "         thetai  =  -90\n",
+      "elif  ((xi==0)&(yi>0)):\n",
+      "         thetai  =  +90\n",
+      "else:\n",
+      "         thetai  =  cmath.phase(complex(xi,yi))*180/math.pi\n",
+      "\n",
+      " #P.d.  across  the  resistor\n",
+      "VR  =  I*R\n",
+      "xr  =  VR.real\n",
+      "yr  =  VR.imag\n",
+      "rr  =  (xr**2  +  yr**2)**0.5\n",
+      "thetar  =  cmath.phase(complex(xr,yr))*180/math.pi\n",
+      " #P.d.  across  the  coil,  VL\n",
+      "VL  =  I*1j*XL\n",
+      "xl  =  VL.real\n",
+      "yl  =  VL.imag\n",
+      "rl  =  (xl**2  +  yl**2)**0.5\n",
+      "thetal  =  cmath.phase(complex(xl,yl))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)impedance  is  \",round(Z.real,2),\"  + \",round(  Z.imag,2),\")i  ohm\"\n",
+      "print  \"\\n  (b)current  flowing  and  its  phase  angle  is lagging the voltage = (\",round(  ri,2),\"/_\",round(  thetai,2),\"deg)  A\"\n",
+      "print  \"\\n  (c)P.d.  across  the  resistor  is  (\",round(rr,2),\"/_\",round(thetar,2),\"deg)  V\"\n",
+      "print  \"\\n  (d)P.d.  across  the  coil,  VL  is  (\",round(rl,2),\"/_\",round(thetal,2),\"deg)  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)impedance  is   32.0   +  47.12 )i  ohm\n",
+        "\n",
+        "  (b)current  flowing  and  its  phase  angle  is lagging the voltage = ( 3.51 /_ -55.82 deg)  A\n",
+        "\n",
+        "  (c)P.d.  across  the  resistor  is  ( 112.36 /_ -55.82 deg)  V\n",
+        "\n",
+        "  (d)P.d.  across  the  coil,  VL  is  ( 165.46 /_ 34.18 deg)  V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 436</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of impedance\n",
+      "#determine the value of the components forming the series circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  120  +  200j;#  in  Volts\n",
+      "f  =  5E6;#  in  Hz\n",
+      "I  =  7  +  16j;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z\n",
+      "Z  =  V/I\n",
+      "R  =  Z.real\n",
+      "X  =  Z.imag  \n",
+      "if  ((R>0)&(X<0)):\n",
+      "    C  =  -1/(2*math.pi*f*X)\n",
+      "#Results\n",
+      "    print  \"\\n\\n  Result  \\n\\n\"\n",
+      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm  \"\n",
+      "    print   \"and  a  capacitor  of  capacitive reactance\", round(X*-1,3),\"ohm and capacitance is\",round(C*1E9,2),\" nFarad\\n\"\n",
+      "elif  ((R>0)&(X>0)):\n",
+      "    L  =  2*math.pi*f*X\n",
+      "#Results\n",
+      "    print  \"\\n\\n  Result  \\n\\n\"\n",
+      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm \"\n",
+      "    print   \" and  a  inductor  of  insuctance  \",round(L*100,2),\" mHenry\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  The  series  circuit  thus  consists  of  a  resistor  of  resistance   13.25   ohm  \n",
+        "and  a  capacitor  of  capacitive reactance 1.705 ohm and capacitance is 18.67  nFarad\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 437</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of impedance Z2.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  70;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "ri  =  3.5;#  in  amperes\n",
+      "thetai  =  -20;#  in  degrees\n",
+      " #z1  consist  of  two  resistance\n",
+      "R1  =  4.36;#  in  ohms\n",
+      "R2  =  -2.1j;#  in  ohms\n",
+      "\n",
+      " #calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #impedance,  Z\n",
+      "Z  =  V/I\n",
+      " #Total  impedance  Z  =  z1  +  z2\n",
+      "Z1  =  R1  +  R2\n",
+      "Z2  =  Z  -  Z1\n",
+      "x  =  Z2.real\n",
+      "y  =  Z2.imag  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  impedance  Z2  is  \",round(x,2),\"  +  (\",round(y,2),\")i  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  impedance  Z2  is   8.5   +  ( 17.42 )i  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 437</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the supply voltage, (b) the voltage across the 90 \u000e resistance, \n",
+      "#(c) the voltage across the inductance, and (d) the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  90;#  in  ohms\n",
+      "XL  =  150;#  in  ohms\n",
+      "ri  =  1.35;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  1j*XL\n",
+      " #Supply  voltage,  V\n",
+      "V  =  I*Z\n",
+      " #Voltage  across  90  ohm\u000e  resistor\n",
+      "VR  =  V.real\n",
+      "#Voltage  across  inductance,  VL\n",
+      "VL  =  V.imag\n",
+      "xv  =  V.real\n",
+      "yv  =  V.imag\n",
+      "rv  =  (xv**2  +  yv**2)**0.5\n",
+      "thetav  = cmath.phase(complex(xv,yv))*180/math.pi\n",
+      "phi  =  thetav  -  thetai\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Supply  voltage,  V  is  \",xv,\"  +  (\",yv,\")i  V\\n\"\n",
+      "print  \"\\n  (b)Voltage  across  90  ohm resistor,  VR  is  \",VR,\"  V\\n\"\n",
+      "print  \"\\n  (c)Voltage  across  inductance,  VL  is  \",VL,\"  V\\n\"\n",
+      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Supply  voltage,  V  is   121.5   +  ( 202.5 )i  V\n",
+        "\n",
+        "\n",
+        "  (b)Voltage  across  90  ohm resistor,  VR  is   121.5   V\n",
+        "\n",
+        "\n",
+        "  (c)Voltage  across  inductance,  VL  is   202.5   V\n",
+        "\n",
+        "\n",
+        "  (d)Circuit  phase  angle  is   59.04 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 438</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the rms value of voltage (in polar form),\n",
+      "#(b) the circuit impedance, (c) the rms current flowing, and\n",
+      "#(d) the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohms\n",
+      "L  =  0.02;#  in  henry\n",
+      "Vm  =  282.8;#  in  volts\n",
+      "w  =  628.4;#  in  rad/sec\n",
+      "phiv  =  math.pi/3;#  phase  angle\n",
+      "\n",
+      "#calculation:\n",
+      " #rms  voltage\n",
+      "Vrms  =  0.707*Vm*math.cos(phiv)  +  0.707*Vm*math.sin(phiv)*1j\n",
+      " #frequency\n",
+      "f  =  w/(2*math.pi)\n",
+      " #Inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  XL*1j\n",
+      " #Rms  current\n",
+      "Irms  =  Vrms/Z\n",
+      "phii  =  cmath.phase(complex(Irms.real, Irms.imag))*180/math.pi\n",
+      "phi  =  phiv*180/math.pi  -  phii\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms.real,2),\"  +  (\",round(  Vrms.imag,2),\")i  V\\n\"\n",
+      "print  \"\\n  (b)the  circuit  impedance  is  \",round(R,2),\"  +  (\",round(  XL,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (c)the  rms  current  flowing  is  \",round(Irms.real,2),\"  +  (\",round(  Irms.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  rms  value  of  voltage  is   99.97   +  ( 173.15 )i  V\n",
+        "\n",
+        "\n",
+        "  (b)the  circuit  impedance  is   25.0   +  ( 12.57 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (c)the  rms  current  flowing  is   5.97   +  ( 3.92 )i  A\n",
+        "\n",
+        "\n",
+        "  (d)Circuit  phase  angle  is   26.69 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 438</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  12;#  in  ohms\n",
+      "L  =  0.10;#  in  henry\n",
+      "C  =  120E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance,  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  1j*(XL  -  Xc)\n",
+      "I  =  V/Z\n",
+      "phii  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phiv  =  0#  in  degrees\n",
+      "phi  =  phiv  -  phii\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  is  \",round(abs(I),1),\"/_\",round(cmath.phase(complex(I.real,I.imag))*180/math.pi,1),\"deg  A\\n\"\n",
+      "print  \"and  Circuit  phase  angle  is  \",round(phi,1),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  is   18.5 /_ -22.2 deg  A\n",
+        "\n",
+        "and  Circuit  phase  angle  is   22.2 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 439</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of R and L, Determine also the voltage across the coil and the voltage across the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  50E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  225;#  in  volts\n",
+      "ri  =  1.5;#  in  Amperes\n",
+      "thetai  =  -30;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  V/I\n",
+      "R  =  Z.real\n",
+      "XL  =  Z.imag  +  Xc\n",
+      " #inductance  L\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      " #Voltage  across  coil\n",
+      "Zcoil  =  R  +  1j*XL\n",
+      "Vcoil  =  I*Zcoil\n",
+      " #Voltage  across  capacitor,\n",
+      "Vc  =  I*(-1j*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  is  \",round(R,2),\"  ohm  and  inductance  is  \",round(  L,3),\" H\\n\"\n",
+      "print  \"\\n  (b)voltage  across  the  coil  is  \",round(Vcoil.real,2),\"  +  (\",round(  Vcoil.imag,2),\")i  V\\n\"\n",
+      "print  \"\\n  (c)voltage  across  the  capacitor  is  \",round(Vc.real,2),\"  +  (\",round(  Vc.imag,2),\")i  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  is   129.9   ohm  and  inductance  is   0.441  H\n",
+        "\n",
+        "\n",
+        "  (b)voltage  across  the  coil  is   272.75   +  ( 82.7 )i  V\n",
+        "\n",
+        "\n",
+        "  (c)voltage  across  the  capacitor  is   -47.75   +  ( -82.7 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 440</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine also the value of the supply voltage V and the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  2.653E-6;#  in  Farads\n",
+      "R1  =  8;#  in  ohms\n",
+      "R2  =  5;#  in  ohms\n",
+      "L  =  0.477E-3;#  in  Henry\n",
+      "f  =  4000;#  in  Hz\n",
+      "ri  =  6;#  in  Amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #impedance  Z1\n",
+      "Z1  =  R1  -  1j*Xc\n",
+      " #inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #impedance  Z2,\n",
+      "Z2  =  R2  +  1j*XL\n",
+      " #voltage  V1\n",
+      "V1  =  I*Z1\n",
+      " #voltage  V2\n",
+      "V2  =  I*Z2\n",
+      " #Supply  voltage,  V\n",
+      "V  =  V1  +  V2\n",
+      "phiv  =  cmath.phase(complex(V.real, V.imag))*180/math.pi\n",
+      "phi  =  phiv  -  thetai\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  voltage  is  \",round(V.real,2),\"  +  (\",round(  V.imag,2),\")i  V\\n\"\n",
+      "print  \"and  Circuit  phase  angle  is  \",round(phi,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  voltage  is   78.0   +  ( -18.06 )i  V\n",
+        "\n",
+        "and  Circuit  phase  angle  is   -13.03 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_2.ipynb
new file mode 100755
index 00000000..e4305467
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_2.ipynb
@@ -0,0 +1,843 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 24: Application of complex numbers to series a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 433</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances:\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z1  =  12  +  5j;\n",
+      "z2  =  -40j;\n",
+      "r3  =  30;\n",
+      "theta3  =  60;#  in  degrees\n",
+      "r4  =  2.20E6;  \n",
+      "theta4  =  -30;#  in  degrees\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #for  an  R-L  series  circuit,  impedance\n",
+      " #  Z  =  R  +  iXL\n",
+      "Ra  =  z1.real\n",
+      "XLa  =  z1.imag\n",
+      "La  =  XLa/(2*math.pi*f)\n",
+      " #for  a  purely  capacitive  circuit,  impedance  Z  =  -iXc\n",
+      "Xcb  =  abs(z2.imag)\n",
+      "Cb  =  1/(2*math.pi*f*Xcb)\n",
+      "z3  =  r3*cmath.cos(theta3*math.pi/180)  +  (r3*cmath.sin(theta3*math.pi/180))*1j\n",
+      "Rc  =  z3.real\n",
+      "XLc  =  z3.imag\n",
+      "Lc  =  XLc/(2*math.pi*f)\n",
+      "z4  =  r4*cmath.cos(theta4*math.pi/180)  +  (r4*cmath.sin(theta4*math.pi/180))*1j\n",
+      "Rd  =  z4.real\n",
+      "Xcd  =  abs(z4.imag)\n",
+      "Cd  =  1/(2*math.pi*f*Xcd)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of  \",round(  Ra,2),\"  ohm  \"\n",
+      "print   \"in  series  with  an  inductance  of  \",round(La*1000,2),\"mH\"\n",
+      "print  \"\\n  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance  \",round(Cb*1E6,2),\"uF\"\n",
+      "print  \"\\n  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of  \",round(Rc,2),\"  ohm  \"\n",
+      "print   \"in  series  with  an  inductance  of  \",round(Lc*1000,2),\"mH\"\n",
+      "print  \"\\n  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of  \",round(Rd/1000,2),\"kohm \"\n",
+      "print   \" in  series  with  a  capacitor  of  capacitance  \",round(Cd*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of   12.0   ohm  \n",
+        "in  series  with  an  inductance  of   15.92 mH\n",
+        "\n",
+        "  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance   79.58 uF\n",
+        "\n",
+        "  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of   15.0   ohm  \n",
+        "in  series  with  an  inductance  of   82.7 mH\n",
+        "\n",
+        "  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of   1905.26 kohm \n",
+        " in  series  with  a  capacitor  of  capacitance   2.89 nF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 434</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.1592  ;#  in  Henry\n",
+      "V  =  250;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  0;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #for  an  R\u00e2\u20ac\u201cL  series  circuit,  impedance\n",
+      " #  Z  =  R  +  iXL\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  R  +  1j*XL\n",
+      "I  =  V/Z\n",
+      "x  =  I.real\n",
+      "y  =  I.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "if  ((x==0)&(y<0)):\n",
+      "    theta  =  -90\n",
+      "elif  ((x==0)&(y>0)):\n",
+      "    theta  =  +90\n",
+      "else:\n",
+      "    theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  is  (\",round(r,2),\"/_\",theta,\"deg)  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  is  ( 5.0 /_ -90 deg)  A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 435</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the supply p.d.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  3E-6  ;#  in  farad\n",
+      "f  =  1000;#  in  Hz\n",
+      "ri  =  2.83;\n",
+      "thetai  =  90;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #Capacitive  reactance  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #  circuit  impedance  Z\n",
+      "Z  =  -1*1j*Xc\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      "V  =  I*Z\n",
+      "x  =  V.real\n",
+      "y  =  V.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  p.d.  is \",round(abs(V),0),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  p.d.  is  150.0 V"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 435</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resistance, (b) the capacitance, \n",
+      "#(c) the modulus of the impedance, and (d) the current flowing and its phase angle,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "Z  =  30  -  50j;\n",
+      "\n",
+      "#calculation:\n",
+      " #Since  impedance  Z  =  30  -  i50,\n",
+      " #resistance\n",
+      "R  =  Z.real\n",
+      " #capacitive  reactance\n",
+      "Xc  =  abs(Z.imag)\n",
+      " #capacitance\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      " #modulus  of  impedance\n",
+      "modZ  =  (R**2  +  Xc**2)**0.5\n",
+      "I  =  V/Z\n",
+      "x  =  I.real\n",
+      "y  =  I.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  is  \",round(  R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)capacitance  is  \",round(C*1E6,2),\"uFarad\"\n",
+      "print  \"\\n  (c)modulus  of  impedance  is  \",round(modZ,2),\"  ohm\"\n",
+      "print  \"\\n  (d)current  flowing  and  its  phase  angle  is  (\",round(  r,2),\"/_\",round(  theta,2),\"deg)  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  is   30.0   ohm\n",
+        "\n",
+        "  (b)capacitance  is   63.66 uFarad\n",
+        "\n",
+        "  (c)modulus  of  impedance  is   58.31   ohm\n",
+        "\n",
+        "  (d)current  flowing  and  its  phase  angle  is  ( 4.12 /_ 59.04 deg)  A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 436</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the impedance of the circuit,\n",
+      "#(b) the current and circuit phase angle, \n",
+      "#(c) the p.d. across the 32 ohm resistor, and (d) the p.d. across the coil\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  200;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  32;#  in  ohms\n",
+      "L  =  0.15;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #impedance,  Z\n",
+      "Z  =  R  +  1j*XL\n",
+      " #Current  I\n",
+      "I  =  V/Z\n",
+      "xi  =  I.real\n",
+      "yi  =  I.imag\n",
+      "ri  =  (xi**2  +  yi**2)**0.5\n",
+      "if  ((xi==0)&(yi<0)):\n",
+      "         thetai  =  -90\n",
+      "elif  ((xi==0)&(yi>0)):\n",
+      "         thetai  =  +90\n",
+      "else:\n",
+      "         thetai  =  cmath.phase(complex(xi,yi))*180/math.pi\n",
+      "\n",
+      " #P.d.  across  the  resistor\n",
+      "VR  =  I*R\n",
+      "xr  =  VR.real\n",
+      "yr  =  VR.imag\n",
+      "rr  =  (xr**2  +  yr**2)**0.5\n",
+      "thetar  =  cmath.phase(complex(xr,yr))*180/math.pi\n",
+      " #P.d.  across  the  coil,  VL\n",
+      "VL  =  I*1j*XL\n",
+      "xl  =  VL.real\n",
+      "yl  =  VL.imag\n",
+      "rl  =  (xl**2  +  yl**2)**0.5\n",
+      "thetal  =  cmath.phase(complex(xl,yl))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)impedance  is  \",round(Z.real,2),\"  + \",round(  Z.imag,2),\")i  ohm\"\n",
+      "print  \"\\n  (b)current  flowing  and  its  phase  angle  is lagging the voltage = (\",round(  ri,2),\"/_\",round(  thetai,2),\"deg)  A\"\n",
+      "print  \"\\n  (c)P.d.  across  the  resistor  is  (\",round(rr,2),\"/_\",round(thetar,2),\"deg)  V\"\n",
+      "print  \"\\n  (d)P.d.  across  the  coil,  VL  is  (\",round(rl,2),\"/_\",round(thetal,2),\"deg)  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)impedance  is   32.0   +  47.12 )i  ohm\n",
+        "\n",
+        "  (b)current  flowing  and  its  phase  angle  is lagging the voltage = ( 3.51 /_ -55.82 deg)  A\n",
+        "\n",
+        "  (c)P.d.  across  the  resistor  is  ( 112.36 /_ -55.82 deg)  V\n",
+        "\n",
+        "  (d)P.d.  across  the  coil,  VL  is  ( 165.46 /_ 34.18 deg)  V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 436</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of impedance\n",
+      "#determine the value of the components forming the series circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  120  +  200j;#  in  Volts\n",
+      "f  =  5E6;#  in  Hz\n",
+      "I  =  7  +  16j;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z\n",
+      "Z  =  V/I\n",
+      "R  =  Z.real\n",
+      "X  =  Z.imag  \n",
+      "if  ((R>0)&(X<0)):\n",
+      "    C  =  -1/(2*math.pi*f*X)\n",
+      "#Results\n",
+      "    print  \"\\n\\n  Result  \\n\\n\"\n",
+      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm  \"\n",
+      "    print   \"and  a  capacitor  of  capacitive reactance\", round(X*-1,3),\"ohm and capacitance is\",round(C*1E9,2),\" nFarad\\n\"\n",
+      "elif  ((R>0)&(X>0)):\n",
+      "    L  =  2*math.pi*f*X\n",
+      "#Results\n",
+      "    print  \"\\n\\n  Result  \\n\\n\"\n",
+      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm \"\n",
+      "    print   \" and  a  inductor  of  insuctance  \",round(L*100,2),\" mHenry\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  The  series  circuit  thus  consists  of  a  resistor  of  resistance   13.25   ohm  \n",
+        "and  a  capacitor  of  capacitive reactance 1.705 ohm and capacitance is 18.67  nFarad\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 437</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of impedance Z2.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  70;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "ri  =  3.5;#  in  amperes\n",
+      "thetai  =  -20;#  in  degrees\n",
+      " #z1  consist  of  two  resistance\n",
+      "R1  =  4.36;#  in  ohms\n",
+      "R2  =  -2.1j;#  in  ohms\n",
+      "\n",
+      " #calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #impedance,  Z\n",
+      "Z  =  V/I\n",
+      " #Total  impedance  Z  =  z1  +  z2\n",
+      "Z1  =  R1  +  R2\n",
+      "Z2  =  Z  -  Z1\n",
+      "x  =  Z2.real\n",
+      "y  =  Z2.imag  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  impedance  Z2  is  \",round(x,2),\"  +  (\",round(y,2),\")i  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  impedance  Z2  is   8.5   +  ( 17.42 )i  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 437</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the supply voltage, (b) the voltage across the 90 \u000e resistance, \n",
+      "#(c) the voltage across the inductance, and (d) the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  90;#  in  ohms\n",
+      "XL  =  150;#  in  ohms\n",
+      "ri  =  1.35;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  1j*XL\n",
+      " #Supply  voltage,  V\n",
+      "V  =  I*Z\n",
+      " #Voltage  across  90  ohm\u000e  resistor\n",
+      "VR  =  V.real\n",
+      "#Voltage  across  inductance,  VL\n",
+      "VL  =  V.imag\n",
+      "xv  =  V.real\n",
+      "yv  =  V.imag\n",
+      "rv  =  (xv**2  +  yv**2)**0.5\n",
+      "thetav  = cmath.phase(complex(xv,yv))*180/math.pi\n",
+      "phi  =  thetav  -  thetai\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Supply  voltage,  V  is  \",xv,\"  +  (\",yv,\")i  V\\n\"\n",
+      "print  \"\\n  (b)Voltage  across  90  ohm resistor,  VR  is  \",VR,\"  V\\n\"\n",
+      "print  \"\\n  (c)Voltage  across  inductance,  VL  is  \",VL,\"  V\\n\"\n",
+      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Supply  voltage,  V  is   121.5   +  ( 202.5 )i  V\n",
+        "\n",
+        "\n",
+        "  (b)Voltage  across  90  ohm resistor,  VR  is   121.5   V\n",
+        "\n",
+        "\n",
+        "  (c)Voltage  across  inductance,  VL  is   202.5   V\n",
+        "\n",
+        "\n",
+        "  (d)Circuit  phase  angle  is   59.04 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 438</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the rms value of voltage (in polar form),\n",
+      "#(b) the circuit impedance, (c) the rms current flowing, and\n",
+      "#(d) the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohms\n",
+      "L  =  0.02;#  in  henry\n",
+      "Vm  =  282.8;#  in  volts\n",
+      "w  =  628.4;#  in  rad/sec\n",
+      "phiv  =  math.pi/3;#  phase  angle\n",
+      "\n",
+      "#calculation:\n",
+      " #rms  voltage\n",
+      "Vrms  =  0.707*Vm*math.cos(phiv)  +  0.707*Vm*math.sin(phiv)*1j\n",
+      " #frequency\n",
+      "f  =  w/(2*math.pi)\n",
+      " #Inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  XL*1j\n",
+      " #Rms  current\n",
+      "Irms  =  Vrms/Z\n",
+      "phii  =  cmath.phase(complex(Irms.real, Irms.imag))*180/math.pi\n",
+      "phi  =  phiv*180/math.pi  -  phii\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms.real,2),\"  +  (\",round(  Vrms.imag,2),\")i  V\\n\"\n",
+      "print  \"\\n  (b)the  circuit  impedance  is  \",round(R,2),\"  +  (\",round(  XL,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (c)the  rms  current  flowing  is  \",round(Irms.real,2),\"  +  (\",round(  Irms.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  rms  value  of  voltage  is   99.97   +  ( 173.15 )i  V\n",
+        "\n",
+        "\n",
+        "  (b)the  circuit  impedance  is   25.0   +  ( 12.57 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (c)the  rms  current  flowing  is   5.97   +  ( 3.92 )i  A\n",
+        "\n",
+        "\n",
+        "  (d)Circuit  phase  angle  is   26.69 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 438</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  12;#  in  ohms\n",
+      "L  =  0.10;#  in  henry\n",
+      "C  =  120E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance,  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  1j*(XL  -  Xc)\n",
+      "I  =  V/Z\n",
+      "phii  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phiv  =  0#  in  degrees\n",
+      "phi  =  phiv  -  phii\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  is  \",round(abs(I),1),\"/_\",round(cmath.phase(complex(I.real,I.imag))*180/math.pi,1),\"deg  A\\n\"\n",
+      "print  \"and  Circuit  phase  angle  is  \",round(phi,1),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  is   18.5 /_ -22.2 deg  A\n",
+        "\n",
+        "and  Circuit  phase  angle  is   22.2 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 439</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of R and L, Determine also the voltage across the coil and the voltage across the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  50E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  225;#  in  volts\n",
+      "ri  =  1.5;#  in  Amperes\n",
+      "thetai  =  -30;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  V/I\n",
+      "R  =  Z.real\n",
+      "XL  =  Z.imag  +  Xc\n",
+      " #inductance  L\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      " #Voltage  across  coil\n",
+      "Zcoil  =  R  +  1j*XL\n",
+      "Vcoil  =  I*Zcoil\n",
+      " #Voltage  across  capacitor,\n",
+      "Vc  =  I*(-1j*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  is  \",round(R,2),\"  ohm  and  inductance  is  \",round(  L,3),\" H\\n\"\n",
+      "print  \"\\n  (b)voltage  across  the  coil  is  \",round(Vcoil.real,2),\"  +  (\",round(  Vcoil.imag,2),\")i  V\\n\"\n",
+      "print  \"\\n  (c)voltage  across  the  capacitor  is  \",round(Vc.real,2),\"  +  (\",round(  Vc.imag,2),\")i  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  is   129.9   ohm  and  inductance  is   0.441  H\n",
+        "\n",
+        "\n",
+        "  (b)voltage  across  the  coil  is   272.75   +  ( 82.7 )i  V\n",
+        "\n",
+        "\n",
+        "  (c)voltage  across  the  capacitor  is   -47.75   +  ( -82.7 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 440</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine also the value of the supply voltage V and the circuit phase angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  2.653E-6;#  in  Farads\n",
+      "R1  =  8;#  in  ohms\n",
+      "R2  =  5;#  in  ohms\n",
+      "L  =  0.477E-3;#  in  Henry\n",
+      "f  =  4000;#  in  Hz\n",
+      "ri  =  6;#  in  Amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #impedance  Z1\n",
+      "Z1  =  R1  -  1j*Xc\n",
+      " #inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #impedance  Z2,\n",
+      "Z2  =  R2  +  1j*XL\n",
+      " #voltage  V1\n",
+      "V1  =  I*Z1\n",
+      " #voltage  V2\n",
+      "V2  =  I*Z2\n",
+      " #Supply  voltage,  V\n",
+      "V  =  V1  +  V2\n",
+      "phiv  =  cmath.phase(complex(V.real, V.imag))*180/math.pi\n",
+      "phi  =  phiv  -  thetai\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  voltage  is  \",round(V.real,2),\"  +  (\",round(  V.imag,2),\")i  V\\n\"\n",
+      "print  \"and  Circuit  phase  angle  is  \",round(phi,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  voltage  is   78.0   +  ( -18.06 )i  V\n",
+        "\n",
+        "and  Circuit  phase  angle  is   -13.03 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_3.ipynb
new file mode 100755
index 00000000..8160e132
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_3.ipynb
@@ -0,0 +1,837 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:829d21461d0ddb81a1efda36f24f3a06a46515329e5239b342aaa3633230388b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 24: Application of complex numbers to series a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 433</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "z1  =  12  +  5j;\n",
+      "z2  =  -40j;\n",
+      "r3  =  30;\n",
+      "theta3  =  60;#  in  degrees\n",
+      "r4  =  2.20E6;  \n",
+      "theta4  =  -30;#  in  degrees\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #for  an  R-L  series  circuit,  impedance\n",
+      " #  Z  =  R  +  iXL\n",
+      "Ra  =  z1.real\n",
+      "XLa  =  z1.imag\n",
+      "La  =  XLa/(2*math.pi*f)\n",
+      " #for  a  purely  capacitive  circuit,  impedance  Z  =  -iXc\n",
+      "Xcb  =  abs(z2.imag)\n",
+      "Cb  =  1/(2*math.pi*f*Xcb)\n",
+      "z3  =  r3*cmath.cos(theta3*math.pi/180)  +  (r3*cmath.sin(theta3*math.pi/180))*1j\n",
+      "Rc  =  z3.real\n",
+      "XLc  =  z3.imag\n",
+      "Lc  =  XLc/(2*math.pi*f)\n",
+      "z4  =  r4*cmath.cos(theta4*math.pi/180)  +  (r4*cmath.sin(theta4*math.pi/180))*1j\n",
+      "Rd  =  z4.real\n",
+      "Xcd  =  abs(z4.imag)\n",
+      "Cd  =  1/(2*math.pi*f*Xcd)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of  \",round(  Ra,2),\"  ohm  \"\n",
+      "print   \"in  series  with  an  inductance  of  \",round(La*1000,2),\"mH\"\n",
+      "print  \"\\n  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance  \",round(Cb*1E6,2),\"uF\"\n",
+      "print  \"\\n  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of  \",round(Rc,2),\"  ohm  \"\n",
+      "print   \"in  series  with  an  inductance  of  \",round(Lc*1000,2),\"mH\"\n",
+      "print  \"\\n  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of  \",round(Rd/1000,2),\"kohm \"\n",
+      "print   \" in  series  with  a  capacitor  of  capacitance  \",round(Cd*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of   12.0   ohm  \n",
+        "in  series  with  an  inductance  of   15.92 mH\n",
+        "\n",
+        "  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance   79.58 uF\n",
+        "\n",
+        "  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of   15.0   ohm  \n",
+        "in  series  with  an  inductance  of   82.7 mH\n",
+        "\n",
+        "  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of   1905.26 kohm \n",
+        " in  series  with  a  capacitor  of  capacitance   2.89 nF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 434</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.1592  ;#  in  Henry\n",
+      "V  =  250;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  0;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #for  an  R\u00e2\u20ac\u201cL  series  circuit,  impedance\n",
+      " #  Z  =  R  +  iXL\n",
+      "XL  =  2*math.pi*f*L\n",
+      "Z  =  R  +  1j*XL\n",
+      "I  =  V/Z\n",
+      "x  =  I.real\n",
+      "y  =  I.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "if  ((x==0)&(y<0)):\n",
+      "    theta  =  -90\n",
+      "elif  ((x==0)&(y>0)):\n",
+      "    theta  =  +90\n",
+      "else:\n",
+      "    theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current  is  (\",round(r,2),\"/_\",theta,\"deg)  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current  is  ( 5.0 /_ -90 deg)  A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 435</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  3E-6  ;#  in  farad\n",
+      "f  =  1000;#  in  Hz\n",
+      "ri  =  2.83;\n",
+      "thetai  =  90;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #Capacitive  reactance  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #  circuit  impedance  Z\n",
+      "Z  =  -1*1j*Xc\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      "V  =  I*Z\n",
+      "x  =  V.real\n",
+      "y  =  V.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  p.d.  is \",round(abs(V),0),\"V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  p.d.  is  150.0 V"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 435</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "Z  =  30  -  50j;\n",
+      "\n",
+      "#calculation:\n",
+      " #Since  impedance  Z  =  30  -  i50,\n",
+      " #resistance\n",
+      "R  =  Z.real\n",
+      " #capacitive  reactance\n",
+      "Xc  =  abs(Z.imag)\n",
+      " #capacitance\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      " #modulus  of  impedance\n",
+      "modZ  =  (R**2  +  Xc**2)**0.5\n",
+      "I  =  V/Z\n",
+      "x  =  I.real\n",
+      "y  =  I.imag\n",
+      "r  =  (x**2  +  y**2)**0.5\n",
+      "theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  is  \",round(  R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)capacitance  is  \",round(C*1E6,2),\"uFarad\"\n",
+      "print  \"\\n  (c)modulus  of  impedance  is  \",round(modZ,2),\"  ohm\"\n",
+      "print  \"\\n  (d)current  flowing  and  its  phase  angle  is  (\",round(  r,2),\"/_\",round(  theta,2),\"deg)  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  is   30.0   ohm\n",
+        "\n",
+        "  (b)capacitance  is   63.66 uFarad\n",
+        "\n",
+        "  (c)modulus  of  impedance  is   58.31   ohm\n",
+        "\n",
+        "  (d)current  flowing  and  its  phase  angle  is  ( 4.12 /_ 59.04 deg)  A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 436</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  200;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  32;#  in  ohms\n",
+      "L  =  0.15;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #impedance,  Z\n",
+      "Z  =  R  +  1j*XL\n",
+      " #Current  I\n",
+      "I  =  V/Z\n",
+      "xi  =  I.real\n",
+      "yi  =  I.imag\n",
+      "ri  =  (xi**2  +  yi**2)**0.5\n",
+      "if  ((xi==0)&(yi<0)):\n",
+      "         thetai  =  -90\n",
+      "elif  ((xi==0)&(yi>0)):\n",
+      "         thetai  =  +90\n",
+      "else:\n",
+      "         thetai  =  cmath.phase(complex(xi,yi))*180/math.pi\n",
+      "\n",
+      " #P.d.  across  the  resistor\n",
+      "VR  =  I*R\n",
+      "xr  =  VR.real\n",
+      "yr  =  VR.imag\n",
+      "rr  =  (xr**2  +  yr**2)**0.5\n",
+      "thetar  =  cmath.phase(complex(xr,yr))*180/math.pi\n",
+      " #P.d.  across  the  coil,  VL\n",
+      "VL  =  I*1j*XL\n",
+      "xl  =  VL.real\n",
+      "yl  =  VL.imag\n",
+      "rl  =  (xl**2  +  yl**2)**0.5\n",
+      "thetal  =  cmath.phase(complex(xl,yl))*180/math.pi\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)impedance  is  \",round(Z.real,2),\"  + \",round(  Z.imag,2),\")i  ohm\"\n",
+      "print  \"\\n  (b)current  flowing  and  its  phase  angle  is lagging the voltage = (\",round(  ri,2),\"/_\",round(  thetai,2),\"deg)  A\"\n",
+      "print  \"\\n  (c)P.d.  across  the  resistor  is  (\",round(rr,2),\"/_\",round(thetar,2),\"deg)  V\"\n",
+      "print  \"\\n  (d)P.d.  across  the  coil,  VL  is  (\",round(rl,2),\"/_\",round(thetal,2),\"deg)  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)impedance  is   32.0   +  47.12 )i  ohm\n",
+        "\n",
+        "  (b)current  flowing  and  its  phase  angle  is lagging the voltage = ( 3.51 /_ -55.82 deg)  A\n",
+        "\n",
+        "  (c)P.d.  across  the  resistor  is  ( 112.36 /_ -55.82 deg)  V\n",
+        "\n",
+        "  (d)P.d.  across  the  coil,  VL  is  ( 165.46 /_ 34.18 deg)  V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 436</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "V  =  120  +  200j;#  in  Volts\n",
+      "f  =  5E6;#  in  Hz\n",
+      "I  =  7  +  16j;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z\n",
+      "Z  =  V/I\n",
+      "R  =  Z.real\n",
+      "X  =  Z.imag  \n",
+      "if  ((R>0)&(X<0)):\n",
+      "    C  =  -1/(2*math.pi*f*X)\n",
+      "#Results\n",
+      "    print  \"\\n\\n  Result  \\n\\n\"\n",
+      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm  \"\n",
+      "    print   \"and  a  capacitor  of  capacitive reactance\", round(X*-1,3),\"ohm and capacitance is\",round(C*1E9,2),\" nFarad\\n\"\n",
+      "elif  ((R>0)&(X>0)):\n",
+      "    L  =  2*math.pi*f*X\n",
+      "#Results\n",
+      "    print  \"\\n\\n  Result  \\n\\n\"\n",
+      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm \"\n",
+      "    print   \" and  a  inductor  of  insuctance  \",round(L*100,2),\" mHenry\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  The  series  circuit  thus  consists  of  a  resistor  of  resistance   13.25   ohm  \n",
+        "and  a  capacitor  of  capacitive reactance 1.705 ohm and capacitance is 18.67  nFarad\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 437</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  70;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "ri  =  3.5;#  in  amperes\n",
+      "thetai  =  -20;#  in  degrees\n",
+      " #z1  consist  of  two  resistance\n",
+      "R1  =  4.36;#  in  ohms\n",
+      "R2  =  -2.1j;#  in  ohms\n",
+      "\n",
+      " #calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #impedance,  Z\n",
+      "Z  =  V/I\n",
+      " #Total  impedance  Z  =  z1  +  z2\n",
+      "Z1  =  R1  +  R2\n",
+      "Z2  =  Z  -  Z1\n",
+      "x  =  Z2.real\n",
+      "y  =  Z2.imag  \n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  impedance  Z2  is  \",round(x,2),\"  +  (\",round(y,2),\")i  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  impedance  Z2  is   8.5   +  ( 17.42 )i  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 437</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  90;#  in  ohms\n",
+      "XL  =  150;#  in  ohms\n",
+      "ri  =  1.35;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  1j*XL\n",
+      " #Supply  voltage,  V\n",
+      "V  =  I*Z\n",
+      " #Voltage  across  90  ohm\u000e  resistor\n",
+      "VR  =  V.real\n",
+      "#Voltage  across  inductance,  VL\n",
+      "VL  =  V.imag\n",
+      "xv  =  V.real\n",
+      "yv  =  V.imag\n",
+      "rv  =  (xv**2  +  yv**2)**0.5\n",
+      "thetav  = cmath.phase(complex(xv,yv))*180/math.pi\n",
+      "phi  =  thetav  -  thetai\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Supply  voltage,  V  is  \",xv,\"  +  (\",yv,\")i  V\\n\"\n",
+      "print  \"\\n  (b)Voltage  across  90  ohm resistor,  VR  is  \",VR,\"  V\\n\"\n",
+      "print  \"\\n  (c)Voltage  across  inductance,  VL  is  \",VL,\"  V\\n\"\n",
+      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Supply  voltage,  V  is   121.5   +  ( 202.5 )i  V\n",
+        "\n",
+        "\n",
+        "  (b)Voltage  across  90  ohm resistor,  VR  is   121.5   V\n",
+        "\n",
+        "\n",
+        "  (c)Voltage  across  inductance,  VL  is   202.5   V\n",
+        "\n",
+        "\n",
+        "  (d)Circuit  phase  angle  is   59.04 deg lagging"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 438</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohms\n",
+      "L  =  0.02;#  in  henry\n",
+      "Vm  =  282.8;#  in  volts\n",
+      "w  =  628.4;#  in  rad/sec\n",
+      "phiv  =  math.pi/3;#  phase  angle\n",
+      "\n",
+      "#calculation:\n",
+      " #rms  voltage\n",
+      "Vrms  =  0.707*Vm*math.cos(phiv)  +  0.707*Vm*math.sin(phiv)*1j\n",
+      " #frequency\n",
+      "f  =  w/(2*math.pi)\n",
+      " #Inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  XL*1j\n",
+      " #Rms  current\n",
+      "Irms  =  Vrms/Z\n",
+      "phii  =  cmath.phase(complex(Irms.real, Irms.imag))*180/math.pi\n",
+      "phi  =  phiv*180/math.pi  -  phii\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms.real,2),\"  +  (\",round(  Vrms.imag,2),\")i  V\\n\"\n",
+      "print  \"\\n  (b)the  circuit  impedance  is  \",round(R,2),\"  +  (\",round(  XL,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (c)the  rms  current  flowing  is  \",round(Irms.real,2),\"  +  (\",round(  Irms.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  rms  value  of  voltage  is   99.97   +  ( 173.15 )i  V\n",
+        "\n",
+        "\n",
+        "  (b)the  circuit  impedance  is   25.0   +  ( 12.57 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (c)the  rms  current  flowing  is   5.97   +  ( 3.92 )i  A\n",
+        "\n",
+        "\n",
+        "  (d)Circuit  phase  angle  is   26.69 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 438</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  12;#  in  ohms\n",
+      "L  =  0.10;#  in  henry\n",
+      "C  =  120E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  240;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance,  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  R  +  1j*(XL  -  Xc)\n",
+      "I  =  V/Z\n",
+      "phii  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phiv  =  0#  in  degrees\n",
+      "phi  =  phiv  -  phii\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  is  \",round(abs(I),1),\"/_\",round(cmath.phase(complex(I.real,I.imag))*180/math.pi,1),\"deg  A\\n\"\n",
+      "print  \"and  Circuit  phase  angle  is  \",round(phi,1),\"deg lagging\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  is   18.5 /_ -22.2 deg  A\n",
+        "\n",
+        "and  Circuit  phase  angle  is   22.2 deg lagging\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 439</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  50E-6;#  in  Farads\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  225;#  in  volts\n",
+      "ri  =  1.5;#  in  Amperes\n",
+      "thetai  =  -30;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Circuit  impedance  Z\n",
+      "Z  =  V/I\n",
+      "R  =  Z.real\n",
+      "XL  =  Z.imag  +  Xc\n",
+      " #inductance  L\n",
+      "L  =  XL/(2*math.pi*f)\n",
+      " #Voltage  across  coil\n",
+      "Zcoil  =  R  +  1j*XL\n",
+      "Vcoil  =  I*Zcoil\n",
+      " #Voltage  across  capacitor,\n",
+      "Vc  =  I*(-1j*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)resistance  is  \",round(R,2),\"  ohm  and  inductance  is  \",round(  L,3),\" H\\n\"\n",
+      "print  \"\\n  (b)voltage  across  the  coil  is  \",round(Vcoil.real,2),\"  +  (\",round(  Vcoil.imag,2),\")i  V\\n\"\n",
+      "print  \"\\n  (c)voltage  across  the  capacitor  is  \",round(Vc.real,2),\"  +  (\",round(  Vc.imag,2),\")i  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)resistance  is   129.9   ohm  and  inductance  is   0.441  H\n",
+        "\n",
+        "\n",
+        "  (b)voltage  across  the  coil  is   272.75   +  ( 82.7 )i  V\n",
+        "\n",
+        "\n",
+        "  (c)voltage  across  the  capacitor  is   -47.75   +  ( -82.7 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 440</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  2.653E-6;#  in  Farads\n",
+      "R1  =  8;#  in  ohms\n",
+      "R2  =  5;#  in  ohms\n",
+      "L  =  0.477E-3;#  in  Henry\n",
+      "f  =  4000;#  in  Hz\n",
+      "ri  =  6;#  in  Amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #impedance  Z1\n",
+      "Z1  =  R1  -  1j*Xc\n",
+      " #inductive  reactance  XL\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #impedance  Z2,\n",
+      "Z2  =  R2  +  1j*XL\n",
+      " #voltage  V1\n",
+      "V1  =  I*Z1\n",
+      " #voltage  V2\n",
+      "V2  =  I*Z2\n",
+      " #Supply  voltage,  V\n",
+      "V  =  V1  +  V2\n",
+      "phiv  =  cmath.phase(complex(V.real, V.imag))*180/math.pi\n",
+      "phi  =  phiv  -  thetai\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  supply  voltage  is  \",round(V.real,2),\"  +  (\",round(  V.imag,2),\")i  V\\n\"\n",
+      "print  \"and  Circuit  phase  angle  is  \",round(phi,2),\"deg leading\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  supply  voltage  is   78.0   +  ( -18.06 )i  V\n",
+        "\n",
+        "and  Circuit  phase  angle  is   -13.03 deg leading"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint.ipynb
new file mode 100755
index 00000000..19e462e5
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint.ipynb
@@ -0,0 +1,647 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 25: Application of complex numbers to parallel a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 446</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the admittance, conductance and susceptance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  0  -  5j;#  in  ohms\n",
+      "Z2  =  25  + 40j;#  in  ohms\n",
+      "Z3  =  3  -  2j;#  in  ohms\n",
+      "r4  =  50;#  in  ohms\n",
+      "theta4  =  40;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #admittance  Y\n",
+      "Y1  =  1/Z1\n",
+      " #conductance,  G\n",
+      "G1  =  Y1.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc1  =  abs(Y1.imag)\n",
+      " #admittance  Y\n",
+      "Y2  =  1/Z2\n",
+      " #conductance,  G\n",
+      "G2  =  Y2.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc2  =  abs(Y2.imag)\n",
+      " #admittance  Y\n",
+      "Y3  =  1/Z3\n",
+      " #conductance,  G\n",
+      "G3  =  Y3.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc3  =  abs(Y3.imag)\n",
+      "Z4  =  r4*math.cos(theta4*math.pi/180)  +  1j*r4*math.sin(theta4*math.pi/180)\n",
+      " #admittance  Y\n",
+      "Y4  =  1/Z4\n",
+      " #conductance,  G\n",
+      "G4  =  Y4.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc4  =  abs(Y4.imag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)admittance  Y  is  (\",round(Y1.real,2),\"  +  (\",round(Y1.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G1,2),\"  S,  susceptance,Bc  is  \",round(Bc1,2),\"  S\\n\"\n",
+      "print  \"\\n  (b)admittance  Y  is  (\",round(Y2.real,2),\"  +  (\",round(Y2.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G2,2),\"  S,  susceptance,Bc  is  \",round(Bc2,2),\"  S\\n\"\n",
+      "print  \"\\n  (c)admittance  Y  is  (\",round(Y3.real,2),\"  +  (\",round(Y3.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G3,2),\"  S,  susceptance,Bc  is  \",round(Bc3,2),\"  S\\n\"\n",
+      "print  \"\\n  (d)admittance  Y  is  (\",round(Y4.real,2),\"  +  (\",round(Y4.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G4,2),\"  S,  susceptance,Bc  is  \",round(Bc4,2),\"  S\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)admittance  Y  is  ( -0.0   +  ( 0.2 )i)  S, \n",
+        " conductance,  G  is   -0.0   S,  susceptance,Bc  is   0.2   S\n",
+        "\n",
+        "\n",
+        "  (b)admittance  Y  is  ( 0.01   +  ( -0.02 )i)  S, \n",
+        " conductance,  G  is   0.01   S,  susceptance,Bc  is   0.02   S\n",
+        "\n",
+        "\n",
+        "  (c)admittance  Y  is  ( 0.23   +  ( 0.15 )i)  S, \n",
+        " conductance,  G  is   0.23   S,  susceptance,Bc  is   0.15   S\n",
+        "\n",
+        "\n",
+        "  (d)admittance  Y  is  ( 0.02   +  ( -0.01 )i)  S, \n",
+        " conductance,  G  is   0.02   S,  susceptance,Bc  is   0.01   S\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 447</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine expressions for the impedance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Y2  =  0.001  -  0.002j;#  in  S\n",
+      "Y3  =  0.05  +  0.08j;#  in  S\n",
+      "r1  =  0.004;#  in  S\n",
+      "theta1  =  30;#  in  degrees\n",
+      "\n",
+      " #calculation:\n",
+      " #impedance,  Z\n",
+      "Z2  =  1/Y2\n",
+      "Z3  =  1/Y3\n",
+      "Y1  =  r1*math.cos(theta1*math.pi/180)  +  1j*r1*math.sin(theta1*math.pi/180)\n",
+      "Z1  =  1/Y1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,Z  is  (\",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i)  ohm\\n\"\n",
+      "print  \"\\n  (b)Impedance,Z  is  (\",round(Z2.real,2),\"  +  (\",round(  Z2.imag,2),\")i)  ohm\\n\"\n",
+      "print  \"\\n  (c)Impedance,Z  is  (\",round(Z3.real,2),\"  +  (\",round(  Z3.imag,2),\")i)  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,Z  is  ( 216.51   +  ( -125.0 )i)  ohm\n",
+        "\n",
+        "\n",
+        "  (b)Impedance,Z  is  ( 200.0   +  ( 400.0 )i)  ohm\n",
+        "\n",
+        "\n",
+        "  (c)Impedance,Z  is  ( 5.62   +  ( -8.99 )i)  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 448</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of the resistance and the capacitive reactance of the circuit if they are connected \n",
+      "#(a) in parallel, (b) in series.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Y  =  0.040  -  1j*0.025;#  in  S\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z\n",
+      "Z  =  1/Y\n",
+      " #conductance,  G\n",
+      "G  =  Y.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc  =  abs(Y.imag)\n",
+      " #parallrl  \n",
+      " #resistance,  R\n",
+      "Rp  =  1/G\n",
+      " #capacitive  reactance\n",
+      "Xcp  =  1/Bc\n",
+      " #series\n",
+      " #resistance,  R\n",
+      "Rs  =  Z.real\n",
+      " #capacitive  reactance\n",
+      "Xcs  =  abs(Z.imag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)for  parallel,  resistance,R  is  \",round(Rp,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcp,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)forseries,  resistance,R  is  \",round(Rs,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcs,2),\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)for  parallel,  resistance,R  is   25.0   ohm  and  capacitive  reactance,  Xc  is   40.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)forseries,  resistance,R  is   17.98   ohm  and  capacitive  reactance,  Xc  is   11.24   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 449</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of currents I, I1 and I2.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohm\n",
+      "R  =  5;#  in  ohm\n",
+      "R2  =  6;#  ohm\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage,V\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #circuit  impedance,  ZT\n",
+      "ZT  =  R  +  (R1*1j*R2/(R1  +  1j*R2))\n",
+      " #Current  I\n",
+      "I  =  V/ZT\n",
+      " #current,I1\n",
+      "I1  =  I*(1j*R2/(R1  +  1j*R2))\n",
+      " #current,  I2\n",
+      "I2  =  I*(R1/(R1  +  1j*R2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A,\"\n",
+      "print   \"current,I1  = \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real, I1.imag))*180/math.pi,2),\"deg  A,  \"\n",
+      "print   \"current,  I2  = \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real, I2.imag))*180/math.pi,2),\"deg  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I  =  5.7 /_ -25.98 deg  A,\n",
+        "current,I1  =  3.42 /_ 27.15 deg  A,  \n",
+        "current,  I2  =  4.56 /_ -62.85 deg  A\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 450</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of supply current I and its phase relative to the 40 V supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  3;#  in  ohm  \n",
+      "R3  =  8;#  ohm\n",
+      "Xc  =  4;#  in  ohms\n",
+      "XL  =  12;#  in  Ohms\n",
+      "V  =  40;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Z1  =  R1  +  1j*XL\n",
+      "Z2  =  R2  -  1j*Xc\n",
+      "Z3  =  R3\n",
+      " #circuit  admittance,  YT  =  1/ZT\n",
+      "YT  =  (1/Z1)  +  (1/Z2)  +  (1/Z3)\n",
+      " #Current  I\n",
+      "I  =  V*YT\n",
+      "I1  =  V/Z1\n",
+      "I2  =  V/Z2\n",
+      "I3  =  V/Z2\n",
+      "thetav  =  0\n",
+      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai  \n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I  is  (\",round(I.real,2),\"  +  (\",round(I.imag,2),\")i)  A,\"\n",
+      "print   \"and  its  phase  relative  to  the  40  V  supply  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I  is  ( 10.98   +  ( 3.56 )i)  A,\n",
+        "and  its  phase  relative  to  the  40  V  supply  is   leading s  by   17.96 deg\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 451</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the total equivalent circuit impedance, (b) the supply current, \n",
+      "#(c) the circuit phase angle, (d) the current in the coil, and (e) the current in the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.07958;#  in  Henry\n",
+      "R  =  18;#  in  ohm\n",
+      "C  =  64.96E-6;#  in  Farad\n",
+      "rv  =  250;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #capacitive  reactance\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #impedance  of  the  coil,\n",
+      "Zcoil  =  R  +  1j*XL\n",
+      " #impedance  presented  by  the  capacitor,\n",
+      "Zc  =  -1j*Xc\n",
+      " #Total  equivalent  circuit  impedance,\n",
+      "ZT  =  Zcoil*Zc/(Zcoil  +  Zc)\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #current,  I\n",
+      "I  =  V/ZT\n",
+      "thetai  =  cmath.phase(complex(I.real,I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai\n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      " #Current  in  the  coil,  ICOIL\n",
+      "Icoil  =  V/Zcoil\n",
+      " #Current  in  the  capacitor,  IC\n",
+      "Ic  =  V/Zc\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (b)supply  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\"\n",
+      "print  \"\\n  (d)current  in  coil,  Icoil  = \",round(abs(Icoil),2),\"/_\",round(cmath.phase(complex(Icoil.real, Icoil.imag))*180/math.pi,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (e)current  in  capacitor,  Ic  = \",round(abs(Ic),2),\"/_\",round(cmath.phase(complex(Ic.real, Ic.imag))*180/math.pi,2),\"deg A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  circuit  impedance  is   48.02   +  ( 15.03 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)supply  current,  I  =  4.97 /_ -17.38 deg  A\n",
+        "\n",
+        "\n",
+        "  (c)circuit  phase  relative  is   lagging s  by   17.38 deg\n",
+        "\n",
+        "\n",
+        "  (d)current  in  coil,  Icoil  =  8.12 /_ -54.25 deg  A\n",
+        "\n",
+        "\n",
+        "  (e)current  in  capacitor,  Ic  =  5.1 /_ 90.0 deg A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 452</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)determine the value of impedance Z1 \n",
+      "#(b) If the supply frequency is 5 kHz,determine the value of the components comprising impedance Z1\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  6j;#  in  ohm\n",
+      "R2  =  8;#  in  ohm\n",
+      "Z3  =  10;#  in  ohm\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "ri  =  31.4;#  in  amperes\n",
+      "thetai  =  52.48;#  in  degrees\n",
+      "f  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z2\n",
+      "Z2  =  R2  +  RL\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #current,  I\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Total  circuit  admittance,\n",
+      "YT  =  I/V\n",
+      " #admittance,  Y3\n",
+      "Y3  =  1/Z3\n",
+      " #admittance,  Y2\n",
+      "Y2  =  1/Z2\n",
+      " #admittance,  Y1\n",
+      "Y1  =  YT  -  Y2  -  Y3\n",
+      " #impedance,  Z1\n",
+      "Z1  =  1/Y1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  impedance  Z1  is  \",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i  ohm\\n\"\n",
+      "\n",
+      " #resistance,  R1\n",
+      "R1  =  Z1.real\n",
+      "X1  =  Z1.imag  \n",
+      "if  ((R1>0)&(X1<0)):\n",
+      "    C1  =  -1/(2*math.pi*f*X1)\n",
+      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm\"\n",
+      "    print   \"  and  a  capacitor  of  capacitance  \",round(C1*1E6,2),\"uFarad\\n\"\n",
+      "elif  ((R1>0)&(X1>0)):\n",
+      "    L1  =  2*math.pi*f*X1\n",
+      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm \"\n",
+      "    print   \" and  a  inductor  of  insuctance  \",round(L1*1000,2),\"mHenry\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  impedance  Z1  is   1.6   +  ( -1.2 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance   1.6   ohm\n",
+        "  and  a  capacitor  of  capacitance   26.55 uFarad\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 453</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the equivalent series circuit impedance,\n",
+      "#(b) the supply current I, (c) the circuit phase angle,\n",
+      "#(d) the values of voltages V1 and V2, and (e) the values of currents IA and IB\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL1  =  1.02j;#  in  ohm\n",
+      "R1  =  1.65;#  in  ohm\n",
+      "RLa  =  7j;#  in  ohm\n",
+      "Ra  =  5;#  in  ohm\n",
+      "Rcb  =  -1j*15;#  in  ohm\n",
+      "Rb  =  4;#  in  ohm\n",
+      "rv  =  91;#  in  volts\n",
+      "thetav  =  0;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #impedance,  Z1\n",
+      "Z1  =  R1  +  RL1\n",
+      " #impedance,  Za\n",
+      "Za  =  Ra  +  RLa\n",
+      " #impedance,  Zb\n",
+      "Zb  =  Rb  +  Rcb\n",
+      " #impedance,  Z,  of  the  two  branches  connected  in  parallel\n",
+      "Z  =  Za*Zb/(Za  +  Zb)\n",
+      " #Total  circuit  impedance\n",
+      "ZT  =  Z1  +  Z\n",
+      " #Supply  current,  I\n",
+      "I  =  V/ZT\n",
+      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai  \n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      " #Voltage  V1\n",
+      "V1  =  I*Z1\n",
+      " #Voltage  V2\n",
+      "V2  =  I*Z\n",
+      " #current  Ia\n",
+      "Ia  =  V2/Za\n",
+      " #Current  Ib\n",
+      "Ib  =  V2/Zb\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)equivalent  series  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (b)supply  current,  I  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"  by  \",round(abs(phi),2),\"deg\\n\"\n",
+      "print  \"\\n  (d)voltage,  V1  is  (\",round(V1.real,2),\"  +  (\",round(V1.imag,2),\")i)  V  and  V2  is(\",round(V2.real,2),\"  +  (\",round(  V2.imag,2),\")i)  V\\n\"\n",
+      "print  \"\\n  (e)current,  Ia  is  (\",round(Ia.real,2),\"  +  (\",round( Ia.imag,2),\")i)  A  and  Ib  is(\",round(Ib.real,2),\"  +  (\",round(  Ib.imag,2),\")i)  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)equivalent  series  circuit  impedance  is   12.0   +  ( 5.0 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)supply  current,  I  is   6.46   +  ( -2.69 )i  A\n",
+        "\n",
+        "\n",
+        "  (c)circuit  phase  relative  is   lagging   by   22.61 deg\n",
+        "\n",
+        "\n",
+        "  (d)voltage,  V1  is  ( 13.41   +  ( 2.15 )i)  V  and  V2  is( 77.59   +  ( -2.15 )i)  V\n",
+        "\n",
+        "\n",
+        "  (e)current,  Ia  is  ( 5.04   +  ( -7.49 )i)  A  and  Ib  is( 1.42   +  ( 4.79 )i)  A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_1.ipynb
new file mode 100755
index 00000000..19e462e5
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_1.ipynb
@@ -0,0 +1,647 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 25: Application of complex numbers to parallel a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 446</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the admittance, conductance and susceptance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  0  -  5j;#  in  ohms\n",
+      "Z2  =  25  + 40j;#  in  ohms\n",
+      "Z3  =  3  -  2j;#  in  ohms\n",
+      "r4  =  50;#  in  ohms\n",
+      "theta4  =  40;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #admittance  Y\n",
+      "Y1  =  1/Z1\n",
+      " #conductance,  G\n",
+      "G1  =  Y1.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc1  =  abs(Y1.imag)\n",
+      " #admittance  Y\n",
+      "Y2  =  1/Z2\n",
+      " #conductance,  G\n",
+      "G2  =  Y2.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc2  =  abs(Y2.imag)\n",
+      " #admittance  Y\n",
+      "Y3  =  1/Z3\n",
+      " #conductance,  G\n",
+      "G3  =  Y3.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc3  =  abs(Y3.imag)\n",
+      "Z4  =  r4*math.cos(theta4*math.pi/180)  +  1j*r4*math.sin(theta4*math.pi/180)\n",
+      " #admittance  Y\n",
+      "Y4  =  1/Z4\n",
+      " #conductance,  G\n",
+      "G4  =  Y4.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc4  =  abs(Y4.imag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)admittance  Y  is  (\",round(Y1.real,2),\"  +  (\",round(Y1.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G1,2),\"  S,  susceptance,Bc  is  \",round(Bc1,2),\"  S\\n\"\n",
+      "print  \"\\n  (b)admittance  Y  is  (\",round(Y2.real,2),\"  +  (\",round(Y2.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G2,2),\"  S,  susceptance,Bc  is  \",round(Bc2,2),\"  S\\n\"\n",
+      "print  \"\\n  (c)admittance  Y  is  (\",round(Y3.real,2),\"  +  (\",round(Y3.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G3,2),\"  S,  susceptance,Bc  is  \",round(Bc3,2),\"  S\\n\"\n",
+      "print  \"\\n  (d)admittance  Y  is  (\",round(Y4.real,2),\"  +  (\",round(Y4.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G4,2),\"  S,  susceptance,Bc  is  \",round(Bc4,2),\"  S\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)admittance  Y  is  ( -0.0   +  ( 0.2 )i)  S, \n",
+        " conductance,  G  is   -0.0   S,  susceptance,Bc  is   0.2   S\n",
+        "\n",
+        "\n",
+        "  (b)admittance  Y  is  ( 0.01   +  ( -0.02 )i)  S, \n",
+        " conductance,  G  is   0.01   S,  susceptance,Bc  is   0.02   S\n",
+        "\n",
+        "\n",
+        "  (c)admittance  Y  is  ( 0.23   +  ( 0.15 )i)  S, \n",
+        " conductance,  G  is   0.23   S,  susceptance,Bc  is   0.15   S\n",
+        "\n",
+        "\n",
+        "  (d)admittance  Y  is  ( 0.02   +  ( -0.01 )i)  S, \n",
+        " conductance,  G  is   0.02   S,  susceptance,Bc  is   0.01   S\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 447</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine expressions for the impedance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Y2  =  0.001  -  0.002j;#  in  S\n",
+      "Y3  =  0.05  +  0.08j;#  in  S\n",
+      "r1  =  0.004;#  in  S\n",
+      "theta1  =  30;#  in  degrees\n",
+      "\n",
+      " #calculation:\n",
+      " #impedance,  Z\n",
+      "Z2  =  1/Y2\n",
+      "Z3  =  1/Y3\n",
+      "Y1  =  r1*math.cos(theta1*math.pi/180)  +  1j*r1*math.sin(theta1*math.pi/180)\n",
+      "Z1  =  1/Y1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,Z  is  (\",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i)  ohm\\n\"\n",
+      "print  \"\\n  (b)Impedance,Z  is  (\",round(Z2.real,2),\"  +  (\",round(  Z2.imag,2),\")i)  ohm\\n\"\n",
+      "print  \"\\n  (c)Impedance,Z  is  (\",round(Z3.real,2),\"  +  (\",round(  Z3.imag,2),\")i)  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,Z  is  ( 216.51   +  ( -125.0 )i)  ohm\n",
+        "\n",
+        "\n",
+        "  (b)Impedance,Z  is  ( 200.0   +  ( 400.0 )i)  ohm\n",
+        "\n",
+        "\n",
+        "  (c)Impedance,Z  is  ( 5.62   +  ( -8.99 )i)  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 448</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of the resistance and the capacitive reactance of the circuit if they are connected \n",
+      "#(a) in parallel, (b) in series.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Y  =  0.040  -  1j*0.025;#  in  S\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z\n",
+      "Z  =  1/Y\n",
+      " #conductance,  G\n",
+      "G  =  Y.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc  =  abs(Y.imag)\n",
+      " #parallrl  \n",
+      " #resistance,  R\n",
+      "Rp  =  1/G\n",
+      " #capacitive  reactance\n",
+      "Xcp  =  1/Bc\n",
+      " #series\n",
+      " #resistance,  R\n",
+      "Rs  =  Z.real\n",
+      " #capacitive  reactance\n",
+      "Xcs  =  abs(Z.imag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)for  parallel,  resistance,R  is  \",round(Rp,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcp,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)forseries,  resistance,R  is  \",round(Rs,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcs,2),\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)for  parallel,  resistance,R  is   25.0   ohm  and  capacitive  reactance,  Xc  is   40.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)forseries,  resistance,R  is   17.98   ohm  and  capacitive  reactance,  Xc  is   11.24   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 449</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of currents I, I1 and I2.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohm\n",
+      "R  =  5;#  in  ohm\n",
+      "R2  =  6;#  ohm\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage,V\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #circuit  impedance,  ZT\n",
+      "ZT  =  R  +  (R1*1j*R2/(R1  +  1j*R2))\n",
+      " #Current  I\n",
+      "I  =  V/ZT\n",
+      " #current,I1\n",
+      "I1  =  I*(1j*R2/(R1  +  1j*R2))\n",
+      " #current,  I2\n",
+      "I2  =  I*(R1/(R1  +  1j*R2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A,\"\n",
+      "print   \"current,I1  = \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real, I1.imag))*180/math.pi,2),\"deg  A,  \"\n",
+      "print   \"current,  I2  = \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real, I2.imag))*180/math.pi,2),\"deg  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I  =  5.7 /_ -25.98 deg  A,\n",
+        "current,I1  =  3.42 /_ 27.15 deg  A,  \n",
+        "current,  I2  =  4.56 /_ -62.85 deg  A\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 450</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of supply current I and its phase relative to the 40 V supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  3;#  in  ohm  \n",
+      "R3  =  8;#  ohm\n",
+      "Xc  =  4;#  in  ohms\n",
+      "XL  =  12;#  in  Ohms\n",
+      "V  =  40;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Z1  =  R1  +  1j*XL\n",
+      "Z2  =  R2  -  1j*Xc\n",
+      "Z3  =  R3\n",
+      " #circuit  admittance,  YT  =  1/ZT\n",
+      "YT  =  (1/Z1)  +  (1/Z2)  +  (1/Z3)\n",
+      " #Current  I\n",
+      "I  =  V*YT\n",
+      "I1  =  V/Z1\n",
+      "I2  =  V/Z2\n",
+      "I3  =  V/Z2\n",
+      "thetav  =  0\n",
+      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai  \n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I  is  (\",round(I.real,2),\"  +  (\",round(I.imag,2),\")i)  A,\"\n",
+      "print   \"and  its  phase  relative  to  the  40  V  supply  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I  is  ( 10.98   +  ( 3.56 )i)  A,\n",
+        "and  its  phase  relative  to  the  40  V  supply  is   leading s  by   17.96 deg\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 451</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the total equivalent circuit impedance, (b) the supply current, \n",
+      "#(c) the circuit phase angle, (d) the current in the coil, and (e) the current in the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.07958;#  in  Henry\n",
+      "R  =  18;#  in  ohm\n",
+      "C  =  64.96E-6;#  in  Farad\n",
+      "rv  =  250;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #capacitive  reactance\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #impedance  of  the  coil,\n",
+      "Zcoil  =  R  +  1j*XL\n",
+      " #impedance  presented  by  the  capacitor,\n",
+      "Zc  =  -1j*Xc\n",
+      " #Total  equivalent  circuit  impedance,\n",
+      "ZT  =  Zcoil*Zc/(Zcoil  +  Zc)\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #current,  I\n",
+      "I  =  V/ZT\n",
+      "thetai  =  cmath.phase(complex(I.real,I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai\n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      " #Current  in  the  coil,  ICOIL\n",
+      "Icoil  =  V/Zcoil\n",
+      " #Current  in  the  capacitor,  IC\n",
+      "Ic  =  V/Zc\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (b)supply  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\"\n",
+      "print  \"\\n  (d)current  in  coil,  Icoil  = \",round(abs(Icoil),2),\"/_\",round(cmath.phase(complex(Icoil.real, Icoil.imag))*180/math.pi,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (e)current  in  capacitor,  Ic  = \",round(abs(Ic),2),\"/_\",round(cmath.phase(complex(Ic.real, Ic.imag))*180/math.pi,2),\"deg A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  circuit  impedance  is   48.02   +  ( 15.03 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)supply  current,  I  =  4.97 /_ -17.38 deg  A\n",
+        "\n",
+        "\n",
+        "  (c)circuit  phase  relative  is   lagging s  by   17.38 deg\n",
+        "\n",
+        "\n",
+        "  (d)current  in  coil,  Icoil  =  8.12 /_ -54.25 deg  A\n",
+        "\n",
+        "\n",
+        "  (e)current  in  capacitor,  Ic  =  5.1 /_ 90.0 deg A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 452</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)determine the value of impedance Z1 \n",
+      "#(b) If the supply frequency is 5 kHz,determine the value of the components comprising impedance Z1\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  6j;#  in  ohm\n",
+      "R2  =  8;#  in  ohm\n",
+      "Z3  =  10;#  in  ohm\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "ri  =  31.4;#  in  amperes\n",
+      "thetai  =  52.48;#  in  degrees\n",
+      "f  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z2\n",
+      "Z2  =  R2  +  RL\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #current,  I\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Total  circuit  admittance,\n",
+      "YT  =  I/V\n",
+      " #admittance,  Y3\n",
+      "Y3  =  1/Z3\n",
+      " #admittance,  Y2\n",
+      "Y2  =  1/Z2\n",
+      " #admittance,  Y1\n",
+      "Y1  =  YT  -  Y2  -  Y3\n",
+      " #impedance,  Z1\n",
+      "Z1  =  1/Y1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  impedance  Z1  is  \",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i  ohm\\n\"\n",
+      "\n",
+      " #resistance,  R1\n",
+      "R1  =  Z1.real\n",
+      "X1  =  Z1.imag  \n",
+      "if  ((R1>0)&(X1<0)):\n",
+      "    C1  =  -1/(2*math.pi*f*X1)\n",
+      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm\"\n",
+      "    print   \"  and  a  capacitor  of  capacitance  \",round(C1*1E6,2),\"uFarad\\n\"\n",
+      "elif  ((R1>0)&(X1>0)):\n",
+      "    L1  =  2*math.pi*f*X1\n",
+      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm \"\n",
+      "    print   \" and  a  inductor  of  insuctance  \",round(L1*1000,2),\"mHenry\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  impedance  Z1  is   1.6   +  ( -1.2 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance   1.6   ohm\n",
+        "  and  a  capacitor  of  capacitance   26.55 uFarad\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 453</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the equivalent series circuit impedance,\n",
+      "#(b) the supply current I, (c) the circuit phase angle,\n",
+      "#(d) the values of voltages V1 and V2, and (e) the values of currents IA and IB\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL1  =  1.02j;#  in  ohm\n",
+      "R1  =  1.65;#  in  ohm\n",
+      "RLa  =  7j;#  in  ohm\n",
+      "Ra  =  5;#  in  ohm\n",
+      "Rcb  =  -1j*15;#  in  ohm\n",
+      "Rb  =  4;#  in  ohm\n",
+      "rv  =  91;#  in  volts\n",
+      "thetav  =  0;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #impedance,  Z1\n",
+      "Z1  =  R1  +  RL1\n",
+      " #impedance,  Za\n",
+      "Za  =  Ra  +  RLa\n",
+      " #impedance,  Zb\n",
+      "Zb  =  Rb  +  Rcb\n",
+      " #impedance,  Z,  of  the  two  branches  connected  in  parallel\n",
+      "Z  =  Za*Zb/(Za  +  Zb)\n",
+      " #Total  circuit  impedance\n",
+      "ZT  =  Z1  +  Z\n",
+      " #Supply  current,  I\n",
+      "I  =  V/ZT\n",
+      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai  \n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      " #Voltage  V1\n",
+      "V1  =  I*Z1\n",
+      " #Voltage  V2\n",
+      "V2  =  I*Z\n",
+      " #current  Ia\n",
+      "Ia  =  V2/Za\n",
+      " #Current  Ib\n",
+      "Ib  =  V2/Zb\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)equivalent  series  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (b)supply  current,  I  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"  by  \",round(abs(phi),2),\"deg\\n\"\n",
+      "print  \"\\n  (d)voltage,  V1  is  (\",round(V1.real,2),\"  +  (\",round(V1.imag,2),\")i)  V  and  V2  is(\",round(V2.real,2),\"  +  (\",round(  V2.imag,2),\")i)  V\\n\"\n",
+      "print  \"\\n  (e)current,  Ia  is  (\",round(Ia.real,2),\"  +  (\",round( Ia.imag,2),\")i)  A  and  Ib  is(\",round(Ib.real,2),\"  +  (\",round(  Ib.imag,2),\")i)  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)equivalent  series  circuit  impedance  is   12.0   +  ( 5.0 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)supply  current,  I  is   6.46   +  ( -2.69 )i  A\n",
+        "\n",
+        "\n",
+        "  (c)circuit  phase  relative  is   lagging   by   22.61 deg\n",
+        "\n",
+        "\n",
+        "  (d)voltage,  V1  is  ( 13.41   +  ( 2.15 )i)  V  and  V2  is( 77.59   +  ( -2.15 )i)  V\n",
+        "\n",
+        "\n",
+        "  (e)current,  Ia  is  ( 5.04   +  ( -7.49 )i)  A  and  Ib  is( 1.42   +  ( 4.79 )i)  A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_2.ipynb
new file mode 100755
index 00000000..19e462e5
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_2.ipynb
@@ -0,0 +1,647 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 25: Application of complex numbers to parallel a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 446</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the admittance, conductance and susceptance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  0  -  5j;#  in  ohms\n",
+      "Z2  =  25  + 40j;#  in  ohms\n",
+      "Z3  =  3  -  2j;#  in  ohms\n",
+      "r4  =  50;#  in  ohms\n",
+      "theta4  =  40;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #admittance  Y\n",
+      "Y1  =  1/Z1\n",
+      " #conductance,  G\n",
+      "G1  =  Y1.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc1  =  abs(Y1.imag)\n",
+      " #admittance  Y\n",
+      "Y2  =  1/Z2\n",
+      " #conductance,  G\n",
+      "G2  =  Y2.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc2  =  abs(Y2.imag)\n",
+      " #admittance  Y\n",
+      "Y3  =  1/Z3\n",
+      " #conductance,  G\n",
+      "G3  =  Y3.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc3  =  abs(Y3.imag)\n",
+      "Z4  =  r4*math.cos(theta4*math.pi/180)  +  1j*r4*math.sin(theta4*math.pi/180)\n",
+      " #admittance  Y\n",
+      "Y4  =  1/Z4\n",
+      " #conductance,  G\n",
+      "G4  =  Y4.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc4  =  abs(Y4.imag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)admittance  Y  is  (\",round(Y1.real,2),\"  +  (\",round(Y1.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G1,2),\"  S,  susceptance,Bc  is  \",round(Bc1,2),\"  S\\n\"\n",
+      "print  \"\\n  (b)admittance  Y  is  (\",round(Y2.real,2),\"  +  (\",round(Y2.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G2,2),\"  S,  susceptance,Bc  is  \",round(Bc2,2),\"  S\\n\"\n",
+      "print  \"\\n  (c)admittance  Y  is  (\",round(Y3.real,2),\"  +  (\",round(Y3.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G3,2),\"  S,  susceptance,Bc  is  \",round(Bc3,2),\"  S\\n\"\n",
+      "print  \"\\n  (d)admittance  Y  is  (\",round(Y4.real,2),\"  +  (\",round(Y4.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G4,2),\"  S,  susceptance,Bc  is  \",round(Bc4,2),\"  S\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)admittance  Y  is  ( -0.0   +  ( 0.2 )i)  S, \n",
+        " conductance,  G  is   -0.0   S,  susceptance,Bc  is   0.2   S\n",
+        "\n",
+        "\n",
+        "  (b)admittance  Y  is  ( 0.01   +  ( -0.02 )i)  S, \n",
+        " conductance,  G  is   0.01   S,  susceptance,Bc  is   0.02   S\n",
+        "\n",
+        "\n",
+        "  (c)admittance  Y  is  ( 0.23   +  ( 0.15 )i)  S, \n",
+        " conductance,  G  is   0.23   S,  susceptance,Bc  is   0.15   S\n",
+        "\n",
+        "\n",
+        "  (d)admittance  Y  is  ( 0.02   +  ( -0.01 )i)  S, \n",
+        " conductance,  G  is   0.02   S,  susceptance,Bc  is   0.01   S\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 447</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine expressions for the impedance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Y2  =  0.001  -  0.002j;#  in  S\n",
+      "Y3  =  0.05  +  0.08j;#  in  S\n",
+      "r1  =  0.004;#  in  S\n",
+      "theta1  =  30;#  in  degrees\n",
+      "\n",
+      " #calculation:\n",
+      " #impedance,  Z\n",
+      "Z2  =  1/Y2\n",
+      "Z3  =  1/Y3\n",
+      "Y1  =  r1*math.cos(theta1*math.pi/180)  +  1j*r1*math.sin(theta1*math.pi/180)\n",
+      "Z1  =  1/Y1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,Z  is  (\",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i)  ohm\\n\"\n",
+      "print  \"\\n  (b)Impedance,Z  is  (\",round(Z2.real,2),\"  +  (\",round(  Z2.imag,2),\")i)  ohm\\n\"\n",
+      "print  \"\\n  (c)Impedance,Z  is  (\",round(Z3.real,2),\"  +  (\",round(  Z3.imag,2),\")i)  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,Z  is  ( 216.51   +  ( -125.0 )i)  ohm\n",
+        "\n",
+        "\n",
+        "  (b)Impedance,Z  is  ( 200.0   +  ( 400.0 )i)  ohm\n",
+        "\n",
+        "\n",
+        "  (c)Impedance,Z  is  ( 5.62   +  ( -8.99 )i)  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 448</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of the resistance and the capacitive reactance of the circuit if they are connected \n",
+      "#(a) in parallel, (b) in series.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Y  =  0.040  -  1j*0.025;#  in  S\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z\n",
+      "Z  =  1/Y\n",
+      " #conductance,  G\n",
+      "G  =  Y.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc  =  abs(Y.imag)\n",
+      " #parallrl  \n",
+      " #resistance,  R\n",
+      "Rp  =  1/G\n",
+      " #capacitive  reactance\n",
+      "Xcp  =  1/Bc\n",
+      " #series\n",
+      " #resistance,  R\n",
+      "Rs  =  Z.real\n",
+      " #capacitive  reactance\n",
+      "Xcs  =  abs(Z.imag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)for  parallel,  resistance,R  is  \",round(Rp,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcp,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)forseries,  resistance,R  is  \",round(Rs,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcs,2),\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)for  parallel,  resistance,R  is   25.0   ohm  and  capacitive  reactance,  Xc  is   40.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)forseries,  resistance,R  is   17.98   ohm  and  capacitive  reactance,  Xc  is   11.24   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 449</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of currents I, I1 and I2.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohm\n",
+      "R  =  5;#  in  ohm\n",
+      "R2  =  6;#  ohm\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage,V\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #circuit  impedance,  ZT\n",
+      "ZT  =  R  +  (R1*1j*R2/(R1  +  1j*R2))\n",
+      " #Current  I\n",
+      "I  =  V/ZT\n",
+      " #current,I1\n",
+      "I1  =  I*(1j*R2/(R1  +  1j*R2))\n",
+      " #current,  I2\n",
+      "I2  =  I*(R1/(R1  +  1j*R2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A,\"\n",
+      "print   \"current,I1  = \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real, I1.imag))*180/math.pi,2),\"deg  A,  \"\n",
+      "print   \"current,  I2  = \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real, I2.imag))*180/math.pi,2),\"deg  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I  =  5.7 /_ -25.98 deg  A,\n",
+        "current,I1  =  3.42 /_ 27.15 deg  A,  \n",
+        "current,  I2  =  4.56 /_ -62.85 deg  A\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 450</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of supply current I and its phase relative to the 40 V supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  3;#  in  ohm  \n",
+      "R3  =  8;#  ohm\n",
+      "Xc  =  4;#  in  ohms\n",
+      "XL  =  12;#  in  Ohms\n",
+      "V  =  40;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Z1  =  R1  +  1j*XL\n",
+      "Z2  =  R2  -  1j*Xc\n",
+      "Z3  =  R3\n",
+      " #circuit  admittance,  YT  =  1/ZT\n",
+      "YT  =  (1/Z1)  +  (1/Z2)  +  (1/Z3)\n",
+      " #Current  I\n",
+      "I  =  V*YT\n",
+      "I1  =  V/Z1\n",
+      "I2  =  V/Z2\n",
+      "I3  =  V/Z2\n",
+      "thetav  =  0\n",
+      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai  \n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I  is  (\",round(I.real,2),\"  +  (\",round(I.imag,2),\")i)  A,\"\n",
+      "print   \"and  its  phase  relative  to  the  40  V  supply  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I  is  ( 10.98   +  ( 3.56 )i)  A,\n",
+        "and  its  phase  relative  to  the  40  V  supply  is   leading s  by   17.96 deg\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 451</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the total equivalent circuit impedance, (b) the supply current, \n",
+      "#(c) the circuit phase angle, (d) the current in the coil, and (e) the current in the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.07958;#  in  Henry\n",
+      "R  =  18;#  in  ohm\n",
+      "C  =  64.96E-6;#  in  Farad\n",
+      "rv  =  250;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #capacitive  reactance\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #impedance  of  the  coil,\n",
+      "Zcoil  =  R  +  1j*XL\n",
+      " #impedance  presented  by  the  capacitor,\n",
+      "Zc  =  -1j*Xc\n",
+      " #Total  equivalent  circuit  impedance,\n",
+      "ZT  =  Zcoil*Zc/(Zcoil  +  Zc)\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #current,  I\n",
+      "I  =  V/ZT\n",
+      "thetai  =  cmath.phase(complex(I.real,I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai\n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      " #Current  in  the  coil,  ICOIL\n",
+      "Icoil  =  V/Zcoil\n",
+      " #Current  in  the  capacitor,  IC\n",
+      "Ic  =  V/Zc\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (b)supply  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\"\n",
+      "print  \"\\n  (d)current  in  coil,  Icoil  = \",round(abs(Icoil),2),\"/_\",round(cmath.phase(complex(Icoil.real, Icoil.imag))*180/math.pi,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (e)current  in  capacitor,  Ic  = \",round(abs(Ic),2),\"/_\",round(cmath.phase(complex(Ic.real, Ic.imag))*180/math.pi,2),\"deg A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  circuit  impedance  is   48.02   +  ( 15.03 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)supply  current,  I  =  4.97 /_ -17.38 deg  A\n",
+        "\n",
+        "\n",
+        "  (c)circuit  phase  relative  is   lagging s  by   17.38 deg\n",
+        "\n",
+        "\n",
+        "  (d)current  in  coil,  Icoil  =  8.12 /_ -54.25 deg  A\n",
+        "\n",
+        "\n",
+        "  (e)current  in  capacitor,  Ic  =  5.1 /_ 90.0 deg A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 452</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)determine the value of impedance Z1 \n",
+      "#(b) If the supply frequency is 5 kHz,determine the value of the components comprising impedance Z1\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  6j;#  in  ohm\n",
+      "R2  =  8;#  in  ohm\n",
+      "Z3  =  10;#  in  ohm\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "ri  =  31.4;#  in  amperes\n",
+      "thetai  =  52.48;#  in  degrees\n",
+      "f  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z2\n",
+      "Z2  =  R2  +  RL\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #current,  I\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Total  circuit  admittance,\n",
+      "YT  =  I/V\n",
+      " #admittance,  Y3\n",
+      "Y3  =  1/Z3\n",
+      " #admittance,  Y2\n",
+      "Y2  =  1/Z2\n",
+      " #admittance,  Y1\n",
+      "Y1  =  YT  -  Y2  -  Y3\n",
+      " #impedance,  Z1\n",
+      "Z1  =  1/Y1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  impedance  Z1  is  \",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i  ohm\\n\"\n",
+      "\n",
+      " #resistance,  R1\n",
+      "R1  =  Z1.real\n",
+      "X1  =  Z1.imag  \n",
+      "if  ((R1>0)&(X1<0)):\n",
+      "    C1  =  -1/(2*math.pi*f*X1)\n",
+      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm\"\n",
+      "    print   \"  and  a  capacitor  of  capacitance  \",round(C1*1E6,2),\"uFarad\\n\"\n",
+      "elif  ((R1>0)&(X1>0)):\n",
+      "    L1  =  2*math.pi*f*X1\n",
+      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm \"\n",
+      "    print   \" and  a  inductor  of  insuctance  \",round(L1*1000,2),\"mHenry\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  impedance  Z1  is   1.6   +  ( -1.2 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance   1.6   ohm\n",
+        "  and  a  capacitor  of  capacitance   26.55 uFarad\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 453</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the equivalent series circuit impedance,\n",
+      "#(b) the supply current I, (c) the circuit phase angle,\n",
+      "#(d) the values of voltages V1 and V2, and (e) the values of currents IA and IB\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL1  =  1.02j;#  in  ohm\n",
+      "R1  =  1.65;#  in  ohm\n",
+      "RLa  =  7j;#  in  ohm\n",
+      "Ra  =  5;#  in  ohm\n",
+      "Rcb  =  -1j*15;#  in  ohm\n",
+      "Rb  =  4;#  in  ohm\n",
+      "rv  =  91;#  in  volts\n",
+      "thetav  =  0;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #impedance,  Z1\n",
+      "Z1  =  R1  +  RL1\n",
+      " #impedance,  Za\n",
+      "Za  =  Ra  +  RLa\n",
+      " #impedance,  Zb\n",
+      "Zb  =  Rb  +  Rcb\n",
+      " #impedance,  Z,  of  the  two  branches  connected  in  parallel\n",
+      "Z  =  Za*Zb/(Za  +  Zb)\n",
+      " #Total  circuit  impedance\n",
+      "ZT  =  Z1  +  Z\n",
+      " #Supply  current,  I\n",
+      "I  =  V/ZT\n",
+      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai  \n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      " #Voltage  V1\n",
+      "V1  =  I*Z1\n",
+      " #Voltage  V2\n",
+      "V2  =  I*Z\n",
+      " #current  Ia\n",
+      "Ia  =  V2/Za\n",
+      " #Current  Ib\n",
+      "Ib  =  V2/Zb\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)equivalent  series  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (b)supply  current,  I  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"  by  \",round(abs(phi),2),\"deg\\n\"\n",
+      "print  \"\\n  (d)voltage,  V1  is  (\",round(V1.real,2),\"  +  (\",round(V1.imag,2),\")i)  V  and  V2  is(\",round(V2.real,2),\"  +  (\",round(  V2.imag,2),\")i)  V\\n\"\n",
+      "print  \"\\n  (e)current,  Ia  is  (\",round(Ia.real,2),\"  +  (\",round( Ia.imag,2),\")i)  A  and  Ib  is(\",round(Ib.real,2),\"  +  (\",round(  Ib.imag,2),\")i)  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)equivalent  series  circuit  impedance  is   12.0   +  ( 5.0 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)supply  current,  I  is   6.46   +  ( -2.69 )i  A\n",
+        "\n",
+        "\n",
+        "  (c)circuit  phase  relative  is   lagging   by   22.61 deg\n",
+        "\n",
+        "\n",
+        "  (d)voltage,  V1  is  ( 13.41   +  ( 2.15 )i)  V  and  V2  is( 77.59   +  ( -2.15 )i)  V\n",
+        "\n",
+        "\n",
+        "  (e)current,  Ia  is  ( 5.04   +  ( -7.49 )i)  A  and  Ib  is( 1.42   +  ( 4.79 )i)  A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_3.ipynb
new file mode 100755
index 00000000..592a4669
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint_3.ipynb
@@ -0,0 +1,643 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:9b0edaefe058f9c7f8efe48336cbbb08a13cafea026561f6301c02956a595ae7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 25: Application of complex numbers to parallel a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 446</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  0  -  5j;#  in  ohms\n",
+      "Z2  =  25  + 40j;#  in  ohms\n",
+      "Z3  =  3  -  2j;#  in  ohms\n",
+      "r4  =  50;#  in  ohms\n",
+      "theta4  =  40;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #admittance  Y\n",
+      "Y1  =  1/Z1\n",
+      " #conductance,  G\n",
+      "G1  =  Y1.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc1  =  abs(Y1.imag)\n",
+      " #admittance  Y\n",
+      "Y2  =  1/Z2\n",
+      " #conductance,  G\n",
+      "G2  =  Y2.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc2  =  abs(Y2.imag)\n",
+      " #admittance  Y\n",
+      "Y3  =  1/Z3\n",
+      " #conductance,  G\n",
+      "G3  =  Y3.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc3  =  abs(Y3.imag)\n",
+      "Z4  =  r4*math.cos(theta4*math.pi/180)  +  1j*r4*math.sin(theta4*math.pi/180)\n",
+      " #admittance  Y\n",
+      "Y4  =  1/Z4\n",
+      " #conductance,  G\n",
+      "G4  =  Y4.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc4  =  abs(Y4.imag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)admittance  Y  is  (\",round(Y1.real,2),\"  +  (\",round(Y1.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G1,2),\"  S,  susceptance,Bc  is  \",round(Bc1,2),\"  S\\n\"\n",
+      "print  \"\\n  (b)admittance  Y  is  (\",round(Y2.real,2),\"  +  (\",round(Y2.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G2,2),\"  S,  susceptance,Bc  is  \",round(Bc2,2),\"  S\\n\"\n",
+      "print  \"\\n  (c)admittance  Y  is  (\",round(Y3.real,2),\"  +  (\",round(Y3.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G3,2),\"  S,  susceptance,Bc  is  \",round(Bc3,2),\"  S\\n\"\n",
+      "print  \"\\n  (d)admittance  Y  is  (\",round(Y4.real,2),\"  +  (\",round(Y4.imag,2),\")i)  S, \"\n",
+      "print   \" conductance,  G  is  \",round(G4,2),\"  S,  susceptance,Bc  is  \",round(Bc4,2),\"  S\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)admittance  Y  is  ( -0.0   +  ( 0.2 )i)  S, \n",
+        " conductance,  G  is   -0.0   S,  susceptance,Bc  is   0.2   S\n",
+        "\n",
+        "\n",
+        "  (b)admittance  Y  is  ( 0.01   +  ( -0.02 )i)  S, \n",
+        " conductance,  G  is   0.01   S,  susceptance,Bc  is   0.02   S\n",
+        "\n",
+        "\n",
+        "  (c)admittance  Y  is  ( 0.23   +  ( 0.15 )i)  S, \n",
+        " conductance,  G  is   0.23   S,  susceptance,Bc  is   0.15   S\n",
+        "\n",
+        "\n",
+        "  (d)admittance  Y  is  ( 0.02   +  ( -0.01 )i)  S, \n",
+        " conductance,  G  is   0.02   S,  susceptance,Bc  is   0.01   S\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 447</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Y2  =  0.001  -  0.002j;#  in  S\n",
+      "Y3  =  0.05  +  0.08j;#  in  S\n",
+      "r1  =  0.004;#  in  S\n",
+      "theta1  =  30;#  in  degrees\n",
+      "\n",
+      " #calculation:\n",
+      " #impedance,  Z\n",
+      "Z2  =  1/Y2\n",
+      "Z3  =  1/Y3\n",
+      "Y1  =  r1*math.cos(theta1*math.pi/180)  +  1j*r1*math.sin(theta1*math.pi/180)\n",
+      "Z1  =  1/Y1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Impedance,Z  is  (\",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i)  ohm\\n\"\n",
+      "print  \"\\n  (b)Impedance,Z  is  (\",round(Z2.real,2),\"  +  (\",round(  Z2.imag,2),\")i)  ohm\\n\"\n",
+      "print  \"\\n  (c)Impedance,Z  is  (\",round(Z3.real,2),\"  +  (\",round(  Z3.imag,2),\")i)  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Impedance,Z  is  ( 216.51   +  ( -125.0 )i)  ohm\n",
+        "\n",
+        "\n",
+        "  (b)Impedance,Z  is  ( 200.0   +  ( 400.0 )i)  ohm\n",
+        "\n",
+        "\n",
+        "  (c)Impedance,Z  is  ( 5.62   +  ( -8.99 )i)  ohm"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 448</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Y  =  0.040  -  1j*0.025;#  in  S\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z\n",
+      "Z  =  1/Y\n",
+      " #conductance,  G\n",
+      "G  =  Y.real\n",
+      " #Suspectance,  Bc\n",
+      "Bc  =  abs(Y.imag)\n",
+      " #parallrl  \n",
+      " #resistance,  R\n",
+      "Rp  =  1/G\n",
+      " #capacitive  reactance\n",
+      "Xcp  =  1/Bc\n",
+      " #series\n",
+      " #resistance,  R\n",
+      "Rs  =  Z.real\n",
+      " #capacitive  reactance\n",
+      "Xcs  =  abs(Z.imag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)for  parallel,  resistance,R  is  \",round(Rp,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcp,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)forseries,  resistance,R  is  \",round(Rs,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcs,2),\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)for  parallel,  resistance,R  is   25.0   ohm  and  capacitive  reactance,  Xc  is   40.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)forseries,  resistance,R  is   17.98   ohm  and  capacitive  reactance,  Xc  is   11.24   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 449</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  8;#  in  ohm\n",
+      "R  =  5;#  in  ohm\n",
+      "R2  =  6;#  ohm\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage,V\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #circuit  impedance,  ZT\n",
+      "ZT  =  R  +  (R1*1j*R2/(R1  +  1j*R2))\n",
+      " #Current  I\n",
+      "I  =  V/ZT\n",
+      " #current,I1\n",
+      "I1  =  I*(1j*R2/(R1  +  1j*R2))\n",
+      " #current,  I2\n",
+      "I2  =  I*(R1/(R1  +  1j*R2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A,\"\n",
+      "print   \"current,I1  = \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real, I1.imag))*180/math.pi,2),\"deg  A,  \"\n",
+      "print   \"current,  I2  = \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real, I2.imag))*180/math.pi,2),\"deg  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I  =  5.7 /_ -25.98 deg  A,\n",
+        "current,I1  =  3.42 /_ 27.15 deg  A,  \n",
+        "current,  I2  =  4.56 /_ -62.85 deg  A\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 450</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  3;#  in  ohm  \n",
+      "R3  =  8;#  ohm\n",
+      "Xc  =  4;#  in  ohms\n",
+      "XL  =  12;#  in  Ohms\n",
+      "V  =  40;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "Z1  =  R1  +  1j*XL\n",
+      "Z2  =  R2  -  1j*Xc\n",
+      "Z3  =  R3\n",
+      " #circuit  admittance,  YT  =  1/ZT\n",
+      "YT  =  (1/Z1)  +  (1/Z2)  +  (1/Z3)\n",
+      " #Current  I\n",
+      "I  =  V*YT\n",
+      "I1  =  V/Z1\n",
+      "I2  =  V/Z2\n",
+      "I3  =  V/Z2\n",
+      "thetav  =  0\n",
+      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai  \n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I  is  (\",round(I.real,2),\"  +  (\",round(I.imag,2),\")i)  A,\"\n",
+      "print   \"and  its  phase  relative  to  the  40  V  supply  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I  is  ( 10.98   +  ( 3.56 )i)  A,\n",
+        "and  its  phase  relative  to  the  40  V  supply  is   leading s  by   17.96 deg\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 451</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.07958;#  in  Henry\n",
+      "R  =  18;#  in  ohm\n",
+      "C  =  64.96E-6;#  in  Farad\n",
+      "rv  =  250;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #Inductive  reactance\n",
+      "XL  =  2*math.pi*f*L\n",
+      " #capacitive  reactance\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #impedance  of  the  coil,\n",
+      "Zcoil  =  R  +  1j*XL\n",
+      " #impedance  presented  by  the  capacitor,\n",
+      "Zc  =  -1j*Xc\n",
+      " #Total  equivalent  circuit  impedance,\n",
+      "ZT  =  Zcoil*Zc/(Zcoil  +  Zc)\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #current,  I\n",
+      "I  =  V/ZT\n",
+      "thetai  =  cmath.phase(complex(I.real,I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai\n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      " #Current  in  the  coil,  ICOIL\n",
+      "Icoil  =  V/Zcoil\n",
+      " #Current  in  the  capacitor,  IC\n",
+      "Ic  =  V/Zc\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (b)supply  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\"\n",
+      "print  \"\\n  (d)current  in  coil,  Icoil  = \",round(abs(Icoil),2),\"/_\",round(cmath.phase(complex(Icoil.real, Icoil.imag))*180/math.pi,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (e)current  in  capacitor,  Ic  = \",round(abs(Ic),2),\"/_\",round(cmath.phase(complex(Ic.real, Ic.imag))*180/math.pi,2),\"deg A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  circuit  impedance  is   48.02   +  ( 15.03 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)supply  current,  I  =  4.97 /_ -17.38 deg  A\n",
+        "\n",
+        "\n",
+        "  (c)circuit  phase  relative  is   lagging s  by   17.38 deg\n",
+        "\n",
+        "\n",
+        "  (d)current  in  coil,  Icoil  =  8.12 /_ -54.25 deg  A\n",
+        "\n",
+        "\n",
+        "  (e)current  in  capacitor,  Ic  =  5.1 /_ 90.0 deg A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 452</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  6j;#  in  ohm\n",
+      "R2  =  8;#  in  ohm\n",
+      "Z3  =  10;#  in  ohm\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "ri  =  31.4;#  in  amperes\n",
+      "thetai  =  52.48;#  in  degrees\n",
+      "f  =  5000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z2\n",
+      "Z2  =  R2  +  RL\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #current,  I\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Total  circuit  admittance,\n",
+      "YT  =  I/V\n",
+      " #admittance,  Y3\n",
+      "Y3  =  1/Z3\n",
+      " #admittance,  Y2\n",
+      "Y2  =  1/Z2\n",
+      " #admittance,  Y1\n",
+      "Y1  =  YT  -  Y2  -  Y3\n",
+      " #impedance,  Z1\n",
+      "Z1  =  1/Y1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  impedance  Z1  is  \",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i  ohm\\n\"\n",
+      "\n",
+      " #resistance,  R1\n",
+      "R1  =  Z1.real\n",
+      "X1  =  Z1.imag  \n",
+      "if  ((R1>0)&(X1<0)):\n",
+      "    C1  =  -1/(2*math.pi*f*X1)\n",
+      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm\"\n",
+      "    print   \"  and  a  capacitor  of  capacitance  \",round(C1*1E6,2),\"uFarad\\n\"\n",
+      "elif  ((R1>0)&(X1>0)):\n",
+      "    L1  =  2*math.pi*f*X1\n",
+      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm \"\n",
+      "    print   \" and  a  inductor  of  insuctance  \",round(L1*1000,2),\"mHenry\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  impedance  Z1  is   1.6   +  ( -1.2 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance   1.6   ohm\n",
+        "  and  a  capacitor  of  capacitance   26.55 uFarad\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 453</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL1  =  1.02j;#  in  ohm\n",
+      "R1  =  1.65;#  in  ohm\n",
+      "RLa  =  7j;#  in  ohm\n",
+      "Ra  =  5;#  in  ohm\n",
+      "Rcb  =  -1j*15;#  in  ohm\n",
+      "Rb  =  4;#  in  ohm\n",
+      "rv  =  91;#  in  volts\n",
+      "thetav  =  0;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #impedance,  Z1\n",
+      "Z1  =  R1  +  RL1\n",
+      " #impedance,  Za\n",
+      "Za  =  Ra  +  RLa\n",
+      " #impedance,  Zb\n",
+      "Zb  =  Rb  +  Rcb\n",
+      " #impedance,  Z,  of  the  two  branches  connected  in  parallel\n",
+      "Z  =  Za*Zb/(Za  +  Zb)\n",
+      " #Total  circuit  impedance\n",
+      "ZT  =  Z1  +  Z\n",
+      " #Supply  current,  I\n",
+      "I  =  V/ZT\n",
+      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
+      "phi  =  thetav  -  thetai  \n",
+      "if  (phi>0):\n",
+      "         a  =  \"lagging\"\n",
+      "else:\n",
+      "         a  =  \"leading\"\n",
+      "\n",
+      " #Voltage  V1\n",
+      "V1  =  I*Z1\n",
+      " #Voltage  V2\n",
+      "V2  =  I*Z\n",
+      " #current  Ia\n",
+      "Ia  =  V2/Za\n",
+      " #Current  Ib\n",
+      "Ib  =  V2/Zb\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)equivalent  series  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
+      "print  \"\\n  (b)supply  current,  I  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"  by  \",round(abs(phi),2),\"deg\\n\"\n",
+      "print  \"\\n  (d)voltage,  V1  is  (\",round(V1.real,2),\"  +  (\",round(V1.imag,2),\")i)  V  and  V2  is(\",round(V2.real,2),\"  +  (\",round(  V2.imag,2),\")i)  V\\n\"\n",
+      "print  \"\\n  (e)current,  Ia  is  (\",round(Ia.real,2),\"  +  (\",round( Ia.imag,2),\")i)  A  and  Ib  is(\",round(Ib.real,2),\"  +  (\",round(  Ib.imag,2),\")i)  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)equivalent  series  circuit  impedance  is   12.0   +  ( 5.0 )i  ohm\n",
+        "\n",
+        "\n",
+        "  (b)supply  current,  I  is   6.46   +  ( -2.69 )i  A\n",
+        "\n",
+        "\n",
+        "  (c)circuit  phase  relative  is   lagging   by   22.61 deg\n",
+        "\n",
+        "\n",
+        "  (d)voltage,  V1  is  ( 13.41   +  ( 2.15 )i)  V  and  V2  is( 77.59   +  ( -2.15 )i)  V\n",
+        "\n",
+        "\n",
+        "  (e)current,  Ia  is  ( 5.04   +  ( -7.49 )i)  A  and  Ib  is( 1.42   +  ( 4.79 )i)  A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint.ipynb
new file mode 100755
index 00000000..4edde932
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint.ipynb
@@ -0,0 +1,496 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 26: Power in a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 466</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the active power in the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  12j;#  in  ohm\n",
+      "R  =  5;#  in  ohm\n",
+      "rv  =  52;#  in  volts\n",
+      "thetav  =  30;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #impedance,  Z\n",
+      "Z  =  R  +  RL\n",
+      " #current\n",
+      "I  =  V/Z\n",
+      " #Active  power,  P\n",
+      "Pa  =  V.real*I.real  +  V.imag*I.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nthe  active  power  in  the  circuit  \",Pa,\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "the  active  power  in  the  circuit   80.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 467</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the active power, and (b) the reactive power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  120  +  200j;#  in  volts\n",
+      "I  =  15  +  8j;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Active  power,  P\n",
+      "Pa  =  V.real*I.real  +  V.imag*I.imag\n",
+      " #Reactive  power,  Q\n",
+      "Q  =  V.imag*I.real  -  V.real*I.imag\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  active  power  in  the  circuit  \",Pa,\"  W\\n\"\n",
+      "print  \"\\n  (b)  the  reactive  power  in  the  circuit  \",Q,\"  var\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  active  power  in  the  circuit   3400.0   W\n",
+        "\n",
+        "\n",
+        "  (b)  the  reactive  power  in  the  circuit   2040.0   var"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 468</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current flowing and its phase, (b) the value of resistance R, and (c) the value of capacitance C.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vm  =  141.4;#  in  volts\n",
+      "w  =  10000;#  in  rad/sec\n",
+      "phiv  =  math.pi/9;#  in  radian\n",
+      "Pd  =  1732;#  in  Watts\n",
+      "pf  =  0.866;#  power  fctr\n",
+      "\n",
+      "#calculation:\n",
+      " #the  rms  voltage,\n",
+      "Vrms  =  0.707*Vm\n",
+      " #Power  P  =  V*I*cos(phi)\n",
+      " #current  magnitude,  Irms\n",
+      "Irms  =  Pd/(Vrms*pf)\n",
+      "phid  =  math.acos(pf)\n",
+      " #current  phase  angle\n",
+      "phii  =  phiv  +  phid\n",
+      "phiid  =  phii*180/math.pi#  in  degrees\n",
+      " #Voltage,  V\n",
+      "V  =  Vrms*math.cos(phiv)  +  1j*Vrms*math.sin(phiv)\n",
+      " #current,  I\n",
+      "I  =  Irms*math.cos(phii)  +  1j*Irms*math.sin(phii)\n",
+      " #Impedance,  Z\n",
+      "Z  =  V/I\n",
+      " #resistance,  R\n",
+      "R  =  Z.real\n",
+      " #capacitive  reactance,  Xc\n",
+      "Xc  =  abs(Z.imag)\n",
+      " #capacitance,  C\n",
+      "C  =  1/  (w*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  and  Circuit  phase  angle  is  \",round(Irms,2),\"/_\",round(phiid,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (b)  the  resistance  is  \",round(R,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (c)  the  capacitance  is  \",round(C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  and  Circuit  phase  angle  is   20.01 /_ 50.0 deg  A\n",
+        "\n",
+        "\n",
+        "  (b)  the  resistance  is   4.33   ohm\n",
+        "\n",
+        "\n",
+        "  (c)  the  capacitance  is   40.02 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 468</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the active power developed between points (a) A and B, (b) C and D, (c) E and F.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R  =  5;#  in  ohm\n",
+      "R1  =  3;#  in  ohms\n",
+      "RL  =  4j;#  in  ohm\n",
+      "Rc  =  -10j;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z1\n",
+      "Z1  =  R1  +  RL\n",
+      " #impedance,  Zc\n",
+      "Zc  =  Rc\n",
+      " #Circuit  impedance,  Z\n",
+      "Z  =  R  +  (Z1*Zc/(Z1  +  Zc))\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)\n",
+      "I  =  V/Z\n",
+      "Imag  =  ((I.real)**2  +  (I.imag)**2)**0.5\n",
+      " #Active  power  developed  between  points  A  and  B\n",
+      "Pab  =  (Imag**2)*R\n",
+      " #Active  power  developed  between  points  C  and  D\n",
+      "Pcd  =  (Imag**2)*Zc.real\n",
+      " #Current,  I1\n",
+      "I1  =  I*Zc/(Zc  +  Z1)\n",
+      "I1mag  =  ((I1.real)**2  +  (I1.imag)**2)**0.5\n",
+      " #active  power  developed  between  points  E  and  F\n",
+      "Pef  =  (I1mag**2)*Z1.real\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Active  power  developed  between  points  A  and  B  is  \",round(Pab,2),\"  W\\n\"\n",
+      "print  \"\\n  (b)Active  power  developed  between  points  C  and  D  is  \",round(Pcd,2),\"  W\\n\"\n",
+      "print  \"\\n  (c)Active  power  developed  between  points  E  and  F  is  \",round(Pef,2),\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Active  power  developed  between  points  A  and  B  is   339.62   W\n",
+        "\n",
+        "\n",
+        "  (b)Active  power  developed  between  points  C  and  D  is   0.0   W\n",
+        "\n",
+        "\n",
+        "  (c)Active  power  developed  between  points  E  and  F  is   452.83   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 469</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the apparent power, (b) the reactive power, \n",
+      "#(c) the value and phase of current I, and (d) the value of impedance Z.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pa  =  400;#  in  Watts\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "R  =  4;#  in  ohm\n",
+      "pf  =  0.766;#  power  factor\n",
+      "\n",
+      " #calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #magnitude  of  apparent  power,S  =  V*I\n",
+      "S  =  Pa/pf\n",
+      "phi  =  math.acos(pf)\n",
+      "theta  =  phi*180/math.pi#  in  degrees\n",
+      " #Reactive  power  Q\n",
+      "Q  =  S*math.sin(phi)\n",
+      " #magnitude  of  current\n",
+      "Imag  =  S/rv\n",
+      "thetai  =  thetav  -  theta\n",
+      "I  =  Imag*math.cos(thetai*math.pi/180)  +  1j*Imag*math.sin(thetai*math.pi/180)\n",
+      " #Total  circuit  impedance  ZT\n",
+      "ZT  =  V/I\n",
+      " #impedance  Z\n",
+      "Z  =  ZT  -  R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)apparent  power  is  \",round(S,2),\"  VA\\n\"\n",
+      "print  \"\\n  (b)reactive  power  is  \",round(Q,1),\"  var lagging\\n\"\n",
+      "print  \"\\n  (c)the  current  flowing  and  Circuit  phase  angle  is  \",round(Imag,2),\"/_\",round(thetai,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (d)impedance,  Z  is  \",round(Z.real,2),\"  +  (\",round(  Z.imag,2),\")i  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)apparent  power  is   522.19   VA\n",
+        "\n",
+        "\n",
+        "  (b)reactive  power  is   335.7   var lagging\n",
+        "\n",
+        "\n",
+        "  (c)the  current  flowing  and  Circuit  phase  angle  is   5.22 /_ -10.0 deg  A\n",
+        "\n",
+        "\n",
+        "  (d)impedance,  Z  is   10.67   +  ( 12.31 )i  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 471</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the rating (in kilovars) of the capacitors required.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "S  =  300000;#  in  VA\n",
+      "pf1  =  0.70;#  in  power  factor\n",
+      "pf2  =  0.90;#  in  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #active  power,  P\n",
+      "Pa  =  S*pf1\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      " #Reactive  power,  Q\n",
+      "Q  =  S*math.sin(phi1)\n",
+      "phi2  =  math.acos(pf2)\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      " #The  capacitor  rating  needed  to  improve  the  power  factor  to  0.90\n",
+      " #the  capacitor  rating,\n",
+      "Pr  =  Q  -  (Pa*math.tan(phi2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  rating  (in  kilovars)  of  the  capacitors  is  \",round((Pr/1E3),2),\"  kvar leading\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  rating  (in  kilovars)  of  the  capacitors  is   112.54   kvar leading"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 471</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the supply current, (b) the active, apparent and reactive power,\n",
+      "#(c) the rating of a capacitor (d) the value of capacitance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z  =  3  +  4j;#  in  ohms\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  30;#  in  Degrees\n",
+      "f  =  1500;#  in  Hz\n",
+      "pf1  =  0.966;#  in  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Supply  current,  I\n",
+      "I  =  V/Z\n",
+      "Istr  =  I.real  -  1j*I.imag\n",
+      " #Apparent  power,  S\n",
+      "S  =  V*Istr\n",
+      " #active  power,  Pa\n",
+      "Pa  =  S.real\n",
+      "#reactive  power,  Q\n",
+      "Q  =  abs(S.imag)\n",
+      " #apparent  power,  S\n",
+      "S  =  (S.real**2  +  S.imag**2)**0.5\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      " #rating  of  the  capacitor  \n",
+      "Pr  =  Q  -  Pa*math.tan(phi1)\n",
+      " #Current  in  capacitor,  Ic\n",
+      "Ic  =  Pr/rv\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  rv/Ic\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)supply  current,  I  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (b)active  power  is  \",round(Pa,2),\"  W,  apparent  power  is  \",round(  S,2),\"  W  \"\n",
+      "print   \"and  reactive  power  is  \",round(  Q,2),\"  W lagging\\n\"\n",
+      "print  \"\\n  (c)the  rating  of  the  capacitors  is  \",round(Pr,2),\"  var leading\\n\"\n",
+      "print  \"\\n  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is  \",round(  C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)supply  current,  I  is   9.2   +  ( -3.93 )i  A\n",
+        "\n",
+        "\n",
+        "  (b)active  power  is   300.0   W,  apparent  power  is   500.0   W  and  reactive  power  is   400.0   W lagging\n",
+        "\n",
+        "\n",
+        "  (c)the  rating  of  the  capacitors  is   319.71   var leading\n",
+        "\n",
+        "\n",
+        "  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is   13.57 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_1.ipynb
new file mode 100755
index 00000000..4edde932
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_1.ipynb
@@ -0,0 +1,496 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 26: Power in a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 466</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the active power in the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  12j;#  in  ohm\n",
+      "R  =  5;#  in  ohm\n",
+      "rv  =  52;#  in  volts\n",
+      "thetav  =  30;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #impedance,  Z\n",
+      "Z  =  R  +  RL\n",
+      " #current\n",
+      "I  =  V/Z\n",
+      " #Active  power,  P\n",
+      "Pa  =  V.real*I.real  +  V.imag*I.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nthe  active  power  in  the  circuit  \",Pa,\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "the  active  power  in  the  circuit   80.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 467</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the active power, and (b) the reactive power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  120  +  200j;#  in  volts\n",
+      "I  =  15  +  8j;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Active  power,  P\n",
+      "Pa  =  V.real*I.real  +  V.imag*I.imag\n",
+      " #Reactive  power,  Q\n",
+      "Q  =  V.imag*I.real  -  V.real*I.imag\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  active  power  in  the  circuit  \",Pa,\"  W\\n\"\n",
+      "print  \"\\n  (b)  the  reactive  power  in  the  circuit  \",Q,\"  var\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  active  power  in  the  circuit   3400.0   W\n",
+        "\n",
+        "\n",
+        "  (b)  the  reactive  power  in  the  circuit   2040.0   var"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 468</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current flowing and its phase, (b) the value of resistance R, and (c) the value of capacitance C.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vm  =  141.4;#  in  volts\n",
+      "w  =  10000;#  in  rad/sec\n",
+      "phiv  =  math.pi/9;#  in  radian\n",
+      "Pd  =  1732;#  in  Watts\n",
+      "pf  =  0.866;#  power  fctr\n",
+      "\n",
+      "#calculation:\n",
+      " #the  rms  voltage,\n",
+      "Vrms  =  0.707*Vm\n",
+      " #Power  P  =  V*I*cos(phi)\n",
+      " #current  magnitude,  Irms\n",
+      "Irms  =  Pd/(Vrms*pf)\n",
+      "phid  =  math.acos(pf)\n",
+      " #current  phase  angle\n",
+      "phii  =  phiv  +  phid\n",
+      "phiid  =  phii*180/math.pi#  in  degrees\n",
+      " #Voltage,  V\n",
+      "V  =  Vrms*math.cos(phiv)  +  1j*Vrms*math.sin(phiv)\n",
+      " #current,  I\n",
+      "I  =  Irms*math.cos(phii)  +  1j*Irms*math.sin(phii)\n",
+      " #Impedance,  Z\n",
+      "Z  =  V/I\n",
+      " #resistance,  R\n",
+      "R  =  Z.real\n",
+      " #capacitive  reactance,  Xc\n",
+      "Xc  =  abs(Z.imag)\n",
+      " #capacitance,  C\n",
+      "C  =  1/  (w*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  and  Circuit  phase  angle  is  \",round(Irms,2),\"/_\",round(phiid,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (b)  the  resistance  is  \",round(R,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (c)  the  capacitance  is  \",round(C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  and  Circuit  phase  angle  is   20.01 /_ 50.0 deg  A\n",
+        "\n",
+        "\n",
+        "  (b)  the  resistance  is   4.33   ohm\n",
+        "\n",
+        "\n",
+        "  (c)  the  capacitance  is   40.02 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 468</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the active power developed between points (a) A and B, (b) C and D, (c) E and F.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R  =  5;#  in  ohm\n",
+      "R1  =  3;#  in  ohms\n",
+      "RL  =  4j;#  in  ohm\n",
+      "Rc  =  -10j;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z1\n",
+      "Z1  =  R1  +  RL\n",
+      " #impedance,  Zc\n",
+      "Zc  =  Rc\n",
+      " #Circuit  impedance,  Z\n",
+      "Z  =  R  +  (Z1*Zc/(Z1  +  Zc))\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)\n",
+      "I  =  V/Z\n",
+      "Imag  =  ((I.real)**2  +  (I.imag)**2)**0.5\n",
+      " #Active  power  developed  between  points  A  and  B\n",
+      "Pab  =  (Imag**2)*R\n",
+      " #Active  power  developed  between  points  C  and  D\n",
+      "Pcd  =  (Imag**2)*Zc.real\n",
+      " #Current,  I1\n",
+      "I1  =  I*Zc/(Zc  +  Z1)\n",
+      "I1mag  =  ((I1.real)**2  +  (I1.imag)**2)**0.5\n",
+      " #active  power  developed  between  points  E  and  F\n",
+      "Pef  =  (I1mag**2)*Z1.real\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Active  power  developed  between  points  A  and  B  is  \",round(Pab,2),\"  W\\n\"\n",
+      "print  \"\\n  (b)Active  power  developed  between  points  C  and  D  is  \",round(Pcd,2),\"  W\\n\"\n",
+      "print  \"\\n  (c)Active  power  developed  between  points  E  and  F  is  \",round(Pef,2),\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Active  power  developed  between  points  A  and  B  is   339.62   W\n",
+        "\n",
+        "\n",
+        "  (b)Active  power  developed  between  points  C  and  D  is   0.0   W\n",
+        "\n",
+        "\n",
+        "  (c)Active  power  developed  between  points  E  and  F  is   452.83   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 469</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the apparent power, (b) the reactive power, \n",
+      "#(c) the value and phase of current I, and (d) the value of impedance Z.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pa  =  400;#  in  Watts\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "R  =  4;#  in  ohm\n",
+      "pf  =  0.766;#  power  factor\n",
+      "\n",
+      " #calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #magnitude  of  apparent  power,S  =  V*I\n",
+      "S  =  Pa/pf\n",
+      "phi  =  math.acos(pf)\n",
+      "theta  =  phi*180/math.pi#  in  degrees\n",
+      " #Reactive  power  Q\n",
+      "Q  =  S*math.sin(phi)\n",
+      " #magnitude  of  current\n",
+      "Imag  =  S/rv\n",
+      "thetai  =  thetav  -  theta\n",
+      "I  =  Imag*math.cos(thetai*math.pi/180)  +  1j*Imag*math.sin(thetai*math.pi/180)\n",
+      " #Total  circuit  impedance  ZT\n",
+      "ZT  =  V/I\n",
+      " #impedance  Z\n",
+      "Z  =  ZT  -  R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)apparent  power  is  \",round(S,2),\"  VA\\n\"\n",
+      "print  \"\\n  (b)reactive  power  is  \",round(Q,1),\"  var lagging\\n\"\n",
+      "print  \"\\n  (c)the  current  flowing  and  Circuit  phase  angle  is  \",round(Imag,2),\"/_\",round(thetai,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (d)impedance,  Z  is  \",round(Z.real,2),\"  +  (\",round(  Z.imag,2),\")i  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)apparent  power  is   522.19   VA\n",
+        "\n",
+        "\n",
+        "  (b)reactive  power  is   335.7   var lagging\n",
+        "\n",
+        "\n",
+        "  (c)the  current  flowing  and  Circuit  phase  angle  is   5.22 /_ -10.0 deg  A\n",
+        "\n",
+        "\n",
+        "  (d)impedance,  Z  is   10.67   +  ( 12.31 )i  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 471</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the rating (in kilovars) of the capacitors required.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "S  =  300000;#  in  VA\n",
+      "pf1  =  0.70;#  in  power  factor\n",
+      "pf2  =  0.90;#  in  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #active  power,  P\n",
+      "Pa  =  S*pf1\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      " #Reactive  power,  Q\n",
+      "Q  =  S*math.sin(phi1)\n",
+      "phi2  =  math.acos(pf2)\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      " #The  capacitor  rating  needed  to  improve  the  power  factor  to  0.90\n",
+      " #the  capacitor  rating,\n",
+      "Pr  =  Q  -  (Pa*math.tan(phi2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  rating  (in  kilovars)  of  the  capacitors  is  \",round((Pr/1E3),2),\"  kvar leading\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  rating  (in  kilovars)  of  the  capacitors  is   112.54   kvar leading"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 471</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the supply current, (b) the active, apparent and reactive power,\n",
+      "#(c) the rating of a capacitor (d) the value of capacitance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z  =  3  +  4j;#  in  ohms\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  30;#  in  Degrees\n",
+      "f  =  1500;#  in  Hz\n",
+      "pf1  =  0.966;#  in  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Supply  current,  I\n",
+      "I  =  V/Z\n",
+      "Istr  =  I.real  -  1j*I.imag\n",
+      " #Apparent  power,  S\n",
+      "S  =  V*Istr\n",
+      " #active  power,  Pa\n",
+      "Pa  =  S.real\n",
+      "#reactive  power,  Q\n",
+      "Q  =  abs(S.imag)\n",
+      " #apparent  power,  S\n",
+      "S  =  (S.real**2  +  S.imag**2)**0.5\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      " #rating  of  the  capacitor  \n",
+      "Pr  =  Q  -  Pa*math.tan(phi1)\n",
+      " #Current  in  capacitor,  Ic\n",
+      "Ic  =  Pr/rv\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  rv/Ic\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)supply  current,  I  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (b)active  power  is  \",round(Pa,2),\"  W,  apparent  power  is  \",round(  S,2),\"  W  \"\n",
+      "print   \"and  reactive  power  is  \",round(  Q,2),\"  W lagging\\n\"\n",
+      "print  \"\\n  (c)the  rating  of  the  capacitors  is  \",round(Pr,2),\"  var leading\\n\"\n",
+      "print  \"\\n  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is  \",round(  C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)supply  current,  I  is   9.2   +  ( -3.93 )i  A\n",
+        "\n",
+        "\n",
+        "  (b)active  power  is   300.0   W,  apparent  power  is   500.0   W  and  reactive  power  is   400.0   W lagging\n",
+        "\n",
+        "\n",
+        "  (c)the  rating  of  the  capacitors  is   319.71   var leading\n",
+        "\n",
+        "\n",
+        "  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is   13.57 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_2.ipynb
new file mode 100755
index 00000000..4edde932
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_2.ipynb
@@ -0,0 +1,496 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 26: Power in a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 466</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the active power in the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  12j;#  in  ohm\n",
+      "R  =  5;#  in  ohm\n",
+      "rv  =  52;#  in  volts\n",
+      "thetav  =  30;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #impedance,  Z\n",
+      "Z  =  R  +  RL\n",
+      " #current\n",
+      "I  =  V/Z\n",
+      " #Active  power,  P\n",
+      "Pa  =  V.real*I.real  +  V.imag*I.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nthe  active  power  in  the  circuit  \",Pa,\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "the  active  power  in  the  circuit   80.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 467</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the active power, and (b) the reactive power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  120  +  200j;#  in  volts\n",
+      "I  =  15  +  8j;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Active  power,  P\n",
+      "Pa  =  V.real*I.real  +  V.imag*I.imag\n",
+      " #Reactive  power,  Q\n",
+      "Q  =  V.imag*I.real  -  V.real*I.imag\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  active  power  in  the  circuit  \",Pa,\"  W\\n\"\n",
+      "print  \"\\n  (b)  the  reactive  power  in  the  circuit  \",Q,\"  var\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  active  power  in  the  circuit   3400.0   W\n",
+        "\n",
+        "\n",
+        "  (b)  the  reactive  power  in  the  circuit   2040.0   var"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 468</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current flowing and its phase, (b) the value of resistance R, and (c) the value of capacitance C.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vm  =  141.4;#  in  volts\n",
+      "w  =  10000;#  in  rad/sec\n",
+      "phiv  =  math.pi/9;#  in  radian\n",
+      "Pd  =  1732;#  in  Watts\n",
+      "pf  =  0.866;#  power  fctr\n",
+      "\n",
+      "#calculation:\n",
+      " #the  rms  voltage,\n",
+      "Vrms  =  0.707*Vm\n",
+      " #Power  P  =  V*I*cos(phi)\n",
+      " #current  magnitude,  Irms\n",
+      "Irms  =  Pd/(Vrms*pf)\n",
+      "phid  =  math.acos(pf)\n",
+      " #current  phase  angle\n",
+      "phii  =  phiv  +  phid\n",
+      "phiid  =  phii*180/math.pi#  in  degrees\n",
+      " #Voltage,  V\n",
+      "V  =  Vrms*math.cos(phiv)  +  1j*Vrms*math.sin(phiv)\n",
+      " #current,  I\n",
+      "I  =  Irms*math.cos(phii)  +  1j*Irms*math.sin(phii)\n",
+      " #Impedance,  Z\n",
+      "Z  =  V/I\n",
+      " #resistance,  R\n",
+      "R  =  Z.real\n",
+      " #capacitive  reactance,  Xc\n",
+      "Xc  =  abs(Z.imag)\n",
+      " #capacitance,  C\n",
+      "C  =  1/  (w*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  and  Circuit  phase  angle  is  \",round(Irms,2),\"/_\",round(phiid,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (b)  the  resistance  is  \",round(R,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (c)  the  capacitance  is  \",round(C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  and  Circuit  phase  angle  is   20.01 /_ 50.0 deg  A\n",
+        "\n",
+        "\n",
+        "  (b)  the  resistance  is   4.33   ohm\n",
+        "\n",
+        "\n",
+        "  (c)  the  capacitance  is   40.02 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 468</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the active power developed between points (a) A and B, (b) C and D, (c) E and F.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R  =  5;#  in  ohm\n",
+      "R1  =  3;#  in  ohms\n",
+      "RL  =  4j;#  in  ohm\n",
+      "Rc  =  -10j;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z1\n",
+      "Z1  =  R1  +  RL\n",
+      " #impedance,  Zc\n",
+      "Zc  =  Rc\n",
+      " #Circuit  impedance,  Z\n",
+      "Z  =  R  +  (Z1*Zc/(Z1  +  Zc))\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)\n",
+      "I  =  V/Z\n",
+      "Imag  =  ((I.real)**2  +  (I.imag)**2)**0.5\n",
+      " #Active  power  developed  between  points  A  and  B\n",
+      "Pab  =  (Imag**2)*R\n",
+      " #Active  power  developed  between  points  C  and  D\n",
+      "Pcd  =  (Imag**2)*Zc.real\n",
+      " #Current,  I1\n",
+      "I1  =  I*Zc/(Zc  +  Z1)\n",
+      "I1mag  =  ((I1.real)**2  +  (I1.imag)**2)**0.5\n",
+      " #active  power  developed  between  points  E  and  F\n",
+      "Pef  =  (I1mag**2)*Z1.real\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Active  power  developed  between  points  A  and  B  is  \",round(Pab,2),\"  W\\n\"\n",
+      "print  \"\\n  (b)Active  power  developed  between  points  C  and  D  is  \",round(Pcd,2),\"  W\\n\"\n",
+      "print  \"\\n  (c)Active  power  developed  between  points  E  and  F  is  \",round(Pef,2),\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Active  power  developed  between  points  A  and  B  is   339.62   W\n",
+        "\n",
+        "\n",
+        "  (b)Active  power  developed  between  points  C  and  D  is   0.0   W\n",
+        "\n",
+        "\n",
+        "  (c)Active  power  developed  between  points  E  and  F  is   452.83   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 469</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the apparent power, (b) the reactive power, \n",
+      "#(c) the value and phase of current I, and (d) the value of impedance Z.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pa  =  400;#  in  Watts\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "R  =  4;#  in  ohm\n",
+      "pf  =  0.766;#  power  factor\n",
+      "\n",
+      " #calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #magnitude  of  apparent  power,S  =  V*I\n",
+      "S  =  Pa/pf\n",
+      "phi  =  math.acos(pf)\n",
+      "theta  =  phi*180/math.pi#  in  degrees\n",
+      " #Reactive  power  Q\n",
+      "Q  =  S*math.sin(phi)\n",
+      " #magnitude  of  current\n",
+      "Imag  =  S/rv\n",
+      "thetai  =  thetav  -  theta\n",
+      "I  =  Imag*math.cos(thetai*math.pi/180)  +  1j*Imag*math.sin(thetai*math.pi/180)\n",
+      " #Total  circuit  impedance  ZT\n",
+      "ZT  =  V/I\n",
+      " #impedance  Z\n",
+      "Z  =  ZT  -  R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)apparent  power  is  \",round(S,2),\"  VA\\n\"\n",
+      "print  \"\\n  (b)reactive  power  is  \",round(Q,1),\"  var lagging\\n\"\n",
+      "print  \"\\n  (c)the  current  flowing  and  Circuit  phase  angle  is  \",round(Imag,2),\"/_\",round(thetai,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (d)impedance,  Z  is  \",round(Z.real,2),\"  +  (\",round(  Z.imag,2),\")i  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)apparent  power  is   522.19   VA\n",
+        "\n",
+        "\n",
+        "  (b)reactive  power  is   335.7   var lagging\n",
+        "\n",
+        "\n",
+        "  (c)the  current  flowing  and  Circuit  phase  angle  is   5.22 /_ -10.0 deg  A\n",
+        "\n",
+        "\n",
+        "  (d)impedance,  Z  is   10.67   +  ( 12.31 )i  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 471</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the rating (in kilovars) of the capacitors required.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "S  =  300000;#  in  VA\n",
+      "pf1  =  0.70;#  in  power  factor\n",
+      "pf2  =  0.90;#  in  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #active  power,  P\n",
+      "Pa  =  S*pf1\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      " #Reactive  power,  Q\n",
+      "Q  =  S*math.sin(phi1)\n",
+      "phi2  =  math.acos(pf2)\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      " #The  capacitor  rating  needed  to  improve  the  power  factor  to  0.90\n",
+      " #the  capacitor  rating,\n",
+      "Pr  =  Q  -  (Pa*math.tan(phi2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  rating  (in  kilovars)  of  the  capacitors  is  \",round((Pr/1E3),2),\"  kvar leading\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  rating  (in  kilovars)  of  the  capacitors  is   112.54   kvar leading"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 471</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the supply current, (b) the active, apparent and reactive power,\n",
+      "#(c) the rating of a capacitor (d) the value of capacitance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z  =  3  +  4j;#  in  ohms\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  30;#  in  Degrees\n",
+      "f  =  1500;#  in  Hz\n",
+      "pf1  =  0.966;#  in  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Supply  current,  I\n",
+      "I  =  V/Z\n",
+      "Istr  =  I.real  -  1j*I.imag\n",
+      " #Apparent  power,  S\n",
+      "S  =  V*Istr\n",
+      " #active  power,  Pa\n",
+      "Pa  =  S.real\n",
+      "#reactive  power,  Q\n",
+      "Q  =  abs(S.imag)\n",
+      " #apparent  power,  S\n",
+      "S  =  (S.real**2  +  S.imag**2)**0.5\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      " #rating  of  the  capacitor  \n",
+      "Pr  =  Q  -  Pa*math.tan(phi1)\n",
+      " #Current  in  capacitor,  Ic\n",
+      "Ic  =  Pr/rv\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  rv/Ic\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)supply  current,  I  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (b)active  power  is  \",round(Pa,2),\"  W,  apparent  power  is  \",round(  S,2),\"  W  \"\n",
+      "print   \"and  reactive  power  is  \",round(  Q,2),\"  W lagging\\n\"\n",
+      "print  \"\\n  (c)the  rating  of  the  capacitors  is  \",round(Pr,2),\"  var leading\\n\"\n",
+      "print  \"\\n  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is  \",round(  C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)supply  current,  I  is   9.2   +  ( -3.93 )i  A\n",
+        "\n",
+        "\n",
+        "  (b)active  power  is   300.0   W,  apparent  power  is   500.0   W  and  reactive  power  is   400.0   W lagging\n",
+        "\n",
+        "\n",
+        "  (c)the  rating  of  the  capacitors  is   319.71   var leading\n",
+        "\n",
+        "\n",
+        "  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is   13.57 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_3.ipynb
new file mode 100755
index 00000000..fe6f0a60
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_26-checkpoint_3.ipynb
@@ -0,0 +1,495 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ef0f04156af110c85f44cbf4f7c9b4a7f78444e3b8eab6fe43fee89327c37d26"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 26: Power in a.c. circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 466</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  12j;#  in  ohm\n",
+      "R  =  5;#  in  ohm\n",
+      "rv  =  52;#  in  volts\n",
+      "thetav  =  30;#  in  degree\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #impedance,  Z\n",
+      "Z  =  R  +  RL\n",
+      " #current\n",
+      "I  =  V/Z\n",
+      " #Active  power,  P\n",
+      "Pa  =  V.real*I.real  +  V.imag*I.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nthe  active  power  in  the  circuit  \",Pa,\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "the  active  power  in  the  circuit   80.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 467</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  120  +  200j;#  in  volts\n",
+      "I  =  15  +  8j;#  in  amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #Active  power,  P\n",
+      "Pa  =  V.real*I.real  +  V.imag*I.imag\n",
+      " #Reactive  power,  Q\n",
+      "Q  =  V.imag*I.real  -  V.real*I.imag\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  active  power  in  the  circuit  \",Pa,\"  W\\n\"\n",
+      "print  \"\\n  (b)  the  reactive  power  in  the  circuit  \",Q,\"  var\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  active  power  in  the  circuit   3400.0   W\n",
+        "\n",
+        "\n",
+        "  (b)  the  reactive  power  in  the  circuit   2040.0   var"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 468</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vm  =  141.4;#  in  volts\n",
+      "w  =  10000;#  in  rad/sec\n",
+      "phiv  =  math.pi/9;#  in  radian\n",
+      "Pd  =  1732;#  in  Watts\n",
+      "pf  =  0.866;#  power  fctr\n",
+      "\n",
+      "#calculation:\n",
+      " #the  rms  voltage,\n",
+      "Vrms  =  0.707*Vm\n",
+      " #Power  P  =  V*I*cos(phi)\n",
+      " #current  magnitude,  Irms\n",
+      "Irms  =  Pd/(Vrms*pf)\n",
+      "phid  =  math.acos(pf)\n",
+      " #current  phase  angle\n",
+      "phii  =  phiv  +  phid\n",
+      "phiid  =  phii*180/math.pi#  in  degrees\n",
+      " #Voltage,  V\n",
+      "V  =  Vrms*math.cos(phiv)  +  1j*Vrms*math.sin(phiv)\n",
+      " #current,  I\n",
+      "I  =  Irms*math.cos(phii)  +  1j*Irms*math.sin(phii)\n",
+      " #Impedance,  Z\n",
+      "Z  =  V/I\n",
+      " #resistance,  R\n",
+      "R  =  Z.real\n",
+      " #capacitive  reactance,  Xc\n",
+      "Xc  =  abs(Z.imag)\n",
+      " #capacitance,  C\n",
+      "C  =  1/  (w*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  and  Circuit  phase  angle  is  \",round(Irms,2),\"/_\",round(phiid,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (b)  the  resistance  is  \",round(R,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (c)  the  capacitance  is  \",round(C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  and  Circuit  phase  angle  is   20.01 /_ 50.0 deg  A\n",
+        "\n",
+        "\n",
+        "  (b)  the  resistance  is   4.33   ohm\n",
+        "\n",
+        "\n",
+        "  (c)  the  capacitance  is   40.02 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 468</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R  =  5;#  in  ohm\n",
+      "R1  =  3;#  in  ohms\n",
+      "RL  =  4j;#  in  ohm\n",
+      "Rc  =  -10j;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #impedance,  Z1\n",
+      "Z1  =  R1  +  RL\n",
+      " #impedance,  Zc\n",
+      "Zc  =  Rc\n",
+      " #Circuit  impedance,  Z\n",
+      "Z  =  R  +  (Z1*Zc/(Z1  +  Zc))\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)\n",
+      "I  =  V/Z\n",
+      "Imag  =  ((I.real)**2  +  (I.imag)**2)**0.5\n",
+      " #Active  power  developed  between  points  A  and  B\n",
+      "Pab  =  (Imag**2)*R\n",
+      " #Active  power  developed  between  points  C  and  D\n",
+      "Pcd  =  (Imag**2)*Zc.real\n",
+      " #Current,  I1\n",
+      "I1  =  I*Zc/(Zc  +  Z1)\n",
+      "I1mag  =  ((I1.real)**2  +  (I1.imag)**2)**0.5\n",
+      " #active  power  developed  between  points  E  and  F\n",
+      "Pef  =  (I1mag**2)*Z1.real\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Active  power  developed  between  points  A  and  B  is  \",round(Pab,2),\"  W\\n\"\n",
+      "print  \"\\n  (b)Active  power  developed  between  points  C  and  D  is  \",round(Pcd,2),\"  W\\n\"\n",
+      "print  \"\\n  (c)Active  power  developed  between  points  E  and  F  is  \",round(Pef,2),\"  W\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Active  power  developed  between  points  A  and  B  is   339.62   W\n",
+        "\n",
+        "\n",
+        "  (b)Active  power  developed  between  points  C  and  D  is   0.0   W\n",
+        "\n",
+        "\n",
+        "  (c)Active  power  developed  between  points  E  and  F  is   452.83   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 469</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pa  =  400;#  in  Watts\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "R  =  4;#  in  ohm\n",
+      "pf  =  0.766;#  power  factor\n",
+      "\n",
+      " #calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #magnitude  of  apparent  power,S  =  V*I\n",
+      "S  =  Pa/pf\n",
+      "phi  =  math.acos(pf)\n",
+      "theta  =  phi*180/math.pi#  in  degrees\n",
+      " #Reactive  power  Q\n",
+      "Q  =  S*math.sin(phi)\n",
+      " #magnitude  of  current\n",
+      "Imag  =  S/rv\n",
+      "thetai  =  thetav  -  theta\n",
+      "I  =  Imag*math.cos(thetai*math.pi/180)  +  1j*Imag*math.sin(thetai*math.pi/180)\n",
+      " #Total  circuit  impedance  ZT\n",
+      "ZT  =  V/I\n",
+      " #impedance  Z\n",
+      "Z  =  ZT  -  R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)apparent  power  is  \",round(S,2),\"  VA\\n\"\n",
+      "print  \"\\n  (b)reactive  power  is  \",round(Q,1),\"  var lagging\\n\"\n",
+      "print  \"\\n  (c)the  current  flowing  and  Circuit  phase  angle  is  \",round(Imag,2),\"/_\",round(thetai,2),\"deg  A\\n\"\n",
+      "print  \"\\n  (d)impedance,  Z  is  \",round(Z.real,2),\"  +  (\",round(  Z.imag,2),\")i  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)apparent  power  is   522.19   VA\n",
+        "\n",
+        "\n",
+        "  (b)reactive  power  is   335.7   var lagging\n",
+        "\n",
+        "\n",
+        "  (c)the  current  flowing  and  Circuit  phase  angle  is   5.22 /_ -10.0 deg  A\n",
+        "\n",
+        "\n",
+        "  (d)impedance,  Z  is   10.67   +  ( 12.31 )i  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 471</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "S  =  300000;#  in  VA\n",
+      "pf1  =  0.70;#  in  power  factor\n",
+      "pf2  =  0.90;#  in  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      " #active  power,  P\n",
+      "Pa  =  S*pf1\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      " #Reactive  power,  Q\n",
+      "Q  =  S*math.sin(phi1)\n",
+      "phi2  =  math.acos(pf2)\n",
+      "phi2d  =  phi2*180/math.pi\n",
+      " #The  capacitor  rating  needed  to  improve  the  power  factor  to  0.90\n",
+      " #the  capacitor  rating,\n",
+      "Pr  =  Q  -  (Pa*math.tan(phi2))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  rating  (in  kilovars)  of  the  capacitors  is  \",round((Pr/1E3),2),\"  kvar leading\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  rating  (in  kilovars)  of  the  capacitors  is   112.54   kvar leading"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 471</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z  =  3  +  4j;#  in  ohms\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  30;#  in  Degrees\n",
+      "f  =  1500;#  in  Hz\n",
+      "pf1  =  0.966;#  in  power  factor\n",
+      "\n",
+      "#calculation:\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Supply  current,  I\n",
+      "I  =  V/Z\n",
+      "Istr  =  I.real  -  1j*I.imag\n",
+      " #Apparent  power,  S\n",
+      "S  =  V*Istr\n",
+      " #active  power,  Pa\n",
+      "Pa  =  S.real\n",
+      "#reactive  power,  Q\n",
+      "Q  =  abs(S.imag)\n",
+      " #apparent  power,  S\n",
+      "S  =  (S.real**2  +  S.imag**2)**0.5\n",
+      "phi1  =  math.acos(pf1)\n",
+      "phi1d  =  phi1*180/math.pi\n",
+      " #rating  of  the  capacitor  \n",
+      "Pr  =  Q  -  Pa*math.tan(phi1)\n",
+      " #Current  in  capacitor,  Ic\n",
+      "Ic  =  Pr/rv\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  rv/Ic\n",
+      "C  =  1/(2*math.pi*f*Xc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)supply  current,  I  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (b)active  power  is  \",round(Pa,2),\"  W,  apparent  power  is  \",round(  S,2),\"  W  \"\n",
+      "print   \"and  reactive  power  is  \",round(  Q,2),\"  W lagging\\n\"\n",
+      "print  \"\\n  (c)the  rating  of  the  capacitors  is  \",round(Pr,2),\"  var leading\\n\"\n",
+      "print  \"\\n  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is  \",round(  C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)supply  current,  I  is   9.2   +  ( -3.93 )i  A\n",
+        "\n",
+        "\n",
+        "  (b)active  power  is   300.0   W,  apparent  power  is   500.0   W  and  reactive  power  is   400.0   W lagging\n",
+        "\n",
+        "\n",
+        "  (c)the  rating  of  the  capacitors  is   319.71   var leading\n",
+        "\n",
+        "\n",
+        "  (d)value  of  capacitance  needed  to  improve  the  power  factor  to  0.966  lagging  is   13.57 uF"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint.ipynb
new file mode 100755
index 00000000..4d37718b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint.ipynb
@@ -0,0 +1,205 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 27: A.c. bridges</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 485</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of Rx and Cx at balance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  2500;#  in  ohms\n",
+      "C2  =  0.2E-6;#  IN  fARADS\n",
+      "R3 = 1;\n",
+      "R4 = 1;\n",
+      "w = 2000*math.pi;\n",
+      "#calculation:\n",
+      "Rx = R4*(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2)\n",
+      "Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  Rx  = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\"\n",
+      "print  \"\\n  (b)at balance Rx = \",round(Rx/1000,2),\"KOhm and Cx = \", round(Cx*1E9,2),\"nF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  Rx  = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
+        "\n",
+        "  (b)at balance Rx =  2.75 KOhm and Cx =  18.4 nF"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 487</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, when the bridge is balanced, (a) the value of resistance R1, and (b) the frequency of the bridge.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  30000;#  in  ohms\n",
+      "R3  =  30000;#  in  ohms\n",
+      "R4  =  1000;#  in  ohms\n",
+      "C2  =  1e-9;#  IN  fARADS\n",
+      "C3  =  1e-9;#  IN  fARADS\n",
+      "\n",
+      "#calculation:\n",
+      " #the  bridge  is  balanced\n",
+      "R1  =  R4/((R3/R2)  +  (C2/C3))\n",
+      " #frequency,  f\n",
+      "f  =  1/(2*math.pi*((C2*C3*R2*R3)**0.5))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  R1  =  \",R1,\"  ohm\\n\"\n",
+      "print  \"\\n  (b)frequency,  f  is  \",round(f,2),\"Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  R1  =   500.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)frequency,  f  is   5305.16 Hz"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 487</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine, when the bridge is balanced, \n",
+      "#(a) the value of resistance Rx, (b) the value of capacitance Cx,\n",
+      "#(c) the phase angle of the unknown arm, (d) the power factor of the unknown arm and (e) its loss angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R3  =  600;#  in  ohms\n",
+      "R4  =  200;#  in  ohms\n",
+      "C2  =  0.2e-6;#  IN  fARADS\n",
+      "C3  =  4000e-12;#  IN  fARADS\n",
+      "f  =  1500;#in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #the  bridge  is  balanced\n",
+      " #Resistance,  Rx\n",
+      "Rx  =  R4*C3/C2\n",
+      " #Capacitance,  Cx\n",
+      "Cx  =  C2*R3/R4\n",
+      " #Phase  angle\n",
+      "phi  =  math.atan(1/(2*math.pi*f*Cx*Rx))\n",
+      "phid  =  phi*180/math.pi#  in  degrees\n",
+      " #Power  factor  of  capacitor\n",
+      "Pc  =  math.cos(phi)\n",
+      " #Loss  angle,\n",
+      "de  =  90  -  phid\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  Rx  =  \",round(Rx,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)capacitance,  Cx  is  \",round(Cx*1E9,2),\"pFarad\\n\"\n",
+      "print  \"\\n  (c)phasor  diagram  =  \",round(phid,2),\"deg lead \"\n",
+      "print  \"\\n  (d)power  factor  is  \",round(Pc,2),\"  \\n\"\n",
+      "print  \"\\n  (e)Loss  angle  =  \",round(de,2),\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  Rx  =   4.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)capacitance,  Cx  is   600.0 pFarad\n",
+        "\n",
+        "\n",
+        "  (c)phasor  diagram  =   88.7 deg lead \n",
+        "\n",
+        "  (d)power  factor  is   0.02   \n",
+        "\n",
+        "\n",
+        "  (e)Loss  angle  =   1.3 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_1.ipynb
new file mode 100755
index 00000000..4d37718b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_1.ipynb
@@ -0,0 +1,205 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 27: A.c. bridges</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 485</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of Rx and Cx at balance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  2500;#  in  ohms\n",
+      "C2  =  0.2E-6;#  IN  fARADS\n",
+      "R3 = 1;\n",
+      "R4 = 1;\n",
+      "w = 2000*math.pi;\n",
+      "#calculation:\n",
+      "Rx = R4*(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2)\n",
+      "Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  Rx  = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\"\n",
+      "print  \"\\n  (b)at balance Rx = \",round(Rx/1000,2),\"KOhm and Cx = \", round(Cx*1E9,2),\"nF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  Rx  = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
+        "\n",
+        "  (b)at balance Rx =  2.75 KOhm and Cx =  18.4 nF"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 487</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, when the bridge is balanced, (a) the value of resistance R1, and (b) the frequency of the bridge.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  30000;#  in  ohms\n",
+      "R3  =  30000;#  in  ohms\n",
+      "R4  =  1000;#  in  ohms\n",
+      "C2  =  1e-9;#  IN  fARADS\n",
+      "C3  =  1e-9;#  IN  fARADS\n",
+      "\n",
+      "#calculation:\n",
+      " #the  bridge  is  balanced\n",
+      "R1  =  R4/((R3/R2)  +  (C2/C3))\n",
+      " #frequency,  f\n",
+      "f  =  1/(2*math.pi*((C2*C3*R2*R3)**0.5))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  R1  =  \",R1,\"  ohm\\n\"\n",
+      "print  \"\\n  (b)frequency,  f  is  \",round(f,2),\"Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  R1  =   500.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)frequency,  f  is   5305.16 Hz"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 487</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine, when the bridge is balanced, \n",
+      "#(a) the value of resistance Rx, (b) the value of capacitance Cx,\n",
+      "#(c) the phase angle of the unknown arm, (d) the power factor of the unknown arm and (e) its loss angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R3  =  600;#  in  ohms\n",
+      "R4  =  200;#  in  ohms\n",
+      "C2  =  0.2e-6;#  IN  fARADS\n",
+      "C3  =  4000e-12;#  IN  fARADS\n",
+      "f  =  1500;#in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #the  bridge  is  balanced\n",
+      " #Resistance,  Rx\n",
+      "Rx  =  R4*C3/C2\n",
+      " #Capacitance,  Cx\n",
+      "Cx  =  C2*R3/R4\n",
+      " #Phase  angle\n",
+      "phi  =  math.atan(1/(2*math.pi*f*Cx*Rx))\n",
+      "phid  =  phi*180/math.pi#  in  degrees\n",
+      " #Power  factor  of  capacitor\n",
+      "Pc  =  math.cos(phi)\n",
+      " #Loss  angle,\n",
+      "de  =  90  -  phid\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  Rx  =  \",round(Rx,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)capacitance,  Cx  is  \",round(Cx*1E9,2),\"pFarad\\n\"\n",
+      "print  \"\\n  (c)phasor  diagram  =  \",round(phid,2),\"deg lead \"\n",
+      "print  \"\\n  (d)power  factor  is  \",round(Pc,2),\"  \\n\"\n",
+      "print  \"\\n  (e)Loss  angle  =  \",round(de,2),\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  Rx  =   4.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)capacitance,  Cx  is   600.0 pFarad\n",
+        "\n",
+        "\n",
+        "  (c)phasor  diagram  =   88.7 deg lead \n",
+        "\n",
+        "  (d)power  factor  is   0.02   \n",
+        "\n",
+        "\n",
+        "  (e)Loss  angle  =   1.3 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb
new file mode 100755
index 00000000..4d37718b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb
@@ -0,0 +1,205 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 27: A.c. bridges</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 485</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of Rx and Cx at balance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  2500;#  in  ohms\n",
+      "C2  =  0.2E-6;#  IN  fARADS\n",
+      "R3 = 1;\n",
+      "R4 = 1;\n",
+      "w = 2000*math.pi;\n",
+      "#calculation:\n",
+      "Rx = R4*(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2)\n",
+      "Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  Rx  = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\"\n",
+      "print  \"\\n  (b)at balance Rx = \",round(Rx/1000,2),\"KOhm and Cx = \", round(Cx*1E9,2),\"nF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  Rx  = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
+        "\n",
+        "  (b)at balance Rx =  2.75 KOhm and Cx =  18.4 nF"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 487</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, when the bridge is balanced, (a) the value of resistance R1, and (b) the frequency of the bridge.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  30000;#  in  ohms\n",
+      "R3  =  30000;#  in  ohms\n",
+      "R4  =  1000;#  in  ohms\n",
+      "C2  =  1e-9;#  IN  fARADS\n",
+      "C3  =  1e-9;#  IN  fARADS\n",
+      "\n",
+      "#calculation:\n",
+      " #the  bridge  is  balanced\n",
+      "R1  =  R4/((R3/R2)  +  (C2/C3))\n",
+      " #frequency,  f\n",
+      "f  =  1/(2*math.pi*((C2*C3*R2*R3)**0.5))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  R1  =  \",R1,\"  ohm\\n\"\n",
+      "print  \"\\n  (b)frequency,  f  is  \",round(f,2),\"Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  R1  =   500.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)frequency,  f  is   5305.16 Hz"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 487</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine, when the bridge is balanced, \n",
+      "#(a) the value of resistance Rx, (b) the value of capacitance Cx,\n",
+      "#(c) the phase angle of the unknown arm, (d) the power factor of the unknown arm and (e) its loss angle.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R3  =  600;#  in  ohms\n",
+      "R4  =  200;#  in  ohms\n",
+      "C2  =  0.2e-6;#  IN  fARADS\n",
+      "C3  =  4000e-12;#  IN  fARADS\n",
+      "f  =  1500;#in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #the  bridge  is  balanced\n",
+      " #Resistance,  Rx\n",
+      "Rx  =  R4*C3/C2\n",
+      " #Capacitance,  Cx\n",
+      "Cx  =  C2*R3/R4\n",
+      " #Phase  angle\n",
+      "phi  =  math.atan(1/(2*math.pi*f*Cx*Rx))\n",
+      "phid  =  phi*180/math.pi#  in  degrees\n",
+      " #Power  factor  of  capacitor\n",
+      "Pc  =  math.cos(phi)\n",
+      " #Loss  angle,\n",
+      "de  =  90  -  phid\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  Rx  =  \",round(Rx,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)capacitance,  Cx  is  \",round(Cx*1E9,2),\"pFarad\\n\"\n",
+      "print  \"\\n  (c)phasor  diagram  =  \",round(phid,2),\"deg lead \"\n",
+      "print  \"\\n  (d)power  factor  is  \",round(Pc,2),\"  \\n\"\n",
+      "print  \"\\n  (e)Loss  angle  =  \",round(de,2),\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  Rx  =   4.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)capacitance,  Cx  is   600.0 pFarad\n",
+        "\n",
+        "\n",
+        "  (c)phasor  diagram  =   88.7 deg lead \n",
+        "\n",
+        "  (d)power  factor  is   0.02   \n",
+        "\n",
+        "\n",
+        "  (e)Loss  angle  =   1.3 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_3.ipynb
new file mode 100755
index 00000000..c141c129
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_3.ipynb
@@ -0,0 +1,204 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:de2147a7b14f8ed39e81d7f13f742b47f7e864d50fa1a5733b8aeda0bb6528ab"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 27: A.c. bridges</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 485</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  2500;#  in  ohms\n",
+      "C2  =  0.2E-6;#  IN  fARADS\n",
+      "R3 = 1;\n",
+      "R4 = 1;\n",
+      "w = 2000*math.pi;\n",
+      "#calculation:\n",
+      "Rx = R4*(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2)\n",
+      "Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  Rx  = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\"\n",
+      "print  \"\\n  (b)at balance Rx = \",round(Rx/1000,2),\"KOhm and Cx = \", round(Cx*1E9,2),\"nF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  Rx  = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n",
+        "\n",
+        "  (b)at balance Rx =  2.75 KOhm and Cx =  18.4 nF"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 487</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  30000;#  in  ohms\n",
+      "R3  =  30000;#  in  ohms\n",
+      "R4  =  1000;#  in  ohms\n",
+      "C2  =  1e-9;#  IN  fARADS\n",
+      "C3  =  1e-9;#  IN  fARADS\n",
+      "\n",
+      "#calculation:\n",
+      " #the  bridge  is  balanced\n",
+      "R1  =  R4/((R3/R2)  +  (C2/C3))\n",
+      " #frequency,  f\n",
+      "f  =  1/(2*math.pi*((C2*C3*R2*R3)**0.5))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  R1  =  \",R1,\"  ohm\\n\"\n",
+      "print  \"\\n  (b)frequency,  f  is  \",round(f,2),\"Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  R1  =   500.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)frequency,  f  is   5305.16 Hz"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 487</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R3  =  600;#  in  ohms\n",
+      "R4  =  200;#  in  ohms\n",
+      "C2  =  0.2e-6;#  IN  fARADS\n",
+      "C3  =  4000e-12;#  IN  fARADS\n",
+      "f  =  1500;#in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #the  bridge  is  balanced\n",
+      " #Resistance,  Rx\n",
+      "Rx  =  R4*C3/C2\n",
+      " #Capacitance,  Cx\n",
+      "Cx  =  C2*R3/R4\n",
+      " #Phase  angle\n",
+      "phi  =  math.atan(1/(2*math.pi*f*Cx*Rx))\n",
+      "phid  =  phi*180/math.pi#  in  degrees\n",
+      " #Power  factor  of  capacitor\n",
+      "Pc  =  math.cos(phi)\n",
+      " #Loss  angle,\n",
+      "de  =  90  -  phid\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resistance  Rx  =  \",round(Rx,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)capacitance,  Cx  is  \",round(Cx*1E9,2),\"pFarad\\n\"\n",
+      "print  \"\\n  (c)phasor  diagram  =  \",round(phid,2),\"deg lead \"\n",
+      "print  \"\\n  (d)power  factor  is  \",round(Pc,2),\"  \\n\"\n",
+      "print  \"\\n  (e)Loss  angle  =  \",round(de,2),\"deg\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resistance  Rx  =   4.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)capacitance,  Cx  is   600.0 pFarad\n",
+        "\n",
+        "\n",
+        "  (c)phasor  diagram  =   88.7 deg lead \n",
+        "\n",
+        "  (d)power  factor  is   0.02   \n",
+        "\n",
+        "\n",
+        "  (e)Loss  angle  =   1.3 deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint.ipynb
new file mode 100755
index 00000000..35880e50
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint.ipynb
@@ -0,0 +1,736 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 28: Series resonance and Q-factor</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 492</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine at what frequency resonance occurs, and (b) the current flowing at resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  40e-6;#  IN  fARADS\n",
+      "L  =  0.075;#  IN  Henry\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #Current  at  resonance,  I\n",
+      "I  =  V/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonant  frequency  =  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Current  at  resonance,  I  is  \",I,\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonant  frequency  =   91.89   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Current  at  resonance,  I  is   20.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 493</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of capacitor C for series resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8;#  in  ohms\n",
+      "L  =  0.010;#  IN  Henry\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #At  resonance\n",
+      " #capacitance  C\n",
+      "C  =  1/(L*(2*math.pi*f)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance,  C  is  \",round(C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance,  C  is   2.53 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 493</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of (a) the stray capacitance CS, and (b) the coil inductance L.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C1  =  1000e-12;#  IN  fARADS\n",
+      "C2  =  500e-12;#  IN  fARADS\n",
+      "fr1  =  92500;#  in  Hz\n",
+      "fr2  =  127800;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  series  R\u2013L\u2013C  circuit  the  resonant  frequency  fr  is  given  by:\n",
+      " #fr  =  1/(2pi*(L*C)**2)\n",
+      "Cs  =  ((C1  -  C2)/((fr2/fr1)**2  -  1))  -  C2\n",
+      "L  =  1/((C1  +  Cs)*(2*math.pi*fr1)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)stray  capacitance,  Cs  is  \",round(Cs*1E12,2),\"pF\\n\"\n",
+      "print  \"\\n  (b)inductance,  L  is  \",round(L*1000,2),\"mH\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)stray  capacitance,  Cs  is   50.13 pF\n",
+        "\n",
+        "\n",
+        "  (b)inductance,  L  is   2.82 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 497</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for this condition (a) the value of inductance L, (b) the p.d. across each component and (c) the Q-factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  5e-6;#  IN  fARADS\n",
+      "rv  =  20;#in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "f  =  318.3;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*f\n",
+      " #The  maximum  voltage  across  the  resistance  occurs  at  resonance  when  the  current  is  a  maximum.  \n",
+      "    #At  resonance,L  =  1/c*wr**2\n",
+      "L  =  1/(C*wr**2)\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  at  resonance  Ir\n",
+      "Ir  =  V/R\n",
+      " #p.d.  across  resistance,  VR\n",
+      "VR  =  Ir*R\n",
+      " #inductive  reactance,  XL\n",
+      "XL  =  wr*L\n",
+      " #p.d.  across  inductance,  VL\n",
+      "VL  =  Ir*(1j*XL)\n",
+      " #capacitive  reactance,  Xc\n",
+      "Xc  =  1/(wr*C)\n",
+      " #p.d.  across  capacitor,  Vc\n",
+      "Vc  =  Ir*(-1j*Xc)\n",
+      " #Q-factor  at  resonance,  Qr\n",
+      "Qr  =  VL.imag/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (b)p.d.  across  resistance,  VR  is  \",VR,\"  V,  p.d.  across  inductance,  VL  \",round( VL.imag,2),\"j V  \"\n",
+      "print   \"and  p.d.  across  capacitor,  VC  \",round(Vc.imag,2),\" V\\n\"\n",
+      "print  \"\\n  (c)Q-factor  at  resonance,  Qr  is  \",round(abs(Qr),2),\"  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)inductance,  L  is   50.0 mH\n",
+        "\n",
+        "\n",
+        "  (b)p.d.  across  resistance,  VR  is   (20+0j)   V,  p.d.  across  inductance,  VL   200.01 j V  \n",
+        "and  p.d.  across  capacitor,  VC   -200.01  V\n",
+        "\n",
+        "\n",
+        "  (c)Q-factor  at  resonance,  Qr  is   10.0   \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 502</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the resonant frequency, (b) the value of the p.d. across the capacitor at the resonant frequency, \n",
+      "#(c) the frequency at which the p.d. across the capacitor is a maximum, and \n",
+      "#(d) the value of the maximum voltage across the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  ohms\n",
+      "C  =  0.4e-6;#  IN  fARADS\n",
+      "L  =  0.020;#  IN  Henry\n",
+      "Vm  =  12;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q  =  wr*L/R\n",
+      "Q  =  wr*L/R\n",
+      "Vc  =  Q*Vm\n",
+      " #the  frequency  f  at  which  VC  is  a  maximum  value,\n",
+      "f  =  fr*(1  -  (1/(2*Q*Q)))**0.5\n",
+      " #the  maximum  value  of  the  p.d.  across  the  capacitor  is  given  by:\n",
+      "Vcm  =  Vc/((1  -  (1/(2*Q*Q)))**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)The  resonant  frequency  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency  \",round(Vc,2),\"  V\\n\"\n",
+      "print  \"\\n  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is  \",round(Vcm,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)The  resonant  frequency  is   1779.41   Hz\n",
+        "\n",
+        "\n",
+        "  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency   33.54   V\n",
+        "\n",
+        "\n",
+        "  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is   1721.52   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is   34.67   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 503</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the overall Q-factor of the circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "QL  =  60;#  Q-factor\n",
+      "Qc  =  390;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "QT  =  QL*Qc/(QL  +  Qc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  overall  Q-factor  is  \",QT"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  overall  Q-factor  is   52.0"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 505</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the bandwidth of the filter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  5;#  in  ohms\n",
+      "L  =  0.010;#  IN  Henry\n",
+      "fr  =  10000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance  is  given  by\n",
+      "Qr  =  wr*L/R\n",
+      " #Since  Qr  =  fr/(f2  -  f1),\n",
+      "bw  =  fr/Qr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  bandwidth  of  the  filter  is  \",round(bw,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  bandwidth  of  the  filter  is   79.58   Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 507</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of (a) the inductance, and (b) the capacitance. Find also (c) the bandwidth,\n",
+      "#(d) the lower and upper half-power frequencies and (e) the value of the circuit impedance at the half-power frequencies\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zr  =  50;#  in  ohms\n",
+      "fr  =  1200;#  in  Hz\n",
+      "Qr  =  30;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      " #At  resonance  the  circuit  impedance,  Z\n",
+      "R  =  Zr\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is\n",
+      "L  =  Qr*R/wr\n",
+      " #At  resonance  r*L  =  1/(wr*C)\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*wr*wr)\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #upper  half-power  frequency,  f2\n",
+      "f2  =  (bw  +  ((bw**2)  +  4*(fr**2))**0.5)/2\n",
+      " #lower  half-power  frequency,  f1\n",
+      "f1  =  f2  -  bw\n",
+      " #At  the  half-power  frequencies,  current  I\n",
+      " #I  =  0.707*Ir\n",
+      " #Hence  impedance\n",
+      "Z  =  (2**0.5)*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (b)capacitance,  C  is  \",round(C*1E9,2),\"nF\\n\"\n",
+      "print  \"\\n  (c)bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  upper  half-power  frequency,  f2  is  \",round(f2,2),\"  Hz \"\n",
+      "print   \" and  the  lower  half-power  frequency,  f1  is  \",round(f1,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (e)impedance  at  the  half-power  frequencies  is  \",round(Z,2),\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)inductance,  L  is   198.94 mH\n",
+        "\n",
+        "\n",
+        "  (b)capacitance,  C  is   88.42 nF\n",
+        "\n",
+        "\n",
+        "  (c)bandwidth  is   40.0   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  upper  half-power  frequency,  f2  is   1220.17   Hz \n",
+        " and  the  lower  half-power  frequency,  f1  is   1180.17   Hz\n",
+        "\n",
+        "\n",
+        "  (e)impedance  at  the  half-power  frequencies  is   70.71   ohm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 508</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of (a) the circuit resistance,\n",
+      "#(b) the circuit inductance, (c) the circuit capacitance, and\n",
+      "#(d) the voltage across the capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  0.2;#  in  Volts\n",
+      "I  =  0.004;#  in  Amperes\n",
+      "fr  =  3000;#  in  Hz\n",
+      "Qr  =  100;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*fr\n",
+      " #At  resonance,  impedance\n",
+      "Z  =  V/I\n",
+      " #At  resonance  the  circuit  impedance,  Z\n",
+      "R  =  Z\n",
+      " #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is\n",
+      "L  =  Qr*R/wr\n",
+      " #At  resonance  r*L  =  1/(wr*C)\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*wr*wr)\n",
+      " #Q-factor  at  resonance  in  a  series  circuit  represents  the  voltage  magnification  Qr  =  Vc/V,  then  Vc  is\n",
+      "Vc  =  Qr*V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  circuit  resistance  is  \",round(R,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (c)capacitance,  C  is  \",round(C*1E9,2),\"nF\\n\"\n",
+      "print  \"\\n  (d)the  voltage  across  the  capacitor  is  \",round(Vc,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  circuit  resistance  is   50.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)inductance,  L  is   265.26 mH\n",
+        "\n",
+        "\n",
+        "  (c)capacitance,  C  is   10.61 nF\n",
+        "\n",
+        "\n",
+        "  (d)the  voltage  across  the  capacitor  is   20.0   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 509</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, (b) the Q-factor at resonance, (c) the bandwidth,\n",
+      "#and (d) the lower and upper -3dB frequencies.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8.84;#  in  ohms\n",
+      "L  =  0.3518;#  IN  Henry\n",
+      "C  =  20e-6;#  IN  fARADS\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #the  lower  \u22123  dB  frequency\n",
+      "f1  =  fr  -  bw/2\n",
+      " #the  upper  \u22123  dB  frequency\n",
+      "f2  =  fr  +  bw/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonant  frequency,  fr  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\"\n",
+      "print  \"\\n  (c)Bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  lower  -3dB  frequency,  f1  is  \",round(f1,2),\"  Hz \"\n",
+      "print   \" and  the  upper  -3dB  frequency,  f2  is  \",round(f2,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonant  frequency,  fr  is   60.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Q-factor  at  resonance  is   15.0 \n",
+        "\n",
+        "\n",
+        "  (c)Bandwidth  is   4.0   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  lower  -3dB  frequency,  f1  is   58.0   Hz \n",
+        " and  the  upper  -3dB  frequency,  f2  is   62.0   Hz\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 511</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the circuit when the input voltage is 7.56/_0\u00b0 V and the frequency is \n",
+      "#(a) the resonant frequency, (b) a frequency 3% above the resonant frequency\n",
+      "#(c) the impedance of the circuit when the frequency is 3% above the resonant frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohms\n",
+      "L  =  0.008;#  IN  Henry\n",
+      "C  =  0.3e-6;#  IN  fARADS\n",
+      "rv  =  7.56;#in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "x  =  0.03;\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #At  resonance,\n",
+      "Zr  =  R\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Zr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #If  the  frequency  is  3%  above    fr,  then\n",
+      "de  =  x\n",
+      "I  =  Ir/(1  +  (2*de*Qr*1j))\n",
+      "Z  =  V/I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  at  resonance,  Ir  is  \",round(abs(Ir),2),\"  A\\n\"\n",
+      "print  \"\\n  (b)current  flowing  in  the  circuit  when  frequency  3  percent\"\n",
+      "print   \"      above  the  resonant  frequency  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent\"\n",
+      "print   \"      above  the  resonant  frequency  is  \",round(Z.real,2),\"  +  (\",round(Z.imag,2),\")i  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  at  resonance,  Ir  is   0.5   A\n",
+        "\n",
+        "\n",
+        "  (b)current  flowing  in  the  circuit  when  frequency  3  percent\n",
+        "      above  the  resonant  frequency  is   0.35   +  ( -0.23 )i  A\n",
+        "\n",
+        "\n",
+        "  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent\n",
+        "      above  the  resonant  frequency  is   15.0   +  ( 9.8 )i  A\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_1.ipynb
new file mode 100755
index 00000000..35880e50
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_1.ipynb
@@ -0,0 +1,736 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 28: Series resonance and Q-factor</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 492</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine at what frequency resonance occurs, and (b) the current flowing at resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  40e-6;#  IN  fARADS\n",
+      "L  =  0.075;#  IN  Henry\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #Current  at  resonance,  I\n",
+      "I  =  V/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonant  frequency  =  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Current  at  resonance,  I  is  \",I,\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonant  frequency  =   91.89   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Current  at  resonance,  I  is   20.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 493</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of capacitor C for series resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8;#  in  ohms\n",
+      "L  =  0.010;#  IN  Henry\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #At  resonance\n",
+      " #capacitance  C\n",
+      "C  =  1/(L*(2*math.pi*f)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance,  C  is  \",round(C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance,  C  is   2.53 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 493</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of (a) the stray capacitance CS, and (b) the coil inductance L.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C1  =  1000e-12;#  IN  fARADS\n",
+      "C2  =  500e-12;#  IN  fARADS\n",
+      "fr1  =  92500;#  in  Hz\n",
+      "fr2  =  127800;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  series  R\u2013L\u2013C  circuit  the  resonant  frequency  fr  is  given  by:\n",
+      " #fr  =  1/(2pi*(L*C)**2)\n",
+      "Cs  =  ((C1  -  C2)/((fr2/fr1)**2  -  1))  -  C2\n",
+      "L  =  1/((C1  +  Cs)*(2*math.pi*fr1)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)stray  capacitance,  Cs  is  \",round(Cs*1E12,2),\"pF\\n\"\n",
+      "print  \"\\n  (b)inductance,  L  is  \",round(L*1000,2),\"mH\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)stray  capacitance,  Cs  is   50.13 pF\n",
+        "\n",
+        "\n",
+        "  (b)inductance,  L  is   2.82 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 497</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for this condition (a) the value of inductance L, (b) the p.d. across each component and (c) the Q-factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  5e-6;#  IN  fARADS\n",
+      "rv  =  20;#in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "f  =  318.3;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*f\n",
+      " #The  maximum  voltage  across  the  resistance  occurs  at  resonance  when  the  current  is  a  maximum.  \n",
+      "    #At  resonance,L  =  1/c*wr**2\n",
+      "L  =  1/(C*wr**2)\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  at  resonance  Ir\n",
+      "Ir  =  V/R\n",
+      " #p.d.  across  resistance,  VR\n",
+      "VR  =  Ir*R\n",
+      " #inductive  reactance,  XL\n",
+      "XL  =  wr*L\n",
+      " #p.d.  across  inductance,  VL\n",
+      "VL  =  Ir*(1j*XL)\n",
+      " #capacitive  reactance,  Xc\n",
+      "Xc  =  1/(wr*C)\n",
+      " #p.d.  across  capacitor,  Vc\n",
+      "Vc  =  Ir*(-1j*Xc)\n",
+      " #Q-factor  at  resonance,  Qr\n",
+      "Qr  =  VL.imag/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (b)p.d.  across  resistance,  VR  is  \",VR,\"  V,  p.d.  across  inductance,  VL  \",round( VL.imag,2),\"j V  \"\n",
+      "print   \"and  p.d.  across  capacitor,  VC  \",round(Vc.imag,2),\" V\\n\"\n",
+      "print  \"\\n  (c)Q-factor  at  resonance,  Qr  is  \",round(abs(Qr),2),\"  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)inductance,  L  is   50.0 mH\n",
+        "\n",
+        "\n",
+        "  (b)p.d.  across  resistance,  VR  is   (20+0j)   V,  p.d.  across  inductance,  VL   200.01 j V  \n",
+        "and  p.d.  across  capacitor,  VC   -200.01  V\n",
+        "\n",
+        "\n",
+        "  (c)Q-factor  at  resonance,  Qr  is   10.0   \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 502</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the resonant frequency, (b) the value of the p.d. across the capacitor at the resonant frequency, \n",
+      "#(c) the frequency at which the p.d. across the capacitor is a maximum, and \n",
+      "#(d) the value of the maximum voltage across the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  ohms\n",
+      "C  =  0.4e-6;#  IN  fARADS\n",
+      "L  =  0.020;#  IN  Henry\n",
+      "Vm  =  12;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q  =  wr*L/R\n",
+      "Q  =  wr*L/R\n",
+      "Vc  =  Q*Vm\n",
+      " #the  frequency  f  at  which  VC  is  a  maximum  value,\n",
+      "f  =  fr*(1  -  (1/(2*Q*Q)))**0.5\n",
+      " #the  maximum  value  of  the  p.d.  across  the  capacitor  is  given  by:\n",
+      "Vcm  =  Vc/((1  -  (1/(2*Q*Q)))**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)The  resonant  frequency  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency  \",round(Vc,2),\"  V\\n\"\n",
+      "print  \"\\n  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is  \",round(Vcm,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)The  resonant  frequency  is   1779.41   Hz\n",
+        "\n",
+        "\n",
+        "  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency   33.54   V\n",
+        "\n",
+        "\n",
+        "  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is   1721.52   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is   34.67   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 503</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the overall Q-factor of the circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "QL  =  60;#  Q-factor\n",
+      "Qc  =  390;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "QT  =  QL*Qc/(QL  +  Qc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  overall  Q-factor  is  \",QT"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  overall  Q-factor  is   52.0"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 505</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the bandwidth of the filter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  5;#  in  ohms\n",
+      "L  =  0.010;#  IN  Henry\n",
+      "fr  =  10000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance  is  given  by\n",
+      "Qr  =  wr*L/R\n",
+      " #Since  Qr  =  fr/(f2  -  f1),\n",
+      "bw  =  fr/Qr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  bandwidth  of  the  filter  is  \",round(bw,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  bandwidth  of  the  filter  is   79.58   Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 507</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of (a) the inductance, and (b) the capacitance. Find also (c) the bandwidth,\n",
+      "#(d) the lower and upper half-power frequencies and (e) the value of the circuit impedance at the half-power frequencies\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zr  =  50;#  in  ohms\n",
+      "fr  =  1200;#  in  Hz\n",
+      "Qr  =  30;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      " #At  resonance  the  circuit  impedance,  Z\n",
+      "R  =  Zr\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is\n",
+      "L  =  Qr*R/wr\n",
+      " #At  resonance  r*L  =  1/(wr*C)\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*wr*wr)\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #upper  half-power  frequency,  f2\n",
+      "f2  =  (bw  +  ((bw**2)  +  4*(fr**2))**0.5)/2\n",
+      " #lower  half-power  frequency,  f1\n",
+      "f1  =  f2  -  bw\n",
+      " #At  the  half-power  frequencies,  current  I\n",
+      " #I  =  0.707*Ir\n",
+      " #Hence  impedance\n",
+      "Z  =  (2**0.5)*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (b)capacitance,  C  is  \",round(C*1E9,2),\"nF\\n\"\n",
+      "print  \"\\n  (c)bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  upper  half-power  frequency,  f2  is  \",round(f2,2),\"  Hz \"\n",
+      "print   \" and  the  lower  half-power  frequency,  f1  is  \",round(f1,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (e)impedance  at  the  half-power  frequencies  is  \",round(Z,2),\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)inductance,  L  is   198.94 mH\n",
+        "\n",
+        "\n",
+        "  (b)capacitance,  C  is   88.42 nF\n",
+        "\n",
+        "\n",
+        "  (c)bandwidth  is   40.0   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  upper  half-power  frequency,  f2  is   1220.17   Hz \n",
+        " and  the  lower  half-power  frequency,  f1  is   1180.17   Hz\n",
+        "\n",
+        "\n",
+        "  (e)impedance  at  the  half-power  frequencies  is   70.71   ohm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 508</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of (a) the circuit resistance,\n",
+      "#(b) the circuit inductance, (c) the circuit capacitance, and\n",
+      "#(d) the voltage across the capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  0.2;#  in  Volts\n",
+      "I  =  0.004;#  in  Amperes\n",
+      "fr  =  3000;#  in  Hz\n",
+      "Qr  =  100;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*fr\n",
+      " #At  resonance,  impedance\n",
+      "Z  =  V/I\n",
+      " #At  resonance  the  circuit  impedance,  Z\n",
+      "R  =  Z\n",
+      " #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is\n",
+      "L  =  Qr*R/wr\n",
+      " #At  resonance  r*L  =  1/(wr*C)\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*wr*wr)\n",
+      " #Q-factor  at  resonance  in  a  series  circuit  represents  the  voltage  magnification  Qr  =  Vc/V,  then  Vc  is\n",
+      "Vc  =  Qr*V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  circuit  resistance  is  \",round(R,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (c)capacitance,  C  is  \",round(C*1E9,2),\"nF\\n\"\n",
+      "print  \"\\n  (d)the  voltage  across  the  capacitor  is  \",round(Vc,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  circuit  resistance  is   50.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)inductance,  L  is   265.26 mH\n",
+        "\n",
+        "\n",
+        "  (c)capacitance,  C  is   10.61 nF\n",
+        "\n",
+        "\n",
+        "  (d)the  voltage  across  the  capacitor  is   20.0   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 509</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, (b) the Q-factor at resonance, (c) the bandwidth,\n",
+      "#and (d) the lower and upper -3dB frequencies.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8.84;#  in  ohms\n",
+      "L  =  0.3518;#  IN  Henry\n",
+      "C  =  20e-6;#  IN  fARADS\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #the  lower  \u22123  dB  frequency\n",
+      "f1  =  fr  -  bw/2\n",
+      " #the  upper  \u22123  dB  frequency\n",
+      "f2  =  fr  +  bw/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonant  frequency,  fr  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\"\n",
+      "print  \"\\n  (c)Bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  lower  -3dB  frequency,  f1  is  \",round(f1,2),\"  Hz \"\n",
+      "print   \" and  the  upper  -3dB  frequency,  f2  is  \",round(f2,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonant  frequency,  fr  is   60.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Q-factor  at  resonance  is   15.0 \n",
+        "\n",
+        "\n",
+        "  (c)Bandwidth  is   4.0   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  lower  -3dB  frequency,  f1  is   58.0   Hz \n",
+        " and  the  upper  -3dB  frequency,  f2  is   62.0   Hz\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 511</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the circuit when the input voltage is 7.56/_0\u00b0 V and the frequency is \n",
+      "#(a) the resonant frequency, (b) a frequency 3% above the resonant frequency\n",
+      "#(c) the impedance of the circuit when the frequency is 3% above the resonant frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohms\n",
+      "L  =  0.008;#  IN  Henry\n",
+      "C  =  0.3e-6;#  IN  fARADS\n",
+      "rv  =  7.56;#in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "x  =  0.03;\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #At  resonance,\n",
+      "Zr  =  R\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Zr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #If  the  frequency  is  3%  above    fr,  then\n",
+      "de  =  x\n",
+      "I  =  Ir/(1  +  (2*de*Qr*1j))\n",
+      "Z  =  V/I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  at  resonance,  Ir  is  \",round(abs(Ir),2),\"  A\\n\"\n",
+      "print  \"\\n  (b)current  flowing  in  the  circuit  when  frequency  3  percent\"\n",
+      "print   \"      above  the  resonant  frequency  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent\"\n",
+      "print   \"      above  the  resonant  frequency  is  \",round(Z.real,2),\"  +  (\",round(Z.imag,2),\")i  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  at  resonance,  Ir  is   0.5   A\n",
+        "\n",
+        "\n",
+        "  (b)current  flowing  in  the  circuit  when  frequency  3  percent\n",
+        "      above  the  resonant  frequency  is   0.35   +  ( -0.23 )i  A\n",
+        "\n",
+        "\n",
+        "  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent\n",
+        "      above  the  resonant  frequency  is   15.0   +  ( 9.8 )i  A\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_2.ipynb
new file mode 100755
index 00000000..35880e50
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_2.ipynb
@@ -0,0 +1,736 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 28: Series resonance and Q-factor</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 492</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine at what frequency resonance occurs, and (b) the current flowing at resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  40e-6;#  IN  fARADS\n",
+      "L  =  0.075;#  IN  Henry\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #Current  at  resonance,  I\n",
+      "I  =  V/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonant  frequency  =  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Current  at  resonance,  I  is  \",I,\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonant  frequency  =   91.89   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Current  at  resonance,  I  is   20.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 493</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of capacitor C for series resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8;#  in  ohms\n",
+      "L  =  0.010;#  IN  Henry\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #At  resonance\n",
+      " #capacitance  C\n",
+      "C  =  1/(L*(2*math.pi*f)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance,  C  is  \",round(C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance,  C  is   2.53 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 493</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the values of (a) the stray capacitance CS, and (b) the coil inductance L.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C1  =  1000e-12;#  IN  fARADS\n",
+      "C2  =  500e-12;#  IN  fARADS\n",
+      "fr1  =  92500;#  in  Hz\n",
+      "fr2  =  127800;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  series  R\u2013L\u2013C  circuit  the  resonant  frequency  fr  is  given  by:\n",
+      " #fr  =  1/(2pi*(L*C)**2)\n",
+      "Cs  =  ((C1  -  C2)/((fr2/fr1)**2  -  1))  -  C2\n",
+      "L  =  1/((C1  +  Cs)*(2*math.pi*fr1)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)stray  capacitance,  Cs  is  \",round(Cs*1E12,2),\"pF\\n\"\n",
+      "print  \"\\n  (b)inductance,  L  is  \",round(L*1000,2),\"mH\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)stray  capacitance,  Cs  is   50.13 pF\n",
+        "\n",
+        "\n",
+        "  (b)inductance,  L  is   2.82 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 497</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for this condition (a) the value of inductance L, (b) the p.d. across each component and (c) the Q-factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  5e-6;#  IN  fARADS\n",
+      "rv  =  20;#in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "f  =  318.3;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*f\n",
+      " #The  maximum  voltage  across  the  resistance  occurs  at  resonance  when  the  current  is  a  maximum.  \n",
+      "    #At  resonance,L  =  1/c*wr**2\n",
+      "L  =  1/(C*wr**2)\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  at  resonance  Ir\n",
+      "Ir  =  V/R\n",
+      " #p.d.  across  resistance,  VR\n",
+      "VR  =  Ir*R\n",
+      " #inductive  reactance,  XL\n",
+      "XL  =  wr*L\n",
+      " #p.d.  across  inductance,  VL\n",
+      "VL  =  Ir*(1j*XL)\n",
+      " #capacitive  reactance,  Xc\n",
+      "Xc  =  1/(wr*C)\n",
+      " #p.d.  across  capacitor,  Vc\n",
+      "Vc  =  Ir*(-1j*Xc)\n",
+      " #Q-factor  at  resonance,  Qr\n",
+      "Qr  =  VL.imag/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (b)p.d.  across  resistance,  VR  is  \",VR,\"  V,  p.d.  across  inductance,  VL  \",round( VL.imag,2),\"j V  \"\n",
+      "print   \"and  p.d.  across  capacitor,  VC  \",round(Vc.imag,2),\" V\\n\"\n",
+      "print  \"\\n  (c)Q-factor  at  resonance,  Qr  is  \",round(abs(Qr),2),\"  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)inductance,  L  is   50.0 mH\n",
+        "\n",
+        "\n",
+        "  (b)p.d.  across  resistance,  VR  is   (20+0j)   V,  p.d.  across  inductance,  VL   200.01 j V  \n",
+        "and  p.d.  across  capacitor,  VC   -200.01  V\n",
+        "\n",
+        "\n",
+        "  (c)Q-factor  at  resonance,  Qr  is   10.0   \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 502</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the resonant frequency, (b) the value of the p.d. across the capacitor at the resonant frequency, \n",
+      "#(c) the frequency at which the p.d. across the capacitor is a maximum, and \n",
+      "#(d) the value of the maximum voltage across the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  ohms\n",
+      "C  =  0.4e-6;#  IN  fARADS\n",
+      "L  =  0.020;#  IN  Henry\n",
+      "Vm  =  12;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q  =  wr*L/R\n",
+      "Q  =  wr*L/R\n",
+      "Vc  =  Q*Vm\n",
+      " #the  frequency  f  at  which  VC  is  a  maximum  value,\n",
+      "f  =  fr*(1  -  (1/(2*Q*Q)))**0.5\n",
+      " #the  maximum  value  of  the  p.d.  across  the  capacitor  is  given  by:\n",
+      "Vcm  =  Vc/((1  -  (1/(2*Q*Q)))**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)The  resonant  frequency  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency  \",round(Vc,2),\"  V\\n\"\n",
+      "print  \"\\n  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is  \",round(Vcm,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)The  resonant  frequency  is   1779.41   Hz\n",
+        "\n",
+        "\n",
+        "  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency   33.54   V\n",
+        "\n",
+        "\n",
+        "  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is   1721.52   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is   34.67   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 503</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the overall Q-factor of the circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "QL  =  60;#  Q-factor\n",
+      "Qc  =  390;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "QT  =  QL*Qc/(QL  +  Qc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  overall  Q-factor  is  \",QT"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  overall  Q-factor  is   52.0"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 505</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the bandwidth of the filter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  5;#  in  ohms\n",
+      "L  =  0.010;#  IN  Henry\n",
+      "fr  =  10000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance  is  given  by\n",
+      "Qr  =  wr*L/R\n",
+      " #Since  Qr  =  fr/(f2  -  f1),\n",
+      "bw  =  fr/Qr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  bandwidth  of  the  filter  is  \",round(bw,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  bandwidth  of  the  filter  is   79.58   Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 507</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of (a) the inductance, and (b) the capacitance. Find also (c) the bandwidth,\n",
+      "#(d) the lower and upper half-power frequencies and (e) the value of the circuit impedance at the half-power frequencies\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zr  =  50;#  in  ohms\n",
+      "fr  =  1200;#  in  Hz\n",
+      "Qr  =  30;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      " #At  resonance  the  circuit  impedance,  Z\n",
+      "R  =  Zr\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is\n",
+      "L  =  Qr*R/wr\n",
+      " #At  resonance  r*L  =  1/(wr*C)\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*wr*wr)\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #upper  half-power  frequency,  f2\n",
+      "f2  =  (bw  +  ((bw**2)  +  4*(fr**2))**0.5)/2\n",
+      " #lower  half-power  frequency,  f1\n",
+      "f1  =  f2  -  bw\n",
+      " #At  the  half-power  frequencies,  current  I\n",
+      " #I  =  0.707*Ir\n",
+      " #Hence  impedance\n",
+      "Z  =  (2**0.5)*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (b)capacitance,  C  is  \",round(C*1E9,2),\"nF\\n\"\n",
+      "print  \"\\n  (c)bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  upper  half-power  frequency,  f2  is  \",round(f2,2),\"  Hz \"\n",
+      "print   \" and  the  lower  half-power  frequency,  f1  is  \",round(f1,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (e)impedance  at  the  half-power  frequencies  is  \",round(Z,2),\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)inductance,  L  is   198.94 mH\n",
+        "\n",
+        "\n",
+        "  (b)capacitance,  C  is   88.42 nF\n",
+        "\n",
+        "\n",
+        "  (c)bandwidth  is   40.0   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  upper  half-power  frequency,  f2  is   1220.17   Hz \n",
+        " and  the  lower  half-power  frequency,  f1  is   1180.17   Hz\n",
+        "\n",
+        "\n",
+        "  (e)impedance  at  the  half-power  frequencies  is   70.71   ohm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 508</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of (a) the circuit resistance,\n",
+      "#(b) the circuit inductance, (c) the circuit capacitance, and\n",
+      "#(d) the voltage across the capacitor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  0.2;#  in  Volts\n",
+      "I  =  0.004;#  in  Amperes\n",
+      "fr  =  3000;#  in  Hz\n",
+      "Qr  =  100;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*fr\n",
+      " #At  resonance,  impedance\n",
+      "Z  =  V/I\n",
+      " #At  resonance  the  circuit  impedance,  Z\n",
+      "R  =  Z\n",
+      " #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is\n",
+      "L  =  Qr*R/wr\n",
+      " #At  resonance  r*L  =  1/(wr*C)\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*wr*wr)\n",
+      " #Q-factor  at  resonance  in  a  series  circuit  represents  the  voltage  magnification  Qr  =  Vc/V,  then  Vc  is\n",
+      "Vc  =  Qr*V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  circuit  resistance  is  \",round(R,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (c)capacitance,  C  is  \",round(C*1E9,2),\"nF\\n\"\n",
+      "print  \"\\n  (d)the  voltage  across  the  capacitor  is  \",round(Vc,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  circuit  resistance  is   50.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)inductance,  L  is   265.26 mH\n",
+        "\n",
+        "\n",
+        "  (c)capacitance,  C  is   10.61 nF\n",
+        "\n",
+        "\n",
+        "  (d)the  voltage  across  the  capacitor  is   20.0   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 509</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, (b) the Q-factor at resonance, (c) the bandwidth,\n",
+      "#and (d) the lower and upper -3dB frequencies.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8.84;#  in  ohms\n",
+      "L  =  0.3518;#  IN  Henry\n",
+      "C  =  20e-6;#  IN  fARADS\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #the  lower  \u22123  dB  frequency\n",
+      "f1  =  fr  -  bw/2\n",
+      " #the  upper  \u22123  dB  frequency\n",
+      "f2  =  fr  +  bw/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonant  frequency,  fr  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\"\n",
+      "print  \"\\n  (c)Bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  lower  -3dB  frequency,  f1  is  \",round(f1,2),\"  Hz \"\n",
+      "print   \" and  the  upper  -3dB  frequency,  f2  is  \",round(f2,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonant  frequency,  fr  is   60.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Q-factor  at  resonance  is   15.0 \n",
+        "\n",
+        "\n",
+        "  (c)Bandwidth  is   4.0   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  lower  -3dB  frequency,  f1  is   58.0   Hz \n",
+        " and  the  upper  -3dB  frequency,  f2  is   62.0   Hz\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 511</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the circuit when the input voltage is 7.56/_0\u00b0 V and the frequency is \n",
+      "#(a) the resonant frequency, (b) a frequency 3% above the resonant frequency\n",
+      "#(c) the impedance of the circuit when the frequency is 3% above the resonant frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohms\n",
+      "L  =  0.008;#  IN  Henry\n",
+      "C  =  0.3e-6;#  IN  fARADS\n",
+      "rv  =  7.56;#in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "x  =  0.03;\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #At  resonance,\n",
+      "Zr  =  R\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Zr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #If  the  frequency  is  3%  above    fr,  then\n",
+      "de  =  x\n",
+      "I  =  Ir/(1  +  (2*de*Qr*1j))\n",
+      "Z  =  V/I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  at  resonance,  Ir  is  \",round(abs(Ir),2),\"  A\\n\"\n",
+      "print  \"\\n  (b)current  flowing  in  the  circuit  when  frequency  3  percent\"\n",
+      "print   \"      above  the  resonant  frequency  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent\"\n",
+      "print   \"      above  the  resonant  frequency  is  \",round(Z.real,2),\"  +  (\",round(Z.imag,2),\")i  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  at  resonance,  Ir  is   0.5   A\n",
+        "\n",
+        "\n",
+        "  (b)current  flowing  in  the  circuit  when  frequency  3  percent\n",
+        "      above  the  resonant  frequency  is   0.35   +  ( -0.23 )i  A\n",
+        "\n",
+        "\n",
+        "  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent\n",
+        "      above  the  resonant  frequency  is   15.0   +  ( 9.8 )i  A\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_3.ipynb
new file mode 100755
index 00000000..100af3b8
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_28-checkpoint_3.ipynb
@@ -0,0 +1,729 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ed4e87165c8ac7092d02719ed97384f22be6f2636ae0697903ffa43695fe04ea"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 28: Series resonance and Q-factor</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 492</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  40e-6;#  IN  fARADS\n",
+      "L  =  0.075;#  IN  Henry\n",
+      "V  =  200;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      " #Current  at  resonance,  I\n",
+      "I  =  V/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonant  frequency  =  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Current  at  resonance,  I  is  \",I,\"  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonant  frequency  =   91.89   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Current  at  resonance,  I  is   20.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 493</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8;#  in  ohms\n",
+      "L  =  0.010;#  IN  Henry\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #At  resonance\n",
+      " #capacitance  C\n",
+      "C  =  1/(L*(2*math.pi*f)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance,  C  is  \",round(C*1E6,2),\"uF\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance,  C  is   2.53 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 493</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C1  =  1000e-12;#  IN  fARADS\n",
+      "C2  =  500e-12;#  IN  fARADS\n",
+      "fr1  =  92500;#  in  Hz\n",
+      "fr2  =  127800;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #For  a  series  R\u2013L\u2013C  circuit  the  resonant  frequency  fr  is  given  by:\n",
+      " #fr  =  1/(2pi*(L*C)**2)\n",
+      "Cs  =  ((C1  -  C2)/((fr2/fr1)**2  -  1))  -  C2\n",
+      "L  =  1/((C1  +  Cs)*(2*math.pi*fr1)**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)stray  capacitance,  Cs  is  \",round(Cs*1E12,2),\"pF\\n\"\n",
+      "print  \"\\n  (b)inductance,  L  is  \",round(L*1000,2),\"mH\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)stray  capacitance,  Cs  is   50.13 pF\n",
+        "\n",
+        "\n",
+        "  (b)inductance,  L  is   2.82 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 497</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "C  =  5e-6;#  IN  fARADS\n",
+      "rv  =  20;#in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "f  =  318.3;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*f\n",
+      " #The  maximum  voltage  across  the  resistance  occurs  at  resonance  when  the  current  is  a  maximum.  \n",
+      "    #At  resonance,L  =  1/c*wr**2\n",
+      "L  =  1/(C*wr**2)\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  at  resonance  Ir\n",
+      "Ir  =  V/R\n",
+      " #p.d.  across  resistance,  VR\n",
+      "VR  =  Ir*R\n",
+      " #inductive  reactance,  XL\n",
+      "XL  =  wr*L\n",
+      " #p.d.  across  inductance,  VL\n",
+      "VL  =  Ir*(1j*XL)\n",
+      " #capacitive  reactance,  Xc\n",
+      "Xc  =  1/(wr*C)\n",
+      " #p.d.  across  capacitor,  Vc\n",
+      "Vc  =  Ir*(-1j*Xc)\n",
+      " #Q-factor  at  resonance,  Qr\n",
+      "Qr  =  VL.imag/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (b)p.d.  across  resistance,  VR  is  \",VR,\"  V,  p.d.  across  inductance,  VL  \",round( VL.imag,2),\"j V  \"\n",
+      "print   \"and  p.d.  across  capacitor,  VC  \",round(Vc.imag,2),\" V\\n\"\n",
+      "print  \"\\n  (c)Q-factor  at  resonance,  Qr  is  \",round(abs(Qr),2),\"  \\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)inductance,  L  is   50.0 mH\n",
+        "\n",
+        "\n",
+        "  (b)p.d.  across  resistance,  VR  is   (20+0j)   V,  p.d.  across  inductance,  VL   200.01 j V  \n",
+        "and  p.d.  across  capacitor,  VC   -200.01  V\n",
+        "\n",
+        "\n",
+        "  (c)Q-factor  at  resonance,  Qr  is   10.0   \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 502</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  ohms\n",
+      "C  =  0.4e-6;#  IN  fARADS\n",
+      "L  =  0.020;#  IN  Henry\n",
+      "Vm  =  12;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q  =  wr*L/R\n",
+      "Q  =  wr*L/R\n",
+      "Vc  =  Q*Vm\n",
+      " #the  frequency  f  at  which  VC  is  a  maximum  value,\n",
+      "f  =  fr*(1  -  (1/(2*Q*Q)))**0.5\n",
+      " #the  maximum  value  of  the  p.d.  across  the  capacitor  is  given  by:\n",
+      "Vcm  =  Vc/((1  -  (1/(2*Q*Q)))**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)The  resonant  frequency  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency  \",round(Vc,2),\"  V\\n\"\n",
+      "print  \"\\n  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is  \",round(f,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is  \",round(Vcm,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)The  resonant  frequency  is   1779.41   Hz\n",
+        "\n",
+        "\n",
+        "  (b)the  value  of  the  p.d.  across  the  capacitor  at  the  resonant  frequency   33.54   V\n",
+        "\n",
+        "\n",
+        "  (c)the  frequency  f  at  which  Vc  is  a  maximum  value,  is   1721.52   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  maximum  value  of  the  p.d.  across  the  capacitor  is   34.67   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 503</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "QL  =  60;#  Q-factor\n",
+      "Qc  =  390;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "QT  =  QL*Qc/(QL  +  Qc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  overall  Q-factor  is  \",QT"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  overall  Q-factor  is   52.0"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 505</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  5;#  in  ohms\n",
+      "L  =  0.010;#  IN  Henry\n",
+      "fr  =  10000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance  is  given  by\n",
+      "Qr  =  wr*L/R\n",
+      " #Since  Qr  =  fr/(f2  -  f1),\n",
+      "bw  =  fr/Qr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  bandwidth  of  the  filter  is  \",round(bw,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  bandwidth  of  the  filter  is   79.58   Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 507</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zr  =  50;#  in  ohms\n",
+      "fr  =  1200;#  in  Hz\n",
+      "Qr  =  30;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      " #At  resonance  the  circuit  impedance,  Z\n",
+      "R  =  Zr\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is\n",
+      "L  =  Qr*R/wr\n",
+      " #At  resonance  r*L  =  1/(wr*C)\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*wr*wr)\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #upper  half-power  frequency,  f2\n",
+      "f2  =  (bw  +  ((bw**2)  +  4*(fr**2))**0.5)/2\n",
+      " #lower  half-power  frequency,  f1\n",
+      "f1  =  f2  -  bw\n",
+      " #At  the  half-power  frequencies,  current  I\n",
+      " #I  =  0.707*Ir\n",
+      " #Hence  impedance\n",
+      "Z  =  (2**0.5)*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (b)capacitance,  C  is  \",round(C*1E9,2),\"nF\\n\"\n",
+      "print  \"\\n  (c)bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  upper  half-power  frequency,  f2  is  \",round(f2,2),\"  Hz \"\n",
+      "print   \" and  the  lower  half-power  frequency,  f1  is  \",round(f1,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (e)impedance  at  the  half-power  frequencies  is  \",round(Z,2),\"  ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)inductance,  L  is   198.94 mH\n",
+        "\n",
+        "\n",
+        "  (b)capacitance,  C  is   88.42 nF\n",
+        "\n",
+        "\n",
+        "  (c)bandwidth  is   40.0   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  upper  half-power  frequency,  f2  is   1220.17   Hz \n",
+        " and  the  lower  half-power  frequency,  f1  is   1180.17   Hz\n",
+        "\n",
+        "\n",
+        "  (e)impedance  at  the  half-power  frequencies  is   70.71   ohm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 508</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  0.2;#  in  Volts\n",
+      "I  =  0.004;#  in  Amperes\n",
+      "fr  =  3000;#  in  Hz\n",
+      "Qr  =  100;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "wr  =  2*math.pi*fr\n",
+      " #At  resonance,  impedance\n",
+      "Z  =  V/I\n",
+      " #At  resonance  the  circuit  impedance,  Z\n",
+      "R  =  Z\n",
+      " #Q-factor  at  resonance  is  given  by  Qr  =  wr*L/R,  then  L  is\n",
+      "L  =  Qr*R/wr\n",
+      " #At  resonance  r*L  =  1/(wr*C)\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*wr*wr)\n",
+      " #Q-factor  at  resonance  in  a  series  circuit  represents  the  voltage  magnification  Qr  =  Vc/V,  then  Vc  is\n",
+      "Vc  =  Qr*V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  circuit  resistance  is  \",round(R,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (b)inductance,  L  is  \",round(L*1000,2),\"mH\\n\"\n",
+      "print  \"\\n  (c)capacitance,  C  is  \",round(C*1E9,2),\"nF\\n\"\n",
+      "print  \"\\n  (d)the  voltage  across  the  capacitor  is  \",round(Vc,2),\"  V\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  circuit  resistance  is   50.0   ohm\n",
+        "\n",
+        "\n",
+        "  (b)inductance,  L  is   265.26 mH\n",
+        "\n",
+        "\n",
+        "  (c)capacitance,  C  is   10.61 nF\n",
+        "\n",
+        "\n",
+        "  (d)the  voltage  across  the  capacitor  is   20.0   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 509</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8.84;#  in  ohms\n",
+      "L  =  0.3518;#  IN  Henry\n",
+      "C  =  20e-6;#  IN  fARADS\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #the  lower  \u22123  dB  frequency\n",
+      "f1  =  fr  -  bw/2\n",
+      " #the  upper  \u22123  dB  frequency\n",
+      "f2  =  fr  +  bw/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonant  frequency,  fr  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\"\n",
+      "print  \"\\n  (c)Bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (d)the  lower  -3dB  frequency,  f1  is  \",round(f1,2),\"  Hz \"\n",
+      "print   \" and  the  upper  -3dB  frequency,  f2  is  \",round(f2,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonant  frequency,  fr  is   60.0   Hz\n",
+        "\n",
+        "\n",
+        "  (b)Q-factor  at  resonance  is   15.0 \n",
+        "\n",
+        "\n",
+        "  (c)Bandwidth  is   4.0   Hz\n",
+        "\n",
+        "\n",
+        "  (d)the  lower  -3dB  frequency,  f1  is   58.0   Hz \n",
+        " and  the  upper  -3dB  frequency,  f2  is   62.0   Hz\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 511</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohms\n",
+      "L  =  0.008;#  IN  Henry\n",
+      "C  =  0.3e-6;#  IN  fARADS\n",
+      "rv  =  7.56;#in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "x  =  0.03;\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,\n",
+      "fr  =  1/(2*math.pi*((L*C)**0.5))\n",
+      "wr  =  2*math.pi*fr\n",
+      " #At  resonance,\n",
+      "Zr  =  R\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Zr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #If  the  frequency  is  3%  above    fr,  then\n",
+      "de  =  x\n",
+      "I  =  Ir/(1  +  (2*de*Qr*1j))\n",
+      "Z  =  V/I\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Current  at  resonance,  Ir  is  \",round(abs(Ir),2),\"  A\\n\"\n",
+      "print  \"\\n  (b)current  flowing  in  the  circuit  when  frequency  3  percent\"\n",
+      "print   \"      above  the  resonant  frequency  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
+      "print  \"\\n  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent\"\n",
+      "print   \"      above  the  resonant  frequency  is  \",round(Z.real,2),\"  +  (\",round(Z.imag,2),\")i  A\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Current  at  resonance,  Ir  is   0.5   A\n",
+        "\n",
+        "\n",
+        "  (b)current  flowing  in  the  circuit  when  frequency  3  percent\n",
+        "      above  the  resonant  frequency  is   0.35   +  ( -0.23 )i  A\n",
+        "\n",
+        "\n",
+        "  (c)impedance  of  the  circuit  when  the  frequency  is  3  percent\n",
+        "      above  the  resonant  frequency  is   15.0   +  ( 9.8 )i  A\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint.ipynb
new file mode 100755
index 00000000..d6dbd493
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint.ipynb
@@ -0,0 +1,449 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 29: parallel resonance and Q-factor</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 521</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, (b) the dynamic resistance, \n",
+      "#(c) the current at resonance, and (d) the circuit Q-factor at resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "L  =  0.005;#  IN  Henry\n",
+      "C  =  0.25e-6;#  IN  fARADS\n",
+      "V  =  50;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,  for  parallel\n",
+      "fr  =  ((1/(L*C)  -  ((R**2)/(L**2)))**0.5)/(2*math.pi)\n",
+      " #dynamic  resistance\n",
+      "Rd  =  L/(C*R)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Rd\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (c)Current  at  resonance,  Ir  is  \",round(Ir,2),\"  A\\n\"\n",
+      "print  \"\\n  (d)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency  is   4490.31   Hz\n",
+        "\n",
+        "\n",
+        "  (b)dynamic  resistance   2000.0   ohm\n",
+        "\n",
+        "\n",
+        "  (c)Current  at  resonance,  Ir  is   0.02   A\n",
+        "\n",
+        "\n",
+        "  (d)Q-factor  at  resonance  is   14.11 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 521</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resonant frequency for the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL1  =  0;#  in  ohms\n",
+      "RL2  =  30;#  in  ohms\n",
+      "L  =  0.100;#  IN  Henry\n",
+      "C  =  40e-6;#  IN  fARADS\n",
+      "V  =  50;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #for  RL1\n",
+      " #Resonant  frequency,\n",
+      "wr1  =  (1/(L*C))**0.5\n",
+      "fr1  =  wr1/(2*math.pi)\n",
+      " #for  RL2\n",
+      " #Resonant  frequency,\n",
+      "wr2  =  (1/(L*C)  -  ((RL2**2)/(L**2)))**0.5\n",
+      "fr2  =  wr2/(2*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency  at  RL  =  0  is  \",round(fr1,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Resonance  frequency  at  RL  =  30  ohm  is  \",round(fr2,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency  at  RL  =  0  is   79.58   Hz\n",
+        "\n",
+        "  (b)Resonance  frequency  at  RL  =  30  ohm  is   63.66   Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 523</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the condition when the supply current is a minimum, \n",
+      "#(a) the capacitance of the capacitor, (b) the dynamic resistance, \n",
+      "#(c) the supply current, (d) the Q-factor, (e) the bandwidth,\n",
+      "#(f) the upper and lower \u00033 dB frequencies, and (g) the value of the circuit impedance at the \u00033 dB frequencies\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  150;#  in  ohms\n",
+      "L  =  0.120;#  IN  Henry\n",
+      "V  =  20;#in  volts\n",
+      "fr  =  4000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*((2*math.pi*fr)**2  +  ((R**2)/(L**2))))\n",
+      "Rd  =  L/(C*R)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Rd\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #upper  half-power  frequency,  f2\n",
+      "f2  =  (bw  +  ((bw**2)  +  4*(fr**2))**0.5)/2\n",
+      " #lower  half-power  frequency,  f1\n",
+      "f1  =  f2  -  bw\n",
+      " #impedance  at  the  \u22123  dB  frequencies\n",
+      "Z  =  Rd/(2**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitance  of  the  capacitor,C  is  \",round(C*1E6,2),\"uF\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"ohm\\n\"\n",
+      "print  \"\\n  (c)Current  at  resonance,  Ir  is  \",round(Ir*1000,2),\"mA\\n\"\n",
+      "print  \"\\n  (d)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\"\n",
+      "print  \"\\n  (e)bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (f)the  upper  half-power  frequency,  f2  is  \",round(f2,2),\"  Hz  and \"\n",
+      "print   \" the  lower  half-power  frequency,  f1  is  \",round(f1,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (g)impedance  at  the  -3  dB  frequencies  is  \",round(Z,2),\" ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitance  of  the  capacitor,C  is   0.01 uF\n",
+        "\n",
+        "  (b)dynamic  resistance   60788.85 ohm\n",
+        "\n",
+        "\n",
+        "  (c)Current  at  resonance,  Ir  is   0.33 mA\n",
+        "\n",
+        "\n",
+        "  (d)Q-factor  at  resonance  is   20.11 \n",
+        "\n",
+        "\n",
+        "  (e)bandwidth  is   198.94   Hz\n",
+        "\n",
+        "\n",
+        "  (f)the  upper  half-power  frequency,  f2  is   4100.71   Hz  and \n",
+        " the  lower  half-power  frequency,  f1  is   3901.76   Hz\n",
+        "\n",
+        "\n",
+        "  (g)impedance  at  the  -3  dB  frequencies  is   42984.21  ohm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 525</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resonant frequency of the network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  5;#  in  ohms\n",
+      "L  =  0.002;#  IN  Henry\n",
+      "C  =  25e-6;#  IN  fARADS\n",
+      "Rc  =  3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,  for  parallel\n",
+      "fr  =  (1/(2*math.pi*((L*C)**0.5)))*((RL**2  -  (L/C))/(Rc**2  -  (L/C)))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  resonant  frequency,  fr  is  \",round(fr,2),\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  resonant  frequency,  fr  is   626.45   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 525</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the parallel network the values of inductance L\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  3;#  in  ohms\n",
+      "fr  =  1000;#  in  Hz\n",
+      "Xc  =  10;#  IN  ohms\n",
+      "Rc  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "XL1  =  (((Rc**2  +  Xc**2)/Xc)  +  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2\n",
+      "XL2  =  (((Rc**2  +  Xc**2)/Xc)  -  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2\n",
+      "wr  =  2*math.pi*fr\n",
+      " #inductance\n",
+      "L1  =  XL1/wr\n",
+      "L2  =  XL2/wr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  is  either  \",round(L1*1000,2),\"mH  or  \",round(L2*1000,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  is  either   1.71 mH  or   0.13 mH"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 526</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the overall Q-factor of the parallel combination.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "QL  =  60;#  Q-factor\n",
+      "Qc  =  300;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "QT  =  QL*Qc/(QL  +  Qc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  overall  Q-factor  is  \",round(QT,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  overall  Q-factor  is   50.0"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 527</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the circuit (a) the Q-factor, (b) the dynamic resistance, and\n",
+      "#(c) the magnitude of the impedance when the supply frequency is 0.4% greater than the tuned frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  10.61E-9;#  in  Farad\n",
+      "bw  =  500;#  in  Hz\n",
+      "fr  =  150000;#  in  Hz\n",
+      "x  =  0.004\n",
+      "\n",
+      "#calculation:\n",
+      " #Q-factor\n",
+      "Q  =  fr/bw\n",
+      "wr  =  2*math.pi*fr\n",
+      " #dynamic  resistance,  RD\n",
+      "Rd =  Q/(C*wr)\n",
+      "de =  x\n",
+      "Z  =  Rd/(1  +  (2*de*Q*1j))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Q-factor  \",round(Q,2),\"\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"ohm\"\n",
+      "print  \"\\n  (c)magnitude  of  the  impedance  \",round(abs(Z),2),\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Q-factor   300.0 \n",
+        "\n",
+        "  (b)dynamic  resistance   30000.93 ohm\n",
+        "\n",
+        "  (c)magnitude  of  the  impedance   11538.82 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_1.ipynb
new file mode 100755
index 00000000..d6dbd493
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_1.ipynb
@@ -0,0 +1,449 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 29: parallel resonance and Q-factor</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 521</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, (b) the dynamic resistance, \n",
+      "#(c) the current at resonance, and (d) the circuit Q-factor at resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "L  =  0.005;#  IN  Henry\n",
+      "C  =  0.25e-6;#  IN  fARADS\n",
+      "V  =  50;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,  for  parallel\n",
+      "fr  =  ((1/(L*C)  -  ((R**2)/(L**2)))**0.5)/(2*math.pi)\n",
+      " #dynamic  resistance\n",
+      "Rd  =  L/(C*R)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Rd\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (c)Current  at  resonance,  Ir  is  \",round(Ir,2),\"  A\\n\"\n",
+      "print  \"\\n  (d)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency  is   4490.31   Hz\n",
+        "\n",
+        "\n",
+        "  (b)dynamic  resistance   2000.0   ohm\n",
+        "\n",
+        "\n",
+        "  (c)Current  at  resonance,  Ir  is   0.02   A\n",
+        "\n",
+        "\n",
+        "  (d)Q-factor  at  resonance  is   14.11 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 521</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resonant frequency for the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL1  =  0;#  in  ohms\n",
+      "RL2  =  30;#  in  ohms\n",
+      "L  =  0.100;#  IN  Henry\n",
+      "C  =  40e-6;#  IN  fARADS\n",
+      "V  =  50;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #for  RL1\n",
+      " #Resonant  frequency,\n",
+      "wr1  =  (1/(L*C))**0.5\n",
+      "fr1  =  wr1/(2*math.pi)\n",
+      " #for  RL2\n",
+      " #Resonant  frequency,\n",
+      "wr2  =  (1/(L*C)  -  ((RL2**2)/(L**2)))**0.5\n",
+      "fr2  =  wr2/(2*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency  at  RL  =  0  is  \",round(fr1,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Resonance  frequency  at  RL  =  30  ohm  is  \",round(fr2,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency  at  RL  =  0  is   79.58   Hz\n",
+        "\n",
+        "  (b)Resonance  frequency  at  RL  =  30  ohm  is   63.66   Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 523</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the condition when the supply current is a minimum, \n",
+      "#(a) the capacitance of the capacitor, (b) the dynamic resistance, \n",
+      "#(c) the supply current, (d) the Q-factor, (e) the bandwidth,\n",
+      "#(f) the upper and lower \u00033 dB frequencies, and (g) the value of the circuit impedance at the \u00033 dB frequencies\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  150;#  in  ohms\n",
+      "L  =  0.120;#  IN  Henry\n",
+      "V  =  20;#in  volts\n",
+      "fr  =  4000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*((2*math.pi*fr)**2  +  ((R**2)/(L**2))))\n",
+      "Rd  =  L/(C*R)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Rd\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #upper  half-power  frequency,  f2\n",
+      "f2  =  (bw  +  ((bw**2)  +  4*(fr**2))**0.5)/2\n",
+      " #lower  half-power  frequency,  f1\n",
+      "f1  =  f2  -  bw\n",
+      " #impedance  at  the  \u22123  dB  frequencies\n",
+      "Z  =  Rd/(2**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitance  of  the  capacitor,C  is  \",round(C*1E6,2),\"uF\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"ohm\\n\"\n",
+      "print  \"\\n  (c)Current  at  resonance,  Ir  is  \",round(Ir*1000,2),\"mA\\n\"\n",
+      "print  \"\\n  (d)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\"\n",
+      "print  \"\\n  (e)bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (f)the  upper  half-power  frequency,  f2  is  \",round(f2,2),\"  Hz  and \"\n",
+      "print   \" the  lower  half-power  frequency,  f1  is  \",round(f1,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (g)impedance  at  the  -3  dB  frequencies  is  \",round(Z,2),\" ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitance  of  the  capacitor,C  is   0.01 uF\n",
+        "\n",
+        "  (b)dynamic  resistance   60788.85 ohm\n",
+        "\n",
+        "\n",
+        "  (c)Current  at  resonance,  Ir  is   0.33 mA\n",
+        "\n",
+        "\n",
+        "  (d)Q-factor  at  resonance  is   20.11 \n",
+        "\n",
+        "\n",
+        "  (e)bandwidth  is   198.94   Hz\n",
+        "\n",
+        "\n",
+        "  (f)the  upper  half-power  frequency,  f2  is   4100.71   Hz  and \n",
+        " the  lower  half-power  frequency,  f1  is   3901.76   Hz\n",
+        "\n",
+        "\n",
+        "  (g)impedance  at  the  -3  dB  frequencies  is   42984.21  ohm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 525</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resonant frequency of the network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  5;#  in  ohms\n",
+      "L  =  0.002;#  IN  Henry\n",
+      "C  =  25e-6;#  IN  fARADS\n",
+      "Rc  =  3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,  for  parallel\n",
+      "fr  =  (1/(2*math.pi*((L*C)**0.5)))*((RL**2  -  (L/C))/(Rc**2  -  (L/C)))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  resonant  frequency,  fr  is  \",round(fr,2),\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  resonant  frequency,  fr  is   626.45   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 525</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the parallel network the values of inductance L\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  3;#  in  ohms\n",
+      "fr  =  1000;#  in  Hz\n",
+      "Xc  =  10;#  IN  ohms\n",
+      "Rc  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "XL1  =  (((Rc**2  +  Xc**2)/Xc)  +  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2\n",
+      "XL2  =  (((Rc**2  +  Xc**2)/Xc)  -  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2\n",
+      "wr  =  2*math.pi*fr\n",
+      " #inductance\n",
+      "L1  =  XL1/wr\n",
+      "L2  =  XL2/wr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  is  either  \",round(L1*1000,2),\"mH  or  \",round(L2*1000,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  is  either   1.71 mH  or   0.13 mH"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 526</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the overall Q-factor of the parallel combination.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "QL  =  60;#  Q-factor\n",
+      "Qc  =  300;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "QT  =  QL*Qc/(QL  +  Qc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  overall  Q-factor  is  \",round(QT,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  overall  Q-factor  is   50.0"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 527</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the circuit (a) the Q-factor, (b) the dynamic resistance, and\n",
+      "#(c) the magnitude of the impedance when the supply frequency is 0.4% greater than the tuned frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  10.61E-9;#  in  Farad\n",
+      "bw  =  500;#  in  Hz\n",
+      "fr  =  150000;#  in  Hz\n",
+      "x  =  0.004\n",
+      "\n",
+      "#calculation:\n",
+      " #Q-factor\n",
+      "Q  =  fr/bw\n",
+      "wr  =  2*math.pi*fr\n",
+      " #dynamic  resistance,  RD\n",
+      "Rd =  Q/(C*wr)\n",
+      "de =  x\n",
+      "Z  =  Rd/(1  +  (2*de*Q*1j))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Q-factor  \",round(Q,2),\"\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"ohm\"\n",
+      "print  \"\\n  (c)magnitude  of  the  impedance  \",round(abs(Z),2),\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Q-factor   300.0 \n",
+        "\n",
+        "  (b)dynamic  resistance   30000.93 ohm\n",
+        "\n",
+        "  (c)magnitude  of  the  impedance   11538.82 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_2.ipynb
new file mode 100755
index 00000000..d6dbd493
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_2.ipynb
@@ -0,0 +1,449 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 29: parallel resonance and Q-factor</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 521</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, (b) the dynamic resistance, \n",
+      "#(c) the current at resonance, and (d) the circuit Q-factor at resonance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "L  =  0.005;#  IN  Henry\n",
+      "C  =  0.25e-6;#  IN  fARADS\n",
+      "V  =  50;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,  for  parallel\n",
+      "fr  =  ((1/(L*C)  -  ((R**2)/(L**2)))**0.5)/(2*math.pi)\n",
+      " #dynamic  resistance\n",
+      "Rd  =  L/(C*R)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Rd\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (c)Current  at  resonance,  Ir  is  \",round(Ir,2),\"  A\\n\"\n",
+      "print  \"\\n  (d)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency  is   4490.31   Hz\n",
+        "\n",
+        "\n",
+        "  (b)dynamic  resistance   2000.0   ohm\n",
+        "\n",
+        "\n",
+        "  (c)Current  at  resonance,  Ir  is   0.02   A\n",
+        "\n",
+        "\n",
+        "  (d)Q-factor  at  resonance  is   14.11 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 521</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resonant frequency for the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL1  =  0;#  in  ohms\n",
+      "RL2  =  30;#  in  ohms\n",
+      "L  =  0.100;#  IN  Henry\n",
+      "C  =  40e-6;#  IN  fARADS\n",
+      "V  =  50;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #for  RL1\n",
+      " #Resonant  frequency,\n",
+      "wr1  =  (1/(L*C))**0.5\n",
+      "fr1  =  wr1/(2*math.pi)\n",
+      " #for  RL2\n",
+      " #Resonant  frequency,\n",
+      "wr2  =  (1/(L*C)  -  ((RL2**2)/(L**2)))**0.5\n",
+      "fr2  =  wr2/(2*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency  at  RL  =  0  is  \",round(fr1,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Resonance  frequency  at  RL  =  30  ohm  is  \",round(fr2,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency  at  RL  =  0  is   79.58   Hz\n",
+        "\n",
+        "  (b)Resonance  frequency  at  RL  =  30  ohm  is   63.66   Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 523</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the condition when the supply current is a minimum, \n",
+      "#(a) the capacitance of the capacitor, (b) the dynamic resistance, \n",
+      "#(c) the supply current, (d) the Q-factor, (e) the bandwidth,\n",
+      "#(f) the upper and lower \u00033 dB frequencies, and (g) the value of the circuit impedance at the \u00033 dB frequencies\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  150;#  in  ohms\n",
+      "L  =  0.120;#  IN  Henry\n",
+      "V  =  20;#in  volts\n",
+      "fr  =  4000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*((2*math.pi*fr)**2  +  ((R**2)/(L**2))))\n",
+      "Rd  =  L/(C*R)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Rd\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #upper  half-power  frequency,  f2\n",
+      "f2  =  (bw  +  ((bw**2)  +  4*(fr**2))**0.5)/2\n",
+      " #lower  half-power  frequency,  f1\n",
+      "f1  =  f2  -  bw\n",
+      " #impedance  at  the  \u22123  dB  frequencies\n",
+      "Z  =  Rd/(2**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitance  of  the  capacitor,C  is  \",round(C*1E6,2),\"uF\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"ohm\\n\"\n",
+      "print  \"\\n  (c)Current  at  resonance,  Ir  is  \",round(Ir*1000,2),\"mA\\n\"\n",
+      "print  \"\\n  (d)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\"\n",
+      "print  \"\\n  (e)bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (f)the  upper  half-power  frequency,  f2  is  \",round(f2,2),\"  Hz  and \"\n",
+      "print   \" the  lower  half-power  frequency,  f1  is  \",round(f1,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (g)impedance  at  the  -3  dB  frequencies  is  \",round(Z,2),\" ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitance  of  the  capacitor,C  is   0.01 uF\n",
+        "\n",
+        "  (b)dynamic  resistance   60788.85 ohm\n",
+        "\n",
+        "\n",
+        "  (c)Current  at  resonance,  Ir  is   0.33 mA\n",
+        "\n",
+        "\n",
+        "  (d)Q-factor  at  resonance  is   20.11 \n",
+        "\n",
+        "\n",
+        "  (e)bandwidth  is   198.94   Hz\n",
+        "\n",
+        "\n",
+        "  (f)the  upper  half-power  frequency,  f2  is   4100.71   Hz  and \n",
+        " the  lower  half-power  frequency,  f1  is   3901.76   Hz\n",
+        "\n",
+        "\n",
+        "  (g)impedance  at  the  -3  dB  frequencies  is   42984.21  ohm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 525</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the resonant frequency of the network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  5;#  in  ohms\n",
+      "L  =  0.002;#  IN  Henry\n",
+      "C  =  25e-6;#  IN  fARADS\n",
+      "Rc  =  3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,  for  parallel\n",
+      "fr  =  (1/(2*math.pi*((L*C)**0.5)))*((RL**2  -  (L/C))/(Rc**2  -  (L/C)))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  resonant  frequency,  fr  is  \",round(fr,2),\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  resonant  frequency,  fr  is   626.45   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 525</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the parallel network the values of inductance L\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  3;#  in  ohms\n",
+      "fr  =  1000;#  in  Hz\n",
+      "Xc  =  10;#  IN  ohms\n",
+      "Rc  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "XL1  =  (((Rc**2  +  Xc**2)/Xc)  +  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2\n",
+      "XL2  =  (((Rc**2  +  Xc**2)/Xc)  -  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2\n",
+      "wr  =  2*math.pi*fr\n",
+      " #inductance\n",
+      "L1  =  XL1/wr\n",
+      "L2  =  XL2/wr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  is  either  \",round(L1*1000,2),\"mH  or  \",round(L2*1000,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  is  either   1.71 mH  or   0.13 mH"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 526</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the overall Q-factor of the parallel combination.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "QL  =  60;#  Q-factor\n",
+      "Qc  =  300;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "QT  =  QL*Qc/(QL  +  Qc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  overall  Q-factor  is  \",round(QT,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  overall  Q-factor  is   50.0"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 527</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the circuit (a) the Q-factor, (b) the dynamic resistance, and\n",
+      "#(c) the magnitude of the impedance when the supply frequency is 0.4% greater than the tuned frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  10.61E-9;#  in  Farad\n",
+      "bw  =  500;#  in  Hz\n",
+      "fr  =  150000;#  in  Hz\n",
+      "x  =  0.004\n",
+      "\n",
+      "#calculation:\n",
+      " #Q-factor\n",
+      "Q  =  fr/bw\n",
+      "wr  =  2*math.pi*fr\n",
+      " #dynamic  resistance,  RD\n",
+      "Rd =  Q/(C*wr)\n",
+      "de =  x\n",
+      "Z  =  Rd/(1  +  (2*de*Q*1j))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Q-factor  \",round(Q,2),\"\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"ohm\"\n",
+      "print  \"\\n  (c)magnitude  of  the  impedance  \",round(abs(Z),2),\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Q-factor   300.0 \n",
+        "\n",
+        "  (b)dynamic  resistance   30000.93 ohm\n",
+        "\n",
+        "  (c)magnitude  of  the  impedance   11538.82 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_3.ipynb
new file mode 100755
index 00000000..169830ec
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_29-checkpoint_3.ipynb
@@ -0,0 +1,445 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:57634c93bf4d4aea76e807a2cbb9e510f617e9cf546f62a2de10f5bf9157d3c1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 29: parallel resonance and Q-factor</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 521</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohms\n",
+      "L  =  0.005;#  IN  Henry\n",
+      "C  =  0.25e-6;#  IN  fARADS\n",
+      "V  =  50;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,  for  parallel\n",
+      "fr  =  ((1/(L*C)  -  ((R**2)/(L**2)))**0.5)/(2*math.pi)\n",
+      " #dynamic  resistance\n",
+      "Rd  =  L/(C*R)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Rd\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency  is  \",round(fr,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"  ohm\\n\"\n",
+      "print  \"\\n  (c)Current  at  resonance,  Ir  is  \",round(Ir,2),\"  A\\n\"\n",
+      "print  \"\\n  (d)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency  is   4490.31   Hz\n",
+        "\n",
+        "\n",
+        "  (b)dynamic  resistance   2000.0   ohm\n",
+        "\n",
+        "\n",
+        "  (c)Current  at  resonance,  Ir  is   0.02   A\n",
+        "\n",
+        "\n",
+        "  (d)Q-factor  at  resonance  is   14.11 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 521</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL1  =  0;#  in  ohms\n",
+      "RL2  =  30;#  in  ohms\n",
+      "L  =  0.100;#  IN  Henry\n",
+      "C  =  40e-6;#  IN  fARADS\n",
+      "V  =  50;#in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #for  RL1\n",
+      " #Resonant  frequency,\n",
+      "wr1  =  (1/(L*C))**0.5\n",
+      "fr1  =  wr1/(2*math.pi)\n",
+      " #for  RL2\n",
+      " #Resonant  frequency,\n",
+      "wr2  =  (1/(L*C)  -  ((RL2**2)/(L**2)))**0.5\n",
+      "fr2  =  wr2/(2*math.pi)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Resonance  frequency  at  RL  =  0  is  \",round(fr1,2),\"  Hz\"\n",
+      "print  \"\\n  (b)Resonance  frequency  at  RL  =  30  ohm  is  \",round(fr2,2),\"  Hz\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Resonance  frequency  at  RL  =  0  is   79.58   Hz\n",
+        "\n",
+        "  (b)Resonance  frequency  at  RL  =  30  ohm  is   63.66   Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 523</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  150;#  in  ohms\n",
+      "L  =  0.120;#  IN  Henry\n",
+      "V  =  20;#in  volts\n",
+      "fr  =  4000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #capacitance,  C\n",
+      "C  =  1/(L*((2*math.pi*fr)**2  +  ((R**2)/(L**2))))\n",
+      "Rd  =  L/(C*R)\n",
+      " #Current  at  resonance\n",
+      "Ir  =  V/Rd\n",
+      "wr  =  2*math.pi*fr\n",
+      " #Q-factor  at  resonance,  Q  =  wr*L/R\n",
+      "Qr  =  wr*L/R\n",
+      " #bandwidth,.(f2  \u2212  f1)\n",
+      "bw  =  fr/Qr\n",
+      " #upper  half-power  frequency,  f2\n",
+      "f2  =  (bw  +  ((bw**2)  +  4*(fr**2))**0.5)/2\n",
+      " #lower  half-power  frequency,  f1\n",
+      "f1  =  f2  -  bw\n",
+      " #impedance  at  the  \u22123  dB  frequencies\n",
+      "Z  =  Rd/(2**0.5)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  capacitance  of  the  capacitor,C  is  \",round(C*1E6,2),\"uF\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"ohm\\n\"\n",
+      "print  \"\\n  (c)Current  at  resonance,  Ir  is  \",round(Ir*1000,2),\"mA\\n\"\n",
+      "print  \"\\n  (d)Q-factor  at  resonance  is  \",round(Qr,2),\"\\n\"\n",
+      "print  \"\\n  (e)bandwidth  is  \",round(bw,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (f)the  upper  half-power  frequency,  f2  is  \",round(f2,2),\"  Hz  and \"\n",
+      "print   \" the  lower  half-power  frequency,  f1  is  \",round(f1,2),\"  Hz\\n\"\n",
+      "print  \"\\n  (g)impedance  at  the  -3  dB  frequencies  is  \",round(Z,2),\" ohm\\n\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  capacitance  of  the  capacitor,C  is   0.01 uF\n",
+        "\n",
+        "  (b)dynamic  resistance   60788.85 ohm\n",
+        "\n",
+        "\n",
+        "  (c)Current  at  resonance,  Ir  is   0.33 mA\n",
+        "\n",
+        "\n",
+        "  (d)Q-factor  at  resonance  is   20.11 \n",
+        "\n",
+        "\n",
+        "  (e)bandwidth  is   198.94   Hz\n",
+        "\n",
+        "\n",
+        "  (f)the  upper  half-power  frequency,  f2  is   4100.71   Hz  and \n",
+        " the  lower  half-power  frequency,  f1  is   3901.76   Hz\n",
+        "\n",
+        "\n",
+        "  (g)impedance  at  the  -3  dB  frequencies  is   42984.21  ohm\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 525</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  5;#  in  ohms\n",
+      "L  =  0.002;#  IN  Henry\n",
+      "C  =  25e-6;#  IN  fARADS\n",
+      "Rc  =  3;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #Resonant  frequency,  for  parallel\n",
+      "fr  =  (1/(2*math.pi*((L*C)**0.5)))*((RL**2  -  (L/C))/(Rc**2  -  (L/C)))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  resonant  frequency,  fr  is  \",round(fr,2),\"  Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  resonant  frequency,  fr  is   626.45   Hz"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 525</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "RL  =  3;#  in  ohms\n",
+      "fr  =  1000;#  in  Hz\n",
+      "Xc  =  10;#  IN  ohms\n",
+      "Rc  =  4;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      "XL1  =  (((Rc**2  +  Xc**2)/Xc)  +  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2\n",
+      "XL2  =  (((Rc**2  +  Xc**2)/Xc)  -  ((((Rc**2  +  Xc**2)/Xc)**2)  -  4*(RL**2))**0.5)/2\n",
+      "wr  =  2*math.pi*fr\n",
+      " #inductance\n",
+      "L1  =  XL1/wr\n",
+      "L2  =  XL2/wr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  is  either  \",round(L1*1000,2),\"mH  or  \",round(L2*1000,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  is  either   1.71 mH  or   0.13 mH"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 526</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "QL  =  60;#  Q-factor\n",
+      "Qc  =  300;#  Q-factor\n",
+      "\n",
+      "#calculation:\n",
+      "QT  =  QL*Qc/(QL  +  Qc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  overall  Q-factor  is  \",round(QT,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  overall  Q-factor  is   50.0"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 527</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  10.61E-9;#  in  Farad\n",
+      "bw  =  500;#  in  Hz\n",
+      "fr  =  150000;#  in  Hz\n",
+      "x  =  0.004\n",
+      "\n",
+      "#calculation:\n",
+      " #Q-factor\n",
+      "Q  =  fr/bw\n",
+      "wr  =  2*math.pi*fr\n",
+      " #dynamic  resistance,  RD\n",
+      "Rd =  Q/(C*wr)\n",
+      "de =  x\n",
+      "Z  =  Rd/(1  +  (2*de*Q*1j))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)Q-factor  \",round(Q,2),\"\"\n",
+      "print  \"\\n  (b)dynamic  resistance  \",round(Rd,2),\"ohm\"\n",
+      "print  \"\\n  (c)magnitude  of  the  impedance  \",round(abs(Z),2),\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)Q-factor   300.0 \n",
+        "\n",
+        "  (b)dynamic  resistance   30000.93 ohm\n",
+        "\n",
+        "  (c)magnitude  of  the  impedance   11538.82 ohm"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint.ipynb
new file mode 100755
index 00000000..b23cbc8e
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint.ipynb
@@ -0,0 +1,303 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 30: Introduction to network analysis</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 536</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the current flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  25;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  branch  currents  and  their  directions  are  labelled  as  shown  in  Figure  30.4\n",
+      " #Two  loops  are  chosen.  loop  ABEF,  and  loop  BCDE\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #two  eqns  obtained\n",
+      " #(R1  +  R2)*I1  +  R2*I2  =  V1\n",
+      " #R2*I1  +  (R2  +  R3)*I2  =  V2\n",
+      "I1  =  (3*V1  -  2*V2)/(3*(R1  +  R2)  -  2*(R2))\n",
+      "I2  =  (V2  -  R2*I1)/(R2  +  R3)\n",
+      "I  =  I1  +  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I1  is  \",round(I1.real,2),\"  +  (\",round( I1.imag,2),\")i  A,  \\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round(  I2.imag,2),\")i  A and \"\n",
+      "print   \" total  current,  I  is  \",round(I.real,2),\"  +  (\",round( I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I1  is   3.16   +  ( -1.05 )i  A,  \n",
+        " current,  I2  is     -2.11   +  ( 2.37 )i  A and \n",
+        " total  current,  I  is   1.05   +  ( 1.32 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 537</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the 2 ohm resistor of the circuit\n",
+      "#find the power dissipated in the 3 ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "#initializing  the  variables:\n",
+      "V  =  8;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  2;#  in  ohm\n",
+      "R3  =  3;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  6;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Currents  and  their  directions  are  assigned  as  shown  in  Figure  30.6.\n",
+      " #Three  loops  are  chosen  since  three  unknown  currents  are  required.  The  choice  of  loop  directions  is  arbitrary.\n",
+      "    #loop  ABCDE,  and  loop  EDGF  and  loop  DCHG\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #R5*I1  +  (R6  +  R4)*I2  -  R4*I3  =  V\n",
+      " #-1*R1*I1  +  (R6  +  R1)*I2  +  R2*I3  =  0\n",
+      " #  R3*I1  -  (R3  +  R4)*I2  +  (R2  +  R3  +  R4)*I3  =  0\n",
+      "#using  determinants\n",
+      "d1  =  [[V, (R6  +  R4), -1*R4],[0, (R6  +  R1),  R2], [0,  (-1*(R3  +  R4)), (R2  +  R3  +  R4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[R5, V,  -1*R4],[-1*R1,  0,  R2],[ R3,  0,  (R2  +  R3  +  R4)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[R5,  (R6  +  R4),  V],[-1*R1,  (R6  +  R1),  0],[ R3,  (-1*(R3  +  R4)),  0]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[R5,  (R6  +  R4),  -1*R4],[-1*R1,  (R6  +  R1),  R2],[ R3,  (-1*(R3  +  R4)),  (R2  +  R3  +  R4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "#Current  in  the  2  ohm  resistance\n",
+      "I  =  I1  -  I2  +  I3\n",
+      "#power  dissipated  in  the  3  ohm  resistance\n",
+      "P3  =  R3*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  through  2  ohm  resistor  is  \",round(I2,3),\"  A\"\n",
+      "print  \"\\n  (b)power  dissipated  in  the  3  ohm  resistor  is  \",round(P3,3),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  through  2  ohm  resistor  is   0.203   A\n",
+        "\n",
+        "  (b)power  dissipated  in  the  3  ohm  resistor  is   1.267 W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 539</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in each branch using Kirchhoff\u2019s laws.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  5  +  0j;#  in  volts\n",
+      "E2  =  2  +  4j;#  in  volts\n",
+      "Z1  =  3  +  4j;#  in  ohm\n",
+      "Z2  =  2  -  5j;#  in  ohm\n",
+      "Z3  =  6  +  8j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Currents  I1  and  I2  with  their  directions  are  shown  in  Figure  30.8.\n",
+      " #Two  loops  are  chosen  with  their  directions  both  clockwise.loop  ABEF  and  loop  BCDE,\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #two  eqns  obtained\n",
+      " #(Z1  +  Z3)*I1  -  Z3*I2  =  E1\n",
+      " #-1*Z3*I1  +  (Z2  +  Z3)*I2  =  E2\n",
+      "I1  =  ((Z2  +  Z3)*E1  +  Z3*E2)/((Z2  +  Z3)*(Z1  +  Z3)  -  Z3*Z3)\n",
+      "I2  =  -1*(E1  -  (Z1  +  Z3)*I1)/Z3\n",
+      "I3  =  I1  -  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current,  I1  is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A,\\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round( I2.imag,2),\")i  A and \"\n",
+      "print   \" current,  I3  is  \",round(I3.real,2),\"  +  (\",round(  I3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current,  I1  is   0.57   +  ( 0.62 )i  A,\n",
+        " current,  I2  is     0.56   +  ( 1.33 )i  A and \n",
+        " current,  I3  is   0.01   +  ( -0.71 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 541</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the current in the (4 + j3)impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  10;#  in  volts\n",
+      "rv2  =  12;#  in  volts\n",
+      "rv3  =  15;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "thetav3  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  -5j;#  in  ohm\n",
+      "R3  =  8;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  3j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      "V3  =  rv3*math.cos(thetav3*math.pi/180)  +  1j*rv3*math.sin(thetav3*math.pi/180)\n",
+      " #Currents  I1,  I2  and  I3  with  their  directions  are  shown  in  Figure  30.10.\n",
+      " #Three  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.  loop  ABGH,  and  loopBCFG  and  loop  CDEF\n",
+      "Z4  =  R4  +  R5\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #R1*I1  +  R2*I2  =  V1  +  V2\n",
+      " #-1*R3*I1  +  (R3  +  R2)*I2  +  R3*I3  =  V2  +  V3\n",
+      " #  -1*R3*I1  +  R3*I2  +  (R3  +  Z4)*I3  =  V3\n",
+      " #using  determinants\n",
+      "d1  =  [[(V1  +  V2),  R2,  0],[(V2  +  V3),  (R3  +  R2),  R3],[V3,  R3,  (R3  +  Z4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[R1,  (V1  +  V2),  0],[-1*R3,  (V2  +  V3),  R3],[-1*R3,  V3,  (R3  +  Z4)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[R1,  R2,  (V1  +  V2)],[-1*R3,  (R3  +  R2),  (V2  +  V3)],[-1*R3,  R3,  V3]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[R1,  R2,  0],[-1*R3,  (R3  +  R2),  R3],[-1*R3,  R3,  (R3  +  Z4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "I3mag  =  abs(I3)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is  \",round(I3mag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is   1.84   A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_1.ipynb
new file mode 100755
index 00000000..b23cbc8e
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_1.ipynb
@@ -0,0 +1,303 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 30: Introduction to network analysis</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 536</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the current flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  25;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  branch  currents  and  their  directions  are  labelled  as  shown  in  Figure  30.4\n",
+      " #Two  loops  are  chosen.  loop  ABEF,  and  loop  BCDE\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #two  eqns  obtained\n",
+      " #(R1  +  R2)*I1  +  R2*I2  =  V1\n",
+      " #R2*I1  +  (R2  +  R3)*I2  =  V2\n",
+      "I1  =  (3*V1  -  2*V2)/(3*(R1  +  R2)  -  2*(R2))\n",
+      "I2  =  (V2  -  R2*I1)/(R2  +  R3)\n",
+      "I  =  I1  +  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I1  is  \",round(I1.real,2),\"  +  (\",round( I1.imag,2),\")i  A,  \\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round(  I2.imag,2),\")i  A and \"\n",
+      "print   \" total  current,  I  is  \",round(I.real,2),\"  +  (\",round( I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I1  is   3.16   +  ( -1.05 )i  A,  \n",
+        " current,  I2  is     -2.11   +  ( 2.37 )i  A and \n",
+        " total  current,  I  is   1.05   +  ( 1.32 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 537</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the 2 ohm resistor of the circuit\n",
+      "#find the power dissipated in the 3 ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "#initializing  the  variables:\n",
+      "V  =  8;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  2;#  in  ohm\n",
+      "R3  =  3;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  6;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Currents  and  their  directions  are  assigned  as  shown  in  Figure  30.6.\n",
+      " #Three  loops  are  chosen  since  three  unknown  currents  are  required.  The  choice  of  loop  directions  is  arbitrary.\n",
+      "    #loop  ABCDE,  and  loop  EDGF  and  loop  DCHG\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #R5*I1  +  (R6  +  R4)*I2  -  R4*I3  =  V\n",
+      " #-1*R1*I1  +  (R6  +  R1)*I2  +  R2*I3  =  0\n",
+      " #  R3*I1  -  (R3  +  R4)*I2  +  (R2  +  R3  +  R4)*I3  =  0\n",
+      "#using  determinants\n",
+      "d1  =  [[V, (R6  +  R4), -1*R4],[0, (R6  +  R1),  R2], [0,  (-1*(R3  +  R4)), (R2  +  R3  +  R4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[R5, V,  -1*R4],[-1*R1,  0,  R2],[ R3,  0,  (R2  +  R3  +  R4)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[R5,  (R6  +  R4),  V],[-1*R1,  (R6  +  R1),  0],[ R3,  (-1*(R3  +  R4)),  0]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[R5,  (R6  +  R4),  -1*R4],[-1*R1,  (R6  +  R1),  R2],[ R3,  (-1*(R3  +  R4)),  (R2  +  R3  +  R4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "#Current  in  the  2  ohm  resistance\n",
+      "I  =  I1  -  I2  +  I3\n",
+      "#power  dissipated  in  the  3  ohm  resistance\n",
+      "P3  =  R3*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  through  2  ohm  resistor  is  \",round(I2,3),\"  A\"\n",
+      "print  \"\\n  (b)power  dissipated  in  the  3  ohm  resistor  is  \",round(P3,3),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  through  2  ohm  resistor  is   0.203   A\n",
+        "\n",
+        "  (b)power  dissipated  in  the  3  ohm  resistor  is   1.267 W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 539</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in each branch using Kirchhoff\u2019s laws.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  5  +  0j;#  in  volts\n",
+      "E2  =  2  +  4j;#  in  volts\n",
+      "Z1  =  3  +  4j;#  in  ohm\n",
+      "Z2  =  2  -  5j;#  in  ohm\n",
+      "Z3  =  6  +  8j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Currents  I1  and  I2  with  their  directions  are  shown  in  Figure  30.8.\n",
+      " #Two  loops  are  chosen  with  their  directions  both  clockwise.loop  ABEF  and  loop  BCDE,\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #two  eqns  obtained\n",
+      " #(Z1  +  Z3)*I1  -  Z3*I2  =  E1\n",
+      " #-1*Z3*I1  +  (Z2  +  Z3)*I2  =  E2\n",
+      "I1  =  ((Z2  +  Z3)*E1  +  Z3*E2)/((Z2  +  Z3)*(Z1  +  Z3)  -  Z3*Z3)\n",
+      "I2  =  -1*(E1  -  (Z1  +  Z3)*I1)/Z3\n",
+      "I3  =  I1  -  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current,  I1  is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A,\\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round( I2.imag,2),\")i  A and \"\n",
+      "print   \" current,  I3  is  \",round(I3.real,2),\"  +  (\",round(  I3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current,  I1  is   0.57   +  ( 0.62 )i  A,\n",
+        " current,  I2  is     0.56   +  ( 1.33 )i  A and \n",
+        " current,  I3  is   0.01   +  ( -0.71 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 541</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the current in the (4 + j3)impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  10;#  in  volts\n",
+      "rv2  =  12;#  in  volts\n",
+      "rv3  =  15;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "thetav3  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  -5j;#  in  ohm\n",
+      "R3  =  8;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  3j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      "V3  =  rv3*math.cos(thetav3*math.pi/180)  +  1j*rv3*math.sin(thetav3*math.pi/180)\n",
+      " #Currents  I1,  I2  and  I3  with  their  directions  are  shown  in  Figure  30.10.\n",
+      " #Three  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.  loop  ABGH,  and  loopBCFG  and  loop  CDEF\n",
+      "Z4  =  R4  +  R5\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #R1*I1  +  R2*I2  =  V1  +  V2\n",
+      " #-1*R3*I1  +  (R3  +  R2)*I2  +  R3*I3  =  V2  +  V3\n",
+      " #  -1*R3*I1  +  R3*I2  +  (R3  +  Z4)*I3  =  V3\n",
+      " #using  determinants\n",
+      "d1  =  [[(V1  +  V2),  R2,  0],[(V2  +  V3),  (R3  +  R2),  R3],[V3,  R3,  (R3  +  Z4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[R1,  (V1  +  V2),  0],[-1*R3,  (V2  +  V3),  R3],[-1*R3,  V3,  (R3  +  Z4)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[R1,  R2,  (V1  +  V2)],[-1*R3,  (R3  +  R2),  (V2  +  V3)],[-1*R3,  R3,  V3]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[R1,  R2,  0],[-1*R3,  (R3  +  R2),  R3],[-1*R3,  R3,  (R3  +  Z4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "I3mag  =  abs(I3)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is  \",round(I3mag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is   1.84   A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_2.ipynb
new file mode 100755
index 00000000..b23cbc8e
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_2.ipynb
@@ -0,0 +1,303 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 30: Introduction to network analysis</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 536</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#find the current flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  25;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  branch  currents  and  their  directions  are  labelled  as  shown  in  Figure  30.4\n",
+      " #Two  loops  are  chosen.  loop  ABEF,  and  loop  BCDE\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #two  eqns  obtained\n",
+      " #(R1  +  R2)*I1  +  R2*I2  =  V1\n",
+      " #R2*I1  +  (R2  +  R3)*I2  =  V2\n",
+      "I1  =  (3*V1  -  2*V2)/(3*(R1  +  R2)  -  2*(R2))\n",
+      "I2  =  (V2  -  R2*I1)/(R2  +  R3)\n",
+      "I  =  I1  +  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I1  is  \",round(I1.real,2),\"  +  (\",round( I1.imag,2),\")i  A,  \\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round(  I2.imag,2),\")i  A and \"\n",
+      "print   \" total  current,  I  is  \",round(I.real,2),\"  +  (\",round( I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I1  is   3.16   +  ( -1.05 )i  A,  \n",
+        " current,  I2  is     -2.11   +  ( 2.37 )i  A and \n",
+        " total  current,  I  is   1.05   +  ( 1.32 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 537</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the current flowing in the 2 ohm resistor of the circuit\n",
+      "#find the power dissipated in the 3 ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "#initializing  the  variables:\n",
+      "V  =  8;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  2;#  in  ohm\n",
+      "R3  =  3;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  6;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Currents  and  their  directions  are  assigned  as  shown  in  Figure  30.6.\n",
+      " #Three  loops  are  chosen  since  three  unknown  currents  are  required.  The  choice  of  loop  directions  is  arbitrary.\n",
+      "    #loop  ABCDE,  and  loop  EDGF  and  loop  DCHG\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #R5*I1  +  (R6  +  R4)*I2  -  R4*I3  =  V\n",
+      " #-1*R1*I1  +  (R6  +  R1)*I2  +  R2*I3  =  0\n",
+      " #  R3*I1  -  (R3  +  R4)*I2  +  (R2  +  R3  +  R4)*I3  =  0\n",
+      "#using  determinants\n",
+      "d1  =  [[V, (R6  +  R4), -1*R4],[0, (R6  +  R1),  R2], [0,  (-1*(R3  +  R4)), (R2  +  R3  +  R4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[R5, V,  -1*R4],[-1*R1,  0,  R2],[ R3,  0,  (R2  +  R3  +  R4)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[R5,  (R6  +  R4),  V],[-1*R1,  (R6  +  R1),  0],[ R3,  (-1*(R3  +  R4)),  0]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[R5,  (R6  +  R4),  -1*R4],[-1*R1,  (R6  +  R1),  R2],[ R3,  (-1*(R3  +  R4)),  (R2  +  R3  +  R4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "#Current  in  the  2  ohm  resistance\n",
+      "I  =  I1  -  I2  +  I3\n",
+      "#power  dissipated  in  the  3  ohm  resistance\n",
+      "P3  =  R3*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  through  2  ohm  resistor  is  \",round(I2,3),\"  A\"\n",
+      "print  \"\\n  (b)power  dissipated  in  the  3  ohm  resistor  is  \",round(P3,3),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  through  2  ohm  resistor  is   0.203   A\n",
+        "\n",
+        "  (b)power  dissipated  in  the  3  ohm  resistor  is   1.267 W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 539</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in each branch using Kirchhoff\u2019s laws.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  5  +  0j;#  in  volts\n",
+      "E2  =  2  +  4j;#  in  volts\n",
+      "Z1  =  3  +  4j;#  in  ohm\n",
+      "Z2  =  2  -  5j;#  in  ohm\n",
+      "Z3  =  6  +  8j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Currents  I1  and  I2  with  their  directions  are  shown  in  Figure  30.8.\n",
+      " #Two  loops  are  chosen  with  their  directions  both  clockwise.loop  ABEF  and  loop  BCDE,\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #two  eqns  obtained\n",
+      " #(Z1  +  Z3)*I1  -  Z3*I2  =  E1\n",
+      " #-1*Z3*I1  +  (Z2  +  Z3)*I2  =  E2\n",
+      "I1  =  ((Z2  +  Z3)*E1  +  Z3*E2)/((Z2  +  Z3)*(Z1  +  Z3)  -  Z3*Z3)\n",
+      "I2  =  -1*(E1  -  (Z1  +  Z3)*I1)/Z3\n",
+      "I3  =  I1  -  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current,  I1  is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A,\\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round( I2.imag,2),\")i  A and \"\n",
+      "print   \" current,  I3  is  \",round(I3.real,2),\"  +  (\",round(  I3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current,  I1  is   0.57   +  ( 0.62 )i  A,\n",
+        " current,  I2  is     0.56   +  ( 1.33 )i  A and \n",
+        " current,  I3  is   0.01   +  ( -0.71 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 541</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the current in the (4 + j3)impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  10;#  in  volts\n",
+      "rv2  =  12;#  in  volts\n",
+      "rv3  =  15;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "thetav3  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  -5j;#  in  ohm\n",
+      "R3  =  8;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  3j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      "V3  =  rv3*math.cos(thetav3*math.pi/180)  +  1j*rv3*math.sin(thetav3*math.pi/180)\n",
+      " #Currents  I1,  I2  and  I3  with  their  directions  are  shown  in  Figure  30.10.\n",
+      " #Three  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.  loop  ABGH,  and  loopBCFG  and  loop  CDEF\n",
+      "Z4  =  R4  +  R5\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #R1*I1  +  R2*I2  =  V1  +  V2\n",
+      " #-1*R3*I1  +  (R3  +  R2)*I2  +  R3*I3  =  V2  +  V3\n",
+      " #  -1*R3*I1  +  R3*I2  +  (R3  +  Z4)*I3  =  V3\n",
+      " #using  determinants\n",
+      "d1  =  [[(V1  +  V2),  R2,  0],[(V2  +  V3),  (R3  +  R2),  R3],[V3,  R3,  (R3  +  Z4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[R1,  (V1  +  V2),  0],[-1*R3,  (V2  +  V3),  R3],[-1*R3,  V3,  (R3  +  Z4)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[R1,  R2,  (V1  +  V2)],[-1*R3,  (R3  +  R2),  (V2  +  V3)],[-1*R3,  R3,  V3]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[R1,  R2,  0],[-1*R3,  (R3  +  R2),  R3],[-1*R3,  R3,  (R3  +  Z4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "I3mag  =  abs(I3)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is  \",round(I3mag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is   1.84   A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb
new file mode 100755
index 00000000..6732d70c
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_30-checkpoint_3.ipynb
@@ -0,0 +1,303 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:cc9bf89f9ef283e2e254a84fd5c837c0ed5af6c8784b676db04bdba61dcf814e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 30: Introduction to network analysis</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 536</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  25;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  branch  currents  and  their  directions  are  labelled  as  shown  in  Figure  30.4\n",
+      " #Two  loops  are  chosen.  loop  ABEF,  and  loop  BCDE\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #two  eqns  obtained\n",
+      " #(R1  +  R2)*I1  +  R2*I2  =  V1\n",
+      " #R2*I1  +  (R2  +  R3)*I2  =  V2\n",
+      "I1  =  (3*V1  -  2*V2)/(3*(R1  +  R2)  -  2*(R2))\n",
+      "I2  =  (V2  -  R2*I1)/(R2  +  R3)\n",
+      "I  =  I1  +  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I1  is  \",round(I1.real,2),\"  +  (\",round( I1.imag,2),\")i  A,  \\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round(  I2.imag,2),\")i  A and \"\n",
+      "print   \" total  current,  I  is  \",round(I.real,2),\"  +  (\",round( I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I1  is   3.16   +  ( -1.05 )i  A,  \n",
+        " current,  I2  is     -2.11   +  ( 2.37 )i  A and \n",
+        " total  current,  I  is   1.05   +  ( 1.32 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 537</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "#initializing  the  variables:\n",
+      "V  =  8;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  2;#  in  ohm\n",
+      "R3  =  3;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  6;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Currents  and  their  directions  are  assigned  as  shown  in  Figure  30.6.\n",
+      " #Three  loops  are  chosen  since  three  unknown  currents  are  required.  The  choice  of  loop  directions  is  arbitrary.\n",
+      "    #loop  ABCDE,  and  loop  EDGF  and  loop  DCHG\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #R5*I1  +  (R6  +  R4)*I2  -  R4*I3  =  V\n",
+      " #-1*R1*I1  +  (R6  +  R1)*I2  +  R2*I3  =  0\n",
+      " #  R3*I1  -  (R3  +  R4)*I2  +  (R2  +  R3  +  R4)*I3  =  0\n",
+      "#using  determinants\n",
+      "d1  =  [[V, (R6  +  R4), -1*R4],[0, (R6  +  R1),  R2], [0,  (-1*(R3  +  R4)), (R2  +  R3  +  R4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[R5, V,  -1*R4],[-1*R1,  0,  R2],[ R3,  0,  (R2  +  R3  +  R4)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[R5,  (R6  +  R4),  V],[-1*R1,  (R6  +  R1),  0],[ R3,  (-1*(R3  +  R4)),  0]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[R5,  (R6  +  R4),  -1*R4],[-1*R1,  (R6  +  R1),  R2],[ R3,  (-1*(R3  +  R4)),  (R2  +  R3  +  R4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "#Current  in  the  2  ohm  resistance\n",
+      "I  =  I1  -  I2  +  I3\n",
+      "#power  dissipated  in  the  3  ohm  resistance\n",
+      "P3  =  R3*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  through  2  ohm  resistor  is  \",round(I2,3),\"  A\"\n",
+      "print  \"\\n  (b)power  dissipated  in  the  3  ohm  resistor  is  \",round(P3,3),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  through  2  ohm  resistor  is   0.203   A\n",
+        "\n",
+        "  (b)power  dissipated  in  the  3  ohm  resistor  is   1.267 W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 539</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  5  +  0j;#  in  volts\n",
+      "E2  =  2  +  4j;#  in  volts\n",
+      "Z1  =  3  +  4j;#  in  ohm\n",
+      "Z2  =  2  -  5j;#  in  ohm\n",
+      "Z3  =  6  +  8j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Currents  I1  and  I2  with  their  directions  are  shown  in  Figure  30.8.\n",
+      " #Two  loops  are  chosen  with  their  directions  both  clockwise.loop  ABEF  and  loop  BCDE,\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #two  eqns  obtained\n",
+      " #(Z1  +  Z3)*I1  -  Z3*I2  =  E1\n",
+      " #-1*Z3*I1  +  (Z2  +  Z3)*I2  =  E2\n",
+      "I1  =  ((Z2  +  Z3)*E1  +  Z3*E2)/((Z2  +  Z3)*(Z1  +  Z3)  -  Z3*Z3)\n",
+      "I2  =  -1*(E1  -  (Z1  +  Z3)*I1)/Z3\n",
+      "I3  =  I1  -  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current,  I1  is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A,\\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round( I2.imag,2),\")i  A and \"\n",
+      "print   \" current,  I3  is  \",round(I3.real,2),\"  +  (\",round(  I3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current,  I1  is   0.57   +  ( 0.62 )i  A,\n",
+        " current,  I2  is     0.56   +  ( 1.33 )i  A and \n",
+        " current,  I3  is   0.01   +  ( -0.71 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 541</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  10;#  in  volts\n",
+      "rv2  =  12;#  in  volts\n",
+      "rv3  =  15;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "thetav3  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  -5j;#  in  ohm\n",
+      "R3  =  8;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  3j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      "V3  =  rv3*math.cos(thetav3*math.pi/180)  +  1j*rv3*math.sin(thetav3*math.pi/180)\n",
+      " #Currents  I1,  I2  and  I3  with  their  directions  are  shown  in  Figure  30.10.\n",
+      " #Three  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.  loop  ABGH,  and  loopBCFG  and  loop  CDEF\n",
+      "Z4  =  R4  +  R5\n",
+      " #using  kirchoff  rule  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #R1*I1  +  R2*I2  =  V1  +  V2\n",
+      " #-1*R3*I1  +  (R3  +  R2)*I2  +  R3*I3  =  V2  +  V3\n",
+      " #  -1*R3*I1  +  R3*I2  +  (R3  +  Z4)*I3  =  V3\n",
+      " #using  determinants\n",
+      "d1  =  [[(V1  +  V2),  R2,  0],[(V2  +  V3),  (R3  +  R2),  R3],[V3,  R3,  (R3  +  Z4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[R1,  (V1  +  V2),  0],[-1*R3,  (V2  +  V3),  R3],[-1*R3,  V3,  (R3  +  Z4)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[R1,  R2,  (V1  +  V2)],[-1*R3,  (R3  +  R2),  (V2  +  V3)],[-1*R3,  R3,  V3]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[R1,  R2,  0],[-1*R3,  (R3  +  R2),  R3],[-1*R3,  R3,  (R3  +  Z4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "I3mag  =  abs(I3)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is  \",round(I3mag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  magnitude  of  the  current  through  (4  +  i3)ohm  impedance  is   1.84   A"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint.ipynb
new file mode 100755
index 00000000..5c1fd9c0
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint.ipynb
@@ -0,0 +1,665 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 31: Mesh-current and nodal analysis</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 546</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use mesh-current analysis to determine the current flowing in \n",
+      "#(a) the 5 ohm resistance, and (b) the 1 ohm resistance of the d.c. circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  4;#  in  volts\n",
+      "V2  =  5;#  in  volts\n",
+      "R1  =  3;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  1;#  in  ohm\n",
+      "R5  =  6;#  in  ohm\n",
+      "R6  =  8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #The  mesh  currents  I1,  I2  and  I3  are  shown  in  Figure  31.2.  Using  Kirchhoff\u2019s  voltage  law  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #(R1  +  R2)*I1  -  R2*I2  =  V1\n",
+      " #-1*R2*I1  +  (R2  +  R3  +  R4  +  R5)*I2  -  R4*I3  =  0\n",
+      " #  -1*R4*I2  +  (R4  +  R6)*I3  =  -1*V2\n",
+      " #using  determinants\n",
+      "d1  =  [[V1,  -1*R2,  0],[0,  (R2  +  R3  +  R4  +  R5),  -1*R4],[-1*V2,  -1*R4,  (R4  +  R6)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R1  +  R2),  V1,  0],[-1*R2,  0,  -1*R4],[0,  -1*V2,  (R4  +  R6)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[(R1  +  R2),  -1*R2,  V1],[-1*R2,  (R2  +  R3  +  R4  +  R5),  0],[0,  -1*R4,  -1*V2]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[(R1  +  R2),  -1*R2,  0],[-1*R2,  (R2  +  R3  +  R4  +  R5),  -1*R4],[0,  -1*R4,  (R4  +  R6)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "IR2  =  I1  -  I2\n",
+      "IR4  =  I2  -  I3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  in  the  5  ohm  resistance  is  \",round(IR2,2),\"  A\"\n",
+      "print  \"\\n  (b)current  in  the  1  ohm  resistance  is  \",round(IR4,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  in  the  5  ohm  resistance  is   0.44   A\n",
+        "\n",
+        "  (b)current  in  the  1  ohm  resistance  is   0.69   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 547</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine, using mesh-current analysis, \n",
+      "#(a) the mesh currents I1 and I2 \n",
+      "#(b) the current flowing in the capacitor,\n",
+      "#(c) the active power delivered by the voltage source.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  -1j*4;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  3j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Currents  I1,  I2  with  their  directions  are  shown  in  Figure  31.03.\n",
+      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
+      " #using  kirchoff  rule  in  2  loops\n",
+      " #two  eqns  obtained\n",
+      " #(R1  +  R2)*I1  -  R2*I2  =  V\n",
+      " #-1*R2*I1  +  (R3  +  R2  +  R4)*I2  =  0\n",
+      " #using  determinants\n",
+      "d1  =  [[V,  -1*R2],[0,  (R3  +  R2  +  R4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R1  +  R2),  V],[-1*R2,  0]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(R1  +  R2),  -1*R2],[-1*R2,  (R3  +  R2  +  R4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I1mag  =  abs(I1)\n",
+      " #Current  flowing  in  capacitor\n",
+      "Ic  =  I1  -  I2\n",
+      " #Source  power  P\n",
+      "phi  =  cmath.phase(complex(I1.real,I1.imag))\n",
+      "P  =  V*I1mag*math.cos(phi)\n",
+      "Icmag  =  abs(Ic)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)current,I1 is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A,  current,  I2  is\",round(I2.real,2),\"  +  (\",round(I2.imag,2),\")i  A\"\n",
+      "print  \"(b)current  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
+      "print  \"(c)Source  power  P  is  \",round(abs(P),2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)current,I1 is   10.17   +  ( 3.55 )i  A,  current,  I2  is 5.73   +  ( -8.74 )i  A\n",
+        "(b)current  in  the  capacitor  is   13.06   A\n",
+        "(c)Source  power  P  is   1017.06   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 548</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the line currents IR, IY and IB using mesh-current analysis\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  415;#  in  volts\n",
+      "rv2  =  415;#  in  volts\n",
+      "thetav1  =  120;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R  =  3  +  4j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Two  mesh  currents  I1  and  I2  are  chosen  as  shown  in  Figure  31.4.\n",
+      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
+      " #using  kirchoff  rule  in  2  loops\n",
+      " #two  eqns  obtained\n",
+      " #2*R*I1  -  R*I2  =  V1\n",
+      " #-1*R*I1  +  2*R*I2  =  V2\n",
+      " #using  determinants\n",
+      "d1  =  [[V1,  -1*R],[V2, 2*R]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[2*R, V1],[-1*R, V2]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[2*R, -1*R],[-1*R,  2*R]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I1mag  =  abs(I1)\n",
+      " #line  current  IR\n",
+      "IR  =  I1\n",
+      " #line  current  IB\n",
+      "IB  =  -1*I2\n",
+      " #line  current  IY\n",
+      "IY  =  I2  -  I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current, IR is\",round(IR.real,2),\"  +  (\",round(  IR.imag,2),\")i  A,  current,  IB  is\",round(IB.real,2),\"  +  (\",round(  IB.imag,2),\")i  A  and  current,  IY  is  \",round(IY.real,2),\"  +  (\",round(IY.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current, IR is 38.34   +  ( 28.75 )i  A,  current,  IB  is -44.07   +  ( 18.82 )i  A  and  current,  IY  is   5.73   +  ( -47.58 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 551</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the voltage VAB, by using nodal analysis.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri  =  20;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "R1  =  10;#  in  ohm\n",
+      "R2  =  3j;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  16;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #current\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Figure  31.8  contains  two  principal  nodes  (at  1  and  B)  and  thus  only  one  nodal  equation  is  required. \n",
+      "    #B  is  taken  as  the  reference  node  and  the  equation  for  node  1  is  obtained  as  follows. \n",
+      "    #Applying  Kirchhoff\u2019s  current  law  to  node  1  gives:\n",
+      " #IX  +  IY  =  I\n",
+      "V1  =  I/((1/R4)  +(1/(R2    +R3)))\n",
+      "IY  =  V1/(R2  +  R3)\n",
+      "VAB  =  IY*R3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VAB  is  \",round(VAB.real,2),\"  +  (\",round(  VAB.imag,2),\")i  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VAB  is   62.59   +  ( -9.39 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 552</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of voltage VXY\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  8;#  in  volts\n",
+      "rv2  =  8;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  6j;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  circuit  contains  no  principal  nodes.  \n",
+      "    #However,  if  point  Y  is  chosen  as  the  reference  node  then  an  equation \n",
+      "    #may  be  written  for  node  X  assuming  that  current  leaves  point  X  by  both  branches\n",
+      "VX  =  ((V1/(R1  +  R3)  +  V2/(R2  +  R4))/(1/(R1  +  R3)  +  1/(R2  +  R4)))\n",
+      "VXY  =  VX\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VXY  = \",round(abs(VXY),2),\"/_\",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),\"deg V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VXY  =  9.12 /_ 52.13 deg V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 553</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use nodal analysis to determine the current flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  25;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #There  are  only  two  principal  nodes  in  Figure  31.10  so  only  one  nodal  equation  is  required.  \n",
+      "    #Node  2  is  taken  as  the  reference  node.\n",
+      " #The  equation  at  node  1  is  I1  +  I2  +  I3  =  0\n",
+      "Vn1  =  ((V1/R1  +  V2/R3)/(1/R1  +  1/R2  +  1/R3))\n",
+      "I1  =  (Vn1  -  V1)/R1\n",
+      "I2  =  Vn1/R2\n",
+      "I3  =  (Vn1  -  V2)/R3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I1  is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A, \\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round( I2.imag,2),\")i  A \\n and  current,  I3  is  \",round(I3.real,2),\"  +  (\",round(I3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I1  is   -3.16   +  ( 1.05 )i  A, \n",
+        " current,  I2  is     1.05   +  ( 1.32 )i  A \n",
+        " and  current,  I3  is   2.11   +  ( -2.37 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 554</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 ohm inductance,\n",
+      "#(c) the current in the 5 ohm resistance, and (d) the magnitude of the active power dissipated in the 2.5ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  25;#  in  volts\n",
+      "rv2  =  25;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  -1j*4;#  in  ohm\n",
+      "R3  =  5;#  in  ohm\n",
+      "R4  =  4j;#  in  ohm\n",
+      "R5  =  2.5;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  equation  at  node  1\n",
+      " #Vn1*(1/R1  +  1/R2  +  1/R3)  -  Vn2/R3  =  V1/R1\n",
+      " #The  equation  at  node  2\n",
+      " #Vn1*(-1/R3)  +  Vn2*(1/R4  +  1/R5  +  1/R3)  =  V2/R5\n",
+      " #using  determinants\n",
+      "d1  =  [[V1/R1,  -1/R3],[V2/R5,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/R1  +  1/R2  +  1/R3),  V1/R1],[-1/R3,  V2/R5]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(1/R1  +  1/R2  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "Vn1  =  D1/D\n",
+      "Vn2  =  D2/D\n",
+      " #current  in  the  j4  ohm  inductance  is  given  by:\n",
+      "I4  =  Vn2/R4\n",
+      " #current  in  the  5  ohm  resistance  is  given  by:\n",
+      "I3  =  (Vn1  -  Vn2)/R3\n",
+      " #active  power  dissipated  in  the  2.5  ohm  resistor  is  given  by\n",
+      "P5  =  R5*((Vn2  -  V2)/R5)**2\n",
+      " #magnitude  of  the  active  power  dissipated\n",
+      "P5mag  =  abs(P5)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  voltage  at  nodes  1  and  2  are \",round(abs(Vn1),1),\"/_\",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),\"deg  V  and \",round(abs(Vn2),1),\"/_\",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),\"deg  V\"\n",
+      "print  \"\\n  (b)the  current  in  the  j4  ohm inductance = \",round(abs(I4),2),\"/_\",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  (c)the  current  in  the  5  ohm resistance  = \",round(abs(I3),2),\"/_\",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is  \",round(P5mag,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  voltage  at  nodes  1  and  2  are  17.1 /_ -5.3 deg  V  and  15.8 /_ 93.2 deg  V\n",
+        "\n",
+        "  (b)the  current  in  the  j4  ohm inductance =  3.95 /_ 3.23 deg  A\n",
+        "\n",
+        "  (c)the  current  in  the  5  ohm resistance  =  4.99 /_ -44.06 deg  A\n",
+        "\n",
+        "  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is   34.4   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 556</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the voltage VXY using nodal analysis.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri  =  25;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  3j;#  in  ohm\n",
+      "R3  =  5;#  in  ohm\n",
+      "R4  =  10j;#  in  ohm\n",
+      "R5  =  20j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #current\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Node  3  is  taken  as  the  reference  node.\n",
+      " #At  node  1,\n",
+      " #V1*(1/(R1  +  R2)  +  1/R3)  -  V2/R3  =  I\n",
+      " #The  equation  at  node  2\n",
+      " #V1*(-1/R3)  +  V2*(1/R4  +  1/R5  +  1/R3)  =  0\n",
+      " #using  determinants\n",
+      "d1  =  [[I,  -1/R3],[0 , (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/(R1  +  R2)  +  1/R3),  I],[-1/R3,  0]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(1/(R1  +  R2)  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "V1  =  D1/D\n",
+      "V2  =  D2/D\n",
+      " #the  voltage  between  point  X  and  node  3  is\n",
+      "VX  =  V1*R2/(R1  +  R2)\n",
+      " #Thus  the  voltage\n",
+      "VY  =  V2\n",
+      "VXY  =  VX  -  VY\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VXY  is  \",round(VXY.real,2),\"  +  (\",round(  VXY.imag,2),\")i  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VXY  is   -16.16   +  ( -15.05 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 557</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use nodal analysis to determine the voltages at nodes 2 and 3 \n",
+      "#determine the current flowing in the 2 ohm resistor and the power dissipated in the 3 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  8;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  2;#  in  ohm\n",
+      "R3  =  3;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  6;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #In  Figure  31.13,  the  reference  node  is  shown  at  point  A.\n",
+      " #At  node  1,\n",
+      " #V1*(1/R1  +  1/R6  +  1/R5)  -  V2/R1  -  V3/R5  =  V/R5\n",
+      " #The  equation  at  node  2\n",
+      " #V1*(-1/R1)  +  V2*(1/R2  +  1/R1  +  1/R3)  -  V3/R3  =  0\n",
+      " #At  node  3\n",
+      " #  -  V1/R5  -  V2/R3  +  V3*(1/R4  +  1/R3  +  1/R5)  =  -1*V/R5\n",
+      "#using  determinants\n",
+      "d1  =  [[V/R5,  -1/R1,  -1/R5],[0, (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1*V/R5, -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/R1  +  1/R6  +  1/R5),  V/R5,  -1/R5],[-1/R1,  0,  -1/R3],[-1/R5,  -1*V/R5,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[(1/R1  +  1/R6  +  1/R5),  -1/R1,  V/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  0],[-1/R5,  -1/R3,  -1*V/R5]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =[[(1/R1  +  1/R6  +  1/R5),  -1/R1,  -1/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1/R5,  -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "Vn1  =  D1/D\n",
+      "Vn2  =  D2/D\n",
+      "Vn3  =  D3/D  \n",
+      " #the  current  in  the  2  ohm  resistor\n",
+      "I2  =  Vn2/R2\n",
+      " #power  dissipated  in  the  3  ohm  resistance\n",
+      "P3  =  R3*((Vn2  -  Vn3)/R3)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  through  2  ohm  resistor  is  \",round(I2,2),\"  A\"\n",
+      "print  \"\\n  (b)power  dissipated  in  the  3  ohm  resistor  is  \",round(P3,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  through  2  ohm  resistor  is   0.19   A\n",
+        "\n",
+        "  (b)power  dissipated  in  the  3  ohm  resistor  is   1.27   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_1.ipynb
new file mode 100755
index 00000000..5c1fd9c0
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_1.ipynb
@@ -0,0 +1,665 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 31: Mesh-current and nodal analysis</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 546</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use mesh-current analysis to determine the current flowing in \n",
+      "#(a) the 5 ohm resistance, and (b) the 1 ohm resistance of the d.c. circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  4;#  in  volts\n",
+      "V2  =  5;#  in  volts\n",
+      "R1  =  3;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  1;#  in  ohm\n",
+      "R5  =  6;#  in  ohm\n",
+      "R6  =  8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #The  mesh  currents  I1,  I2  and  I3  are  shown  in  Figure  31.2.  Using  Kirchhoff\u2019s  voltage  law  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #(R1  +  R2)*I1  -  R2*I2  =  V1\n",
+      " #-1*R2*I1  +  (R2  +  R3  +  R4  +  R5)*I2  -  R4*I3  =  0\n",
+      " #  -1*R4*I2  +  (R4  +  R6)*I3  =  -1*V2\n",
+      " #using  determinants\n",
+      "d1  =  [[V1,  -1*R2,  0],[0,  (R2  +  R3  +  R4  +  R5),  -1*R4],[-1*V2,  -1*R4,  (R4  +  R6)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R1  +  R2),  V1,  0],[-1*R2,  0,  -1*R4],[0,  -1*V2,  (R4  +  R6)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[(R1  +  R2),  -1*R2,  V1],[-1*R2,  (R2  +  R3  +  R4  +  R5),  0],[0,  -1*R4,  -1*V2]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[(R1  +  R2),  -1*R2,  0],[-1*R2,  (R2  +  R3  +  R4  +  R5),  -1*R4],[0,  -1*R4,  (R4  +  R6)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "IR2  =  I1  -  I2\n",
+      "IR4  =  I2  -  I3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  in  the  5  ohm  resistance  is  \",round(IR2,2),\"  A\"\n",
+      "print  \"\\n  (b)current  in  the  1  ohm  resistance  is  \",round(IR4,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  in  the  5  ohm  resistance  is   0.44   A\n",
+        "\n",
+        "  (b)current  in  the  1  ohm  resistance  is   0.69   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 547</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine, using mesh-current analysis, \n",
+      "#(a) the mesh currents I1 and I2 \n",
+      "#(b) the current flowing in the capacitor,\n",
+      "#(c) the active power delivered by the voltage source.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  -1j*4;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  3j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Currents  I1,  I2  with  their  directions  are  shown  in  Figure  31.03.\n",
+      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
+      " #using  kirchoff  rule  in  2  loops\n",
+      " #two  eqns  obtained\n",
+      " #(R1  +  R2)*I1  -  R2*I2  =  V\n",
+      " #-1*R2*I1  +  (R3  +  R2  +  R4)*I2  =  0\n",
+      " #using  determinants\n",
+      "d1  =  [[V,  -1*R2],[0,  (R3  +  R2  +  R4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R1  +  R2),  V],[-1*R2,  0]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(R1  +  R2),  -1*R2],[-1*R2,  (R3  +  R2  +  R4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I1mag  =  abs(I1)\n",
+      " #Current  flowing  in  capacitor\n",
+      "Ic  =  I1  -  I2\n",
+      " #Source  power  P\n",
+      "phi  =  cmath.phase(complex(I1.real,I1.imag))\n",
+      "P  =  V*I1mag*math.cos(phi)\n",
+      "Icmag  =  abs(Ic)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)current,I1 is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A,  current,  I2  is\",round(I2.real,2),\"  +  (\",round(I2.imag,2),\")i  A\"\n",
+      "print  \"(b)current  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
+      "print  \"(c)Source  power  P  is  \",round(abs(P),2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)current,I1 is   10.17   +  ( 3.55 )i  A,  current,  I2  is 5.73   +  ( -8.74 )i  A\n",
+        "(b)current  in  the  capacitor  is   13.06   A\n",
+        "(c)Source  power  P  is   1017.06   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 548</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the line currents IR, IY and IB using mesh-current analysis\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  415;#  in  volts\n",
+      "rv2  =  415;#  in  volts\n",
+      "thetav1  =  120;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R  =  3  +  4j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Two  mesh  currents  I1  and  I2  are  chosen  as  shown  in  Figure  31.4.\n",
+      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
+      " #using  kirchoff  rule  in  2  loops\n",
+      " #two  eqns  obtained\n",
+      " #2*R*I1  -  R*I2  =  V1\n",
+      " #-1*R*I1  +  2*R*I2  =  V2\n",
+      " #using  determinants\n",
+      "d1  =  [[V1,  -1*R],[V2, 2*R]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[2*R, V1],[-1*R, V2]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[2*R, -1*R],[-1*R,  2*R]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I1mag  =  abs(I1)\n",
+      " #line  current  IR\n",
+      "IR  =  I1\n",
+      " #line  current  IB\n",
+      "IB  =  -1*I2\n",
+      " #line  current  IY\n",
+      "IY  =  I2  -  I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current, IR is\",round(IR.real,2),\"  +  (\",round(  IR.imag,2),\")i  A,  current,  IB  is\",round(IB.real,2),\"  +  (\",round(  IB.imag,2),\")i  A  and  current,  IY  is  \",round(IY.real,2),\"  +  (\",round(IY.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current, IR is 38.34   +  ( 28.75 )i  A,  current,  IB  is -44.07   +  ( 18.82 )i  A  and  current,  IY  is   5.73   +  ( -47.58 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 551</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the voltage VAB, by using nodal analysis.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri  =  20;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "R1  =  10;#  in  ohm\n",
+      "R2  =  3j;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  16;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #current\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Figure  31.8  contains  two  principal  nodes  (at  1  and  B)  and  thus  only  one  nodal  equation  is  required. \n",
+      "    #B  is  taken  as  the  reference  node  and  the  equation  for  node  1  is  obtained  as  follows. \n",
+      "    #Applying  Kirchhoff\u2019s  current  law  to  node  1  gives:\n",
+      " #IX  +  IY  =  I\n",
+      "V1  =  I/((1/R4)  +(1/(R2    +R3)))\n",
+      "IY  =  V1/(R2  +  R3)\n",
+      "VAB  =  IY*R3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VAB  is  \",round(VAB.real,2),\"  +  (\",round(  VAB.imag,2),\")i  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VAB  is   62.59   +  ( -9.39 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 552</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of voltage VXY\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  8;#  in  volts\n",
+      "rv2  =  8;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  6j;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  circuit  contains  no  principal  nodes.  \n",
+      "    #However,  if  point  Y  is  chosen  as  the  reference  node  then  an  equation \n",
+      "    #may  be  written  for  node  X  assuming  that  current  leaves  point  X  by  both  branches\n",
+      "VX  =  ((V1/(R1  +  R3)  +  V2/(R2  +  R4))/(1/(R1  +  R3)  +  1/(R2  +  R4)))\n",
+      "VXY  =  VX\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VXY  = \",round(abs(VXY),2),\"/_\",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),\"deg V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VXY  =  9.12 /_ 52.13 deg V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 553</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use nodal analysis to determine the current flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  25;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #There  are  only  two  principal  nodes  in  Figure  31.10  so  only  one  nodal  equation  is  required.  \n",
+      "    #Node  2  is  taken  as  the  reference  node.\n",
+      " #The  equation  at  node  1  is  I1  +  I2  +  I3  =  0\n",
+      "Vn1  =  ((V1/R1  +  V2/R3)/(1/R1  +  1/R2  +  1/R3))\n",
+      "I1  =  (Vn1  -  V1)/R1\n",
+      "I2  =  Vn1/R2\n",
+      "I3  =  (Vn1  -  V2)/R3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I1  is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A, \\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round( I2.imag,2),\")i  A \\n and  current,  I3  is  \",round(I3.real,2),\"  +  (\",round(I3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I1  is   -3.16   +  ( 1.05 )i  A, \n",
+        " current,  I2  is     1.05   +  ( 1.32 )i  A \n",
+        " and  current,  I3  is   2.11   +  ( -2.37 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 554</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 ohm inductance,\n",
+      "#(c) the current in the 5 ohm resistance, and (d) the magnitude of the active power dissipated in the 2.5ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  25;#  in  volts\n",
+      "rv2  =  25;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  -1j*4;#  in  ohm\n",
+      "R3  =  5;#  in  ohm\n",
+      "R4  =  4j;#  in  ohm\n",
+      "R5  =  2.5;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  equation  at  node  1\n",
+      " #Vn1*(1/R1  +  1/R2  +  1/R3)  -  Vn2/R3  =  V1/R1\n",
+      " #The  equation  at  node  2\n",
+      " #Vn1*(-1/R3)  +  Vn2*(1/R4  +  1/R5  +  1/R3)  =  V2/R5\n",
+      " #using  determinants\n",
+      "d1  =  [[V1/R1,  -1/R3],[V2/R5,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/R1  +  1/R2  +  1/R3),  V1/R1],[-1/R3,  V2/R5]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(1/R1  +  1/R2  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "Vn1  =  D1/D\n",
+      "Vn2  =  D2/D\n",
+      " #current  in  the  j4  ohm  inductance  is  given  by:\n",
+      "I4  =  Vn2/R4\n",
+      " #current  in  the  5  ohm  resistance  is  given  by:\n",
+      "I3  =  (Vn1  -  Vn2)/R3\n",
+      " #active  power  dissipated  in  the  2.5  ohm  resistor  is  given  by\n",
+      "P5  =  R5*((Vn2  -  V2)/R5)**2\n",
+      " #magnitude  of  the  active  power  dissipated\n",
+      "P5mag  =  abs(P5)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  voltage  at  nodes  1  and  2  are \",round(abs(Vn1),1),\"/_\",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),\"deg  V  and \",round(abs(Vn2),1),\"/_\",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),\"deg  V\"\n",
+      "print  \"\\n  (b)the  current  in  the  j4  ohm inductance = \",round(abs(I4),2),\"/_\",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  (c)the  current  in  the  5  ohm resistance  = \",round(abs(I3),2),\"/_\",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is  \",round(P5mag,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  voltage  at  nodes  1  and  2  are  17.1 /_ -5.3 deg  V  and  15.8 /_ 93.2 deg  V\n",
+        "\n",
+        "  (b)the  current  in  the  j4  ohm inductance =  3.95 /_ 3.23 deg  A\n",
+        "\n",
+        "  (c)the  current  in  the  5  ohm resistance  =  4.99 /_ -44.06 deg  A\n",
+        "\n",
+        "  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is   34.4   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 556</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the voltage VXY using nodal analysis.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri  =  25;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  3j;#  in  ohm\n",
+      "R3  =  5;#  in  ohm\n",
+      "R4  =  10j;#  in  ohm\n",
+      "R5  =  20j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #current\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Node  3  is  taken  as  the  reference  node.\n",
+      " #At  node  1,\n",
+      " #V1*(1/(R1  +  R2)  +  1/R3)  -  V2/R3  =  I\n",
+      " #The  equation  at  node  2\n",
+      " #V1*(-1/R3)  +  V2*(1/R4  +  1/R5  +  1/R3)  =  0\n",
+      " #using  determinants\n",
+      "d1  =  [[I,  -1/R3],[0 , (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/(R1  +  R2)  +  1/R3),  I],[-1/R3,  0]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(1/(R1  +  R2)  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "V1  =  D1/D\n",
+      "V2  =  D2/D\n",
+      " #the  voltage  between  point  X  and  node  3  is\n",
+      "VX  =  V1*R2/(R1  +  R2)\n",
+      " #Thus  the  voltage\n",
+      "VY  =  V2\n",
+      "VXY  =  VX  -  VY\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VXY  is  \",round(VXY.real,2),\"  +  (\",round(  VXY.imag,2),\")i  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VXY  is   -16.16   +  ( -15.05 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 557</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use nodal analysis to determine the voltages at nodes 2 and 3 \n",
+      "#determine the current flowing in the 2 ohm resistor and the power dissipated in the 3 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  8;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  2;#  in  ohm\n",
+      "R3  =  3;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  6;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #In  Figure  31.13,  the  reference  node  is  shown  at  point  A.\n",
+      " #At  node  1,\n",
+      " #V1*(1/R1  +  1/R6  +  1/R5)  -  V2/R1  -  V3/R5  =  V/R5\n",
+      " #The  equation  at  node  2\n",
+      " #V1*(-1/R1)  +  V2*(1/R2  +  1/R1  +  1/R3)  -  V3/R3  =  0\n",
+      " #At  node  3\n",
+      " #  -  V1/R5  -  V2/R3  +  V3*(1/R4  +  1/R3  +  1/R5)  =  -1*V/R5\n",
+      "#using  determinants\n",
+      "d1  =  [[V/R5,  -1/R1,  -1/R5],[0, (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1*V/R5, -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/R1  +  1/R6  +  1/R5),  V/R5,  -1/R5],[-1/R1,  0,  -1/R3],[-1/R5,  -1*V/R5,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[(1/R1  +  1/R6  +  1/R5),  -1/R1,  V/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  0],[-1/R5,  -1/R3,  -1*V/R5]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =[[(1/R1  +  1/R6  +  1/R5),  -1/R1,  -1/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1/R5,  -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "Vn1  =  D1/D\n",
+      "Vn2  =  D2/D\n",
+      "Vn3  =  D3/D  \n",
+      " #the  current  in  the  2  ohm  resistor\n",
+      "I2  =  Vn2/R2\n",
+      " #power  dissipated  in  the  3  ohm  resistance\n",
+      "P3  =  R3*((Vn2  -  Vn3)/R3)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  through  2  ohm  resistor  is  \",round(I2,2),\"  A\"\n",
+      "print  \"\\n  (b)power  dissipated  in  the  3  ohm  resistor  is  \",round(P3,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  through  2  ohm  resistor  is   0.19   A\n",
+        "\n",
+        "  (b)power  dissipated  in  the  3  ohm  resistor  is   1.27   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_2.ipynb
new file mode 100755
index 00000000..5c1fd9c0
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_2.ipynb
@@ -0,0 +1,665 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 31: Mesh-current and nodal analysis</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 546</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use mesh-current analysis to determine the current flowing in \n",
+      "#(a) the 5 ohm resistance, and (b) the 1 ohm resistance of the d.c. circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  4;#  in  volts\n",
+      "V2  =  5;#  in  volts\n",
+      "R1  =  3;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  1;#  in  ohm\n",
+      "R5  =  6;#  in  ohm\n",
+      "R6  =  8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #The  mesh  currents  I1,  I2  and  I3  are  shown  in  Figure  31.2.  Using  Kirchhoff\u2019s  voltage  law  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #(R1  +  R2)*I1  -  R2*I2  =  V1\n",
+      " #-1*R2*I1  +  (R2  +  R3  +  R4  +  R5)*I2  -  R4*I3  =  0\n",
+      " #  -1*R4*I2  +  (R4  +  R6)*I3  =  -1*V2\n",
+      " #using  determinants\n",
+      "d1  =  [[V1,  -1*R2,  0],[0,  (R2  +  R3  +  R4  +  R5),  -1*R4],[-1*V2,  -1*R4,  (R4  +  R6)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R1  +  R2),  V1,  0],[-1*R2,  0,  -1*R4],[0,  -1*V2,  (R4  +  R6)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[(R1  +  R2),  -1*R2,  V1],[-1*R2,  (R2  +  R3  +  R4  +  R5),  0],[0,  -1*R4,  -1*V2]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[(R1  +  R2),  -1*R2,  0],[-1*R2,  (R2  +  R3  +  R4  +  R5),  -1*R4],[0,  -1*R4,  (R4  +  R6)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "IR2  =  I1  -  I2\n",
+      "IR4  =  I2  -  I3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  in  the  5  ohm  resistance  is  \",round(IR2,2),\"  A\"\n",
+      "print  \"\\n  (b)current  in  the  1  ohm  resistance  is  \",round(IR4,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  in  the  5  ohm  resistance  is   0.44   A\n",
+        "\n",
+        "  (b)current  in  the  1  ohm  resistance  is   0.69   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 547</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine, using mesh-current analysis, \n",
+      "#(a) the mesh currents I1 and I2 \n",
+      "#(b) the current flowing in the capacitor,\n",
+      "#(c) the active power delivered by the voltage source.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  -1j*4;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  3j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Currents  I1,  I2  with  their  directions  are  shown  in  Figure  31.03.\n",
+      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
+      " #using  kirchoff  rule  in  2  loops\n",
+      " #two  eqns  obtained\n",
+      " #(R1  +  R2)*I1  -  R2*I2  =  V\n",
+      " #-1*R2*I1  +  (R3  +  R2  +  R4)*I2  =  0\n",
+      " #using  determinants\n",
+      "d1  =  [[V,  -1*R2],[0,  (R3  +  R2  +  R4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R1  +  R2),  V],[-1*R2,  0]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(R1  +  R2),  -1*R2],[-1*R2,  (R3  +  R2  +  R4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I1mag  =  abs(I1)\n",
+      " #Current  flowing  in  capacitor\n",
+      "Ic  =  I1  -  I2\n",
+      " #Source  power  P\n",
+      "phi  =  cmath.phase(complex(I1.real,I1.imag))\n",
+      "P  =  V*I1mag*math.cos(phi)\n",
+      "Icmag  =  abs(Ic)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)current,I1 is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A,  current,  I2  is\",round(I2.real,2),\"  +  (\",round(I2.imag,2),\")i  A\"\n",
+      "print  \"(b)current  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
+      "print  \"(c)Source  power  P  is  \",round(abs(P),2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)current,I1 is   10.17   +  ( 3.55 )i  A,  current,  I2  is 5.73   +  ( -8.74 )i  A\n",
+        "(b)current  in  the  capacitor  is   13.06   A\n",
+        "(c)Source  power  P  is   1017.06   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 548</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the line currents IR, IY and IB using mesh-current analysis\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  415;#  in  volts\n",
+      "rv2  =  415;#  in  volts\n",
+      "thetav1  =  120;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R  =  3  +  4j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Two  mesh  currents  I1  and  I2  are  chosen  as  shown  in  Figure  31.4.\n",
+      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
+      " #using  kirchoff  rule  in  2  loops\n",
+      " #two  eqns  obtained\n",
+      " #2*R*I1  -  R*I2  =  V1\n",
+      " #-1*R*I1  +  2*R*I2  =  V2\n",
+      " #using  determinants\n",
+      "d1  =  [[V1,  -1*R],[V2, 2*R]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[2*R, V1],[-1*R, V2]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[2*R, -1*R],[-1*R,  2*R]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I1mag  =  abs(I1)\n",
+      " #line  current  IR\n",
+      "IR  =  I1\n",
+      " #line  current  IB\n",
+      "IB  =  -1*I2\n",
+      " #line  current  IY\n",
+      "IY  =  I2  -  I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current, IR is\",round(IR.real,2),\"  +  (\",round(  IR.imag,2),\")i  A,  current,  IB  is\",round(IB.real,2),\"  +  (\",round(  IB.imag,2),\")i  A  and  current,  IY  is  \",round(IY.real,2),\"  +  (\",round(IY.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current, IR is 38.34   +  ( 28.75 )i  A,  current,  IB  is -44.07   +  ( 18.82 )i  A  and  current,  IY  is   5.73   +  ( -47.58 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 551</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the voltage VAB, by using nodal analysis.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri  =  20;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "R1  =  10;#  in  ohm\n",
+      "R2  =  3j;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  16;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #current\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Figure  31.8  contains  two  principal  nodes  (at  1  and  B)  and  thus  only  one  nodal  equation  is  required. \n",
+      "    #B  is  taken  as  the  reference  node  and  the  equation  for  node  1  is  obtained  as  follows. \n",
+      "    #Applying  Kirchhoff\u2019s  current  law  to  node  1  gives:\n",
+      " #IX  +  IY  =  I\n",
+      "V1  =  I/((1/R4)  +(1/(R2    +R3)))\n",
+      "IY  =  V1/(R2  +  R3)\n",
+      "VAB  =  IY*R3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VAB  is  \",round(VAB.real,2),\"  +  (\",round(  VAB.imag,2),\")i  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VAB  is   62.59   +  ( -9.39 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 552</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of voltage VXY\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  8;#  in  volts\n",
+      "rv2  =  8;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  6j;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  circuit  contains  no  principal  nodes.  \n",
+      "    #However,  if  point  Y  is  chosen  as  the  reference  node  then  an  equation \n",
+      "    #may  be  written  for  node  X  assuming  that  current  leaves  point  X  by  both  branches\n",
+      "VX  =  ((V1/(R1  +  R3)  +  V2/(R2  +  R4))/(1/(R1  +  R3)  +  1/(R2  +  R4)))\n",
+      "VXY  =  VX\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VXY  = \",round(abs(VXY),2),\"/_\",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),\"deg V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VXY  =  9.12 /_ 52.13 deg V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 553</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use nodal analysis to determine the current flowing in each branch of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  25;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #There  are  only  two  principal  nodes  in  Figure  31.10  so  only  one  nodal  equation  is  required.  \n",
+      "    #Node  2  is  taken  as  the  reference  node.\n",
+      " #The  equation  at  node  1  is  I1  +  I2  +  I3  =  0\n",
+      "Vn1  =  ((V1/R1  +  V2/R3)/(1/R1  +  1/R2  +  1/R3))\n",
+      "I1  =  (Vn1  -  V1)/R1\n",
+      "I2  =  Vn1/R2\n",
+      "I3  =  (Vn1  -  V2)/R3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I1  is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A, \\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round( I2.imag,2),\")i  A \\n and  current,  I3  is  \",round(I3.real,2),\"  +  (\",round(I3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I1  is   -3.16   +  ( 1.05 )i  A, \n",
+        " current,  I2  is     1.05   +  ( 1.32 )i  A \n",
+        " and  current,  I3  is   2.11   +  ( -2.37 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 554</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 ohm inductance,\n",
+      "#(c) the current in the 5 ohm resistance, and (d) the magnitude of the active power dissipated in the 2.5ohm resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  25;#  in  volts\n",
+      "rv2  =  25;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  -1j*4;#  in  ohm\n",
+      "R3  =  5;#  in  ohm\n",
+      "R4  =  4j;#  in  ohm\n",
+      "R5  =  2.5;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  equation  at  node  1\n",
+      " #Vn1*(1/R1  +  1/R2  +  1/R3)  -  Vn2/R3  =  V1/R1\n",
+      " #The  equation  at  node  2\n",
+      " #Vn1*(-1/R3)  +  Vn2*(1/R4  +  1/R5  +  1/R3)  =  V2/R5\n",
+      " #using  determinants\n",
+      "d1  =  [[V1/R1,  -1/R3],[V2/R5,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/R1  +  1/R2  +  1/R3),  V1/R1],[-1/R3,  V2/R5]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(1/R1  +  1/R2  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "Vn1  =  D1/D\n",
+      "Vn2  =  D2/D\n",
+      " #current  in  the  j4  ohm  inductance  is  given  by:\n",
+      "I4  =  Vn2/R4\n",
+      " #current  in  the  5  ohm  resistance  is  given  by:\n",
+      "I3  =  (Vn1  -  Vn2)/R3\n",
+      " #active  power  dissipated  in  the  2.5  ohm  resistor  is  given  by\n",
+      "P5  =  R5*((Vn2  -  V2)/R5)**2\n",
+      " #magnitude  of  the  active  power  dissipated\n",
+      "P5mag  =  abs(P5)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  voltage  at  nodes  1  and  2  are \",round(abs(Vn1),1),\"/_\",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),\"deg  V  and \",round(abs(Vn2),1),\"/_\",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),\"deg  V\"\n",
+      "print  \"\\n  (b)the  current  in  the  j4  ohm inductance = \",round(abs(I4),2),\"/_\",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  (c)the  current  in  the  5  ohm resistance  = \",round(abs(I3),2),\"/_\",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is  \",round(P5mag,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  voltage  at  nodes  1  and  2  are  17.1 /_ -5.3 deg  V  and  15.8 /_ 93.2 deg  V\n",
+        "\n",
+        "  (b)the  current  in  the  j4  ohm inductance =  3.95 /_ 3.23 deg  A\n",
+        "\n",
+        "  (c)the  current  in  the  5  ohm resistance  =  4.99 /_ -44.06 deg  A\n",
+        "\n",
+        "  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is   34.4   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 556</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the voltage VXY using nodal analysis.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri  =  25;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  3j;#  in  ohm\n",
+      "R3  =  5;#  in  ohm\n",
+      "R4  =  10j;#  in  ohm\n",
+      "R5  =  20j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #current\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Node  3  is  taken  as  the  reference  node.\n",
+      " #At  node  1,\n",
+      " #V1*(1/(R1  +  R2)  +  1/R3)  -  V2/R3  =  I\n",
+      " #The  equation  at  node  2\n",
+      " #V1*(-1/R3)  +  V2*(1/R4  +  1/R5  +  1/R3)  =  0\n",
+      " #using  determinants\n",
+      "d1  =  [[I,  -1/R3],[0 , (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/(R1  +  R2)  +  1/R3),  I],[-1/R3,  0]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(1/(R1  +  R2)  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "V1  =  D1/D\n",
+      "V2  =  D2/D\n",
+      " #the  voltage  between  point  X  and  node  3  is\n",
+      "VX  =  V1*R2/(R1  +  R2)\n",
+      " #Thus  the  voltage\n",
+      "VY  =  V2\n",
+      "VXY  =  VX  -  VY\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VXY  is  \",round(VXY.real,2),\"  +  (\",round(  VXY.imag,2),\")i  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VXY  is   -16.16   +  ( -15.05 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 557</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use nodal analysis to determine the voltages at nodes 2 and 3 \n",
+      "#determine the current flowing in the 2 ohm resistor and the power dissipated in the 3 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  8;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  2;#  in  ohm\n",
+      "R3  =  3;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  6;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #In  Figure  31.13,  the  reference  node  is  shown  at  point  A.\n",
+      " #At  node  1,\n",
+      " #V1*(1/R1  +  1/R6  +  1/R5)  -  V2/R1  -  V3/R5  =  V/R5\n",
+      " #The  equation  at  node  2\n",
+      " #V1*(-1/R1)  +  V2*(1/R2  +  1/R1  +  1/R3)  -  V3/R3  =  0\n",
+      " #At  node  3\n",
+      " #  -  V1/R5  -  V2/R3  +  V3*(1/R4  +  1/R3  +  1/R5)  =  -1*V/R5\n",
+      "#using  determinants\n",
+      "d1  =  [[V/R5,  -1/R1,  -1/R5],[0, (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1*V/R5, -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/R1  +  1/R6  +  1/R5),  V/R5,  -1/R5],[-1/R1,  0,  -1/R3],[-1/R5,  -1*V/R5,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[(1/R1  +  1/R6  +  1/R5),  -1/R1,  V/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  0],[-1/R5,  -1/R3,  -1*V/R5]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =[[(1/R1  +  1/R6  +  1/R5),  -1/R1,  -1/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1/R5,  -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "Vn1  =  D1/D\n",
+      "Vn2  =  D2/D\n",
+      "Vn3  =  D3/D  \n",
+      " #the  current  in  the  2  ohm  resistor\n",
+      "I2  =  Vn2/R2\n",
+      " #power  dissipated  in  the  3  ohm  resistance\n",
+      "P3  =  R3*((Vn2  -  Vn3)/R3)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  through  2  ohm  resistor  is  \",round(I2,2),\"  A\"\n",
+      "print  \"\\n  (b)power  dissipated  in  the  3  ohm  resistor  is  \",round(P3,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  through  2  ohm  resistor  is   0.19   A\n",
+        "\n",
+        "  (b)power  dissipated  in  the  3  ohm  resistor  is   1.27   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_3.ipynb
new file mode 100755
index 00000000..bbaa441d
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_31-checkpoint_3.ipynb
@@ -0,0 +1,660 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2c3da80fc1d074df861310f0cc84dab7f94d8474cb6362025e07b853bd81e258"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 31: Mesh-current and nodal analysis</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 546</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  4;#  in  volts\n",
+      "V2  =  5;#  in  volts\n",
+      "R1  =  3;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  1;#  in  ohm\n",
+      "R5  =  6;#  in  ohm\n",
+      "R6  =  8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #The  mesh  currents  I1,  I2  and  I3  are  shown  in  Figure  31.2.  Using  Kirchhoff\u2019s  voltage  law  in  3  loops\n",
+      " #three  eqns  obtained\n",
+      " #(R1  +  R2)*I1  -  R2*I2  =  V1\n",
+      " #-1*R2*I1  +  (R2  +  R3  +  R4  +  R5)*I2  -  R4*I3  =  0\n",
+      " #  -1*R4*I2  +  (R4  +  R6)*I3  =  -1*V2\n",
+      " #using  determinants\n",
+      "d1  =  [[V1,  -1*R2,  0],[0,  (R2  +  R3  +  R4  +  R5),  -1*R4],[-1*V2,  -1*R4,  (R4  +  R6)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R1  +  R2),  V1,  0],[-1*R2,  0,  -1*R4],[0,  -1*V2,  (R4  +  R6)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[(R1  +  R2),  -1*R2,  V1],[-1*R2,  (R2  +  R3  +  R4  +  R5),  0],[0,  -1*R4,  -1*V2]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =  [[(R1  +  R2),  -1*R2,  0],[-1*R2,  (R2  +  R3  +  R4  +  R5),  -1*R4],[0,  -1*R4,  (R4  +  R6)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I3  =  D3/D  \n",
+      "IR2  =  I1  -  I2\n",
+      "IR4  =  I2  -  I3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  in  the  5  ohm  resistance  is  \",round(IR2,2),\"  A\"\n",
+      "print  \"\\n  (b)current  in  the  1  ohm  resistance  is  \",round(IR4,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  in  the  5  ohm  resistance  is   0.44   A\n",
+        "\n",
+        "  (b)current  in  the  1  ohm  resistance  is   0.69   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 547</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  -1j*4;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  3j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Currents  I1,  I2  with  their  directions  are  shown  in  Figure  31.03.\n",
+      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
+      " #using  kirchoff  rule  in  2  loops\n",
+      " #two  eqns  obtained\n",
+      " #(R1  +  R2)*I1  -  R2*I2  =  V\n",
+      " #-1*R2*I1  +  (R3  +  R2  +  R4)*I2  =  0\n",
+      " #using  determinants\n",
+      "d1  =  [[V,  -1*R2],[0,  (R3  +  R2  +  R4)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R1  +  R2),  V],[-1*R2,  0]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(R1  +  R2),  -1*R2],[-1*R2,  (R3  +  R2  +  R4)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I1mag  =  abs(I1)\n",
+      " #Current  flowing  in  capacitor\n",
+      "Ic  =  I1  -  I2\n",
+      " #Source  power  P\n",
+      "phi  =  cmath.phase(complex(I1.real,I1.imag))\n",
+      "P  =  V*I1mag*math.cos(phi)\n",
+      "Icmag  =  abs(Ic)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)current,I1 is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A,  current,  I2  is\",round(I2.real,2),\"  +  (\",round(I2.imag,2),\")i  A\"\n",
+      "print  \"(b)current  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
+      "print  \"(c)Source  power  P  is  \",round(abs(P),2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)current,I1 is   10.17   +  ( 3.55 )i  A,  current,  I2  is 5.73   +  ( -8.74 )i  A\n",
+        "(b)current  in  the  capacitor  is   13.06   A\n",
+        "(c)Source  power  P  is   1017.06   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 548</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  415;#  in  volts\n",
+      "rv2  =  415;#  in  volts\n",
+      "thetav1  =  120;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R  =  3  +  4j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Two  mesh  currents  I1  and  I2  are  chosen  as  shown  in  Figure  31.4.\n",
+      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
+      " #using  kirchoff  rule  in  2  loops\n",
+      " #two  eqns  obtained\n",
+      " #2*R*I1  -  R*I2  =  V1\n",
+      " #-1*R*I1  +  2*R*I2  =  V2\n",
+      " #using  determinants\n",
+      "d1  =  [[V1,  -1*R],[V2, 2*R]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[2*R, V1],[-1*R, V2]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[2*R, -1*R],[-1*R,  2*R]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      "I1mag  =  abs(I1)\n",
+      " #line  current  IR\n",
+      "IR  =  I1\n",
+      " #line  current  IB\n",
+      "IB  =  -1*I2\n",
+      " #line  current  IY\n",
+      "IY  =  I2  -  I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current, IR is\",round(IR.real,2),\"  +  (\",round(  IR.imag,2),\")i  A,  current,  IB  is\",round(IB.real,2),\"  +  (\",round(  IB.imag,2),\")i  A  and  current,  IY  is  \",round(IY.real,2),\"  +  (\",round(IY.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current, IR is 38.34   +  ( 28.75 )i  A,  current,  IB  is -44.07   +  ( 18.82 )i  A  and  current,  IY  is   5.73   +  ( -47.58 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 551</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri  =  20;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "R1  =  10;#  in  ohm\n",
+      "R2  =  3j;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  16;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #current\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Figure  31.8  contains  two  principal  nodes  (at  1  and  B)  and  thus  only  one  nodal  equation  is  required. \n",
+      "    #B  is  taken  as  the  reference  node  and  the  equation  for  node  1  is  obtained  as  follows. \n",
+      "    #Applying  Kirchhoff\u2019s  current  law  to  node  1  gives:\n",
+      " #IX  +  IY  =  I\n",
+      "V1  =  I/((1/R4)  +(1/(R2    +R3)))\n",
+      "IY  =  V1/(R2  +  R3)\n",
+      "VAB  =  IY*R3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VAB  is  \",round(VAB.real,2),\"  +  (\",round(  VAB.imag,2),\")i  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VAB  is   62.59   +  ( -9.39 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 552</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  8;#  in  volts\n",
+      "rv2  =  8;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  6j;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  circuit  contains  no  principal  nodes.  \n",
+      "    #However,  if  point  Y  is  chosen  as  the  reference  node  then  an  equation \n",
+      "    #may  be  written  for  node  X  assuming  that  current  leaves  point  X  by  both  branches\n",
+      "VX  =  ((V1/(R1  +  R3)  +  V2/(R2  +  R4))/(1/(R1  +  R3)  +  1/(R2  +  R4)))\n",
+      "VXY  =  VX\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VXY  = \",round(abs(VXY),2),\"/_\",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),\"deg V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VXY  =  9.12 /_ 52.13 deg V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 553</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  25;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #There  are  only  two  principal  nodes  in  Figure  31.10  so  only  one  nodal  equation  is  required.  \n",
+      "    #Node  2  is  taken  as  the  reference  node.\n",
+      " #The  equation  at  node  1  is  I1  +  I2  +  I3  =  0\n",
+      "Vn1  =  ((V1/R1  +  V2/R3)/(1/R1  +  1/R2  +  1/R3))\n",
+      "I1  =  (Vn1  -  V1)/R1\n",
+      "I2  =  Vn1/R2\n",
+      "I3  =  (Vn1  -  V2)/R3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  current,  I1  is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A, \\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round( I2.imag,2),\")i  A \\n and  current,  I3  is  \",round(I3.real,2),\"  +  (\",round(I3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  current,  I1  is   -3.16   +  ( 1.05 )i  A, \n",
+        " current,  I2  is     1.05   +  ( 1.32 )i  A \n",
+        " and  current,  I3  is   2.11   +  ( -2.37 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 554</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  25;#  in  volts\n",
+      "rv2  =  25;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  -1j*4;#  in  ohm\n",
+      "R3  =  5;#  in  ohm\n",
+      "R4  =  4j;#  in  ohm\n",
+      "R5  =  2.5;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltages\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  equation  at  node  1\n",
+      " #Vn1*(1/R1  +  1/R2  +  1/R3)  -  Vn2/R3  =  V1/R1\n",
+      " #The  equation  at  node  2\n",
+      " #Vn1*(-1/R3)  +  Vn2*(1/R4  +  1/R5  +  1/R3)  =  V2/R5\n",
+      " #using  determinants\n",
+      "d1  =  [[V1/R1,  -1/R3],[V2/R5,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/R1  +  1/R2  +  1/R3),  V1/R1],[-1/R3,  V2/R5]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(1/R1  +  1/R2  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "Vn1  =  D1/D\n",
+      "Vn2  =  D2/D\n",
+      " #current  in  the  j4  ohm  inductance  is  given  by:\n",
+      "I4  =  Vn2/R4\n",
+      " #current  in  the  5  ohm  resistance  is  given  by:\n",
+      "I3  =  (Vn1  -  Vn2)/R3\n",
+      " #active  power  dissipated  in  the  2.5  ohm  resistor  is  given  by\n",
+      "P5  =  R5*((Vn2  -  V2)/R5)**2\n",
+      " #magnitude  of  the  active  power  dissipated\n",
+      "P5mag  =  abs(P5)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  voltage  at  nodes  1  and  2  are \",round(abs(Vn1),1),\"/_\",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),\"deg  V  and \",round(abs(Vn2),1),\"/_\",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),\"deg  V\"\n",
+      "print  \"\\n  (b)the  current  in  the  j4  ohm inductance = \",round(abs(I4),2),\"/_\",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  (c)the  current  in  the  5  ohm resistance  = \",round(abs(I3),2),\"/_\",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is  \",round(P5mag,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  voltage  at  nodes  1  and  2  are  17.1 /_ -5.3 deg  V  and  15.8 /_ 93.2 deg  V\n",
+        "\n",
+        "  (b)the  current  in  the  j4  ohm inductance =  3.95 /_ 3.23 deg  A\n",
+        "\n",
+        "  (c)the  current  in  the  5  ohm resistance  =  4.99 /_ -44.06 deg  A\n",
+        "\n",
+        "  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is   34.4   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 556</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri  =  25;#  in  amperes\n",
+      "thetai  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  3j;#  in  ohm\n",
+      "R3  =  5;#  in  ohm\n",
+      "R4  =  10j;#  in  ohm\n",
+      "R5  =  20j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #current\n",
+      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
+      " #Node  3  is  taken  as  the  reference  node.\n",
+      " #At  node  1,\n",
+      " #V1*(1/(R1  +  R2)  +  1/R3)  -  V2/R3  =  I\n",
+      " #The  equation  at  node  2\n",
+      " #V1*(-1/R3)  +  V2*(1/R4  +  1/R5  +  1/R3)  =  0\n",
+      " #using  determinants\n",
+      "d1  =  [[I,  -1/R3],[0 , (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/(R1  +  R2)  +  1/R3),  I],[-1/R3,  0]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(1/(R1  +  R2)  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "V1  =  D1/D\n",
+      "V2  =  D2/D\n",
+      " #the  voltage  between  point  X  and  node  3  is\n",
+      "VX  =  V1*R2/(R1  +  R2)\n",
+      " #Thus  the  voltage\n",
+      "VY  =  V2\n",
+      "VXY  =  VX  -  VY\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  VXY  is  \",round(VXY.real,2),\"  +  (\",round(  VXY.imag,2),\")i  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  VXY  is   -16.16   +  ( -15.05 )i  V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 557</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  8;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  2;#  in  ohm\n",
+      "R3  =  3;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  6;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #In  Figure  31.13,  the  reference  node  is  shown  at  point  A.\n",
+      " #At  node  1,\n",
+      " #V1*(1/R1  +  1/R6  +  1/R5)  -  V2/R1  -  V3/R5  =  V/R5\n",
+      " #The  equation  at  node  2\n",
+      " #V1*(-1/R1)  +  V2*(1/R2  +  1/R1  +  1/R3)  -  V3/R3  =  0\n",
+      " #At  node  3\n",
+      " #  -  V1/R5  -  V2/R3  +  V3*(1/R4  +  1/R3  +  1/R5)  =  -1*V/R5\n",
+      "#using  determinants\n",
+      "d1  =  [[V/R5,  -1/R1,  -1/R5],[0, (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1*V/R5, -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(1/R1  +  1/R6  +  1/R5),  V/R5,  -1/R5],[-1/R1,  0,  -1/R3],[-1/R5,  -1*V/R5,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d3  =  [[(1/R1  +  1/R6  +  1/R5),  -1/R1,  V/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  0],[-1/R5,  -1/R3,  -1*V/R5]]\n",
+      "D3  =  numpy.linalg.det(d3)\n",
+      "d  =[[(1/R1  +  1/R6  +  1/R5),  -1/R1,  -1/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1/R5,  -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "Vn1  =  D1/D\n",
+      "Vn2  =  D2/D\n",
+      "Vn3  =  D3/D  \n",
+      " #the  current  in  the  2  ohm  resistor\n",
+      "I2  =  Vn2/R2\n",
+      " #power  dissipated  in  the  3  ohm  resistance\n",
+      "P3  =  R3*((Vn2  -  Vn3)/R3)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  through  2  ohm  resistor  is  \",round(I2,2),\"  A\"\n",
+      "print  \"\\n  (b)power  dissipated  in  the  3  ohm  resistor  is  \",round(P3,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  through  2  ohm  resistor  is   0.19   A\n",
+        "\n",
+        "  (b)power  dissipated  in  the  3  ohm  resistor  is   1.27   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint.ipynb
new file mode 100755
index 00000000..985a6ce1
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint.ipynb
@@ -0,0 +1,436 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 32: The superposition theorem</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 564</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine using the superposition theorem, the current in the 20 ohm load and the current in each voltage source.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "r1  =  25;#  in  ohm\n",
+      "R  =  20;#  in  ohm\n",
+      "r2  =  10;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  circuit  diagram  is  shown  in  Figure  32.7.  Following  the  above  procedure:\n",
+      " #The  network  is  redrawn  with  the  50/_90\u00b0  V  source  removed  as  shown  in  Figure  32.8\n",
+      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.8.\n",
+      "I1  =  V1/(r1  +  r2*R/(R  +  r2))\n",
+      "I2  =  (r2/(r2  +  R))*I1\n",
+      "I3  =  (R/(r2  +  R))*I1\n",
+      " #The  network  is  redrawn  with  the  100/_0\u00b0  V  source  removed  as  shown  in  Figure  32.9\n",
+      " #Currents  I4,  I5  and  I6  are  labelled  as  shown  in  Figure  32.9.\n",
+      "I4  =  V2/(r2  +  r1*R/(r1  +  R))\n",
+      "I5  =  (r1/(r1  +  R))*I4\n",
+      "I6  =  (R/(r1  +  R))*I4\n",
+      " #Figure  32.10  shows  Figure  32.9  superimposed  on  Figure  32.8,  giving  the  currents  shown.\n",
+      " #Current  in  the  20  ohm  load,\n",
+      "I20  =  I2  +  I5\n",
+      " #Current  in  the  100/_0\u00b0  V  source\n",
+      "IV1  =  I1  -  I6\n",
+      " #Current  in  the  50/_90\u00b0  V  source\n",
+      "IV2  =  I4  -  I3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  in  the  20  ohm  load  is  \",round(I20.real,2),\"  +  (\",round(I20.imag,2),\")i  A\"\n",
+      "print  \"\\n  (b)Current  in  the  100/_0deg  V  source  is  \",round(IV1.real,2),\"  +  (\",round(IV1.imag,2),\")i  A\"\n",
+      "print  \"\\n  (b)Current  in  the  50/_90deg  V  source  is  \",round(IV2.real,2),\"  +  (\",round(IV2.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  in  the  20  ohm  load  is   1.05   +  ( 1.32 )i  A\n",
+        "\n",
+        "  (b)Current  in  the  100/_0deg  V  source  is   3.16   +  ( -1.05 )i  A\n",
+        "\n",
+        "  (b)Current  in  the  50/_90deg  V  source  is   -2.11   +  ( 2.37 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 566</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use the superposition theorem to determine the current in the 4 ohm resistor of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  12;#  in  volts\n",
+      "V2  =  20;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  2.5;#  in  ohm\n",
+      "R4  =  6;#  in  ohm\n",
+      "R5  =  2;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Removing  the  20  V  source  gives  the  network  shown  in  Figure  32.12.\n",
+      " #Currents  I1  and  I2  are  shown  labelled  in  Figure  32.12\n",
+      "Re1  =  (R4*R5/(R4  +  R5))  +  R3\n",
+      "Re2  =  Re1*R2/(Re1    +  R2)  +  R1\n",
+      "I1  =  V1/Re2\n",
+      "I2  =  (R2/(Re1  +  R2))*I1\n",
+      " #Removing  the  12  V  source  from  the  original  network  gives  the  network  shown  in  Figure  32.14.\n",
+      " #Currents  I3,  I4  and  I5  are  shown  labelled  in  Figure  32.14.\n",
+      "Re3  =  (R1*R2/(R1  +  R2))  +  R3\n",
+      "Re4  =  Re3*R4/(Re3  +  R4)  +  R5\n",
+      "I3  =  V2/Re4\n",
+      "I4  =  (R4/(Re3  +  R4))*I3\n",
+      "I5  =  (R1/(R1  +  R2))*I4\n",
+      " #Superimposing  Figure  32.14  on  Figure  32.12  shows  that  the  current  flowing  in  the  4  ohm  resistor  is  given  by\n",
+      "Ir4  =  I5  -  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncurrent  in  the  4  ohm  resistor  of  the  network  is  \",round(Ir4,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "current  in  the  4  ohm  resistor  of  the  network  is   0.48   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 567</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use the superposition theorem to obtain the current flowing in the \u0005(4 + j3) ohm impedance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  30;#  in  volts\n",
+      "rv2  =  30;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  -45;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  1j*3;#  in  ohm\n",
+      "R4  =  -1j*10;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  network  is  redrawn  with  V2  removed,  as  shown  in  Figure  32.17.\n",
+      " #Current  I1  and  I2  are  shown  in  Figure  32.17.  From  Figure  32.17,\n",
+      "Re1  =  R4*(R2  +  R3)/(R4  +  R3  +  R2)\n",
+      "Re2  =  Re1  +  R1\n",
+      " #current\n",
+      "I1  =  V1/Re2\n",
+      "I2  =  (R4/(R2  +  R3  +  R4))*I1\n",
+      " #The  original  network  is  redrawn  with  V1  removed,  as  shown  in  Figure  32.18\n",
+      " #Currents  I3  and  I4  are  shown  in  Figure  32.18.  From  Figure  32.18,\n",
+      "Re3  =  R1*(R2  +  R3)/(R1  +  R3  +  R2)\n",
+      "Re4  =  Re3  +  R4\n",
+      "I3  =  V2/Re4\n",
+      "I4  =  (R1/(R2  +  R3  +  R1))*I3\n",
+      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
+      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
+      "Ir4i3  =  I2  -  I4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current  in  (4  +  i3)  ohm  impedance  of  the  network  is  \",round(Ir4i3.real,2),\"  +  (\",round(  Ir4i3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current  in  (4  +  i3)  ohm  impedance  of  the  network  is   2.15   +  ( 0.42 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 568</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine, using the superposition theorem, \n",
+      "#(a) the current in each branch,\n",
+      "#(b) the magnitude of the voltage across the \u0005(6 + j8) ohm\u0006 impedance,\n",
+      "#and (c) the total active power delivered to the network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  5  +  0j;#  in  volts\n",
+      "E2  =  2  +  4j;#  in  volts\n",
+      "Z1  =  3  +  4j;#  in  ohm\n",
+      "Z2  =  2  -  5j;#  in  ohm\n",
+      "Z3  =  6  +  8j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #The  original  network  is  redrawn  with  E2  removed,  as  shown  in  Figure  32.20.\n",
+      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.20.\n",
+      "Ze1  =  Z3*Z2/(Z3  +  Z2)\n",
+      "Ze2  =  Ze1  +  Z1\n",
+      " #current\n",
+      "I1  =  E1/Ze2\n",
+      "I2  =  (Z2/(Z3  +  Z2))*I1\n",
+      "I3  =  (Z3/(Z3  +  Z2))*I1\n",
+      " #The  original  network  is  redrawn  with  E1  removed,  as  shown  in  Figure  32.22\n",
+      " #Currents  I4,  I5  and  I6  are  shown  labelled  in  Figure  32.22  \n",
+      "    #with  I4  flowing  away  from  the  positive  terminal  of  the  E2  source.\n",
+      "Ze3  =  Z3*Z1/(Z3  +  Z1)\n",
+      "Ze4  =  Ze3  +  Z2\n",
+      "I4  =  E2/Ze4\n",
+      "I5  =  (Z1/(Z3  +  Z1))*I4\n",
+      "I6  =  (Z3/(Z3  +  Z1))*I4\n",
+      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
+      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
+      "i1  =  I1  +  I6\n",
+      "i2  =  I3  +  I4\n",
+      "i3  =  I2  -  I5\n",
+      " #magnitude\n",
+      "i1mag  =  abs(i1)\n",
+      "i2mag  =  abs(i2)\n",
+      "E1mag  =  abs(E1)\n",
+      "E2mag  =  abs(E2)\n",
+      " #phase\n",
+      "phi1  =  cmath.phase(complex(i1.real,i1.imag))\n",
+      "phi2  =  cmath.phase(complex(i2.real,i2.imag))\n",
+      " #voltage  across  the(6  +  i8)  ohm  impedance\n",
+      "V6i8  =  i3*Z3\n",
+      "V6i8m  =  abs(V6i8)\n",
+      " #power\n",
+      "P  =  (E1mag*i1mag*math.cos(phi1))  +  (E2mag*i2mag*math.cos(phi2  -  cmath.phase(complex(E2.real,E2.imag))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)currents  are: \\n \",round(i1.real,2),\"  +  (\",round(  i1.imag,2),\")i  A, \\n \",round(i2.real,2),\"  +  (\",round(i2.imag,2),\")i  A \\n and  \",round(i3.real,2),\"  +  (\",round(i3.imag,2),\")i  A\"\n",
+      "print  \"\\n(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is  \",round(V6i8m,2),\"  V\"\n",
+      "print  \"\\n(c)the  total  active  power  delivered  to  the  network  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)currents  are: \n",
+        "  0.57   +  ( 0.62 )i  A, \n",
+        "  0.56   +  ( 1.33 )i  A \n",
+        " and   0.01   +  ( -0.71 )i  A\n",
+        "\n",
+        "(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is   7.09   V\n",
+        "\n",
+        "(c)the  total  active  power  delivered  to  the  network  is   9.29   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 571</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine,(a) the magnitude of the current flowing in the capacitor, \n",
+      "#(b) the p.d. across the 5 ohm resistance, (c) the active power dissipated in the 20 ohm resistance and \n",
+      "#(d) the total active power taken from the supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  50;#  in  volts\n",
+      "rv2  =  30;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  20;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  -1j*3;#  in  ohm\n",
+      "R4  =  8;#  in  ohm\n",
+      "R5  =  8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  network  is  redrawn  with  the  V2  source  removed,  as  shown  in  Figure  32.26.\n",
+      " #Currents  I1  to  I5  are  shown  labelled  in  Figure  32.26.  \n",
+      " #current\n",
+      "Re1  =  R4*R5/(R5  +  R4)  +  R3\n",
+      "Re2  =  Re1*R2/(R2  +  Re1)\n",
+      "I1  =  V1/(Re2  +  R1)\n",
+      "I2  =  (Re1/(R2  +  Re1))*I1\n",
+      "I3  =  (R2/(Re1  +  R2))*I1\n",
+      "I4  =  (R4/(R4  +  R5))*I3\n",
+      "I5  =  I3  -  I4\n",
+      " #The  original  network  is  redrawn  with  the  V1  source  removed,  as  shown  in  Figure  32.27.\n",
+      " #Currents  I6  to  I10  are  shown  labelled  in  Figure  32.27\n",
+      "Re3  =  R1*R2/(R1  +  R2)\n",
+      "Re4  =  Re3  +  R3\n",
+      "Re5  =  Re4*R4/(Re4  +  R4)\n",
+      "Re6  =  Re5    +  R5\n",
+      "I6  =  V2/Re6\n",
+      "I7  =  (Re4/(Re4  +  R4))*I6\n",
+      "I8  =  (R4/(Re4  +  R4))*I6\n",
+      "I9  =  (R1/(R1  +  R2))*I8\n",
+      "I10  =  (R2/(R1  +  R2))*I8\n",
+      " #current  flowing  in  the  capacitor  is  given  by\n",
+      "Ic  =  I3  -  I8\n",
+      " #magnitude  of  the  current  in  the  capacitor\n",
+      "Icmag  =  abs(Ic)\n",
+      "\n",
+      "i1  =  I2  +  I9\n",
+      "i1mag  =  abs(i1)\n",
+      " #magnitude  of  the  p.d.  across  the  5  ohm  resistance  is  given  by\n",
+      "Vr5m  =  i1mag*R2\n",
+      " #Active  power  dissipated  in  the  20  ohm  resistance  is  given  by\n",
+      "i2  =  I1  -  I10\n",
+      "i2mag  =  abs(i2)\n",
+      "phii2  =  cmath.phase(complex(i2.real,i2.imag))\n",
+      "Pr20  =  R1*(i2mag)**2\n",
+      " #Active  power  developed  by  the  V1\n",
+      "P1  =  rv1*i2mag*math.cos(phii2)\n",
+      " #Active  power  developed  by  V2  source\n",
+      "i3  =  I6  -  I5\n",
+      "i3mag  =  abs(i3)\n",
+      "phii3  =  cmath.phase(complex(i3.real,i3.imag))\n",
+      "P2  =  rv2*i3mag*math.cos(phii3  -  (thetav2*math.pi/180))\n",
+      " #Total  power  developed\n",
+      "P  =  P1  +  P2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
+      "print  \"\\n(b)  the  p.d.  across  the  5  ohm  resistance  is  \",round(Vr5m,2),\"  V\"\n",
+      "print  \"\\n(c)the  active  power  dissipated  in  the  20  ohm  resistance  is  \",round(Pr20,0),\"  W\"\n",
+      "print  \"\\n(d)the  total  active  power  taken  from  the  supply  is  \",round(P,1),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is   2.11   A\n",
+        "\n",
+        "(b)  the  p.d.  across  the  5  ohm  resistance  is   5.85   V\n",
+        "\n",
+        "(c)the  active  power  dissipated  in  the  20  ohm  resistance  is   111.0   W\n",
+        "\n",
+        "(d)the  total  active  power  taken  from  the  supply  is   191.9   W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_1.ipynb
new file mode 100755
index 00000000..985a6ce1
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_1.ipynb
@@ -0,0 +1,436 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 32: The superposition theorem</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 564</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine using the superposition theorem, the current in the 20 ohm load and the current in each voltage source.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "r1  =  25;#  in  ohm\n",
+      "R  =  20;#  in  ohm\n",
+      "r2  =  10;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  circuit  diagram  is  shown  in  Figure  32.7.  Following  the  above  procedure:\n",
+      " #The  network  is  redrawn  with  the  50/_90\u00b0  V  source  removed  as  shown  in  Figure  32.8\n",
+      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.8.\n",
+      "I1  =  V1/(r1  +  r2*R/(R  +  r2))\n",
+      "I2  =  (r2/(r2  +  R))*I1\n",
+      "I3  =  (R/(r2  +  R))*I1\n",
+      " #The  network  is  redrawn  with  the  100/_0\u00b0  V  source  removed  as  shown  in  Figure  32.9\n",
+      " #Currents  I4,  I5  and  I6  are  labelled  as  shown  in  Figure  32.9.\n",
+      "I4  =  V2/(r2  +  r1*R/(r1  +  R))\n",
+      "I5  =  (r1/(r1  +  R))*I4\n",
+      "I6  =  (R/(r1  +  R))*I4\n",
+      " #Figure  32.10  shows  Figure  32.9  superimposed  on  Figure  32.8,  giving  the  currents  shown.\n",
+      " #Current  in  the  20  ohm  load,\n",
+      "I20  =  I2  +  I5\n",
+      " #Current  in  the  100/_0\u00b0  V  source\n",
+      "IV1  =  I1  -  I6\n",
+      " #Current  in  the  50/_90\u00b0  V  source\n",
+      "IV2  =  I4  -  I3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  in  the  20  ohm  load  is  \",round(I20.real,2),\"  +  (\",round(I20.imag,2),\")i  A\"\n",
+      "print  \"\\n  (b)Current  in  the  100/_0deg  V  source  is  \",round(IV1.real,2),\"  +  (\",round(IV1.imag,2),\")i  A\"\n",
+      "print  \"\\n  (b)Current  in  the  50/_90deg  V  source  is  \",round(IV2.real,2),\"  +  (\",round(IV2.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  in  the  20  ohm  load  is   1.05   +  ( 1.32 )i  A\n",
+        "\n",
+        "  (b)Current  in  the  100/_0deg  V  source  is   3.16   +  ( -1.05 )i  A\n",
+        "\n",
+        "  (b)Current  in  the  50/_90deg  V  source  is   -2.11   +  ( 2.37 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 566</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use the superposition theorem to determine the current in the 4 ohm resistor of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  12;#  in  volts\n",
+      "V2  =  20;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  2.5;#  in  ohm\n",
+      "R4  =  6;#  in  ohm\n",
+      "R5  =  2;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Removing  the  20  V  source  gives  the  network  shown  in  Figure  32.12.\n",
+      " #Currents  I1  and  I2  are  shown  labelled  in  Figure  32.12\n",
+      "Re1  =  (R4*R5/(R4  +  R5))  +  R3\n",
+      "Re2  =  Re1*R2/(Re1    +  R2)  +  R1\n",
+      "I1  =  V1/Re2\n",
+      "I2  =  (R2/(Re1  +  R2))*I1\n",
+      " #Removing  the  12  V  source  from  the  original  network  gives  the  network  shown  in  Figure  32.14.\n",
+      " #Currents  I3,  I4  and  I5  are  shown  labelled  in  Figure  32.14.\n",
+      "Re3  =  (R1*R2/(R1  +  R2))  +  R3\n",
+      "Re4  =  Re3*R4/(Re3  +  R4)  +  R5\n",
+      "I3  =  V2/Re4\n",
+      "I4  =  (R4/(Re3  +  R4))*I3\n",
+      "I5  =  (R1/(R1  +  R2))*I4\n",
+      " #Superimposing  Figure  32.14  on  Figure  32.12  shows  that  the  current  flowing  in  the  4  ohm  resistor  is  given  by\n",
+      "Ir4  =  I5  -  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncurrent  in  the  4  ohm  resistor  of  the  network  is  \",round(Ir4,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "current  in  the  4  ohm  resistor  of  the  network  is   0.48   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 567</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use the superposition theorem to obtain the current flowing in the \u0005(4 + j3) ohm impedance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  30;#  in  volts\n",
+      "rv2  =  30;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  -45;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  1j*3;#  in  ohm\n",
+      "R4  =  -1j*10;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  network  is  redrawn  with  V2  removed,  as  shown  in  Figure  32.17.\n",
+      " #Current  I1  and  I2  are  shown  in  Figure  32.17.  From  Figure  32.17,\n",
+      "Re1  =  R4*(R2  +  R3)/(R4  +  R3  +  R2)\n",
+      "Re2  =  Re1  +  R1\n",
+      " #current\n",
+      "I1  =  V1/Re2\n",
+      "I2  =  (R4/(R2  +  R3  +  R4))*I1\n",
+      " #The  original  network  is  redrawn  with  V1  removed,  as  shown  in  Figure  32.18\n",
+      " #Currents  I3  and  I4  are  shown  in  Figure  32.18.  From  Figure  32.18,\n",
+      "Re3  =  R1*(R2  +  R3)/(R1  +  R3  +  R2)\n",
+      "Re4  =  Re3  +  R4\n",
+      "I3  =  V2/Re4\n",
+      "I4  =  (R1/(R2  +  R3  +  R1))*I3\n",
+      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
+      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
+      "Ir4i3  =  I2  -  I4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current  in  (4  +  i3)  ohm  impedance  of  the  network  is  \",round(Ir4i3.real,2),\"  +  (\",round(  Ir4i3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current  in  (4  +  i3)  ohm  impedance  of  the  network  is   2.15   +  ( 0.42 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 568</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine, using the superposition theorem, \n",
+      "#(a) the current in each branch,\n",
+      "#(b) the magnitude of the voltage across the \u0005(6 + j8) ohm\u0006 impedance,\n",
+      "#and (c) the total active power delivered to the network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  5  +  0j;#  in  volts\n",
+      "E2  =  2  +  4j;#  in  volts\n",
+      "Z1  =  3  +  4j;#  in  ohm\n",
+      "Z2  =  2  -  5j;#  in  ohm\n",
+      "Z3  =  6  +  8j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #The  original  network  is  redrawn  with  E2  removed,  as  shown  in  Figure  32.20.\n",
+      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.20.\n",
+      "Ze1  =  Z3*Z2/(Z3  +  Z2)\n",
+      "Ze2  =  Ze1  +  Z1\n",
+      " #current\n",
+      "I1  =  E1/Ze2\n",
+      "I2  =  (Z2/(Z3  +  Z2))*I1\n",
+      "I3  =  (Z3/(Z3  +  Z2))*I1\n",
+      " #The  original  network  is  redrawn  with  E1  removed,  as  shown  in  Figure  32.22\n",
+      " #Currents  I4,  I5  and  I6  are  shown  labelled  in  Figure  32.22  \n",
+      "    #with  I4  flowing  away  from  the  positive  terminal  of  the  E2  source.\n",
+      "Ze3  =  Z3*Z1/(Z3  +  Z1)\n",
+      "Ze4  =  Ze3  +  Z2\n",
+      "I4  =  E2/Ze4\n",
+      "I5  =  (Z1/(Z3  +  Z1))*I4\n",
+      "I6  =  (Z3/(Z3  +  Z1))*I4\n",
+      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
+      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
+      "i1  =  I1  +  I6\n",
+      "i2  =  I3  +  I4\n",
+      "i3  =  I2  -  I5\n",
+      " #magnitude\n",
+      "i1mag  =  abs(i1)\n",
+      "i2mag  =  abs(i2)\n",
+      "E1mag  =  abs(E1)\n",
+      "E2mag  =  abs(E2)\n",
+      " #phase\n",
+      "phi1  =  cmath.phase(complex(i1.real,i1.imag))\n",
+      "phi2  =  cmath.phase(complex(i2.real,i2.imag))\n",
+      " #voltage  across  the(6  +  i8)  ohm  impedance\n",
+      "V6i8  =  i3*Z3\n",
+      "V6i8m  =  abs(V6i8)\n",
+      " #power\n",
+      "P  =  (E1mag*i1mag*math.cos(phi1))  +  (E2mag*i2mag*math.cos(phi2  -  cmath.phase(complex(E2.real,E2.imag))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)currents  are: \\n \",round(i1.real,2),\"  +  (\",round(  i1.imag,2),\")i  A, \\n \",round(i2.real,2),\"  +  (\",round(i2.imag,2),\")i  A \\n and  \",round(i3.real,2),\"  +  (\",round(i3.imag,2),\")i  A\"\n",
+      "print  \"\\n(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is  \",round(V6i8m,2),\"  V\"\n",
+      "print  \"\\n(c)the  total  active  power  delivered  to  the  network  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)currents  are: \n",
+        "  0.57   +  ( 0.62 )i  A, \n",
+        "  0.56   +  ( 1.33 )i  A \n",
+        " and   0.01   +  ( -0.71 )i  A\n",
+        "\n",
+        "(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is   7.09   V\n",
+        "\n",
+        "(c)the  total  active  power  delivered  to  the  network  is   9.29   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 571</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine,(a) the magnitude of the current flowing in the capacitor, \n",
+      "#(b) the p.d. across the 5 ohm resistance, (c) the active power dissipated in the 20 ohm resistance and \n",
+      "#(d) the total active power taken from the supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  50;#  in  volts\n",
+      "rv2  =  30;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  20;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  -1j*3;#  in  ohm\n",
+      "R4  =  8;#  in  ohm\n",
+      "R5  =  8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  network  is  redrawn  with  the  V2  source  removed,  as  shown  in  Figure  32.26.\n",
+      " #Currents  I1  to  I5  are  shown  labelled  in  Figure  32.26.  \n",
+      " #current\n",
+      "Re1  =  R4*R5/(R5  +  R4)  +  R3\n",
+      "Re2  =  Re1*R2/(R2  +  Re1)\n",
+      "I1  =  V1/(Re2  +  R1)\n",
+      "I2  =  (Re1/(R2  +  Re1))*I1\n",
+      "I3  =  (R2/(Re1  +  R2))*I1\n",
+      "I4  =  (R4/(R4  +  R5))*I3\n",
+      "I5  =  I3  -  I4\n",
+      " #The  original  network  is  redrawn  with  the  V1  source  removed,  as  shown  in  Figure  32.27.\n",
+      " #Currents  I6  to  I10  are  shown  labelled  in  Figure  32.27\n",
+      "Re3  =  R1*R2/(R1  +  R2)\n",
+      "Re4  =  Re3  +  R3\n",
+      "Re5  =  Re4*R4/(Re4  +  R4)\n",
+      "Re6  =  Re5    +  R5\n",
+      "I6  =  V2/Re6\n",
+      "I7  =  (Re4/(Re4  +  R4))*I6\n",
+      "I8  =  (R4/(Re4  +  R4))*I6\n",
+      "I9  =  (R1/(R1  +  R2))*I8\n",
+      "I10  =  (R2/(R1  +  R2))*I8\n",
+      " #current  flowing  in  the  capacitor  is  given  by\n",
+      "Ic  =  I3  -  I8\n",
+      " #magnitude  of  the  current  in  the  capacitor\n",
+      "Icmag  =  abs(Ic)\n",
+      "\n",
+      "i1  =  I2  +  I9\n",
+      "i1mag  =  abs(i1)\n",
+      " #magnitude  of  the  p.d.  across  the  5  ohm  resistance  is  given  by\n",
+      "Vr5m  =  i1mag*R2\n",
+      " #Active  power  dissipated  in  the  20  ohm  resistance  is  given  by\n",
+      "i2  =  I1  -  I10\n",
+      "i2mag  =  abs(i2)\n",
+      "phii2  =  cmath.phase(complex(i2.real,i2.imag))\n",
+      "Pr20  =  R1*(i2mag)**2\n",
+      " #Active  power  developed  by  the  V1\n",
+      "P1  =  rv1*i2mag*math.cos(phii2)\n",
+      " #Active  power  developed  by  V2  source\n",
+      "i3  =  I6  -  I5\n",
+      "i3mag  =  abs(i3)\n",
+      "phii3  =  cmath.phase(complex(i3.real,i3.imag))\n",
+      "P2  =  rv2*i3mag*math.cos(phii3  -  (thetav2*math.pi/180))\n",
+      " #Total  power  developed\n",
+      "P  =  P1  +  P2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
+      "print  \"\\n(b)  the  p.d.  across  the  5  ohm  resistance  is  \",round(Vr5m,2),\"  V\"\n",
+      "print  \"\\n(c)the  active  power  dissipated  in  the  20  ohm  resistance  is  \",round(Pr20,0),\"  W\"\n",
+      "print  \"\\n(d)the  total  active  power  taken  from  the  supply  is  \",round(P,1),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is   2.11   A\n",
+        "\n",
+        "(b)  the  p.d.  across  the  5  ohm  resistance  is   5.85   V\n",
+        "\n",
+        "(c)the  active  power  dissipated  in  the  20  ohm  resistance  is   111.0   W\n",
+        "\n",
+        "(d)the  total  active  power  taken  from  the  supply  is   191.9   W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_2.ipynb
new file mode 100755
index 00000000..985a6ce1
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_2.ipynb
@@ -0,0 +1,436 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 32: The superposition theorem</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 564</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine using the superposition theorem, the current in the 20 ohm load and the current in each voltage source.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "r1  =  25;#  in  ohm\n",
+      "R  =  20;#  in  ohm\n",
+      "r2  =  10;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  circuit  diagram  is  shown  in  Figure  32.7.  Following  the  above  procedure:\n",
+      " #The  network  is  redrawn  with  the  50/_90\u00b0  V  source  removed  as  shown  in  Figure  32.8\n",
+      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.8.\n",
+      "I1  =  V1/(r1  +  r2*R/(R  +  r2))\n",
+      "I2  =  (r2/(r2  +  R))*I1\n",
+      "I3  =  (R/(r2  +  R))*I1\n",
+      " #The  network  is  redrawn  with  the  100/_0\u00b0  V  source  removed  as  shown  in  Figure  32.9\n",
+      " #Currents  I4,  I5  and  I6  are  labelled  as  shown  in  Figure  32.9.\n",
+      "I4  =  V2/(r2  +  r1*R/(r1  +  R))\n",
+      "I5  =  (r1/(r1  +  R))*I4\n",
+      "I6  =  (R/(r1  +  R))*I4\n",
+      " #Figure  32.10  shows  Figure  32.9  superimposed  on  Figure  32.8,  giving  the  currents  shown.\n",
+      " #Current  in  the  20  ohm  load,\n",
+      "I20  =  I2  +  I5\n",
+      " #Current  in  the  100/_0\u00b0  V  source\n",
+      "IV1  =  I1  -  I6\n",
+      " #Current  in  the  50/_90\u00b0  V  source\n",
+      "IV2  =  I4  -  I3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  in  the  20  ohm  load  is  \",round(I20.real,2),\"  +  (\",round(I20.imag,2),\")i  A\"\n",
+      "print  \"\\n  (b)Current  in  the  100/_0deg  V  source  is  \",round(IV1.real,2),\"  +  (\",round(IV1.imag,2),\")i  A\"\n",
+      "print  \"\\n  (b)Current  in  the  50/_90deg  V  source  is  \",round(IV2.real,2),\"  +  (\",round(IV2.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  in  the  20  ohm  load  is   1.05   +  ( 1.32 )i  A\n",
+        "\n",
+        "  (b)Current  in  the  100/_0deg  V  source  is   3.16   +  ( -1.05 )i  A\n",
+        "\n",
+        "  (b)Current  in  the  50/_90deg  V  source  is   -2.11   +  ( 2.37 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 566</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use the superposition theorem to determine the current in the 4 ohm resistor of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  12;#  in  volts\n",
+      "V2  =  20;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  2.5;#  in  ohm\n",
+      "R4  =  6;#  in  ohm\n",
+      "R5  =  2;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Removing  the  20  V  source  gives  the  network  shown  in  Figure  32.12.\n",
+      " #Currents  I1  and  I2  are  shown  labelled  in  Figure  32.12\n",
+      "Re1  =  (R4*R5/(R4  +  R5))  +  R3\n",
+      "Re2  =  Re1*R2/(Re1    +  R2)  +  R1\n",
+      "I1  =  V1/Re2\n",
+      "I2  =  (R2/(Re1  +  R2))*I1\n",
+      " #Removing  the  12  V  source  from  the  original  network  gives  the  network  shown  in  Figure  32.14.\n",
+      " #Currents  I3,  I4  and  I5  are  shown  labelled  in  Figure  32.14.\n",
+      "Re3  =  (R1*R2/(R1  +  R2))  +  R3\n",
+      "Re4  =  Re3*R4/(Re3  +  R4)  +  R5\n",
+      "I3  =  V2/Re4\n",
+      "I4  =  (R4/(Re3  +  R4))*I3\n",
+      "I5  =  (R1/(R1  +  R2))*I4\n",
+      " #Superimposing  Figure  32.14  on  Figure  32.12  shows  that  the  current  flowing  in  the  4  ohm  resistor  is  given  by\n",
+      "Ir4  =  I5  -  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncurrent  in  the  4  ohm  resistor  of  the  network  is  \",round(Ir4,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "current  in  the  4  ohm  resistor  of  the  network  is   0.48   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 567</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Use the superposition theorem to obtain the current flowing in the \u0005(4 + j3) ohm impedance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  30;#  in  volts\n",
+      "rv2  =  30;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  -45;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  1j*3;#  in  ohm\n",
+      "R4  =  -1j*10;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  network  is  redrawn  with  V2  removed,  as  shown  in  Figure  32.17.\n",
+      " #Current  I1  and  I2  are  shown  in  Figure  32.17.  From  Figure  32.17,\n",
+      "Re1  =  R4*(R2  +  R3)/(R4  +  R3  +  R2)\n",
+      "Re2  =  Re1  +  R1\n",
+      " #current\n",
+      "I1  =  V1/Re2\n",
+      "I2  =  (R4/(R2  +  R3  +  R4))*I1\n",
+      " #The  original  network  is  redrawn  with  V1  removed,  as  shown  in  Figure  32.18\n",
+      " #Currents  I3  and  I4  are  shown  in  Figure  32.18.  From  Figure  32.18,\n",
+      "Re3  =  R1*(R2  +  R3)/(R1  +  R3  +  R2)\n",
+      "Re4  =  Re3  +  R4\n",
+      "I3  =  V2/Re4\n",
+      "I4  =  (R1/(R2  +  R3  +  R1))*I3\n",
+      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
+      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
+      "Ir4i3  =  I2  -  I4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current  in  (4  +  i3)  ohm  impedance  of  the  network  is  \",round(Ir4i3.real,2),\"  +  (\",round(  Ir4i3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current  in  (4  +  i3)  ohm  impedance  of  the  network  is   2.15   +  ( 0.42 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 568</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine, using the superposition theorem, \n",
+      "#(a) the current in each branch,\n",
+      "#(b) the magnitude of the voltage across the \u0005(6 + j8) ohm\u0006 impedance,\n",
+      "#and (c) the total active power delivered to the network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  5  +  0j;#  in  volts\n",
+      "E2  =  2  +  4j;#  in  volts\n",
+      "Z1  =  3  +  4j;#  in  ohm\n",
+      "Z2  =  2  -  5j;#  in  ohm\n",
+      "Z3  =  6  +  8j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #The  original  network  is  redrawn  with  E2  removed,  as  shown  in  Figure  32.20.\n",
+      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.20.\n",
+      "Ze1  =  Z3*Z2/(Z3  +  Z2)\n",
+      "Ze2  =  Ze1  +  Z1\n",
+      " #current\n",
+      "I1  =  E1/Ze2\n",
+      "I2  =  (Z2/(Z3  +  Z2))*I1\n",
+      "I3  =  (Z3/(Z3  +  Z2))*I1\n",
+      " #The  original  network  is  redrawn  with  E1  removed,  as  shown  in  Figure  32.22\n",
+      " #Currents  I4,  I5  and  I6  are  shown  labelled  in  Figure  32.22  \n",
+      "    #with  I4  flowing  away  from  the  positive  terminal  of  the  E2  source.\n",
+      "Ze3  =  Z3*Z1/(Z3  +  Z1)\n",
+      "Ze4  =  Ze3  +  Z2\n",
+      "I4  =  E2/Ze4\n",
+      "I5  =  (Z1/(Z3  +  Z1))*I4\n",
+      "I6  =  (Z3/(Z3  +  Z1))*I4\n",
+      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
+      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
+      "i1  =  I1  +  I6\n",
+      "i2  =  I3  +  I4\n",
+      "i3  =  I2  -  I5\n",
+      " #magnitude\n",
+      "i1mag  =  abs(i1)\n",
+      "i2mag  =  abs(i2)\n",
+      "E1mag  =  abs(E1)\n",
+      "E2mag  =  abs(E2)\n",
+      " #phase\n",
+      "phi1  =  cmath.phase(complex(i1.real,i1.imag))\n",
+      "phi2  =  cmath.phase(complex(i2.real,i2.imag))\n",
+      " #voltage  across  the(6  +  i8)  ohm  impedance\n",
+      "V6i8  =  i3*Z3\n",
+      "V6i8m  =  abs(V6i8)\n",
+      " #power\n",
+      "P  =  (E1mag*i1mag*math.cos(phi1))  +  (E2mag*i2mag*math.cos(phi2  -  cmath.phase(complex(E2.real,E2.imag))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)currents  are: \\n \",round(i1.real,2),\"  +  (\",round(  i1.imag,2),\")i  A, \\n \",round(i2.real,2),\"  +  (\",round(i2.imag,2),\")i  A \\n and  \",round(i3.real,2),\"  +  (\",round(i3.imag,2),\")i  A\"\n",
+      "print  \"\\n(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is  \",round(V6i8m,2),\"  V\"\n",
+      "print  \"\\n(c)the  total  active  power  delivered  to  the  network  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)currents  are: \n",
+        "  0.57   +  ( 0.62 )i  A, \n",
+        "  0.56   +  ( 1.33 )i  A \n",
+        " and   0.01   +  ( -0.71 )i  A\n",
+        "\n",
+        "(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is   7.09   V\n",
+        "\n",
+        "(c)the  total  active  power  delivered  to  the  network  is   9.29   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 571</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine,(a) the magnitude of the current flowing in the capacitor, \n",
+      "#(b) the p.d. across the 5 ohm resistance, (c) the active power dissipated in the 20 ohm resistance and \n",
+      "#(d) the total active power taken from the supply.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  50;#  in  volts\n",
+      "rv2  =  30;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  20;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  -1j*3;#  in  ohm\n",
+      "R4  =  8;#  in  ohm\n",
+      "R5  =  8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  network  is  redrawn  with  the  V2  source  removed,  as  shown  in  Figure  32.26.\n",
+      " #Currents  I1  to  I5  are  shown  labelled  in  Figure  32.26.  \n",
+      " #current\n",
+      "Re1  =  R4*R5/(R5  +  R4)  +  R3\n",
+      "Re2  =  Re1*R2/(R2  +  Re1)\n",
+      "I1  =  V1/(Re2  +  R1)\n",
+      "I2  =  (Re1/(R2  +  Re1))*I1\n",
+      "I3  =  (R2/(Re1  +  R2))*I1\n",
+      "I4  =  (R4/(R4  +  R5))*I3\n",
+      "I5  =  I3  -  I4\n",
+      " #The  original  network  is  redrawn  with  the  V1  source  removed,  as  shown  in  Figure  32.27.\n",
+      " #Currents  I6  to  I10  are  shown  labelled  in  Figure  32.27\n",
+      "Re3  =  R1*R2/(R1  +  R2)\n",
+      "Re4  =  Re3  +  R3\n",
+      "Re5  =  Re4*R4/(Re4  +  R4)\n",
+      "Re6  =  Re5    +  R5\n",
+      "I6  =  V2/Re6\n",
+      "I7  =  (Re4/(Re4  +  R4))*I6\n",
+      "I8  =  (R4/(Re4  +  R4))*I6\n",
+      "I9  =  (R1/(R1  +  R2))*I8\n",
+      "I10  =  (R2/(R1  +  R2))*I8\n",
+      " #current  flowing  in  the  capacitor  is  given  by\n",
+      "Ic  =  I3  -  I8\n",
+      " #magnitude  of  the  current  in  the  capacitor\n",
+      "Icmag  =  abs(Ic)\n",
+      "\n",
+      "i1  =  I2  +  I9\n",
+      "i1mag  =  abs(i1)\n",
+      " #magnitude  of  the  p.d.  across  the  5  ohm  resistance  is  given  by\n",
+      "Vr5m  =  i1mag*R2\n",
+      " #Active  power  dissipated  in  the  20  ohm  resistance  is  given  by\n",
+      "i2  =  I1  -  I10\n",
+      "i2mag  =  abs(i2)\n",
+      "phii2  =  cmath.phase(complex(i2.real,i2.imag))\n",
+      "Pr20  =  R1*(i2mag)**2\n",
+      " #Active  power  developed  by  the  V1\n",
+      "P1  =  rv1*i2mag*math.cos(phii2)\n",
+      " #Active  power  developed  by  V2  source\n",
+      "i3  =  I6  -  I5\n",
+      "i3mag  =  abs(i3)\n",
+      "phii3  =  cmath.phase(complex(i3.real,i3.imag))\n",
+      "P2  =  rv2*i3mag*math.cos(phii3  -  (thetav2*math.pi/180))\n",
+      " #Total  power  developed\n",
+      "P  =  P1  +  P2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
+      "print  \"\\n(b)  the  p.d.  across  the  5  ohm  resistance  is  \",round(Vr5m,2),\"  V\"\n",
+      "print  \"\\n(c)the  active  power  dissipated  in  the  20  ohm  resistance  is  \",round(Pr20,0),\"  W\"\n",
+      "print  \"\\n(d)the  total  active  power  taken  from  the  supply  is  \",round(P,1),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is   2.11   A\n",
+        "\n",
+        "(b)  the  p.d.  across  the  5  ohm  resistance  is   5.85   V\n",
+        "\n",
+        "(c)the  active  power  dissipated  in  the  20  ohm  resistance  is   111.0   W\n",
+        "\n",
+        "(d)the  total  active  power  taken  from  the  supply  is   191.9   W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_3.ipynb
new file mode 100755
index 00000000..a2070ae8
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_32-checkpoint_3.ipynb
@@ -0,0 +1,432 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:5db7f11d16032beee814462a087253a6b78ac4c2f805e28898833a5c147fee39"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 32: The superposition theorem</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 564</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  100;#  in  volts\n",
+      "rv2  =  50;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "r1  =  25;#  in  ohm\n",
+      "R  =  20;#  in  ohm\n",
+      "r2  =  10;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  circuit  diagram  is  shown  in  Figure  32.7.  Following  the  above  procedure:\n",
+      " #The  network  is  redrawn  with  the  50/_90\u00b0  V  source  removed  as  shown  in  Figure  32.8\n",
+      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.8.\n",
+      "I1  =  V1/(r1  +  r2*R/(R  +  r2))\n",
+      "I2  =  (r2/(r2  +  R))*I1\n",
+      "I3  =  (R/(r2  +  R))*I1\n",
+      " #The  network  is  redrawn  with  the  100/_0\u00b0  V  source  removed  as  shown  in  Figure  32.9\n",
+      " #Currents  I4,  I5  and  I6  are  labelled  as  shown  in  Figure  32.9.\n",
+      "I4  =  V2/(r2  +  r1*R/(r1  +  R))\n",
+      "I5  =  (r1/(r1  +  R))*I4\n",
+      "I6  =  (R/(r1  +  R))*I4\n",
+      " #Figure  32.10  shows  Figure  32.9  superimposed  on  Figure  32.8,  giving  the  currents  shown.\n",
+      " #Current  in  the  20  ohm  load,\n",
+      "I20  =  I2  +  I5\n",
+      " #Current  in  the  100/_0\u00b0  V  source\n",
+      "IV1  =  I1  -  I6\n",
+      " #Current  in  the  50/_90\u00b0  V  source\n",
+      "IV2  =  I4  -  I3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)current  in  the  20  ohm  load  is  \",round(I20.real,2),\"  +  (\",round(I20.imag,2),\")i  A\"\n",
+      "print  \"\\n  (b)Current  in  the  100/_0deg  V  source  is  \",round(IV1.real,2),\"  +  (\",round(IV1.imag,2),\")i  A\"\n",
+      "print  \"\\n  (b)Current  in  the  50/_90deg  V  source  is  \",round(IV2.real,2),\"  +  (\",round(IV2.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)current  in  the  20  ohm  load  is   1.05   +  ( 1.32 )i  A\n",
+        "\n",
+        "  (b)Current  in  the  100/_0deg  V  source  is   3.16   +  ( -1.05 )i  A\n",
+        "\n",
+        "  (b)Current  in  the  50/_90deg  V  source  is   -2.11   +  ( 2.37 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 566</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  12;#  in  volts\n",
+      "V2  =  20;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  2.5;#  in  ohm\n",
+      "R4  =  6;#  in  ohm\n",
+      "R5  =  2;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #Removing  the  20  V  source  gives  the  network  shown  in  Figure  32.12.\n",
+      " #Currents  I1  and  I2  are  shown  labelled  in  Figure  32.12\n",
+      "Re1  =  (R4*R5/(R4  +  R5))  +  R3\n",
+      "Re2  =  Re1*R2/(Re1    +  R2)  +  R1\n",
+      "I1  =  V1/Re2\n",
+      "I2  =  (R2/(Re1  +  R2))*I1\n",
+      " #Removing  the  12  V  source  from  the  original  network  gives  the  network  shown  in  Figure  32.14.\n",
+      " #Currents  I3,  I4  and  I5  are  shown  labelled  in  Figure  32.14.\n",
+      "Re3  =  (R1*R2/(R1  +  R2))  +  R3\n",
+      "Re4  =  Re3*R4/(Re3  +  R4)  +  R5\n",
+      "I3  =  V2/Re4\n",
+      "I4  =  (R4/(Re3  +  R4))*I3\n",
+      "I5  =  (R1/(R1  +  R2))*I4\n",
+      " #Superimposing  Figure  32.14  on  Figure  32.12  shows  that  the  current  flowing  in  the  4  ohm  resistor  is  given  by\n",
+      "Ir4  =  I5  -  I2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncurrent  in  the  4  ohm  resistor  of  the  network  is  \",round(Ir4,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "current  in  the  4  ohm  resistor  of  the  network  is   0.48   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 567</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  30;#  in  volts\n",
+      "rv2  =  30;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  -45;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  1j*3;#  in  ohm\n",
+      "R4  =  -1j*10;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  network  is  redrawn  with  V2  removed,  as  shown  in  Figure  32.17.\n",
+      " #Current  I1  and  I2  are  shown  in  Figure  32.17.  From  Figure  32.17,\n",
+      "Re1  =  R4*(R2  +  R3)/(R4  +  R3  +  R2)\n",
+      "Re2  =  Re1  +  R1\n",
+      " #current\n",
+      "I1  =  V1/Re2\n",
+      "I2  =  (R4/(R2  +  R3  +  R4))*I1\n",
+      " #The  original  network  is  redrawn  with  V1  removed,  as  shown  in  Figure  32.18\n",
+      " #Currents  I3  and  I4  are  shown  in  Figure  32.18.  From  Figure  32.18,\n",
+      "Re3  =  R1*(R2  +  R3)/(R1  +  R3  +  R2)\n",
+      "Re4  =  Re3  +  R4\n",
+      "I3  =  V2/Re4\n",
+      "I4  =  (R1/(R2  +  R3  +  R1))*I3\n",
+      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
+      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
+      "Ir4i3  =  I2  -  I4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"current  in  (4  +  i3)  ohm  impedance  of  the  network  is  \",round(Ir4i3.real,2),\"  +  (\",round(  Ir4i3.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "current  in  (4  +  i3)  ohm  impedance  of  the  network  is   2.15   +  ( 0.42 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 568</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  5  +  0j;#  in  volts\n",
+      "E2  =  2  +  4j;#  in  volts\n",
+      "Z1  =  3  +  4j;#  in  ohm\n",
+      "Z2  =  2  -  5j;#  in  ohm\n",
+      "Z3  =  6  +  8j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #The  original  network  is  redrawn  with  E2  removed,  as  shown  in  Figure  32.20.\n",
+      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.20.\n",
+      "Ze1  =  Z3*Z2/(Z3  +  Z2)\n",
+      "Ze2  =  Ze1  +  Z1\n",
+      " #current\n",
+      "I1  =  E1/Ze2\n",
+      "I2  =  (Z2/(Z3  +  Z2))*I1\n",
+      "I3  =  (Z3/(Z3  +  Z2))*I1\n",
+      " #The  original  network  is  redrawn  with  E1  removed,  as  shown  in  Figure  32.22\n",
+      " #Currents  I4,  I5  and  I6  are  shown  labelled  in  Figure  32.22  \n",
+      "    #with  I4  flowing  away  from  the  positive  terminal  of  the  E2  source.\n",
+      "Ze3  =  Z3*Z1/(Z3  +  Z1)\n",
+      "Ze4  =  Ze3  +  Z2\n",
+      "I4  =  E2/Ze4\n",
+      "I5  =  (Z1/(Z3  +  Z1))*I4\n",
+      "I6  =  (Z3/(Z3  +  Z1))*I4\n",
+      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
+      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
+      "i1  =  I1  +  I6\n",
+      "i2  =  I3  +  I4\n",
+      "i3  =  I2  -  I5\n",
+      " #magnitude\n",
+      "i1mag  =  abs(i1)\n",
+      "i2mag  =  abs(i2)\n",
+      "E1mag  =  abs(E1)\n",
+      "E2mag  =  abs(E2)\n",
+      " #phase\n",
+      "phi1  =  cmath.phase(complex(i1.real,i1.imag))\n",
+      "phi2  =  cmath.phase(complex(i2.real,i2.imag))\n",
+      " #voltage  across  the(6  +  i8)  ohm  impedance\n",
+      "V6i8  =  i3*Z3\n",
+      "V6i8m  =  abs(V6i8)\n",
+      " #power\n",
+      "P  =  (E1mag*i1mag*math.cos(phi1))  +  (E2mag*i2mag*math.cos(phi2  -  cmath.phase(complex(E2.real,E2.imag))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)currents  are: \\n \",round(i1.real,2),\"  +  (\",round(  i1.imag,2),\")i  A, \\n \",round(i2.real,2),\"  +  (\",round(i2.imag,2),\")i  A \\n and  \",round(i3.real,2),\"  +  (\",round(i3.imag,2),\")i  A\"\n",
+      "print  \"\\n(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is  \",round(V6i8m,2),\"  V\"\n",
+      "print  \"\\n(c)the  total  active  power  delivered  to  the  network  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)currents  are: \n",
+        "  0.57   +  ( 0.62 )i  A, \n",
+        "  0.56   +  ( 1.33 )i  A \n",
+        " and   0.01   +  ( -0.71 )i  A\n",
+        "\n",
+        "(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is   7.09   V\n",
+        "\n",
+        "(c)the  total  active  power  delivered  to  the  network  is   9.29   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 571</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  50;#  in  volts\n",
+      "rv2  =  30;#  in  volts\n",
+      "thetav1  =  0;#  in  degrees\n",
+      "thetav2  =  90;#  in  degrees\n",
+      "R1  =  20;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  -1j*3;#  in  ohm\n",
+      "R4  =  8;#  in  ohm\n",
+      "R5  =  8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #The  network  is  redrawn  with  the  V2  source  removed,  as  shown  in  Figure  32.26.\n",
+      " #Currents  I1  to  I5  are  shown  labelled  in  Figure  32.26.  \n",
+      " #current\n",
+      "Re1  =  R4*R5/(R5  +  R4)  +  R3\n",
+      "Re2  =  Re1*R2/(R2  +  Re1)\n",
+      "I1  =  V1/(Re2  +  R1)\n",
+      "I2  =  (Re1/(R2  +  Re1))*I1\n",
+      "I3  =  (R2/(Re1  +  R2))*I1\n",
+      "I4  =  (R4/(R4  +  R5))*I3\n",
+      "I5  =  I3  -  I4\n",
+      " #The  original  network  is  redrawn  with  the  V1  source  removed,  as  shown  in  Figure  32.27.\n",
+      " #Currents  I6  to  I10  are  shown  labelled  in  Figure  32.27\n",
+      "Re3  =  R1*R2/(R1  +  R2)\n",
+      "Re4  =  Re3  +  R3\n",
+      "Re5  =  Re4*R4/(Re4  +  R4)\n",
+      "Re6  =  Re5    +  R5\n",
+      "I6  =  V2/Re6\n",
+      "I7  =  (Re4/(Re4  +  R4))*I6\n",
+      "I8  =  (R4/(Re4  +  R4))*I6\n",
+      "I9  =  (R1/(R1  +  R2))*I8\n",
+      "I10  =  (R2/(R1  +  R2))*I8\n",
+      " #current  flowing  in  the  capacitor  is  given  by\n",
+      "Ic  =  I3  -  I8\n",
+      " #magnitude  of  the  current  in  the  capacitor\n",
+      "Icmag  =  abs(Ic)\n",
+      "\n",
+      "i1  =  I2  +  I9\n",
+      "i1mag  =  abs(i1)\n",
+      " #magnitude  of  the  p.d.  across  the  5  ohm  resistance  is  given  by\n",
+      "Vr5m  =  i1mag*R2\n",
+      " #Active  power  dissipated  in  the  20  ohm  resistance  is  given  by\n",
+      "i2  =  I1  -  I10\n",
+      "i2mag  =  abs(i2)\n",
+      "phii2  =  cmath.phase(complex(i2.real,i2.imag))\n",
+      "Pr20  =  R1*(i2mag)**2\n",
+      " #Active  power  developed  by  the  V1\n",
+      "P1  =  rv1*i2mag*math.cos(phii2)\n",
+      " #Active  power  developed  by  V2  source\n",
+      "i3  =  I6  -  I5\n",
+      "i3mag  =  abs(i3)\n",
+      "phii3  =  cmath.phase(complex(i3.real,i3.imag))\n",
+      "P2  =  rv2*i3mag*math.cos(phii3  -  (thetav2*math.pi/180))\n",
+      " #Total  power  developed\n",
+      "P  =  P1  +  P2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
+      "print  \"\\n(b)  the  p.d.  across  the  5  ohm  resistance  is  \",round(Vr5m,2),\"  V\"\n",
+      "print  \"\\n(c)the  active  power  dissipated  in  the  20  ohm  resistance  is  \",round(Pr20,0),\"  W\"\n",
+      "print  \"\\n(d)the  total  active  power  taken  from  the  supply  is  \",round(P,1),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is   2.11   A\n",
+        "\n",
+        "(b)  the  p.d.  across  the  5  ohm  resistance  is   5.85   V\n",
+        "\n",
+        "(c)the  active  power  dissipated  in  the  20  ohm  resistance  is   111.0   W\n",
+        "\n",
+        "(d)the  total  active  power  taken  from  the  supply  is   191.9   W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint.ipynb
new file mode 100755
index 00000000..55e4e2b7
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint.ipynb
@@ -0,0 +1,1103 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 33: Thevenin\u2019s and Norton\u2019s theorems</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 578</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current flowing in the capacitor, and (b) the p.d. across the 150 k\u0006ohmresistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  200;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  5000;#  in  ohm\n",
+      "R2  =  20000;#  in  ohm\n",
+      "R3  =  -1j*120000;#  in  ohm\n",
+      "R4  =  150000;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Initially  the  (150-i120)kohm\u0006  impedance  is  removed  from  the    circuit  as  shown  in  Figure  33.13.\n",
+      " #Note  that,  to  find  the  current  in  the  capacitor,  \n",
+      "    #only  the  capacitor  need  have  been  initially  removed  from  the  circuit.  \n",
+      "    #However,  removing  each  of  the  components  from  the  branch  through  \n",
+      "    #which  the  current  is  required  will  often  result  in  a  simpler  solution.  \n",
+      " #From  Figure  33.13,\n",
+      " #current,  I1  \n",
+      "I1  =  V/(R1  +  R2)\n",
+      " #The  open-circuit  e.m.f.  E  is  equal  to  the  p.d.  across  the  20  k\u0006ohm  resistor,  i.e.\n",
+      "E  =  I1*R2\n",
+      " #Removing  the  V1  source  gives  the  network  shown  in  Figure  33.14.\n",
+      " #The  impedance,  z,  looking  in at  the  open-circuited  terminals  is  given  by\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.15,  where  current  iL  is  given  by\n",
+      "ZL  =  R3  +  R4\n",
+      "IL  =  E/(ZL  +  z)\n",
+      "ILmag  =  abs(IL)\n",
+      " #current  flowing  in  the  capacitor\n",
+      "Ic  =  ILmag\n",
+      " #P.d.  across  the  150  kohm  resistor,\n",
+      "Vr150  =  ILmag*R4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  current  flowing  in  the  capacitor  is  \",round(Ic*1000,2),\" mA\"\n",
+      "print  \"\\n(b)  the  p.d.  across  the  150  ohm  resistance  is  \",round(Vr150,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  current  flowing  in  the  capacitor  is   0.82  mA\n",
+        "\n",
+        "(b)  the  p.d.  across  the  150  ohm  resistance  is   122.93   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 579</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of current I.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  20;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  1.5;#  in  ohm\n",
+      "L  =  235E-6;#  in  Henry\n",
+      "R4  =  3;#  in  ohm\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #The  impedance  through  which  current  I  is  flowing  is  initially  removed  from  the  network,  as  shown  in  Figure  33.17.\n",
+      " #From  Figure  33.17,\n",
+      " #current,  I1  \n",
+      "I1  =  (V1  -  V2)/(R1  +  R4)\n",
+      " #the  open  circuit  e.m.f.  E\n",
+      "E  =  V1  -  I1*R1\n",
+      " #When  the  sources  of  e.m.f.  are  removed  from  the  circuit,  \n",
+      "    #the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
+      "z  =  R1*R4/(R1  +  R4)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.18,  where  inductive  reactance,\n",
+      "XL  =  2*math.pi*f*L\n",
+      "R3  =  1j*XL\n",
+      " #Hence  current\n",
+      "I  =  E/(R2  +  R3  +  z)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  I  is  \",round(I.real,2),\"  +  (\",round(I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  I  is   2.7   +  ( -2.95 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 580</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power dissipated in the 48 ohm resistor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  -1j*400;#  in  ohm\n",
+      "R2  =  300;#  in  ohm\n",
+      "R3  =  144j;#  in  ohm\n",
+      "R4  =  48;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  R3  and  R4  impedance  is  initially  removed  from  the  network  as  shown  in  Figure  33.20.\n",
+      " #From  Figure  33.20,\n",
+      " #current,  I\n",
+      "i  =  V/(R1  +  R2)\n",
+      " #the  open  circuit  e.m.f.  E\n",
+      "E  =  i*R2\n",
+      " #When  the  V  is  removed  from  the  circuit,  the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.21  connected  to  R#  and  R4,\n",
+      " #Hence  current\n",
+      "I  =  E/(R4  +  R3  +  z)\n",
+      "Imag  =  abs(I)\n",
+      " #the  power  dissipated  in  the  48  ohm  resistor\n",
+      "Pr48  =  R4*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  48  ohm  resistor  is  \",round(Pr48,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  48  ohm  resistor  is   0.75   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 581</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the 80 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  46;#  in  ohm\n",
+      "R4  =  50;#  in  ohm\n",
+      "R5  =  15;#  in  ohm\n",
+      "R6  =  60;#  in  ohm\n",
+      "R7  =  16;#  in  ohm\n",
+      "R8  =  80;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #One  method  of  analysing  a  multi-branch  network  as  shown  in  Figure  33.22 \n",
+      "    #is  to  use  Th\u00b4evenin\u2019s  theorem  on  one  part  of  the  network  at  a  time.  \n",
+      "    #For  example,  the  part  of  the  circuit  to  the  left  of  AA  may  be  reduced  to  a  Th\u00b4evenin  equivalent  circuit.\n",
+      " #From  Figure  33.23,\n",
+      "E1  =  (R2/(R1  +  R2))*V\n",
+      "z1  =  R1*R2/(R1  +  R2)\n",
+      " #Thus  the  network  of  Figure  33.22  reduces  to  that  of  Figure  33.24.  \n",
+      "    #The  part  of  the  network  shown  in  Figure  33.24  to  the  left  of  BB  may  be  reduced \n",
+      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
+      "E2  =  (R4/(R3  +  R4  +  z1))*E1\n",
+      "z2  =  R4*(z1  +  R3)/(R4  +  z1  +  R3)\n",
+      " #Thus  the  original  network  reduces  to  that  shown  in  Figure  33.25.  \n",
+      "    #The  part  of  the  network  shown  in  Figure  33.25  to  the  left  of  CC  may  be  reduced  \n",
+      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
+      "E3  =  (R6/(R5  +  R6  +  z2))*E2\n",
+      "z3  =  R6*(z2  +  R5)/(R5  +  z2  +  R6)\n",
+      " #Thus  the  original  network  reduces  to  that  of  Figure  33.26, \n",
+      "    #from  which  the  current  in  the  80  ohm  resistor  is  given  by\n",
+      "I  =  E3/(z3  +  R7  +  R8)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  in  the  80  ohm  resistor  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  in  the  80  ohm  resistor  is   0.2   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 582</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the magnitude of the current flowing in a \t(3.75 + j11) ohmimpedance connected across terminals AB, and\n",
+      "#(b) the magnitude of the p.d. across the (3.75 + j11) ohmimpedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  24;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  -1j*3;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  I1  shown  in  Figure  33.27  is  given  by\n",
+      "I1  =  V/(R1  +  R2  +  R3)\n",
+      " #The  Th\u00b4evenin  equivalent  voltage,  i.e.,  the  open-circuit  voltage  across  terminals  AB,  is  given  by\n",
+      "E  =  I1*(R2  +  R3)\n",
+      " #When  the  voltage  source  is  removed,  the  impedance  z  \u2018looking  in\u2019  at  AB  is  given  by\n",
+      "z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)\n",
+      " #Thus  the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.28.\n",
+      " #when  (3.75  +  i11)  ohm  impedance  connected  across  terminals  AB, \n",
+      "    #the  current  I  flowing  in  the  impedance  is  given  by\n",
+      "R  =  3.75  +  11j;#  in  ohms\n",
+      "I  =  E/(R  +  z)\n",
+      "Imag  =  abs(I)\n",
+      " #the  p.d.  across  the(  3.75  +  i11)ohm  impedance.\n",
+      "VR  =  I*R\n",
+      "VRmag  =  abs(VR)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is  \",round(Imag,2),\"  A\"\n",
+      "print  \"\\n  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is  \",round(VRmag,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is   3.0   A\n",
+        "\n",
+        "  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is   34.86   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 583</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the capacitor of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  16.55;#  in  volts\n",
+      "thetav  =  -22.62;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  1j*2;#  in  ohm\n",
+      "R3  =  1j*6;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  -1*1j*8;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  capacitor  is  removed  from  branch  AB,  as  shown  in  Figure  33.30.\n",
+      " #Impedance,  Z\n",
+      "Z1  =  R3  +  R4  +  R5\n",
+      "Z  =  R1  +  (Z1*R2/(R2  +  Z1))\n",
+      "I1  =  V/Z\n",
+      "I2  =  (R2/(R2  +Z1))*I1\n",
+      " #The  open-circuit  voltage,  E\n",
+      "E  =  I2*R5\n",
+      " #If  the  voltage  source  is  removed  from  Figure  33.30,  the  impedance,  z,  \u2018looking  in\u2019  at  AB  is  given  by\n",
+      "z  =  R5*((R1*R2/(R1  +  R2))  +  R3  +  R4)/(R5  +  ((R1*R2/(R1  +  R2))  +  R3  +  R4))\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.31,  \n",
+      "    #where  the  current  flowing  in  the  capacitor,  I,  is  given  by\n",
+      "I  =  E/(z  +  R6)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  in  the  capacitor  of  the  network  is  \",round(Imag,2),\"    A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  in  the  capacitor  of  the  network  is   0.43     A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 584</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power dissipated by a 2 ohm resistor connected across PQ.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  5;#  in  volts\n",
+      "rv2  =  10;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R1  =  8;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Current  I1  shown  in  Figure  33.32  is  given  by\n",
+      "I1  =  V2/(R2  +  R3  +  R4)\n",
+      " #Hence  the  voltage  drop  across  the  5  ohm\u0006  resistor  is  given  by  VX  is  in  the  direction  shown  in  Figure  33.32,\n",
+      "Vx  =  I1*R2\n",
+      " #The  open-circuit  voltage  E  across  PQ  is  the  phasor  sum  of  V1,  Vx  and  V2,  as  shown  in  Figure  33.33.\n",
+      "E  =  V2  -  V1  -  Vx\n",
+      " #The  impedance,  z,  \u2018looking  in\u2019  at  terminals  PQ  with  the  voltage  sources  removed  is  given  by\n",
+      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.34  with  the  2  ohm  resistance  connected  across  terminals  PQ.\n",
+      " #The  current  flowing  in  the  2  ohm\u0006  resistance  is  given  by\n",
+      "R  =  2;#  in  ohms\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      " #power  P  dissipated  in  the  2  ohm\u0006  resistor  is  given  by\n",
+      "Pr2  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  P  dissipated  in  the  2  ohm  resistor  is  \",round(Pr2,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  P  dissipated  in  the  2  ohm  resistor  is   0.07   W"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 585</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the capacitor, and its direction\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  30;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  15;#  in  ohm\n",
+      "R2  =  40;#  in  ohm\n",
+      "R3  =  20j;#  in  ohm\n",
+      "R4  =  20;#  in  ohm\n",
+      "R5  =  5j;#  in  ohm\n",
+      "R6  =  5;#  in  ohm\n",
+      "R7  =  -1j*25;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  R7\u0006  is  initially  removed  from  the  network,  as  shown  in  Figure  33.36\n",
+      "Z1  =  R1\n",
+      "Z2  =  R2\n",
+      "Z3  =  R3  +  R4\n",
+      "Z4  =  R5  +  R6\n",
+      " #P.d.  between  A  and  C,\n",
+      "Vac  =  (Z1/(Z1  +  Z4))*V\n",
+      " #P.d.  between  B  and  C,\n",
+      "Vbc  =  (Z2/(Z2  +  Z3))*V\n",
+      " #Assuming  that  point  A  is  at  a  higher  potential  than  point  B,  then  the  p.d.  between  A  and  B  is\n",
+      "Vab  =  Vac  -  Vbc\n",
+      " #the  open-circuit  voltage  across  AB  is  given  by\n",
+      "E  =  Vab\n",
+      " #Point  C  is  at  a  potential  of  V  .  Between  C  and  A  is  a  volt  drop  of  Vac.  Hence  the  voltage  at  point  A  is\n",
+      "Va  =  V  -  Vac\n",
+      " #Between  points  C  and  B  is  a  voltage  drop  of  Vbc.  Hence  the  voltage  at  point  B\n",
+      "Vb  =  V  -  Vbc\n",
+      " #Replacing  the  V  source  with  a  short-circuit  (i.e.,  zero  internal  impedance) \n",
+      "    #gives  the  network  shown  in  Figure  33.37(a).  The  network  is  shown  redrawn  in  Figure  33.37(b)  \n",
+      "    #and  simplified  in  Figure  33.37(c).  Hence  the  impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
+      "z  =  Z1*Z4/(Z1  +  Z4)  +  Z2*Z3/(Z2  +  Z3)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.38,  where  current  I  is  given  by\n",
+      "I  =  E/(z  +  R7)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(Imag,2),\"  A  in  direction  from  B  to  A.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  capacitor  is   0.13   A  in  direction  from  B  to  A."
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 589</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of current I in the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  5;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  3;#  in  ohm\n",
+      "R3  =  -1j*3;#  in  ohm\n",
+      "R4  =  2.8;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #The  branch  containing  the  R4  is  short-circuited,  as  shown  in  Figure  33.48.\n",
+      " #The  R2  in  parallel  with  a  short-circuit  is  the  same  as  R2  in  parallel  with  0  ohm \n",
+      "    #giving  an  equivalent  impedance  of\n",
+      "Z1  =  R2*0/(R3  +  0)\n",
+      " #Hence  the  network  reduces  to  that  shown  in  Figure  33.49,  where\n",
+      "Isc  =  V/R1\n",
+      " #If  the  Voltage  source  is  removed  from  the  network  the  input  impedance,  z,  \u2018looking-in\u2019  \n",
+      "    #at  a  break  made  in  AB  of  Figure  33.48  gives\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.51,  where  current  I  is  given  by\n",
+      "I  =  (z/(z  +  R4  +  R3))*Isc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(I.real,2),\"  +  (\", round(I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  capacitor  is   0.48   +  ( 0.36 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 589</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  20;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  1.5;#  in  ohm\n",
+      "R3  =  2.95j;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #The  inductive  branch  is  initially  short-circuited,  as  shown  in  Figure  33.53.\n",
+      " #From  Figure  33.53,\n",
+      "I1  =  V1/R1\n",
+      "I2  =  V2/R4\n",
+      "Isc  =  I1  +  I2\n",
+      " #If  the  voltage  sources  are  removed,  the  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  in  AB  is  given  by\n",
+      "z  =  R1*R4/(R1  +  R4)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.54,  where  current  I  is  given  by\n",
+      "I  =  (z/(z  +  R2  +  R3))*Isc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  inductive  branch  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  inductive  branch  is   2.7   +  ( -2.95 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 590</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  -2j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R3\n",
+      "z = 1/(1/R2 + 1/R3 + 1/R4)\n",
+      "I = (z/(R1 + z))*Isc\n",
+      "pd1 = abs(I)*R1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the magnitude of the p.d. across the 1 ohm resistor is \",round(pd1, 2),\"V\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the magnitude of the p.d. across the 1 ohm resistor is  1.58 V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 591</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power dissipated in a 5 ohm resistor connected between A and B.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  20;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  -1j*3;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Terminals  AB  are  initially  short-circuited,  as  shown  in  Figure  33.61.\n",
+      " #The  circuit  impedance  Z  presented  to  the  voltage  source  is  given  by\n",
+      "Z  =  R1  +  R4*(R2  +  R3)/(R2  +  R3  +  R4)\n",
+      " #Thus  current  I  in  Figure  33.61  is  given  by\n",
+      "I  =  V/Z\n",
+      "Isc  =  ((R2  +  R3)/(R2  +  R3  +  R4))*I\n",
+      " #Removing  the  voltage  source  of  Figure  33.60  gives  the  network  Figure  33.62  of  Figure  33.62. \n",
+      "    #Impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
+      "z  =  R4  +  R1*(R2  +  R3)/(R2  +  R3  +  R1)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.63.\n",
+      "R  =  5;#  in  ohms\n",
+      " #Current  IL\n",
+      "IL  =  (z/(z  +  R))*Isc\n",
+      "ILmag  =  abs(IL)\n",
+      " #the  power  dissipated  in  the  5  ohm  resistor  is\n",
+      "Pr5  =  R*ILmag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  5  ohm  resistor  is  \",round(Pr5,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  5  ohm  resistor  is   22.54   W"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 592</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the current flowing in a 2 ohm resistor connected across PQ.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  5;#  in  volts\n",
+      "rv2  =  10;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R1  =  8;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Terminals  PQ  are  initially  short-circuited,  as  shown  in  Figure  33.65.\n",
+      " #Currents  I1  and  I2  are  shown  labelled.  Kirchhoff\u2019s  laws  are  used.\n",
+      " #For  loop  ABCD,  and  moving  anticlockwise,\n",
+      " #I1*(R2  +  R3  +  R4)  +  I2*(R3  +  R4)  =  V2\n",
+      " #For  loop  DPQC,  and  moving  clockwise,\n",
+      " #R2*I1  -  R1*I2  =  V2  -  V1\n",
+      " #Solving  Equations  by  using  determinants  gives\n",
+      "d1  =  [[V2,  (R3  +  R4)],[(V2  -  V1),  -1*R1]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R2  +  R3  +  R4),  V2],[R2,  (V2  -  V1)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(R2  +  R3  +  R4),  (R3  +  R4)],[R2,  -1*R1]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      " #the  short-circuit  current  Isc\n",
+      "Isc  =  I2\n",
+      " #The  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  between  P  and  Q  is  given  by\n",
+      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
+      " #The  Norton  equivalent  circuit  is  shown  in  Figure  33.66,  where  current  I  is  given  by\n",
+      "R  =  2;#in  ohm\n",
+      "I  =  (z/(z  +  R))*Isc\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  current  flowing  5  ohm  resistor  is  \",round(Imag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  current  flowing  5  ohm  resistor  is   0.19   A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 595</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnitude of the current flowing in the \t(1.8 + j4) ohm impedance connected between terminals A and B\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  12;#  in  volts\n",
+      "E2  =  24;#  in  volts\n",
+      "Z1  =  3;#  in  ohm\n",
+      "Z2  =  2;#  in  ohm\n",
+      "R1  =  4j;#  in  ohm\n",
+      "R2  =  1.8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Z3  =  R1  +  R2\n",
+      " #For  the  branch  containing  the  E1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
+      "Isc1  =  E1/Z1\n",
+      " #For  the  branch  containing  the  E2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
+      "Isc2  =  E2/Z2\n",
+      " #Thus  Figure  33.73  shows  a  network  equivalent  to  Figure  33.72.  From  Figure  33.73,  \n",
+      "    #the  total  short-circuit  current\n",
+      "Isc  =  Isc1  +  Isc2\n",
+      " #the  total  impedance  is  given  by\n",
+      "z  =  Z1*Z2/(Z1  +  Z2)\n",
+      " #Thus  Figure  33.73  simplifies  to  Figure  33.74.\n",
+      " #The  open-circuit  voltage  across  AB  of  Figure  33.74,  E\n",
+      "E  =  Isc*z\n",
+      " #the  impedance  \u2018looking  in\u2019  at  AB,is  z\n",
+      " #the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.75.\n",
+      "R  =  1.8  +  4j;#  in  ohm\n",
+      " #when  R  impedance  is  connected  to  terminals  AB  of  Figure  33.75,  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is  \",round(Imag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is   3.84   A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 596</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the current flowing in the capacitive branch connected to terminals AB.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  5;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "i  =  0.001;#  in  Amperes\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  4000;#  in  ohm\n",
+      "R3  =  2000;#  in  ohm\n",
+      "R4  =  200;#  in  ohm\n",
+      "R5  =  -1j*4000;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #For  the  branch  containing  the  V1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
+      "Isc1  =  V1/R1\n",
+      "z1  =  R1\n",
+      " #For  the  branch  containing  the  V2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
+      "Isc2  =  V2/R2\n",
+      "z2  =  R2\n",
+      " #Thus  the  circuit  of  Figure  33.76  converts  to  that  of  Figure  33.77.\n",
+      " #The  above  two  Norton  equivalent  networks  shown  in  Figure  33.77  may  be  combined,  \n",
+      "    #since  the  total  short-circuit  current  is\n",
+      "Isc  =  Isc1  +  Isc2\n",
+      " #the  total  impedance  is  given  by\n",
+      "Z1  =  z1*z2/(z1  +  z2)\n",
+      " #Both  of  the  Norton  equivalent  networks  shown  in  Figure  33.78  may  be  converted  to  Th\u00b4evenin  equivalent  circuits. \n",
+      "    #Open-circuit  voltage  across  CD  is\n",
+      "Ecd  =  Isc*Z1\n",
+      " #the  impedance  \u2018looking  in\u2019  at  CD  is  Z1\n",
+      " #Open-circuit  voltage  across  EF\n",
+      "Eef  =  i*R3\n",
+      " #the  impedance  \u2018looking  in\u2019  Figure  33.79  at  EF\n",
+      "Z2  =  R3\n",
+      " #Thus  Figure  33.78  converts  to  Figure  33.79.\n",
+      " #Combining  the  two  Th\u00b4evenin  circuits  gives  e.m.f.\n",
+      "E  =  Ecd  -  Eef\n",
+      " #impedance  z\n",
+      "z  =  Z1  +  Z2\n",
+      " #the  Th\u00b4evenin  equivalent  circuit  for  terminals  AB  of  Figure  33.76  is  as  shown  in  Figure  33.80.\n",
+      "Z3  =  R4  +  R5\n",
+      " #If  an  impedance  Z3  is  connected  across  terminals  AB,  then  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  Z3)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  in  the  capacitive  branch  is  \",  Imag*1000,\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  in  the  capacitive  branch  is   0.8 mA"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 597</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated in a \t(600 - j800) ohm impedance connected between A and B\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  5;#  in  volts\n",
+      "i  =  0.004;#  in  Amperes\n",
+      "R1  =  2000;#  in  ohm\n",
+      "R2  =  1000j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #Converting  the  Th\u00b4evenin  circuit  to  a  Norton  network  gives\n",
+      "Isc1  =  V/R2\n",
+      " #Thus  Figure  33.81  converts  to  that  shown  in  Figure  33.82.  \n",
+      "    #The  two  Norton  equivalent  networks  may  be  combined,  giving\n",
+      "Isc  =  Isc1  +  i\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #This  results  in  the  equivalent  network  shown  in  Figure  33.83. \n",
+      "    #Converting  to  an  equivalent  Th\u00b4evenin  circuit  gives  open  circuit  e.m.f.  across  AB,\n",
+      "E  =  Isc*z\n",
+      " #Thus  the  The\u00b4venin  equivalent  circuit  is  as  shown  in  Figure  33.84.\n",
+      "R  =  600  -  800j;#  in  ohms\n",
+      " #When  a  R  impedance  is  connected  across  AB,  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      " #the  power  dissipated  in  the  R  resistor  is\n",
+      "PR  =  R.real*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is  \",round(PR*1000,2),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is   19.68 mW"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_1.ipynb
new file mode 100755
index 00000000..55e4e2b7
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_1.ipynb
@@ -0,0 +1,1103 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 33: Thevenin\u2019s and Norton\u2019s theorems</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 578</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current flowing in the capacitor, and (b) the p.d. across the 150 k\u0006ohmresistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  200;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  5000;#  in  ohm\n",
+      "R2  =  20000;#  in  ohm\n",
+      "R3  =  -1j*120000;#  in  ohm\n",
+      "R4  =  150000;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Initially  the  (150-i120)kohm\u0006  impedance  is  removed  from  the    circuit  as  shown  in  Figure  33.13.\n",
+      " #Note  that,  to  find  the  current  in  the  capacitor,  \n",
+      "    #only  the  capacitor  need  have  been  initially  removed  from  the  circuit.  \n",
+      "    #However,  removing  each  of  the  components  from  the  branch  through  \n",
+      "    #which  the  current  is  required  will  often  result  in  a  simpler  solution.  \n",
+      " #From  Figure  33.13,\n",
+      " #current,  I1  \n",
+      "I1  =  V/(R1  +  R2)\n",
+      " #The  open-circuit  e.m.f.  E  is  equal  to  the  p.d.  across  the  20  k\u0006ohm  resistor,  i.e.\n",
+      "E  =  I1*R2\n",
+      " #Removing  the  V1  source  gives  the  network  shown  in  Figure  33.14.\n",
+      " #The  impedance,  z,  looking  in at  the  open-circuited  terminals  is  given  by\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.15,  where  current  iL  is  given  by\n",
+      "ZL  =  R3  +  R4\n",
+      "IL  =  E/(ZL  +  z)\n",
+      "ILmag  =  abs(IL)\n",
+      " #current  flowing  in  the  capacitor\n",
+      "Ic  =  ILmag\n",
+      " #P.d.  across  the  150  kohm  resistor,\n",
+      "Vr150  =  ILmag*R4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  current  flowing  in  the  capacitor  is  \",round(Ic*1000,2),\" mA\"\n",
+      "print  \"\\n(b)  the  p.d.  across  the  150  ohm  resistance  is  \",round(Vr150,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  current  flowing  in  the  capacitor  is   0.82  mA\n",
+        "\n",
+        "(b)  the  p.d.  across  the  150  ohm  resistance  is   122.93   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 579</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of current I.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  20;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  1.5;#  in  ohm\n",
+      "L  =  235E-6;#  in  Henry\n",
+      "R4  =  3;#  in  ohm\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #The  impedance  through  which  current  I  is  flowing  is  initially  removed  from  the  network,  as  shown  in  Figure  33.17.\n",
+      " #From  Figure  33.17,\n",
+      " #current,  I1  \n",
+      "I1  =  (V1  -  V2)/(R1  +  R4)\n",
+      " #the  open  circuit  e.m.f.  E\n",
+      "E  =  V1  -  I1*R1\n",
+      " #When  the  sources  of  e.m.f.  are  removed  from  the  circuit,  \n",
+      "    #the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
+      "z  =  R1*R4/(R1  +  R4)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.18,  where  inductive  reactance,\n",
+      "XL  =  2*math.pi*f*L\n",
+      "R3  =  1j*XL\n",
+      " #Hence  current\n",
+      "I  =  E/(R2  +  R3  +  z)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  I  is  \",round(I.real,2),\"  +  (\",round(I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  I  is   2.7   +  ( -2.95 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 580</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power dissipated in the 48 ohm resistor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  -1j*400;#  in  ohm\n",
+      "R2  =  300;#  in  ohm\n",
+      "R3  =  144j;#  in  ohm\n",
+      "R4  =  48;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  R3  and  R4  impedance  is  initially  removed  from  the  network  as  shown  in  Figure  33.20.\n",
+      " #From  Figure  33.20,\n",
+      " #current,  I\n",
+      "i  =  V/(R1  +  R2)\n",
+      " #the  open  circuit  e.m.f.  E\n",
+      "E  =  i*R2\n",
+      " #When  the  V  is  removed  from  the  circuit,  the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.21  connected  to  R#  and  R4,\n",
+      " #Hence  current\n",
+      "I  =  E/(R4  +  R3  +  z)\n",
+      "Imag  =  abs(I)\n",
+      " #the  power  dissipated  in  the  48  ohm  resistor\n",
+      "Pr48  =  R4*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  48  ohm  resistor  is  \",round(Pr48,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  48  ohm  resistor  is   0.75   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 581</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the 80 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  46;#  in  ohm\n",
+      "R4  =  50;#  in  ohm\n",
+      "R5  =  15;#  in  ohm\n",
+      "R6  =  60;#  in  ohm\n",
+      "R7  =  16;#  in  ohm\n",
+      "R8  =  80;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #One  method  of  analysing  a  multi-branch  network  as  shown  in  Figure  33.22 \n",
+      "    #is  to  use  Th\u00b4evenin\u2019s  theorem  on  one  part  of  the  network  at  a  time.  \n",
+      "    #For  example,  the  part  of  the  circuit  to  the  left  of  AA  may  be  reduced  to  a  Th\u00b4evenin  equivalent  circuit.\n",
+      " #From  Figure  33.23,\n",
+      "E1  =  (R2/(R1  +  R2))*V\n",
+      "z1  =  R1*R2/(R1  +  R2)\n",
+      " #Thus  the  network  of  Figure  33.22  reduces  to  that  of  Figure  33.24.  \n",
+      "    #The  part  of  the  network  shown  in  Figure  33.24  to  the  left  of  BB  may  be  reduced \n",
+      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
+      "E2  =  (R4/(R3  +  R4  +  z1))*E1\n",
+      "z2  =  R4*(z1  +  R3)/(R4  +  z1  +  R3)\n",
+      " #Thus  the  original  network  reduces  to  that  shown  in  Figure  33.25.  \n",
+      "    #The  part  of  the  network  shown  in  Figure  33.25  to  the  left  of  CC  may  be  reduced  \n",
+      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
+      "E3  =  (R6/(R5  +  R6  +  z2))*E2\n",
+      "z3  =  R6*(z2  +  R5)/(R5  +  z2  +  R6)\n",
+      " #Thus  the  original  network  reduces  to  that  of  Figure  33.26, \n",
+      "    #from  which  the  current  in  the  80  ohm  resistor  is  given  by\n",
+      "I  =  E3/(z3  +  R7  +  R8)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  in  the  80  ohm  resistor  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  in  the  80  ohm  resistor  is   0.2   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 582</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the magnitude of the current flowing in a \t(3.75 + j11) ohmimpedance connected across terminals AB, and\n",
+      "#(b) the magnitude of the p.d. across the (3.75 + j11) ohmimpedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  24;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  -1j*3;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  I1  shown  in  Figure  33.27  is  given  by\n",
+      "I1  =  V/(R1  +  R2  +  R3)\n",
+      " #The  Th\u00b4evenin  equivalent  voltage,  i.e.,  the  open-circuit  voltage  across  terminals  AB,  is  given  by\n",
+      "E  =  I1*(R2  +  R3)\n",
+      " #When  the  voltage  source  is  removed,  the  impedance  z  \u2018looking  in\u2019  at  AB  is  given  by\n",
+      "z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)\n",
+      " #Thus  the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.28.\n",
+      " #when  (3.75  +  i11)  ohm  impedance  connected  across  terminals  AB, \n",
+      "    #the  current  I  flowing  in  the  impedance  is  given  by\n",
+      "R  =  3.75  +  11j;#  in  ohms\n",
+      "I  =  E/(R  +  z)\n",
+      "Imag  =  abs(I)\n",
+      " #the  p.d.  across  the(  3.75  +  i11)ohm  impedance.\n",
+      "VR  =  I*R\n",
+      "VRmag  =  abs(VR)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is  \",round(Imag,2),\"  A\"\n",
+      "print  \"\\n  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is  \",round(VRmag,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is   3.0   A\n",
+        "\n",
+        "  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is   34.86   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 583</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the capacitor of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  16.55;#  in  volts\n",
+      "thetav  =  -22.62;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  1j*2;#  in  ohm\n",
+      "R3  =  1j*6;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  -1*1j*8;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  capacitor  is  removed  from  branch  AB,  as  shown  in  Figure  33.30.\n",
+      " #Impedance,  Z\n",
+      "Z1  =  R3  +  R4  +  R5\n",
+      "Z  =  R1  +  (Z1*R2/(R2  +  Z1))\n",
+      "I1  =  V/Z\n",
+      "I2  =  (R2/(R2  +Z1))*I1\n",
+      " #The  open-circuit  voltage,  E\n",
+      "E  =  I2*R5\n",
+      " #If  the  voltage  source  is  removed  from  Figure  33.30,  the  impedance,  z,  \u2018looking  in\u2019  at  AB  is  given  by\n",
+      "z  =  R5*((R1*R2/(R1  +  R2))  +  R3  +  R4)/(R5  +  ((R1*R2/(R1  +  R2))  +  R3  +  R4))\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.31,  \n",
+      "    #where  the  current  flowing  in  the  capacitor,  I,  is  given  by\n",
+      "I  =  E/(z  +  R6)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  in  the  capacitor  of  the  network  is  \",round(Imag,2),\"    A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  in  the  capacitor  of  the  network  is   0.43     A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 584</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power dissipated by a 2 ohm resistor connected across PQ.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  5;#  in  volts\n",
+      "rv2  =  10;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R1  =  8;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Current  I1  shown  in  Figure  33.32  is  given  by\n",
+      "I1  =  V2/(R2  +  R3  +  R4)\n",
+      " #Hence  the  voltage  drop  across  the  5  ohm\u0006  resistor  is  given  by  VX  is  in  the  direction  shown  in  Figure  33.32,\n",
+      "Vx  =  I1*R2\n",
+      " #The  open-circuit  voltage  E  across  PQ  is  the  phasor  sum  of  V1,  Vx  and  V2,  as  shown  in  Figure  33.33.\n",
+      "E  =  V2  -  V1  -  Vx\n",
+      " #The  impedance,  z,  \u2018looking  in\u2019  at  terminals  PQ  with  the  voltage  sources  removed  is  given  by\n",
+      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.34  with  the  2  ohm  resistance  connected  across  terminals  PQ.\n",
+      " #The  current  flowing  in  the  2  ohm\u0006  resistance  is  given  by\n",
+      "R  =  2;#  in  ohms\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      " #power  P  dissipated  in  the  2  ohm\u0006  resistor  is  given  by\n",
+      "Pr2  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  P  dissipated  in  the  2  ohm  resistor  is  \",round(Pr2,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  P  dissipated  in  the  2  ohm  resistor  is   0.07   W"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 585</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the capacitor, and its direction\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  30;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  15;#  in  ohm\n",
+      "R2  =  40;#  in  ohm\n",
+      "R3  =  20j;#  in  ohm\n",
+      "R4  =  20;#  in  ohm\n",
+      "R5  =  5j;#  in  ohm\n",
+      "R6  =  5;#  in  ohm\n",
+      "R7  =  -1j*25;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  R7\u0006  is  initially  removed  from  the  network,  as  shown  in  Figure  33.36\n",
+      "Z1  =  R1\n",
+      "Z2  =  R2\n",
+      "Z3  =  R3  +  R4\n",
+      "Z4  =  R5  +  R6\n",
+      " #P.d.  between  A  and  C,\n",
+      "Vac  =  (Z1/(Z1  +  Z4))*V\n",
+      " #P.d.  between  B  and  C,\n",
+      "Vbc  =  (Z2/(Z2  +  Z3))*V\n",
+      " #Assuming  that  point  A  is  at  a  higher  potential  than  point  B,  then  the  p.d.  between  A  and  B  is\n",
+      "Vab  =  Vac  -  Vbc\n",
+      " #the  open-circuit  voltage  across  AB  is  given  by\n",
+      "E  =  Vab\n",
+      " #Point  C  is  at  a  potential  of  V  .  Between  C  and  A  is  a  volt  drop  of  Vac.  Hence  the  voltage  at  point  A  is\n",
+      "Va  =  V  -  Vac\n",
+      " #Between  points  C  and  B  is  a  voltage  drop  of  Vbc.  Hence  the  voltage  at  point  B\n",
+      "Vb  =  V  -  Vbc\n",
+      " #Replacing  the  V  source  with  a  short-circuit  (i.e.,  zero  internal  impedance) \n",
+      "    #gives  the  network  shown  in  Figure  33.37(a).  The  network  is  shown  redrawn  in  Figure  33.37(b)  \n",
+      "    #and  simplified  in  Figure  33.37(c).  Hence  the  impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
+      "z  =  Z1*Z4/(Z1  +  Z4)  +  Z2*Z3/(Z2  +  Z3)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.38,  where  current  I  is  given  by\n",
+      "I  =  E/(z  +  R7)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(Imag,2),\"  A  in  direction  from  B  to  A.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  capacitor  is   0.13   A  in  direction  from  B  to  A."
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 589</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of current I in the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  5;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  3;#  in  ohm\n",
+      "R3  =  -1j*3;#  in  ohm\n",
+      "R4  =  2.8;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #The  branch  containing  the  R4  is  short-circuited,  as  shown  in  Figure  33.48.\n",
+      " #The  R2  in  parallel  with  a  short-circuit  is  the  same  as  R2  in  parallel  with  0  ohm \n",
+      "    #giving  an  equivalent  impedance  of\n",
+      "Z1  =  R2*0/(R3  +  0)\n",
+      " #Hence  the  network  reduces  to  that  shown  in  Figure  33.49,  where\n",
+      "Isc  =  V/R1\n",
+      " #If  the  Voltage  source  is  removed  from  the  network  the  input  impedance,  z,  \u2018looking-in\u2019  \n",
+      "    #at  a  break  made  in  AB  of  Figure  33.48  gives\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.51,  where  current  I  is  given  by\n",
+      "I  =  (z/(z  +  R4  +  R3))*Isc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(I.real,2),\"  +  (\", round(I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  capacitor  is   0.48   +  ( 0.36 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 589</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  20;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  1.5;#  in  ohm\n",
+      "R3  =  2.95j;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #The  inductive  branch  is  initially  short-circuited,  as  shown  in  Figure  33.53.\n",
+      " #From  Figure  33.53,\n",
+      "I1  =  V1/R1\n",
+      "I2  =  V2/R4\n",
+      "Isc  =  I1  +  I2\n",
+      " #If  the  voltage  sources  are  removed,  the  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  in  AB  is  given  by\n",
+      "z  =  R1*R4/(R1  +  R4)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.54,  where  current  I  is  given  by\n",
+      "I  =  (z/(z  +  R2  +  R3))*Isc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  inductive  branch  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  inductive  branch  is   2.7   +  ( -2.95 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 590</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  -2j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R3\n",
+      "z = 1/(1/R2 + 1/R3 + 1/R4)\n",
+      "I = (z/(R1 + z))*Isc\n",
+      "pd1 = abs(I)*R1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the magnitude of the p.d. across the 1 ohm resistor is \",round(pd1, 2),\"V\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the magnitude of the p.d. across the 1 ohm resistor is  1.58 V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 591</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power dissipated in a 5 ohm resistor connected between A and B.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  20;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  -1j*3;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Terminals  AB  are  initially  short-circuited,  as  shown  in  Figure  33.61.\n",
+      " #The  circuit  impedance  Z  presented  to  the  voltage  source  is  given  by\n",
+      "Z  =  R1  +  R4*(R2  +  R3)/(R2  +  R3  +  R4)\n",
+      " #Thus  current  I  in  Figure  33.61  is  given  by\n",
+      "I  =  V/Z\n",
+      "Isc  =  ((R2  +  R3)/(R2  +  R3  +  R4))*I\n",
+      " #Removing  the  voltage  source  of  Figure  33.60  gives  the  network  Figure  33.62  of  Figure  33.62. \n",
+      "    #Impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
+      "z  =  R4  +  R1*(R2  +  R3)/(R2  +  R3  +  R1)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.63.\n",
+      "R  =  5;#  in  ohms\n",
+      " #Current  IL\n",
+      "IL  =  (z/(z  +  R))*Isc\n",
+      "ILmag  =  abs(IL)\n",
+      " #the  power  dissipated  in  the  5  ohm  resistor  is\n",
+      "Pr5  =  R*ILmag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  5  ohm  resistor  is  \",round(Pr5,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  5  ohm  resistor  is   22.54   W"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 592</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the current flowing in a 2 ohm resistor connected across PQ.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  5;#  in  volts\n",
+      "rv2  =  10;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R1  =  8;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Terminals  PQ  are  initially  short-circuited,  as  shown  in  Figure  33.65.\n",
+      " #Currents  I1  and  I2  are  shown  labelled.  Kirchhoff\u2019s  laws  are  used.\n",
+      " #For  loop  ABCD,  and  moving  anticlockwise,\n",
+      " #I1*(R2  +  R3  +  R4)  +  I2*(R3  +  R4)  =  V2\n",
+      " #For  loop  DPQC,  and  moving  clockwise,\n",
+      " #R2*I1  -  R1*I2  =  V2  -  V1\n",
+      " #Solving  Equations  by  using  determinants  gives\n",
+      "d1  =  [[V2,  (R3  +  R4)],[(V2  -  V1),  -1*R1]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R2  +  R3  +  R4),  V2],[R2,  (V2  -  V1)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(R2  +  R3  +  R4),  (R3  +  R4)],[R2,  -1*R1]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      " #the  short-circuit  current  Isc\n",
+      "Isc  =  I2\n",
+      " #The  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  between  P  and  Q  is  given  by\n",
+      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
+      " #The  Norton  equivalent  circuit  is  shown  in  Figure  33.66,  where  current  I  is  given  by\n",
+      "R  =  2;#in  ohm\n",
+      "I  =  (z/(z  +  R))*Isc\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  current  flowing  5  ohm  resistor  is  \",round(Imag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  current  flowing  5  ohm  resistor  is   0.19   A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 595</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnitude of the current flowing in the \t(1.8 + j4) ohm impedance connected between terminals A and B\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  12;#  in  volts\n",
+      "E2  =  24;#  in  volts\n",
+      "Z1  =  3;#  in  ohm\n",
+      "Z2  =  2;#  in  ohm\n",
+      "R1  =  4j;#  in  ohm\n",
+      "R2  =  1.8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Z3  =  R1  +  R2\n",
+      " #For  the  branch  containing  the  E1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
+      "Isc1  =  E1/Z1\n",
+      " #For  the  branch  containing  the  E2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
+      "Isc2  =  E2/Z2\n",
+      " #Thus  Figure  33.73  shows  a  network  equivalent  to  Figure  33.72.  From  Figure  33.73,  \n",
+      "    #the  total  short-circuit  current\n",
+      "Isc  =  Isc1  +  Isc2\n",
+      " #the  total  impedance  is  given  by\n",
+      "z  =  Z1*Z2/(Z1  +  Z2)\n",
+      " #Thus  Figure  33.73  simplifies  to  Figure  33.74.\n",
+      " #The  open-circuit  voltage  across  AB  of  Figure  33.74,  E\n",
+      "E  =  Isc*z\n",
+      " #the  impedance  \u2018looking  in\u2019  at  AB,is  z\n",
+      " #the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.75.\n",
+      "R  =  1.8  +  4j;#  in  ohm\n",
+      " #when  R  impedance  is  connected  to  terminals  AB  of  Figure  33.75,  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is  \",round(Imag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is   3.84   A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 596</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the current flowing in the capacitive branch connected to terminals AB.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  5;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "i  =  0.001;#  in  Amperes\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  4000;#  in  ohm\n",
+      "R3  =  2000;#  in  ohm\n",
+      "R4  =  200;#  in  ohm\n",
+      "R5  =  -1j*4000;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #For  the  branch  containing  the  V1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
+      "Isc1  =  V1/R1\n",
+      "z1  =  R1\n",
+      " #For  the  branch  containing  the  V2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
+      "Isc2  =  V2/R2\n",
+      "z2  =  R2\n",
+      " #Thus  the  circuit  of  Figure  33.76  converts  to  that  of  Figure  33.77.\n",
+      " #The  above  two  Norton  equivalent  networks  shown  in  Figure  33.77  may  be  combined,  \n",
+      "    #since  the  total  short-circuit  current  is\n",
+      "Isc  =  Isc1  +  Isc2\n",
+      " #the  total  impedance  is  given  by\n",
+      "Z1  =  z1*z2/(z1  +  z2)\n",
+      " #Both  of  the  Norton  equivalent  networks  shown  in  Figure  33.78  may  be  converted  to  Th\u00b4evenin  equivalent  circuits. \n",
+      "    #Open-circuit  voltage  across  CD  is\n",
+      "Ecd  =  Isc*Z1\n",
+      " #the  impedance  \u2018looking  in\u2019  at  CD  is  Z1\n",
+      " #Open-circuit  voltage  across  EF\n",
+      "Eef  =  i*R3\n",
+      " #the  impedance  \u2018looking  in\u2019  Figure  33.79  at  EF\n",
+      "Z2  =  R3\n",
+      " #Thus  Figure  33.78  converts  to  Figure  33.79.\n",
+      " #Combining  the  two  Th\u00b4evenin  circuits  gives  e.m.f.\n",
+      "E  =  Ecd  -  Eef\n",
+      " #impedance  z\n",
+      "z  =  Z1  +  Z2\n",
+      " #the  Th\u00b4evenin  equivalent  circuit  for  terminals  AB  of  Figure  33.76  is  as  shown  in  Figure  33.80.\n",
+      "Z3  =  R4  +  R5\n",
+      " #If  an  impedance  Z3  is  connected  across  terminals  AB,  then  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  Z3)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  in  the  capacitive  branch  is  \",  Imag*1000,\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  in  the  capacitive  branch  is   0.8 mA"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 597</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated in a \t(600 - j800) ohm impedance connected between A and B\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  5;#  in  volts\n",
+      "i  =  0.004;#  in  Amperes\n",
+      "R1  =  2000;#  in  ohm\n",
+      "R2  =  1000j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #Converting  the  Th\u00b4evenin  circuit  to  a  Norton  network  gives\n",
+      "Isc1  =  V/R2\n",
+      " #Thus  Figure  33.81  converts  to  that  shown  in  Figure  33.82.  \n",
+      "    #The  two  Norton  equivalent  networks  may  be  combined,  giving\n",
+      "Isc  =  Isc1  +  i\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #This  results  in  the  equivalent  network  shown  in  Figure  33.83. \n",
+      "    #Converting  to  an  equivalent  Th\u00b4evenin  circuit  gives  open  circuit  e.m.f.  across  AB,\n",
+      "E  =  Isc*z\n",
+      " #Thus  the  The\u00b4venin  equivalent  circuit  is  as  shown  in  Figure  33.84.\n",
+      "R  =  600  -  800j;#  in  ohms\n",
+      " #When  a  R  impedance  is  connected  across  AB,  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      " #the  power  dissipated  in  the  R  resistor  is\n",
+      "PR  =  R.real*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is  \",round(PR*1000,2),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is   19.68 mW"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_2.ipynb
new file mode 100755
index 00000000..55e4e2b7
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_2.ipynb
@@ -0,0 +1,1103 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 33: Thevenin\u2019s and Norton\u2019s theorems</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 578</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current flowing in the capacitor, and (b) the p.d. across the 150 k\u0006ohmresistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  200;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  5000;#  in  ohm\n",
+      "R2  =  20000;#  in  ohm\n",
+      "R3  =  -1j*120000;#  in  ohm\n",
+      "R4  =  150000;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Initially  the  (150-i120)kohm\u0006  impedance  is  removed  from  the    circuit  as  shown  in  Figure  33.13.\n",
+      " #Note  that,  to  find  the  current  in  the  capacitor,  \n",
+      "    #only  the  capacitor  need  have  been  initially  removed  from  the  circuit.  \n",
+      "    #However,  removing  each  of  the  components  from  the  branch  through  \n",
+      "    #which  the  current  is  required  will  often  result  in  a  simpler  solution.  \n",
+      " #From  Figure  33.13,\n",
+      " #current,  I1  \n",
+      "I1  =  V/(R1  +  R2)\n",
+      " #The  open-circuit  e.m.f.  E  is  equal  to  the  p.d.  across  the  20  k\u0006ohm  resistor,  i.e.\n",
+      "E  =  I1*R2\n",
+      " #Removing  the  V1  source  gives  the  network  shown  in  Figure  33.14.\n",
+      " #The  impedance,  z,  looking  in at  the  open-circuited  terminals  is  given  by\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.15,  where  current  iL  is  given  by\n",
+      "ZL  =  R3  +  R4\n",
+      "IL  =  E/(ZL  +  z)\n",
+      "ILmag  =  abs(IL)\n",
+      " #current  flowing  in  the  capacitor\n",
+      "Ic  =  ILmag\n",
+      " #P.d.  across  the  150  kohm  resistor,\n",
+      "Vr150  =  ILmag*R4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  current  flowing  in  the  capacitor  is  \",round(Ic*1000,2),\" mA\"\n",
+      "print  \"\\n(b)  the  p.d.  across  the  150  ohm  resistance  is  \",round(Vr150,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  current  flowing  in  the  capacitor  is   0.82  mA\n",
+        "\n",
+        "(b)  the  p.d.  across  the  150  ohm  resistance  is   122.93   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 579</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of current I.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  20;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  1.5;#  in  ohm\n",
+      "L  =  235E-6;#  in  Henry\n",
+      "R4  =  3;#  in  ohm\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #The  impedance  through  which  current  I  is  flowing  is  initially  removed  from  the  network,  as  shown  in  Figure  33.17.\n",
+      " #From  Figure  33.17,\n",
+      " #current,  I1  \n",
+      "I1  =  (V1  -  V2)/(R1  +  R4)\n",
+      " #the  open  circuit  e.m.f.  E\n",
+      "E  =  V1  -  I1*R1\n",
+      " #When  the  sources  of  e.m.f.  are  removed  from  the  circuit,  \n",
+      "    #the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
+      "z  =  R1*R4/(R1  +  R4)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.18,  where  inductive  reactance,\n",
+      "XL  =  2*math.pi*f*L\n",
+      "R3  =  1j*XL\n",
+      " #Hence  current\n",
+      "I  =  E/(R2  +  R3  +  z)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  I  is  \",round(I.real,2),\"  +  (\",round(I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  I  is   2.7   +  ( -2.95 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 580</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power dissipated in the 48 ohm resistor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  -1j*400;#  in  ohm\n",
+      "R2  =  300;#  in  ohm\n",
+      "R3  =  144j;#  in  ohm\n",
+      "R4  =  48;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  R3  and  R4  impedance  is  initially  removed  from  the  network  as  shown  in  Figure  33.20.\n",
+      " #From  Figure  33.20,\n",
+      " #current,  I\n",
+      "i  =  V/(R1  +  R2)\n",
+      " #the  open  circuit  e.m.f.  E\n",
+      "E  =  i*R2\n",
+      " #When  the  V  is  removed  from  the  circuit,  the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.21  connected  to  R#  and  R4,\n",
+      " #Hence  current\n",
+      "I  =  E/(R4  +  R3  +  z)\n",
+      "Imag  =  abs(I)\n",
+      " #the  power  dissipated  in  the  48  ohm  resistor\n",
+      "Pr48  =  R4*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  48  ohm  resistor  is  \",round(Pr48,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  48  ohm  resistor  is   0.75   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 581</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the 80 ohm resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  46;#  in  ohm\n",
+      "R4  =  50;#  in  ohm\n",
+      "R5  =  15;#  in  ohm\n",
+      "R6  =  60;#  in  ohm\n",
+      "R7  =  16;#  in  ohm\n",
+      "R8  =  80;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #One  method  of  analysing  a  multi-branch  network  as  shown  in  Figure  33.22 \n",
+      "    #is  to  use  Th\u00b4evenin\u2019s  theorem  on  one  part  of  the  network  at  a  time.  \n",
+      "    #For  example,  the  part  of  the  circuit  to  the  left  of  AA  may  be  reduced  to  a  Th\u00b4evenin  equivalent  circuit.\n",
+      " #From  Figure  33.23,\n",
+      "E1  =  (R2/(R1  +  R2))*V\n",
+      "z1  =  R1*R2/(R1  +  R2)\n",
+      " #Thus  the  network  of  Figure  33.22  reduces  to  that  of  Figure  33.24.  \n",
+      "    #The  part  of  the  network  shown  in  Figure  33.24  to  the  left  of  BB  may  be  reduced \n",
+      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
+      "E2  =  (R4/(R3  +  R4  +  z1))*E1\n",
+      "z2  =  R4*(z1  +  R3)/(R4  +  z1  +  R3)\n",
+      " #Thus  the  original  network  reduces  to  that  shown  in  Figure  33.25.  \n",
+      "    #The  part  of  the  network  shown  in  Figure  33.25  to  the  left  of  CC  may  be  reduced  \n",
+      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
+      "E3  =  (R6/(R5  +  R6  +  z2))*E2\n",
+      "z3  =  R6*(z2  +  R5)/(R5  +  z2  +  R6)\n",
+      " #Thus  the  original  network  reduces  to  that  of  Figure  33.26, \n",
+      "    #from  which  the  current  in  the  80  ohm  resistor  is  given  by\n",
+      "I  =  E3/(z3  +  R7  +  R8)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  in  the  80  ohm  resistor  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  in  the  80  ohm  resistor  is   0.2   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 582</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the magnitude of the current flowing in a \t(3.75 + j11) ohmimpedance connected across terminals AB, and\n",
+      "#(b) the magnitude of the p.d. across the (3.75 + j11) ohmimpedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  24;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  -1j*3;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  I1  shown  in  Figure  33.27  is  given  by\n",
+      "I1  =  V/(R1  +  R2  +  R3)\n",
+      " #The  Th\u00b4evenin  equivalent  voltage,  i.e.,  the  open-circuit  voltage  across  terminals  AB,  is  given  by\n",
+      "E  =  I1*(R2  +  R3)\n",
+      " #When  the  voltage  source  is  removed,  the  impedance  z  \u2018looking  in\u2019  at  AB  is  given  by\n",
+      "z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)\n",
+      " #Thus  the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.28.\n",
+      " #when  (3.75  +  i11)  ohm  impedance  connected  across  terminals  AB, \n",
+      "    #the  current  I  flowing  in  the  impedance  is  given  by\n",
+      "R  =  3.75  +  11j;#  in  ohms\n",
+      "I  =  E/(R  +  z)\n",
+      "Imag  =  abs(I)\n",
+      " #the  p.d.  across  the(  3.75  +  i11)ohm  impedance.\n",
+      "VR  =  I*R\n",
+      "VRmag  =  abs(VR)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is  \",round(Imag,2),\"  A\"\n",
+      "print  \"\\n  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is  \",round(VRmag,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is   3.0   A\n",
+        "\n",
+        "  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is   34.86   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 583</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the capacitor of the network\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  16.55;#  in  volts\n",
+      "thetav  =  -22.62;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  1j*2;#  in  ohm\n",
+      "R3  =  1j*6;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  -1*1j*8;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  capacitor  is  removed  from  branch  AB,  as  shown  in  Figure  33.30.\n",
+      " #Impedance,  Z\n",
+      "Z1  =  R3  +  R4  +  R5\n",
+      "Z  =  R1  +  (Z1*R2/(R2  +  Z1))\n",
+      "I1  =  V/Z\n",
+      "I2  =  (R2/(R2  +Z1))*I1\n",
+      " #The  open-circuit  voltage,  E\n",
+      "E  =  I2*R5\n",
+      " #If  the  voltage  source  is  removed  from  Figure  33.30,  the  impedance,  z,  \u2018looking  in\u2019  at  AB  is  given  by\n",
+      "z  =  R5*((R1*R2/(R1  +  R2))  +  R3  +  R4)/(R5  +  ((R1*R2/(R1  +  R2))  +  R3  +  R4))\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.31,  \n",
+      "    #where  the  current  flowing  in  the  capacitor,  I,  is  given  by\n",
+      "I  =  E/(z  +  R6)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  in  the  capacitor  of  the  network  is  \",round(Imag,2),\"    A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  in  the  capacitor  of  the  network  is   0.43     A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 584</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power dissipated by a 2 ohm resistor connected across PQ.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  5;#  in  volts\n",
+      "rv2  =  10;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R1  =  8;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Current  I1  shown  in  Figure  33.32  is  given  by\n",
+      "I1  =  V2/(R2  +  R3  +  R4)\n",
+      " #Hence  the  voltage  drop  across  the  5  ohm\u0006  resistor  is  given  by  VX  is  in  the  direction  shown  in  Figure  33.32,\n",
+      "Vx  =  I1*R2\n",
+      " #The  open-circuit  voltage  E  across  PQ  is  the  phasor  sum  of  V1,  Vx  and  V2,  as  shown  in  Figure  33.33.\n",
+      "E  =  V2  -  V1  -  Vx\n",
+      " #The  impedance,  z,  \u2018looking  in\u2019  at  terminals  PQ  with  the  voltage  sources  removed  is  given  by\n",
+      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.34  with  the  2  ohm  resistance  connected  across  terminals  PQ.\n",
+      " #The  current  flowing  in  the  2  ohm\u0006  resistance  is  given  by\n",
+      "R  =  2;#  in  ohms\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      " #power  P  dissipated  in  the  2  ohm\u0006  resistor  is  given  by\n",
+      "Pr2  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  P  dissipated  in  the  2  ohm  resistor  is  \",round(Pr2,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  P  dissipated  in  the  2  ohm  resistor  is   0.07   W"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 585</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the capacitor, and its direction\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  30;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  15;#  in  ohm\n",
+      "R2  =  40;#  in  ohm\n",
+      "R3  =  20j;#  in  ohm\n",
+      "R4  =  20;#  in  ohm\n",
+      "R5  =  5j;#  in  ohm\n",
+      "R6  =  5;#  in  ohm\n",
+      "R7  =  -1j*25;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  R7\u0006  is  initially  removed  from  the  network,  as  shown  in  Figure  33.36\n",
+      "Z1  =  R1\n",
+      "Z2  =  R2\n",
+      "Z3  =  R3  +  R4\n",
+      "Z4  =  R5  +  R6\n",
+      " #P.d.  between  A  and  C,\n",
+      "Vac  =  (Z1/(Z1  +  Z4))*V\n",
+      " #P.d.  between  B  and  C,\n",
+      "Vbc  =  (Z2/(Z2  +  Z3))*V\n",
+      " #Assuming  that  point  A  is  at  a  higher  potential  than  point  B,  then  the  p.d.  between  A  and  B  is\n",
+      "Vab  =  Vac  -  Vbc\n",
+      " #the  open-circuit  voltage  across  AB  is  given  by\n",
+      "E  =  Vab\n",
+      " #Point  C  is  at  a  potential  of  V  .  Between  C  and  A  is  a  volt  drop  of  Vac.  Hence  the  voltage  at  point  A  is\n",
+      "Va  =  V  -  Vac\n",
+      " #Between  points  C  and  B  is  a  voltage  drop  of  Vbc.  Hence  the  voltage  at  point  B\n",
+      "Vb  =  V  -  Vbc\n",
+      " #Replacing  the  V  source  with  a  short-circuit  (i.e.,  zero  internal  impedance) \n",
+      "    #gives  the  network  shown  in  Figure  33.37(a).  The  network  is  shown  redrawn  in  Figure  33.37(b)  \n",
+      "    #and  simplified  in  Figure  33.37(c).  Hence  the  impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
+      "z  =  Z1*Z4/(Z1  +  Z4)  +  Z2*Z3/(Z2  +  Z3)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.38,  where  current  I  is  given  by\n",
+      "I  =  E/(z  +  R7)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(Imag,2),\"  A  in  direction  from  B  to  A.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  capacitor  is   0.13   A  in  direction  from  B  to  A."
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 589</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of current I in the circuit\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  5;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  3;#  in  ohm\n",
+      "R3  =  -1j*3;#  in  ohm\n",
+      "R4  =  2.8;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #The  branch  containing  the  R4  is  short-circuited,  as  shown  in  Figure  33.48.\n",
+      " #The  R2  in  parallel  with  a  short-circuit  is  the  same  as  R2  in  parallel  with  0  ohm \n",
+      "    #giving  an  equivalent  impedance  of\n",
+      "Z1  =  R2*0/(R3  +  0)\n",
+      " #Hence  the  network  reduces  to  that  shown  in  Figure  33.49,  where\n",
+      "Isc  =  V/R1\n",
+      " #If  the  Voltage  source  is  removed  from  the  network  the  input  impedance,  z,  \u2018looking-in\u2019  \n",
+      "    #at  a  break  made  in  AB  of  Figure  33.48  gives\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.51,  where  current  I  is  given  by\n",
+      "I  =  (z/(z  +  R4  +  R3))*Isc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(I.real,2),\"  +  (\", round(I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  capacitor  is   0.48   +  ( 0.36 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 589</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  20;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  1.5;#  in  ohm\n",
+      "R3  =  2.95j;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #The  inductive  branch  is  initially  short-circuited,  as  shown  in  Figure  33.53.\n",
+      " #From  Figure  33.53,\n",
+      "I1  =  V1/R1\n",
+      "I2  =  V2/R4\n",
+      "Isc  =  I1  +  I2\n",
+      " #If  the  voltage  sources  are  removed,  the  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  in  AB  is  given  by\n",
+      "z  =  R1*R4/(R1  +  R4)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.54,  where  current  I  is  given  by\n",
+      "I  =  (z/(z  +  R2  +  R3))*Isc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  inductive  branch  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  inductive  branch  is   2.7   +  ( -2.95 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 590</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  -2j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R3\n",
+      "z = 1/(1/R2 + 1/R3 + 1/R4)\n",
+      "I = (z/(R1 + z))*Isc\n",
+      "pd1 = abs(I)*R1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the magnitude of the p.d. across the 1 ohm resistor is \",round(pd1, 2),\"V\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the magnitude of the p.d. across the 1 ohm resistor is  1.58 V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 591</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the power dissipated in a 5 ohm resistor connected between A and B.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  20;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  -1j*3;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Terminals  AB  are  initially  short-circuited,  as  shown  in  Figure  33.61.\n",
+      " #The  circuit  impedance  Z  presented  to  the  voltage  source  is  given  by\n",
+      "Z  =  R1  +  R4*(R2  +  R3)/(R2  +  R3  +  R4)\n",
+      " #Thus  current  I  in  Figure  33.61  is  given  by\n",
+      "I  =  V/Z\n",
+      "Isc  =  ((R2  +  R3)/(R2  +  R3  +  R4))*I\n",
+      " #Removing  the  voltage  source  of  Figure  33.60  gives  the  network  Figure  33.62  of  Figure  33.62. \n",
+      "    #Impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
+      "z  =  R4  +  R1*(R2  +  R3)/(R2  +  R3  +  R1)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.63.\n",
+      "R  =  5;#  in  ohms\n",
+      " #Current  IL\n",
+      "IL  =  (z/(z  +  R))*Isc\n",
+      "ILmag  =  abs(IL)\n",
+      " #the  power  dissipated  in  the  5  ohm  resistor  is\n",
+      "Pr5  =  R*ILmag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  5  ohm  resistor  is  \",round(Pr5,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  5  ohm  resistor  is   22.54   W"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 592</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the current flowing in a 2 ohm resistor connected across PQ.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  5;#  in  volts\n",
+      "rv2  =  10;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R1  =  8;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Terminals  PQ  are  initially  short-circuited,  as  shown  in  Figure  33.65.\n",
+      " #Currents  I1  and  I2  are  shown  labelled.  Kirchhoff\u2019s  laws  are  used.\n",
+      " #For  loop  ABCD,  and  moving  anticlockwise,\n",
+      " #I1*(R2  +  R3  +  R4)  +  I2*(R3  +  R4)  =  V2\n",
+      " #For  loop  DPQC,  and  moving  clockwise,\n",
+      " #R2*I1  -  R1*I2  =  V2  -  V1\n",
+      " #Solving  Equations  by  using  determinants  gives\n",
+      "d1  =  [[V2,  (R3  +  R4)],[(V2  -  V1),  -1*R1]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R2  +  R3  +  R4),  V2],[R2,  (V2  -  V1)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(R2  +  R3  +  R4),  (R3  +  R4)],[R2,  -1*R1]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      " #the  short-circuit  current  Isc\n",
+      "Isc  =  I2\n",
+      " #The  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  between  P  and  Q  is  given  by\n",
+      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
+      " #The  Norton  equivalent  circuit  is  shown  in  Figure  33.66,  where  current  I  is  given  by\n",
+      "R  =  2;#in  ohm\n",
+      "I  =  (z/(z  +  R))*Isc\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  current  flowing  5  ohm  resistor  is  \",round(Imag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  current  flowing  5  ohm  resistor  is   0.19   A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 595</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnitude of the current flowing in the \t(1.8 + j4) ohm impedance connected between terminals A and B\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  12;#  in  volts\n",
+      "E2  =  24;#  in  volts\n",
+      "Z1  =  3;#  in  ohm\n",
+      "Z2  =  2;#  in  ohm\n",
+      "R1  =  4j;#  in  ohm\n",
+      "R2  =  1.8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Z3  =  R1  +  R2\n",
+      " #For  the  branch  containing  the  E1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
+      "Isc1  =  E1/Z1\n",
+      " #For  the  branch  containing  the  E2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
+      "Isc2  =  E2/Z2\n",
+      " #Thus  Figure  33.73  shows  a  network  equivalent  to  Figure  33.72.  From  Figure  33.73,  \n",
+      "    #the  total  short-circuit  current\n",
+      "Isc  =  Isc1  +  Isc2\n",
+      " #the  total  impedance  is  given  by\n",
+      "z  =  Z1*Z2/(Z1  +  Z2)\n",
+      " #Thus  Figure  33.73  simplifies  to  Figure  33.74.\n",
+      " #The  open-circuit  voltage  across  AB  of  Figure  33.74,  E\n",
+      "E  =  Isc*z\n",
+      " #the  impedance  \u2018looking  in\u2019  at  AB,is  z\n",
+      " #the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.75.\n",
+      "R  =  1.8  +  4j;#  in  ohm\n",
+      " #when  R  impedance  is  connected  to  terminals  AB  of  Figure  33.75,  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is  \",round(Imag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is   3.84   A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 596</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the current flowing in the capacitive branch connected to terminals AB.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  5;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "i  =  0.001;#  in  Amperes\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  4000;#  in  ohm\n",
+      "R3  =  2000;#  in  ohm\n",
+      "R4  =  200;#  in  ohm\n",
+      "R5  =  -1j*4000;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #For  the  branch  containing  the  V1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
+      "Isc1  =  V1/R1\n",
+      "z1  =  R1\n",
+      " #For  the  branch  containing  the  V2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
+      "Isc2  =  V2/R2\n",
+      "z2  =  R2\n",
+      " #Thus  the  circuit  of  Figure  33.76  converts  to  that  of  Figure  33.77.\n",
+      " #The  above  two  Norton  equivalent  networks  shown  in  Figure  33.77  may  be  combined,  \n",
+      "    #since  the  total  short-circuit  current  is\n",
+      "Isc  =  Isc1  +  Isc2\n",
+      " #the  total  impedance  is  given  by\n",
+      "Z1  =  z1*z2/(z1  +  z2)\n",
+      " #Both  of  the  Norton  equivalent  networks  shown  in  Figure  33.78  may  be  converted  to  Th\u00b4evenin  equivalent  circuits. \n",
+      "    #Open-circuit  voltage  across  CD  is\n",
+      "Ecd  =  Isc*Z1\n",
+      " #the  impedance  \u2018looking  in\u2019  at  CD  is  Z1\n",
+      " #Open-circuit  voltage  across  EF\n",
+      "Eef  =  i*R3\n",
+      " #the  impedance  \u2018looking  in\u2019  Figure  33.79  at  EF\n",
+      "Z2  =  R3\n",
+      " #Thus  Figure  33.78  converts  to  Figure  33.79.\n",
+      " #Combining  the  two  Th\u00b4evenin  circuits  gives  e.m.f.\n",
+      "E  =  Ecd  -  Eef\n",
+      " #impedance  z\n",
+      "z  =  Z1  +  Z2\n",
+      " #the  Th\u00b4evenin  equivalent  circuit  for  terminals  AB  of  Figure  33.76  is  as  shown  in  Figure  33.80.\n",
+      "Z3  =  R4  +  R5\n",
+      " #If  an  impedance  Z3  is  connected  across  terminals  AB,  then  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  Z3)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  in  the  capacitive  branch  is  \",  Imag*1000,\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  in  the  capacitive  branch  is   0.8 mA"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 597</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the power dissipated in a \t(600 - j800) ohm impedance connected between A and B\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  5;#  in  volts\n",
+      "i  =  0.004;#  in  Amperes\n",
+      "R1  =  2000;#  in  ohm\n",
+      "R2  =  1000j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #Converting  the  Th\u00b4evenin  circuit  to  a  Norton  network  gives\n",
+      "Isc1  =  V/R2\n",
+      " #Thus  Figure  33.81  converts  to  that  shown  in  Figure  33.82.  \n",
+      "    #The  two  Norton  equivalent  networks  may  be  combined,  giving\n",
+      "Isc  =  Isc1  +  i\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #This  results  in  the  equivalent  network  shown  in  Figure  33.83. \n",
+      "    #Converting  to  an  equivalent  Th\u00b4evenin  circuit  gives  open  circuit  e.m.f.  across  AB,\n",
+      "E  =  Isc*z\n",
+      " #Thus  the  The\u00b4venin  equivalent  circuit  is  as  shown  in  Figure  33.84.\n",
+      "R  =  600  -  800j;#  in  ohms\n",
+      " #When  a  R  impedance  is  connected  across  AB,  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      " #the  power  dissipated  in  the  R  resistor  is\n",
+      "PR  =  R.real*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is  \",round(PR*1000,2),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is   19.68 mW"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_3.ipynb
new file mode 100755
index 00000000..d0fda89c
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint_3.ipynb
@@ -0,0 +1,1103 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f149fa9435f5a9b6282f5eb58b9d4797c5c882108a3bdb7bb56b34160f170ab8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 33: Thevenin\u2019s and Norton\u2019s theorems</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 578</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  200;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  5000;#  in  ohm\n",
+      "R2  =  20000;#  in  ohm\n",
+      "R3  =  -1j*120000;#  in  ohm\n",
+      "R4  =  150000;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Initially  the  (150-i120)kohm\u0006  impedance  is  removed  from  the    circuit  as  shown  in  Figure  33.13.\n",
+      " #Note  that,  to  find  the  current  in  the  capacitor,  \n",
+      "    #only  the  capacitor  need  have  been  initially  removed  from  the  circuit.  \n",
+      "    #However,  removing  each  of  the  components  from  the  branch  through  \n",
+      "    #which  the  current  is  required  will  often  result  in  a  simpler  solution.  \n",
+      " #From  Figure  33.13,\n",
+      " #current,  I1  \n",
+      "I1  =  V/(R1  +  R2)\n",
+      " #The  open-circuit  e.m.f.  E  is  equal  to  the  p.d.  across  the  20  k\u0006ohm  resistor,  i.e.\n",
+      "E  =  I1*R2\n",
+      " #Removing  the  V1  source  gives  the  network  shown  in  Figure  33.14.\n",
+      " #The  impedance,  z,  looking  in at  the  open-circuited  terminals  is  given  by\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.15,  where  current  iL  is  given  by\n",
+      "ZL  =  R3  +  R4\n",
+      "IL  =  E/(ZL  +  z)\n",
+      "ILmag  =  abs(IL)\n",
+      " #current  flowing  in  the  capacitor\n",
+      "Ic  =  ILmag\n",
+      " #P.d.  across  the  150  kohm  resistor,\n",
+      "Vr150  =  ILmag*R4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  current  flowing  in  the  capacitor  is  \",round(Ic*1000,2),\" mA\"\n",
+      "print  \"\\n(b)  the  p.d.  across  the  150  ohm  resistance  is  \",round(Vr150,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  current  flowing  in  the  capacitor  is   0.82  mA\n",
+        "\n",
+        "(b)  the  p.d.  across  the  150  ohm  resistance  is   122.93   V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 579</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  20;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  1.5;#  in  ohm\n",
+      "L  =  235E-6;#  in  Henry\n",
+      "R4  =  3;#  in  ohm\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #The  impedance  through  which  current  I  is  flowing  is  initially  removed  from  the  network,  as  shown  in  Figure  33.17.\n",
+      " #From  Figure  33.17,\n",
+      " #current,  I1  \n",
+      "I1  =  (V1  -  V2)/(R1  +  R4)\n",
+      " #the  open  circuit  e.m.f.  E\n",
+      "E  =  V1  -  I1*R1\n",
+      " #When  the  sources  of  e.m.f.  are  removed  from  the  circuit,  \n",
+      "    #the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
+      "z  =  R1*R4/(R1  +  R4)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.18,  where  inductive  reactance,\n",
+      "XL  =  2*math.pi*f*L\n",
+      "R3  =  1j*XL\n",
+      " #Hence  current\n",
+      "I  =  E/(R2  +  R3  +  z)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  I  is  \",round(I.real,2),\"  +  (\",round(I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  I  is   2.7   +  ( -2.95 )i  A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 580</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  50;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  -1j*400;#  in  ohm\n",
+      "R2  =  300;#  in  ohm\n",
+      "R3  =  144j;#  in  ohm\n",
+      "R4  =  48;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  R3  and  R4  impedance  is  initially  removed  from  the  network  as  shown  in  Figure  33.20.\n",
+      " #From  Figure  33.20,\n",
+      " #current,  I\n",
+      "i  =  V/(R1  +  R2)\n",
+      " #the  open  circuit  e.m.f.  E\n",
+      "E  =  i*R2\n",
+      " #When  the  V  is  removed  from  the  circuit,  the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.21  connected  to  R#  and  R4,\n",
+      " #Hence  current\n",
+      "I  =  E/(R4  +  R3  +  z)\n",
+      "Imag  =  abs(I)\n",
+      " #the  power  dissipated  in  the  48  ohm  resistor\n",
+      "Pr48  =  R4*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  48  ohm  resistor  is  \",round(Pr48,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  48  ohm  resistor  is   0.75   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 581</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  100;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  20;#  in  ohm\n",
+      "R3  =  46;#  in  ohm\n",
+      "R4  =  50;#  in  ohm\n",
+      "R5  =  15;#  in  ohm\n",
+      "R6  =  60;#  in  ohm\n",
+      "R7  =  16;#  in  ohm\n",
+      "R8  =  80;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #One  method  of  analysing  a  multi-branch  network  as  shown  in  Figure  33.22 \n",
+      "    #is  to  use  Th\u00b4evenin\u2019s  theorem  on  one  part  of  the  network  at  a  time.  \n",
+      "    #For  example,  the  part  of  the  circuit  to  the  left  of  AA  may  be  reduced  to  a  Th\u00b4evenin  equivalent  circuit.\n",
+      " #From  Figure  33.23,\n",
+      "E1  =  (R2/(R1  +  R2))*V\n",
+      "z1  =  R1*R2/(R1  +  R2)\n",
+      " #Thus  the  network  of  Figure  33.22  reduces  to  that  of  Figure  33.24.  \n",
+      "    #The  part  of  the  network  shown  in  Figure  33.24  to  the  left  of  BB  may  be  reduced \n",
+      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
+      "E2  =  (R4/(R3  +  R4  +  z1))*E1\n",
+      "z2  =  R4*(z1  +  R3)/(R4  +  z1  +  R3)\n",
+      " #Thus  the  original  network  reduces  to  that  shown  in  Figure  33.25.  \n",
+      "    #The  part  of  the  network  shown  in  Figure  33.25  to  the  left  of  CC  may  be  reduced  \n",
+      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
+      "E3  =  (R6/(R5  +  R6  +  z2))*E2\n",
+      "z3  =  R6*(z2  +  R5)/(R5  +  z2  +  R6)\n",
+      " #Thus  the  original  network  reduces  to  that  of  Figure  33.26, \n",
+      "    #from  which  the  current  in  the  80  ohm  resistor  is  given  by\n",
+      "I  =  E3/(z3  +  R7  +  R8)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  in  the  80  ohm  resistor  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  in  the  80  ohm  resistor  is   0.2   A"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 582</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  24;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  -1j*3;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Current  I1  shown  in  Figure  33.27  is  given  by\n",
+      "I1  =  V/(R1  +  R2  +  R3)\n",
+      " #The  Th\u00b4evenin  equivalent  voltage,  i.e.,  the  open-circuit  voltage  across  terminals  AB,  is  given  by\n",
+      "E  =  I1*(R2  +  R3)\n",
+      " #When  the  voltage  source  is  removed,  the  impedance  z  \u2018looking  in\u2019  at  AB  is  given  by\n",
+      "z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)\n",
+      " #Thus  the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.28.\n",
+      " #when  (3.75  +  i11)  ohm  impedance  connected  across  terminals  AB, \n",
+      "    #the  current  I  flowing  in  the  impedance  is  given  by\n",
+      "R  =  3.75  +  11j;#  in  ohms\n",
+      "I  =  E/(R  +  z)\n",
+      "Imag  =  abs(I)\n",
+      " #the  p.d.  across  the(  3.75  +  i11)ohm  impedance.\n",
+      "VR  =  I*R\n",
+      "VRmag  =  abs(VR)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is  \",round(Imag,2),\"  A\"\n",
+      "print  \"\\n  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is  \",round(VRmag,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is   3.0   A\n",
+        "\n",
+        "  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is   34.86   V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 583</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  16.55;#  in  volts\n",
+      "thetav  =  -22.62;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  1j*2;#  in  ohm\n",
+      "R3  =  1j*6;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "R5  =  5;#  in  ohm\n",
+      "R6  =  -1*1j*8;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  capacitor  is  removed  from  branch  AB,  as  shown  in  Figure  33.30.\n",
+      " #Impedance,  Z\n",
+      "Z1  =  R3  +  R4  +  R5\n",
+      "Z  =  R1  +  (Z1*R2/(R2  +  Z1))\n",
+      "I1  =  V/Z\n",
+      "I2  =  (R2/(R2  +Z1))*I1\n",
+      " #The  open-circuit  voltage,  E\n",
+      "E  =  I2*R5\n",
+      " #If  the  voltage  source  is  removed  from  Figure  33.30,  the  impedance,  z,  \u2018looking  in\u2019  at  AB  is  given  by\n",
+      "z  =  R5*((R1*R2/(R1  +  R2))  +  R3  +  R4)/(R5  +  ((R1*R2/(R1  +  R2))  +  R3  +  R4))\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.31,  \n",
+      "    #where  the  current  flowing  in  the  capacitor,  I,  is  given  by\n",
+      "I  =  E/(z  +  R6)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  flowing  in  the  capacitor  of  the  network  is  \",round(Imag,2),\"    A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  flowing  in  the  capacitor  of  the  network  is   0.43     A"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 584</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  5;#  in  volts\n",
+      "rv2  =  10;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R1  =  8;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Current  I1  shown  in  Figure  33.32  is  given  by\n",
+      "I1  =  V2/(R2  +  R3  +  R4)\n",
+      " #Hence  the  voltage  drop  across  the  5  ohm\u0006  resistor  is  given  by  VX  is  in  the  direction  shown  in  Figure  33.32,\n",
+      "Vx  =  I1*R2\n",
+      " #The  open-circuit  voltage  E  across  PQ  is  the  phasor  sum  of  V1,  Vx  and  V2,  as  shown  in  Figure  33.33.\n",
+      "E  =  V2  -  V1  -  Vx\n",
+      " #The  impedance,  z,  \u2018looking  in\u2019  at  terminals  PQ  with  the  voltage  sources  removed  is  given  by\n",
+      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.34  with  the  2  ohm  resistance  connected  across  terminals  PQ.\n",
+      " #The  current  flowing  in  the  2  ohm\u0006  resistance  is  given  by\n",
+      "R  =  2;#  in  ohms\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      " #power  P  dissipated  in  the  2  ohm\u0006  resistor  is  given  by\n",
+      "Pr2  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  P  dissipated  in  the  2  ohm  resistor  is  \",round(Pr2,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  P  dissipated  in  the  2  ohm  resistor  is   0.07   W"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 585</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  30;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  15;#  in  ohm\n",
+      "R2  =  40;#  in  ohm\n",
+      "R3  =  20j;#  in  ohm\n",
+      "R4  =  20;#  in  ohm\n",
+      "R5  =  5j;#  in  ohm\n",
+      "R6  =  5;#  in  ohm\n",
+      "R7  =  -1j*25;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  R7\u0006  is  initially  removed  from  the  network,  as  shown  in  Figure  33.36\n",
+      "Z1  =  R1\n",
+      "Z2  =  R2\n",
+      "Z3  =  R3  +  R4\n",
+      "Z4  =  R5  +  R6\n",
+      " #P.d.  between  A  and  C,\n",
+      "Vac  =  (Z1/(Z1  +  Z4))*V\n",
+      " #P.d.  between  B  and  C,\n",
+      "Vbc  =  (Z2/(Z2  +  Z3))*V\n",
+      " #Assuming  that  point  A  is  at  a  higher  potential  than  point  B,  then  the  p.d.  between  A  and  B  is\n",
+      "Vab  =  Vac  -  Vbc\n",
+      " #the  open-circuit  voltage  across  AB  is  given  by\n",
+      "E  =  Vab\n",
+      " #Point  C  is  at  a  potential  of  V  .  Between  C  and  A  is  a  volt  drop  of  Vac.  Hence  the  voltage  at  point  A  is\n",
+      "Va  =  V  -  Vac\n",
+      " #Between  points  C  and  B  is  a  voltage  drop  of  Vbc.  Hence  the  voltage  at  point  B\n",
+      "Vb  =  V  -  Vbc\n",
+      " #Replacing  the  V  source  with  a  short-circuit  (i.e.,  zero  internal  impedance) \n",
+      "    #gives  the  network  shown  in  Figure  33.37(a).  The  network  is  shown  redrawn  in  Figure  33.37(b)  \n",
+      "    #and  simplified  in  Figure  33.37(c).  Hence  the  impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
+      "z  =  Z1*Z4/(Z1  +  Z4)  +  Z2*Z3/(Z2  +  Z3)\n",
+      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.38,  where  current  I  is  given  by\n",
+      "I  =  E/(z  +  R7)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(Imag,2),\"  A  in  direction  from  B  to  A.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  capacitor  is   0.13   A  in  direction  from  B  to  A."
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 589</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  5;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  3;#  in  ohm\n",
+      "R3  =  -1j*3;#  in  ohm\n",
+      "R4  =  2.8;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #The  branch  containing  the  R4  is  short-circuited,  as  shown  in  Figure  33.48.\n",
+      " #The  R2  in  parallel  with  a  short-circuit  is  the  same  as  R2  in  parallel  with  0  ohm \n",
+      "    #giving  an  equivalent  impedance  of\n",
+      "Z1  =  R2*0/(R3  +  0)\n",
+      " #Hence  the  network  reduces  to  that  shown  in  Figure  33.49,  where\n",
+      "Isc  =  V/R1\n",
+      " #If  the  Voltage  source  is  removed  from  the  network  the  input  impedance,  z,  \u2018looking-in\u2019  \n",
+      "    #at  a  break  made  in  AB  of  Figure  33.48  gives\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.51,  where  current  I  is  given  by\n",
+      "I  =  (z/(z  +  R4  +  R3))*Isc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(I.real,2),\"  +  (\", round(I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  capacitor  is   0.48   +  ( 0.36 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 589</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  20;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  1.5;#  in  ohm\n",
+      "R3  =  2.95j;#  in  ohm\n",
+      "R4  =  3;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #The  inductive  branch  is  initially  short-circuited,  as  shown  in  Figure  33.53.\n",
+      " #From  Figure  33.53,\n",
+      "I1  =  V1/R1\n",
+      "I2  =  V2/R4\n",
+      "Isc  =  I1  +  I2\n",
+      " #If  the  voltage  sources  are  removed,  the  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  in  AB  is  given  by\n",
+      "z  =  R1*R4/(R1  +  R4)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.54,  where  current  I  is  given  by\n",
+      "I  =  (z/(z  +  R2  +  R3))*Isc\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    the  current  flowing  in  the  inductive  branch  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    the  current  flowing  in  the  inductive  branch  is   2.7   +  ( -2.95 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 590</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  10;#  in  volts\n",
+      "R1  =  1;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  4;#  in  ohm\n",
+      "R4  =  -2j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Isc = V/R3\n",
+      "z = 1/(1/R2 + 1/R3 + 1/R4)\n",
+      "I = (z/(R1 + z))*Isc\n",
+      "pd1 = abs(I)*R1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n the magnitude of the p.d. across the 1 ohm resistor is \",round(pd1, 2),\"V\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " the magnitude of the p.d. across the 1 ohm resistor is  1.58 V"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 591</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  20;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  2;#  in  ohm\n",
+      "R2  =  4;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  -1j*3;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Terminals  AB  are  initially  short-circuited,  as  shown  in  Figure  33.61.\n",
+      " #The  circuit  impedance  Z  presented  to  the  voltage  source  is  given  by\n",
+      "Z  =  R1  +  R4*(R2  +  R3)/(R2  +  R3  +  R4)\n",
+      " #Thus  current  I  in  Figure  33.61  is  given  by\n",
+      "I  =  V/Z\n",
+      "Isc  =  ((R2  +  R3)/(R2  +  R3  +  R4))*I\n",
+      " #Removing  the  voltage  source  of  Figure  33.60  gives  the  network  Figure  33.62  of  Figure  33.62. \n",
+      "    #Impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
+      "z  =  R4  +  R1*(R2  +  R3)/(R2  +  R3  +  R1)\n",
+      " #The  Norton  equivalent  network  is  shown  in  Figure  33.63.\n",
+      "R  =  5;#  in  ohms\n",
+      " #Current  IL\n",
+      "IL  =  (z/(z  +  R))*Isc\n",
+      "ILmag  =  abs(IL)\n",
+      " #the  power  dissipated  in  the  5  ohm  resistor  is\n",
+      "Pr5  =  R*ILmag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  5  ohm  resistor  is  \",round(Pr5,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  5  ohm  resistor  is   22.54   W"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 592</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import numpy\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv1  =  5;#  in  volts\n",
+      "rv2  =  10;#  in  volts\n",
+      "thetav1  =  45;#  in  degrees\n",
+      "thetav2  =  0;#  in  degrees\n",
+      "R1  =  8;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  3j;#  in  ohm\n",
+      "R4  =  4;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
+      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
+      " #Terminals  PQ  are  initially  short-circuited,  as  shown  in  Figure  33.65.\n",
+      " #Currents  I1  and  I2  are  shown  labelled.  Kirchhoff\u2019s  laws  are  used.\n",
+      " #For  loop  ABCD,  and  moving  anticlockwise,\n",
+      " #I1*(R2  +  R3  +  R4)  +  I2*(R3  +  R4)  =  V2\n",
+      " #For  loop  DPQC,  and  moving  clockwise,\n",
+      " #R2*I1  -  R1*I2  =  V2  -  V1\n",
+      " #Solving  Equations  by  using  determinants  gives\n",
+      "d1  =  [[V2,  (R3  +  R4)],[(V2  -  V1),  -1*R1]]\n",
+      "D1  =  numpy.linalg.det(d1)\n",
+      "d2  =  [[(R2  +  R3  +  R4),  V2],[R2,  (V2  -  V1)]]\n",
+      "D2  =  numpy.linalg.det(d2)\n",
+      "d  =  [[(R2  +  R3  +  R4),  (R3  +  R4)],[R2,  -1*R1]]\n",
+      "D  =  numpy.linalg.det(d)\n",
+      "I1  =  D1/D\n",
+      "I2  =  D2/D\n",
+      " #the  short-circuit  current  Isc\n",
+      "Isc  =  I2\n",
+      " #The  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  between  P  and  Q  is  given  by\n",
+      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
+      " #The  Norton  equivalent  circuit  is  shown  in  Figure  33.66,  where  current  I  is  given  by\n",
+      "R  =  2;#in  ohm\n",
+      "I  =  (z/(z  +  R))*Isc\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  current  flowing  5  ohm  resistor  is  \",round(Imag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  current  flowing  5  ohm  resistor  is   0.19   A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 595</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  12;#  in  volts\n",
+      "E2  =  24;#  in  volts\n",
+      "Z1  =  3;#  in  ohm\n",
+      "Z2  =  2;#  in  ohm\n",
+      "R1  =  4j;#  in  ohm\n",
+      "R2  =  1.8;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Z3  =  R1  +  R2\n",
+      " #For  the  branch  containing  the  E1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
+      "Isc1  =  E1/Z1\n",
+      " #For  the  branch  containing  the  E2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
+      "Isc2  =  E2/Z2\n",
+      " #Thus  Figure  33.73  shows  a  network  equivalent  to  Figure  33.72.  From  Figure  33.73,  \n",
+      "    #the  total  short-circuit  current\n",
+      "Isc  =  Isc1  +  Isc2\n",
+      " #the  total  impedance  is  given  by\n",
+      "z  =  Z1*Z2/(Z1  +  Z2)\n",
+      " #Thus  Figure  33.73  simplifies  to  Figure  33.74.\n",
+      " #The  open-circuit  voltage  across  AB  of  Figure  33.74,  E\n",
+      "E  =  Isc*z\n",
+      " #the  impedance  \u2018looking  in\u2019  at  AB,is  z\n",
+      " #the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.75.\n",
+      "R  =  1.8  +  4j;#  in  ohm\n",
+      " #when  R  impedance  is  connected  to  terminals  AB  of  Figure  33.75,  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is  \",round(Imag,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is   3.84   A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 596</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  5;#  in  volts\n",
+      "V2  =  10;#  in  volts\n",
+      "i  =  0.001;#  in  Amperes\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  4000;#  in  ohm\n",
+      "R3  =  2000;#  in  ohm\n",
+      "R4  =  200;#  in  ohm\n",
+      "R5  =  -1j*4000;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #For  the  branch  containing  the  V1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
+      "Isc1  =  V1/R1\n",
+      "z1  =  R1\n",
+      " #For  the  branch  containing  the  V2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
+      "Isc2  =  V2/R2\n",
+      "z2  =  R2\n",
+      " #Thus  the  circuit  of  Figure  33.76  converts  to  that  of  Figure  33.77.\n",
+      " #The  above  two  Norton  equivalent  networks  shown  in  Figure  33.77  may  be  combined,  \n",
+      "    #since  the  total  short-circuit  current  is\n",
+      "Isc  =  Isc1  +  Isc2\n",
+      " #the  total  impedance  is  given  by\n",
+      "Z1  =  z1*z2/(z1  +  z2)\n",
+      " #Both  of  the  Norton  equivalent  networks  shown  in  Figure  33.78  may  be  converted  to  Th\u00b4evenin  equivalent  circuits. \n",
+      "    #Open-circuit  voltage  across  CD  is\n",
+      "Ecd  =  Isc*Z1\n",
+      " #the  impedance  \u2018looking  in\u2019  at  CD  is  Z1\n",
+      " #Open-circuit  voltage  across  EF\n",
+      "Eef  =  i*R3\n",
+      " #the  impedance  \u2018looking  in\u2019  Figure  33.79  at  EF\n",
+      "Z2  =  R3\n",
+      " #Thus  Figure  33.78  converts  to  Figure  33.79.\n",
+      " #Combining  the  two  Th\u00b4evenin  circuits  gives  e.m.f.\n",
+      "E  =  Ecd  -  Eef\n",
+      " #impedance  z\n",
+      "z  =  Z1  +  Z2\n",
+      " #the  Th\u00b4evenin  equivalent  circuit  for  terminals  AB  of  Figure  33.76  is  as  shown  in  Figure  33.80.\n",
+      "Z3  =  R4  +  R5\n",
+      " #If  an  impedance  Z3  is  connected  across  terminals  AB,  then  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  Z3)\n",
+      "Imag  =  abs(I)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  current  in  the  capacitive  branch  is  \",  Imag*1000,\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  current  in  the  capacitive  branch  is   0.8 mA"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 597</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  5;#  in  volts\n",
+      "i  =  0.004;#  in  Amperes\n",
+      "R1  =  2000;#  in  ohm\n",
+      "R2  =  1000j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #Converting  the  Th\u00b4evenin  circuit  to  a  Norton  network  gives\n",
+      "Isc1  =  V/R2\n",
+      " #Thus  Figure  33.81  converts  to  that  shown  in  Figure  33.82.  \n",
+      "    #The  two  Norton  equivalent  networks  may  be  combined,  giving\n",
+      "Isc  =  Isc1  +  i\n",
+      "z  =  R1*R2/(R1  +  R2)\n",
+      " #This  results  in  the  equivalent  network  shown  in  Figure  33.83. \n",
+      "    #Converting  to  an  equivalent  Th\u00b4evenin  circuit  gives  open  circuit  e.m.f.  across  AB,\n",
+      "E  =  Isc*z\n",
+      " #Thus  the  The\u00b4venin  equivalent  circuit  is  as  shown  in  Figure  33.84.\n",
+      "R  =  600  -  800j;#  in  ohms\n",
+      " #When  a  R  impedance  is  connected  across  AB,  the  current  I  flowing  is  given  by\n",
+      "I  =  E/(z  +  R)\n",
+      "Imag  =  abs(I)\n",
+      " #the  power  dissipated  in  the  R  resistor  is\n",
+      "PR  =  R.real*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is  \",round(PR*1000,2),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is   19.68 mW"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint.ipynb
new file mode 100755
index 00000000..ce641b05
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint.ipynb
@@ -0,0 +1,449 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 34: Delta-star and star-delta transformations</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 605</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Replace the delta-connected network by an equivalent star connection.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ZA  =  20;#  in  ohm\n",
+      "ZB  =  10 + 10j;#  in  ohm\n",
+      "ZC  =  -20j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Z1 = ZA*ZB/(ZA+ZB+ZC)\n",
+      "Z2 = ZB*ZC/(ZA+ZB+ZC)\n",
+      "Z3 = ZA*ZC/(ZA+ZB+ZC)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent star connection are, Z1 =\",round(Z1.real,2),\"  +  (\",round(Z1.imag,2),\")i  ohmn, Z2 =\",round(Z2.real,2),\"  +  (\",round(Z2.imag,2),\")i  Ohm and Z3 =\",round(Z3.real,2),\"  +  (\",round(Z3.imag,2),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent star connection are, Z1 = 4.0   +  ( 8.0 )i  ohmn, Z2 = 8.0   +  ( -4.0 )i  Ohm and Z3 = 4.0   +  ( -12.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 606</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the equivalent circuit impedance across terminals AB,\n",
+      "#(b) supply current I and (c) the power dissipated in the 10 ohm resistor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  40;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "ZA  =  10j;#  in  ohm\n",
+      "ZB  =  15j;#  in  ohm\n",
+      "ZC  =  25j;#  in  ohm\n",
+      "ZD  =  -8j;#  in  ohm\n",
+      "ZE  =  10;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  network  of  Figure  34.7  is  redrawn,  as  in  Figure  34.8,  \n",
+      "    #showing  more  clearly  the  part  of  the  network  1,  2,  3  forming  a  delta  connection   \n",
+      "    #This  may  he  transformed  into  a  star  connection  as  shown  in  Figure  34.9.\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  equivalent  network  is  shown  in  Figure  34.10  and  is  further  simplified  in  Figure  34.11\n",
+      " #(ZE  +  Z3)  in  parallel  with  (Z1  +  ZD)  gives  an  equivalent  impedance  of\n",
+      "z  =  (ZE  +  Z3)*(Z1  +  ZD)/(Z1  +  ZD  +  ZE  +  Z3)\n",
+      " #Hence  the  total  circuit  equivalent  impedance  across  terminals  AB  is  given  by\n",
+      "Zab  =  z  +  Z2\n",
+      " #Supply  current  I\n",
+      "I  =  V/Zab\n",
+      "I1  =  ((Z1  +  ZD)/(Z1  +  ZD  +  ZE  +  Z3))*I\n",
+      "I1mag  =  abs(I1)\n",
+      " #Power  P  dissipated  in  the  10  ohm  resistance  of  Figure  34.7  is  given  by\n",
+      "Pr10  =  ZE*I1mag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  circuit  impedance  across  terminals  AB  is  \",round(Zab.real,2),\"  +  (\",round(Zab.imag,2),\")i  ohm\"\n",
+      "print  \"\\n  (b)supply  current  I  is  \",round(I.real,2),\"  +  (\",round(I.imag,2),\")i  A\"\n",
+      "print  \"\\n  (c)power  P  dissipated  in  the  10  ohm  resistor  is  \",round(Pr10,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  circuit  impedance  across  terminals  AB  is   2.5   +  ( 2.5 )i  ohm\n",
+        "\n",
+        "  (b)supply  current  I  is   8.0   +  ( -8.0 )i  A\n",
+        "\n",
+        "  (c)power  P  dissipated  in  the  10  ohm  resistor  is   320.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 607</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) the value of the single equivalent resistance that replaces the network between terminals A and B, \n",
+      "#(b) the current supplied by the 52 V source, and (c) the current flowing in the 8 ohm resistance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  52;#  in  volts\n",
+      "ZA  =  8;#  in  ohm\n",
+      "ZB  =  16;#  in  ohm\n",
+      "ZC  =  40;#  in  ohm\n",
+      "ZD  =  1;#  in  ohm\n",
+      "ZE  =  4;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #In  Figure  34.12,  no  resistances  are  directly  in  parallel  or  directly  in  series  with  each  other.  \n",
+      "    #However,  ACD  and  BCD  are  both  delta  connections  and  either  may  be  converted  into  an  equivalent  star  connection.  The  delta  network  BCD  is  redrawn  in  Figure  34.13(a)  and  is  transformed  into  an  equivalent  star  connection  as  shown  in  Figure  34.13(b),  where\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  network  of  Figure  34.12  may  thus  be  redrawn  as  shown  in  Figure  34.14.  \n",
+      "    #The  Z1  and  ZE  are  in  series  with  each  other,  as  are  the  ZD  and  Z3  resistors.  \n",
+      "    #Hence  the  equivalent  network  is  as  shown  in  Figure  34.15.  \n",
+      "    #The  total  equivalent  resistance  across  terminals  A  and  B  is  given  by\n",
+      "Zab  =  (Z1  +  ZE)*(ZD  +  Z3)/(Z1  +  ZE  +  ZD  +  Z3)  +  Z2\n",
+      " #Current  supplied  by  the  source,  i.e.,  current  I  in  Figure  34.15,  is  given  by\n",
+      "I  =  V/Zab\n",
+      " #From  Figure  34.15,  current  I1\n",
+      "I1  =  ((ZD  +  Z3)/(Z1  +  ZE  +  ZD  +  Z3))*I\n",
+      " #current  I2\n",
+      "I2  =  I  -  I1\n",
+      " #From  Figure  34.14,  p.d.  across  AC,\n",
+      "Vac  =  I1*ZE\n",
+      " #p.d.  across  AD\n",
+      "Vad  =  I2*ZD\n",
+      " #Hence  p.d.  between  C  and  D  is  given\n",
+      "Vcd  =  Vac  -  Vad\n",
+      " #current  in  the  8  ohm  resistance\n",
+      "Ir8  =  Vcd/ZA\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  circuit  impedance  across  terminals  AB  is  \",round(Zab,2),\"ohm\"\n",
+      "print  \"\\n  (b)the  current  supplied  by  the  52  V  source  is  \",round(I,2),\"    A\"\n",
+      "print  \"\\n  (c)the  current  flowing  in  the  8  ohm  resistance  is  \",round(Ir8,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  circuit  impedance  across  terminals  AB  is   13.0 ohm\n",
+        "\n",
+        "  (b)the  current  supplied  by  the  52  V  source  is   4.0     A\n",
+        "\n",
+        "  (c)the  current  flowing  in  the  8  ohm  resistance  is   0.75   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 608</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(b)determine the values of RX and LX at balance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  1000;#  in  ohm\n",
+      "R3  =  1000;#  in  ohm\n",
+      "R4  =  500;#  in  ohm\n",
+      "R5  =  200;#  in  ohm\n",
+      "C = 2E-6; # in Farad\n",
+      "\n",
+      "#calculation:\n",
+      "Rx = R2*R4/R3\n",
+      "Lx = R2*C*(R4 + R5 + R4*R5/R3)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (b)Rx is  \",round(Rx,2),\"ohm, Lx is  \",round(Lx,2),\"H\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (b)Rx is   500.0 ohm, Lx is   1.6 H"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 610</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current flowing in the j10 ohm impedance, and \n",
+      "#(b) the power dissipated in the \u000720 ohm impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "ZA  =  25  - 5j;#  in  ohm\n",
+      "ZB  =  15  + 10j;#  in  ohm\n",
+      "ZC  =  20  - 30j;#  in  ohm\n",
+      "ZD  =  20  + 0j;#  in  ohm\n",
+      "ZE  =  0  +  10j;#  in  ohm\n",
+      "ZF  =  2.5  - 5j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  network  may  initially  be  simplified  by  transforming  the  delta  PQR  to  its  equivalent  star  connection  \n",
+      "    #as  represented  by  impedances  Z1,  Z2  and  Z3  in  Figure  34.21.  From  equation  (34.7),\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  network  is  shown  redrawn  in  Figure  34.22  and  further  simplified  in  Figure  34.23,  from  which,\n",
+      "Zab  =  ((Z3  +  ZE)*(ZD  +  Z2)/(Z2  +  ZE  +  ZD  +  Z3))  +  (Z1  +  ZF)\n",
+      " #Current  I1\n",
+      "I1  =  V/Zab\n",
+      " #current  I2\n",
+      "I2  =  ((ZE  +  Z3)/(Z2  +  ZE  +  ZD  +  Z3))*I1\n",
+      " #current  I3\n",
+      "I3  =  I1  -  I2\n",
+      " #The  power  P  dissipated  in  the  ZD  impedance  of  Figure  34.20  is  given  by\n",
+      "Pzd  =  ZD*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  in  the  (0+i10)  ohm  impedance  is  \",round(abs(I3),2),\"    A\"\n",
+      "print  \"\\n  (b)  the  power  dissipated  in  the  (20  +  i0)  ohm  impedance  is  \",round(abs(Pzd),2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  in  the  (0+i10)  ohm  impedance  is   6.0     A\n",
+        "\n",
+        "  (b)  the  power  dissipated  in  the  (20  +  i0)  ohm  impedance  is   80.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 613</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the delta-connected equivalent network for the star-connected impedances.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  10;#  in  ohm\n",
+      "Z2  =  20;#  in  ohm\n",
+      "Z3  =  5j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
+      "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
+      "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,2),\"  +  (\",round(ZA.imag,2),\")i  ohmn, ZB =\",round(ZB.real,2),\"  +  (\",round(ZB.imag,2),\")i  Ohm and ZC =\",round(ZC.real,2),\"  +  (\",round(ZC.imag,2),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent Delta connection are, ZA = 10.0   +  ( 7.5 )i  ohmn, ZB = 30.0   +  ( -40.0 )i  Ohm and ZC = 20.0   +  ( 15.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 613</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert the star to an equivalent delta connection\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1r  =  100;#  in  ohm\n",
+      "Z1a = 0; # in Deg\n",
+      "Z2r  =  63.25;#  in  ohm\n",
+      "Z2a = 18.43; # in Deg\n",
+      "Z3r  =  100;#  in  ohm\n",
+      "Z3a = -90; # in Deg\n",
+      "\n",
+      "#calculation\n",
+      "Z1  =  Z1r*math.cos(Z1a*math.pi/180)  +  1j*Z1r*math.sin(Z1a*math.pi/180)\n",
+      "Z2  =  Z2r*math.cos(Z2a*math.pi/180)  +  1j*Z2r*math.sin(Z2a*math.pi/180)\n",
+      "Z3  =  Z3r*math.cos(Z3a*math.pi/180)  +  1j*Z3r*math.sin(Z3a*math.pi/180)\n",
+      "\n",
+      "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
+      "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
+      "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,0),\"  +  (\",round(ZA.imag,0),\")i  ohmn, ZB =\",round(ZB.real,0),\"  +  (\",round(ZB.imag,0),\")i  Ohm and ZC =\",round(ZC.real,0),\"  - (\",round(ZC.imag,0),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent Delta connection are, ZA = 50.0   +  ( -250.0 )i  ohmn, ZB = 140.0   +  ( 80.0 )i  Ohm and ZC = 80.0   - ( -140.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_1.ipynb
new file mode 100755
index 00000000..ce641b05
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_1.ipynb
@@ -0,0 +1,449 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 34: Delta-star and star-delta transformations</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 605</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Replace the delta-connected network by an equivalent star connection.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ZA  =  20;#  in  ohm\n",
+      "ZB  =  10 + 10j;#  in  ohm\n",
+      "ZC  =  -20j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Z1 = ZA*ZB/(ZA+ZB+ZC)\n",
+      "Z2 = ZB*ZC/(ZA+ZB+ZC)\n",
+      "Z3 = ZA*ZC/(ZA+ZB+ZC)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent star connection are, Z1 =\",round(Z1.real,2),\"  +  (\",round(Z1.imag,2),\")i  ohmn, Z2 =\",round(Z2.real,2),\"  +  (\",round(Z2.imag,2),\")i  Ohm and Z3 =\",round(Z3.real,2),\"  +  (\",round(Z3.imag,2),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent star connection are, Z1 = 4.0   +  ( 8.0 )i  ohmn, Z2 = 8.0   +  ( -4.0 )i  Ohm and Z3 = 4.0   +  ( -12.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 606</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the equivalent circuit impedance across terminals AB,\n",
+      "#(b) supply current I and (c) the power dissipated in the 10 ohm resistor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  40;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "ZA  =  10j;#  in  ohm\n",
+      "ZB  =  15j;#  in  ohm\n",
+      "ZC  =  25j;#  in  ohm\n",
+      "ZD  =  -8j;#  in  ohm\n",
+      "ZE  =  10;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  network  of  Figure  34.7  is  redrawn,  as  in  Figure  34.8,  \n",
+      "    #showing  more  clearly  the  part  of  the  network  1,  2,  3  forming  a  delta  connection   \n",
+      "    #This  may  he  transformed  into  a  star  connection  as  shown  in  Figure  34.9.\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  equivalent  network  is  shown  in  Figure  34.10  and  is  further  simplified  in  Figure  34.11\n",
+      " #(ZE  +  Z3)  in  parallel  with  (Z1  +  ZD)  gives  an  equivalent  impedance  of\n",
+      "z  =  (ZE  +  Z3)*(Z1  +  ZD)/(Z1  +  ZD  +  ZE  +  Z3)\n",
+      " #Hence  the  total  circuit  equivalent  impedance  across  terminals  AB  is  given  by\n",
+      "Zab  =  z  +  Z2\n",
+      " #Supply  current  I\n",
+      "I  =  V/Zab\n",
+      "I1  =  ((Z1  +  ZD)/(Z1  +  ZD  +  ZE  +  Z3))*I\n",
+      "I1mag  =  abs(I1)\n",
+      " #Power  P  dissipated  in  the  10  ohm  resistance  of  Figure  34.7  is  given  by\n",
+      "Pr10  =  ZE*I1mag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  circuit  impedance  across  terminals  AB  is  \",round(Zab.real,2),\"  +  (\",round(Zab.imag,2),\")i  ohm\"\n",
+      "print  \"\\n  (b)supply  current  I  is  \",round(I.real,2),\"  +  (\",round(I.imag,2),\")i  A\"\n",
+      "print  \"\\n  (c)power  P  dissipated  in  the  10  ohm  resistor  is  \",round(Pr10,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  circuit  impedance  across  terminals  AB  is   2.5   +  ( 2.5 )i  ohm\n",
+        "\n",
+        "  (b)supply  current  I  is   8.0   +  ( -8.0 )i  A\n",
+        "\n",
+        "  (c)power  P  dissipated  in  the  10  ohm  resistor  is   320.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 607</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) the value of the single equivalent resistance that replaces the network between terminals A and B, \n",
+      "#(b) the current supplied by the 52 V source, and (c) the current flowing in the 8 ohm resistance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  52;#  in  volts\n",
+      "ZA  =  8;#  in  ohm\n",
+      "ZB  =  16;#  in  ohm\n",
+      "ZC  =  40;#  in  ohm\n",
+      "ZD  =  1;#  in  ohm\n",
+      "ZE  =  4;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #In  Figure  34.12,  no  resistances  are  directly  in  parallel  or  directly  in  series  with  each  other.  \n",
+      "    #However,  ACD  and  BCD  are  both  delta  connections  and  either  may  be  converted  into  an  equivalent  star  connection.  The  delta  network  BCD  is  redrawn  in  Figure  34.13(a)  and  is  transformed  into  an  equivalent  star  connection  as  shown  in  Figure  34.13(b),  where\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  network  of  Figure  34.12  may  thus  be  redrawn  as  shown  in  Figure  34.14.  \n",
+      "    #The  Z1  and  ZE  are  in  series  with  each  other,  as  are  the  ZD  and  Z3  resistors.  \n",
+      "    #Hence  the  equivalent  network  is  as  shown  in  Figure  34.15.  \n",
+      "    #The  total  equivalent  resistance  across  terminals  A  and  B  is  given  by\n",
+      "Zab  =  (Z1  +  ZE)*(ZD  +  Z3)/(Z1  +  ZE  +  ZD  +  Z3)  +  Z2\n",
+      " #Current  supplied  by  the  source,  i.e.,  current  I  in  Figure  34.15,  is  given  by\n",
+      "I  =  V/Zab\n",
+      " #From  Figure  34.15,  current  I1\n",
+      "I1  =  ((ZD  +  Z3)/(Z1  +  ZE  +  ZD  +  Z3))*I\n",
+      " #current  I2\n",
+      "I2  =  I  -  I1\n",
+      " #From  Figure  34.14,  p.d.  across  AC,\n",
+      "Vac  =  I1*ZE\n",
+      " #p.d.  across  AD\n",
+      "Vad  =  I2*ZD\n",
+      " #Hence  p.d.  between  C  and  D  is  given\n",
+      "Vcd  =  Vac  -  Vad\n",
+      " #current  in  the  8  ohm  resistance\n",
+      "Ir8  =  Vcd/ZA\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  circuit  impedance  across  terminals  AB  is  \",round(Zab,2),\"ohm\"\n",
+      "print  \"\\n  (b)the  current  supplied  by  the  52  V  source  is  \",round(I,2),\"    A\"\n",
+      "print  \"\\n  (c)the  current  flowing  in  the  8  ohm  resistance  is  \",round(Ir8,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  circuit  impedance  across  terminals  AB  is   13.0 ohm\n",
+        "\n",
+        "  (b)the  current  supplied  by  the  52  V  source  is   4.0     A\n",
+        "\n",
+        "  (c)the  current  flowing  in  the  8  ohm  resistance  is   0.75   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 608</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(b)determine the values of RX and LX at balance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  1000;#  in  ohm\n",
+      "R3  =  1000;#  in  ohm\n",
+      "R4  =  500;#  in  ohm\n",
+      "R5  =  200;#  in  ohm\n",
+      "C = 2E-6; # in Farad\n",
+      "\n",
+      "#calculation:\n",
+      "Rx = R2*R4/R3\n",
+      "Lx = R2*C*(R4 + R5 + R4*R5/R3)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (b)Rx is  \",round(Rx,2),\"ohm, Lx is  \",round(Lx,2),\"H\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (b)Rx is   500.0 ohm, Lx is   1.6 H"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 610</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current flowing in the j10 ohm impedance, and \n",
+      "#(b) the power dissipated in the \u000720 ohm impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "ZA  =  25  - 5j;#  in  ohm\n",
+      "ZB  =  15  + 10j;#  in  ohm\n",
+      "ZC  =  20  - 30j;#  in  ohm\n",
+      "ZD  =  20  + 0j;#  in  ohm\n",
+      "ZE  =  0  +  10j;#  in  ohm\n",
+      "ZF  =  2.5  - 5j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  network  may  initially  be  simplified  by  transforming  the  delta  PQR  to  its  equivalent  star  connection  \n",
+      "    #as  represented  by  impedances  Z1,  Z2  and  Z3  in  Figure  34.21.  From  equation  (34.7),\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  network  is  shown  redrawn  in  Figure  34.22  and  further  simplified  in  Figure  34.23,  from  which,\n",
+      "Zab  =  ((Z3  +  ZE)*(ZD  +  Z2)/(Z2  +  ZE  +  ZD  +  Z3))  +  (Z1  +  ZF)\n",
+      " #Current  I1\n",
+      "I1  =  V/Zab\n",
+      " #current  I2\n",
+      "I2  =  ((ZE  +  Z3)/(Z2  +  ZE  +  ZD  +  Z3))*I1\n",
+      " #current  I3\n",
+      "I3  =  I1  -  I2\n",
+      " #The  power  P  dissipated  in  the  ZD  impedance  of  Figure  34.20  is  given  by\n",
+      "Pzd  =  ZD*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  in  the  (0+i10)  ohm  impedance  is  \",round(abs(I3),2),\"    A\"\n",
+      "print  \"\\n  (b)  the  power  dissipated  in  the  (20  +  i0)  ohm  impedance  is  \",round(abs(Pzd),2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  in  the  (0+i10)  ohm  impedance  is   6.0     A\n",
+        "\n",
+        "  (b)  the  power  dissipated  in  the  (20  +  i0)  ohm  impedance  is   80.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 613</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the delta-connected equivalent network for the star-connected impedances.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  10;#  in  ohm\n",
+      "Z2  =  20;#  in  ohm\n",
+      "Z3  =  5j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
+      "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
+      "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,2),\"  +  (\",round(ZA.imag,2),\")i  ohmn, ZB =\",round(ZB.real,2),\"  +  (\",round(ZB.imag,2),\")i  Ohm and ZC =\",round(ZC.real,2),\"  +  (\",round(ZC.imag,2),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent Delta connection are, ZA = 10.0   +  ( 7.5 )i  ohmn, ZB = 30.0   +  ( -40.0 )i  Ohm and ZC = 20.0   +  ( 15.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 613</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert the star to an equivalent delta connection\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1r  =  100;#  in  ohm\n",
+      "Z1a = 0; # in Deg\n",
+      "Z2r  =  63.25;#  in  ohm\n",
+      "Z2a = 18.43; # in Deg\n",
+      "Z3r  =  100;#  in  ohm\n",
+      "Z3a = -90; # in Deg\n",
+      "\n",
+      "#calculation\n",
+      "Z1  =  Z1r*math.cos(Z1a*math.pi/180)  +  1j*Z1r*math.sin(Z1a*math.pi/180)\n",
+      "Z2  =  Z2r*math.cos(Z2a*math.pi/180)  +  1j*Z2r*math.sin(Z2a*math.pi/180)\n",
+      "Z3  =  Z3r*math.cos(Z3a*math.pi/180)  +  1j*Z3r*math.sin(Z3a*math.pi/180)\n",
+      "\n",
+      "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
+      "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
+      "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,0),\"  +  (\",round(ZA.imag,0),\")i  ohmn, ZB =\",round(ZB.real,0),\"  +  (\",round(ZB.imag,0),\")i  Ohm and ZC =\",round(ZC.real,0),\"  - (\",round(ZC.imag,0),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent Delta connection are, ZA = 50.0   +  ( -250.0 )i  ohmn, ZB = 140.0   +  ( 80.0 )i  Ohm and ZC = 80.0   - ( -140.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_2.ipynb
new file mode 100755
index 00000000..ce641b05
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_2.ipynb
@@ -0,0 +1,449 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 34: Delta-star and star-delta transformations</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 605</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Replace the delta-connected network by an equivalent star connection.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ZA  =  20;#  in  ohm\n",
+      "ZB  =  10 + 10j;#  in  ohm\n",
+      "ZC  =  -20j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Z1 = ZA*ZB/(ZA+ZB+ZC)\n",
+      "Z2 = ZB*ZC/(ZA+ZB+ZC)\n",
+      "Z3 = ZA*ZC/(ZA+ZB+ZC)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent star connection are, Z1 =\",round(Z1.real,2),\"  +  (\",round(Z1.imag,2),\")i  ohmn, Z2 =\",round(Z2.real,2),\"  +  (\",round(Z2.imag,2),\")i  Ohm and Z3 =\",round(Z3.real,2),\"  +  (\",round(Z3.imag,2),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent star connection are, Z1 = 4.0   +  ( 8.0 )i  ohmn, Z2 = 8.0   +  ( -4.0 )i  Ohm and Z3 = 4.0   +  ( -12.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 606</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the equivalent circuit impedance across terminals AB,\n",
+      "#(b) supply current I and (c) the power dissipated in the 10 ohm resistor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  40;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "ZA  =  10j;#  in  ohm\n",
+      "ZB  =  15j;#  in  ohm\n",
+      "ZC  =  25j;#  in  ohm\n",
+      "ZD  =  -8j;#  in  ohm\n",
+      "ZE  =  10;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  network  of  Figure  34.7  is  redrawn,  as  in  Figure  34.8,  \n",
+      "    #showing  more  clearly  the  part  of  the  network  1,  2,  3  forming  a  delta  connection   \n",
+      "    #This  may  he  transformed  into  a  star  connection  as  shown  in  Figure  34.9.\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  equivalent  network  is  shown  in  Figure  34.10  and  is  further  simplified  in  Figure  34.11\n",
+      " #(ZE  +  Z3)  in  parallel  with  (Z1  +  ZD)  gives  an  equivalent  impedance  of\n",
+      "z  =  (ZE  +  Z3)*(Z1  +  ZD)/(Z1  +  ZD  +  ZE  +  Z3)\n",
+      " #Hence  the  total  circuit  equivalent  impedance  across  terminals  AB  is  given  by\n",
+      "Zab  =  z  +  Z2\n",
+      " #Supply  current  I\n",
+      "I  =  V/Zab\n",
+      "I1  =  ((Z1  +  ZD)/(Z1  +  ZD  +  ZE  +  Z3))*I\n",
+      "I1mag  =  abs(I1)\n",
+      " #Power  P  dissipated  in  the  10  ohm  resistance  of  Figure  34.7  is  given  by\n",
+      "Pr10  =  ZE*I1mag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  circuit  impedance  across  terminals  AB  is  \",round(Zab.real,2),\"  +  (\",round(Zab.imag,2),\")i  ohm\"\n",
+      "print  \"\\n  (b)supply  current  I  is  \",round(I.real,2),\"  +  (\",round(I.imag,2),\")i  A\"\n",
+      "print  \"\\n  (c)power  P  dissipated  in  the  10  ohm  resistor  is  \",round(Pr10,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  circuit  impedance  across  terminals  AB  is   2.5   +  ( 2.5 )i  ohm\n",
+        "\n",
+        "  (b)supply  current  I  is   8.0   +  ( -8.0 )i  A\n",
+        "\n",
+        "  (c)power  P  dissipated  in  the  10  ohm  resistor  is   320.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 607</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) the value of the single equivalent resistance that replaces the network between terminals A and B, \n",
+      "#(b) the current supplied by the 52 V source, and (c) the current flowing in the 8 ohm resistance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  52;#  in  volts\n",
+      "ZA  =  8;#  in  ohm\n",
+      "ZB  =  16;#  in  ohm\n",
+      "ZC  =  40;#  in  ohm\n",
+      "ZD  =  1;#  in  ohm\n",
+      "ZE  =  4;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #In  Figure  34.12,  no  resistances  are  directly  in  parallel  or  directly  in  series  with  each  other.  \n",
+      "    #However,  ACD  and  BCD  are  both  delta  connections  and  either  may  be  converted  into  an  equivalent  star  connection.  The  delta  network  BCD  is  redrawn  in  Figure  34.13(a)  and  is  transformed  into  an  equivalent  star  connection  as  shown  in  Figure  34.13(b),  where\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  network  of  Figure  34.12  may  thus  be  redrawn  as  shown  in  Figure  34.14.  \n",
+      "    #The  Z1  and  ZE  are  in  series  with  each  other,  as  are  the  ZD  and  Z3  resistors.  \n",
+      "    #Hence  the  equivalent  network  is  as  shown  in  Figure  34.15.  \n",
+      "    #The  total  equivalent  resistance  across  terminals  A  and  B  is  given  by\n",
+      "Zab  =  (Z1  +  ZE)*(ZD  +  Z3)/(Z1  +  ZE  +  ZD  +  Z3)  +  Z2\n",
+      " #Current  supplied  by  the  source,  i.e.,  current  I  in  Figure  34.15,  is  given  by\n",
+      "I  =  V/Zab\n",
+      " #From  Figure  34.15,  current  I1\n",
+      "I1  =  ((ZD  +  Z3)/(Z1  +  ZE  +  ZD  +  Z3))*I\n",
+      " #current  I2\n",
+      "I2  =  I  -  I1\n",
+      " #From  Figure  34.14,  p.d.  across  AC,\n",
+      "Vac  =  I1*ZE\n",
+      " #p.d.  across  AD\n",
+      "Vad  =  I2*ZD\n",
+      " #Hence  p.d.  between  C  and  D  is  given\n",
+      "Vcd  =  Vac  -  Vad\n",
+      " #current  in  the  8  ohm  resistance\n",
+      "Ir8  =  Vcd/ZA\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  circuit  impedance  across  terminals  AB  is  \",round(Zab,2),\"ohm\"\n",
+      "print  \"\\n  (b)the  current  supplied  by  the  52  V  source  is  \",round(I,2),\"    A\"\n",
+      "print  \"\\n  (c)the  current  flowing  in  the  8  ohm  resistance  is  \",round(Ir8,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  circuit  impedance  across  terminals  AB  is   13.0 ohm\n",
+        "\n",
+        "  (b)the  current  supplied  by  the  52  V  source  is   4.0     A\n",
+        "\n",
+        "  (c)the  current  flowing  in  the  8  ohm  resistance  is   0.75   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 608</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(b)determine the values of RX and LX at balance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  1000;#  in  ohm\n",
+      "R3  =  1000;#  in  ohm\n",
+      "R4  =  500;#  in  ohm\n",
+      "R5  =  200;#  in  ohm\n",
+      "C = 2E-6; # in Farad\n",
+      "\n",
+      "#calculation:\n",
+      "Rx = R2*R4/R3\n",
+      "Lx = R2*C*(R4 + R5 + R4*R5/R3)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (b)Rx is  \",round(Rx,2),\"ohm, Lx is  \",round(Lx,2),\"H\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (b)Rx is   500.0 ohm, Lx is   1.6 H"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 610</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current flowing in the j10 ohm impedance, and \n",
+      "#(b) the power dissipated in the \u000720 ohm impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "ZA  =  25  - 5j;#  in  ohm\n",
+      "ZB  =  15  + 10j;#  in  ohm\n",
+      "ZC  =  20  - 30j;#  in  ohm\n",
+      "ZD  =  20  + 0j;#  in  ohm\n",
+      "ZE  =  0  +  10j;#  in  ohm\n",
+      "ZF  =  2.5  - 5j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  network  may  initially  be  simplified  by  transforming  the  delta  PQR  to  its  equivalent  star  connection  \n",
+      "    #as  represented  by  impedances  Z1,  Z2  and  Z3  in  Figure  34.21.  From  equation  (34.7),\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  network  is  shown  redrawn  in  Figure  34.22  and  further  simplified  in  Figure  34.23,  from  which,\n",
+      "Zab  =  ((Z3  +  ZE)*(ZD  +  Z2)/(Z2  +  ZE  +  ZD  +  Z3))  +  (Z1  +  ZF)\n",
+      " #Current  I1\n",
+      "I1  =  V/Zab\n",
+      " #current  I2\n",
+      "I2  =  ((ZE  +  Z3)/(Z2  +  ZE  +  ZD  +  Z3))*I1\n",
+      " #current  I3\n",
+      "I3  =  I1  -  I2\n",
+      " #The  power  P  dissipated  in  the  ZD  impedance  of  Figure  34.20  is  given  by\n",
+      "Pzd  =  ZD*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  in  the  (0+i10)  ohm  impedance  is  \",round(abs(I3),2),\"    A\"\n",
+      "print  \"\\n  (b)  the  power  dissipated  in  the  (20  +  i0)  ohm  impedance  is  \",round(abs(Pzd),2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  in  the  (0+i10)  ohm  impedance  is   6.0     A\n",
+        "\n",
+        "  (b)  the  power  dissipated  in  the  (20  +  i0)  ohm  impedance  is   80.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 613</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the delta-connected equivalent network for the star-connected impedances.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  10;#  in  ohm\n",
+      "Z2  =  20;#  in  ohm\n",
+      "Z3  =  5j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
+      "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
+      "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,2),\"  +  (\",round(ZA.imag,2),\")i  ohmn, ZB =\",round(ZB.real,2),\"  +  (\",round(ZB.imag,2),\")i  Ohm and ZC =\",round(ZC.real,2),\"  +  (\",round(ZC.imag,2),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent Delta connection are, ZA = 10.0   +  ( 7.5 )i  ohmn, ZB = 30.0   +  ( -40.0 )i  Ohm and ZC = 20.0   +  ( 15.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 613</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert the star to an equivalent delta connection\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1r  =  100;#  in  ohm\n",
+      "Z1a = 0; # in Deg\n",
+      "Z2r  =  63.25;#  in  ohm\n",
+      "Z2a = 18.43; # in Deg\n",
+      "Z3r  =  100;#  in  ohm\n",
+      "Z3a = -90; # in Deg\n",
+      "\n",
+      "#calculation\n",
+      "Z1  =  Z1r*math.cos(Z1a*math.pi/180)  +  1j*Z1r*math.sin(Z1a*math.pi/180)\n",
+      "Z2  =  Z2r*math.cos(Z2a*math.pi/180)  +  1j*Z2r*math.sin(Z2a*math.pi/180)\n",
+      "Z3  =  Z3r*math.cos(Z3a*math.pi/180)  +  1j*Z3r*math.sin(Z3a*math.pi/180)\n",
+      "\n",
+      "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
+      "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
+      "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,0),\"  +  (\",round(ZA.imag,0),\")i  ohmn, ZB =\",round(ZB.real,0),\"  +  (\",round(ZB.imag,0),\")i  Ohm and ZC =\",round(ZC.real,0),\"  - (\",round(ZC.imag,0),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent Delta connection are, ZA = 50.0   +  ( -250.0 )i  ohmn, ZB = 140.0   +  ( 80.0 )i  Ohm and ZC = 80.0   - ( -140.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_3.ipynb
new file mode 100755
index 00000000..4c4d28d9
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_34-checkpoint_3.ipynb
@@ -0,0 +1,447 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:27ff567c9a2bfcc7567440cb0e3cbbe64500e9c89ec5f91d53c57a7d3ff6f06f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 34: Delta-star and star-delta transformations</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 605</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ZA  =  20;#  in  ohm\n",
+      "ZB  =  10 + 10j;#  in  ohm\n",
+      "ZC  =  -20j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Z1 = ZA*ZB/(ZA+ZB+ZC)\n",
+      "Z2 = ZB*ZC/(ZA+ZB+ZC)\n",
+      "Z3 = ZA*ZC/(ZA+ZB+ZC)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent star connection are, Z1 =\",round(Z1.real,2),\"  +  (\",round(Z1.imag,2),\")i  ohmn, Z2 =\",round(Z2.real,2),\"  +  (\",round(Z2.imag,2),\")i  Ohm and Z3 =\",round(Z3.real,2),\"  +  (\",round(Z3.imag,2),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent star connection are, Z1 = 4.0   +  ( 8.0 )i  ohmn, Z2 = 8.0   +  ( -4.0 )i  Ohm and Z3 = 4.0   +  ( -12.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 606</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  40;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "ZA  =  10j;#  in  ohm\n",
+      "ZB  =  15j;#  in  ohm\n",
+      "ZC  =  25j;#  in  ohm\n",
+      "ZD  =  -8j;#  in  ohm\n",
+      "ZE  =  10;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  network  of  Figure  34.7  is  redrawn,  as  in  Figure  34.8,  \n",
+      "    #showing  more  clearly  the  part  of  the  network  1,  2,  3  forming  a  delta  connection   \n",
+      "    #This  may  he  transformed  into  a  star  connection  as  shown  in  Figure  34.9.\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  equivalent  network  is  shown  in  Figure  34.10  and  is  further  simplified  in  Figure  34.11\n",
+      " #(ZE  +  Z3)  in  parallel  with  (Z1  +  ZD)  gives  an  equivalent  impedance  of\n",
+      "z  =  (ZE  +  Z3)*(Z1  +  ZD)/(Z1  +  ZD  +  ZE  +  Z3)\n",
+      " #Hence  the  total  circuit  equivalent  impedance  across  terminals  AB  is  given  by\n",
+      "Zab  =  z  +  Z2\n",
+      " #Supply  current  I\n",
+      "I  =  V/Zab\n",
+      "I1  =  ((Z1  +  ZD)/(Z1  +  ZD  +  ZE  +  Z3))*I\n",
+      "I1mag  =  abs(I1)\n",
+      " #Power  P  dissipated  in  the  10  ohm  resistance  of  Figure  34.7  is  given  by\n",
+      "Pr10  =  ZE*I1mag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  circuit  impedance  across  terminals  AB  is  \",round(Zab.real,2),\"  +  (\",round(Zab.imag,2),\")i  ohm\"\n",
+      "print  \"\\n  (b)supply  current  I  is  \",round(I.real,2),\"  +  (\",round(I.imag,2),\")i  A\"\n",
+      "print  \"\\n  (c)power  P  dissipated  in  the  10  ohm  resistor  is  \",round(Pr10,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  circuit  impedance  across  terminals  AB  is   2.5   +  ( 2.5 )i  ohm\n",
+        "\n",
+        "  (b)supply  current  I  is   8.0   +  ( -8.0 )i  A\n",
+        "\n",
+        "  (c)power  P  dissipated  in  the  10  ohm  resistor  is   320.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 607</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  52;#  in  volts\n",
+      "ZA  =  8;#  in  ohm\n",
+      "ZB  =  16;#  in  ohm\n",
+      "ZC  =  40;#  in  ohm\n",
+      "ZD  =  1;#  in  ohm\n",
+      "ZE  =  4;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #In  Figure  34.12,  no  resistances  are  directly  in  parallel  or  directly  in  series  with  each  other.  \n",
+      "    #However,  ACD  and  BCD  are  both  delta  connections  and  either  may  be  converted  into  an  equivalent  star  connection.  The  delta  network  BCD  is  redrawn  in  Figure  34.13(a)  and  is  transformed  into  an  equivalent  star  connection  as  shown  in  Figure  34.13(b),  where\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  network  of  Figure  34.12  may  thus  be  redrawn  as  shown  in  Figure  34.14.  \n",
+      "    #The  Z1  and  ZE  are  in  series  with  each  other,  as  are  the  ZD  and  Z3  resistors.  \n",
+      "    #Hence  the  equivalent  network  is  as  shown  in  Figure  34.15.  \n",
+      "    #The  total  equivalent  resistance  across  terminals  A  and  B  is  given  by\n",
+      "Zab  =  (Z1  +  ZE)*(ZD  +  Z3)/(Z1  +  ZE  +  ZD  +  Z3)  +  Z2\n",
+      " #Current  supplied  by  the  source,  i.e.,  current  I  in  Figure  34.15,  is  given  by\n",
+      "I  =  V/Zab\n",
+      " #From  Figure  34.15,  current  I1\n",
+      "I1  =  ((ZD  +  Z3)/(Z1  +  ZE  +  ZD  +  Z3))*I\n",
+      " #current  I2\n",
+      "I2  =  I  -  I1\n",
+      " #From  Figure  34.14,  p.d.  across  AC,\n",
+      "Vac  =  I1*ZE\n",
+      " #p.d.  across  AD\n",
+      "Vad  =  I2*ZD\n",
+      " #Hence  p.d.  between  C  and  D  is  given\n",
+      "Vcd  =  Vac  -  Vad\n",
+      " #current  in  the  8  ohm  resistance\n",
+      "Ir8  =  Vcd/ZA\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  equivalent  circuit  impedance  across  terminals  AB  is  \",round(Zab,2),\"ohm\"\n",
+      "print  \"\\n  (b)the  current  supplied  by  the  52  V  source  is  \",round(I,2),\"    A\"\n",
+      "print  \"\\n  (c)the  current  flowing  in  the  8  ohm  resistance  is  \",round(Ir8,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  equivalent  circuit  impedance  across  terminals  AB  is   13.0 ohm\n",
+        "\n",
+        "  (b)the  current  supplied  by  the  52  V  source  is   4.0     A\n",
+        "\n",
+        "  (c)the  current  flowing  in  the  8  ohm  resistance  is   0.75   A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 608</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R2  =  1000;#  in  ohm\n",
+      "R3  =  1000;#  in  ohm\n",
+      "R4  =  500;#  in  ohm\n",
+      "R5  =  200;#  in  ohm\n",
+      "C = 2E-6; # in Farad\n",
+      "\n",
+      "#calculation:\n",
+      "Rx = R2*R4/R3\n",
+      "Lx = R2*C*(R4 + R5 + R4*R5/R3)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (b)Rx is  \",round(Rx,2),\"ohm, Lx is  \",round(Lx,2),\"H\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (b)Rx is   500.0 ohm, Lx is   1.6 H"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 610</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "ZA  =  25  - 5j;#  in  ohm\n",
+      "ZB  =  15  + 10j;#  in  ohm\n",
+      "ZC  =  20  - 30j;#  in  ohm\n",
+      "ZD  =  20  + 0j;#  in  ohm\n",
+      "ZE  =  0  +  10j;#  in  ohm\n",
+      "ZF  =  2.5  - 5j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #The  network  may  initially  be  simplified  by  transforming  the  delta  PQR  to  its  equivalent  star  connection  \n",
+      "    #as  represented  by  impedances  Z1,  Z2  and  Z3  in  Figure  34.21.  From  equation  (34.7),\n",
+      "Z1  =  ZA*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z2  =  ZC*ZB/(ZA  +  ZB  +  ZC)\n",
+      "Z3  =  ZA*ZC/(ZA  +  ZB  +  ZC)\n",
+      " #The  network  is  shown  redrawn  in  Figure  34.22  and  further  simplified  in  Figure  34.23,  from  which,\n",
+      "Zab  =  ((Z3  +  ZE)*(ZD  +  Z2)/(Z2  +  ZE  +  ZD  +  Z3))  +  (Z1  +  ZF)\n",
+      " #Current  I1\n",
+      "I1  =  V/Zab\n",
+      " #current  I2\n",
+      "I2  =  ((ZE  +  Z3)/(Z2  +  ZE  +  ZD  +  Z3))*I1\n",
+      " #current  I3\n",
+      "I3  =  I1  -  I2\n",
+      " #The  power  P  dissipated  in  the  ZD  impedance  of  Figure  34.20  is  given  by\n",
+      "Pzd  =  ZD*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  current  flowing  in  the  (0+i10)  ohm  impedance  is  \",round(abs(I3),2),\"    A\"\n",
+      "print  \"\\n  (b)  the  power  dissipated  in  the  (20  +  i0)  ohm  impedance  is  \",round(abs(Pzd),2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  current  flowing  in  the  (0+i10)  ohm  impedance  is   6.0     A\n",
+        "\n",
+        "  (b)  the  power  dissipated  in  the  (20  +  i0)  ohm  impedance  is   80.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 613</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1  =  10;#  in  ohm\n",
+      "Z2  =  20;#  in  ohm\n",
+      "Z3  =  5j;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
+      "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
+      "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,2),\"  +  (\",round(ZA.imag,2),\")i  ohmn, ZB =\",round(ZB.real,2),\"  +  (\",round(ZB.imag,2),\")i  Ohm and ZC =\",round(ZC.real,2),\"  +  (\",round(ZC.imag,2),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent Delta connection are, ZA = 10.0   +  ( 7.5 )i  ohmn, ZB = 30.0   +  ( -40.0 )i  Ohm and ZC = 20.0   +  ( 15.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 613</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Z1r  =  100;#  in  ohm\n",
+      "Z1a = 0; # in Deg\n",
+      "Z2r  =  63.25;#  in  ohm\n",
+      "Z2a = 18.43; # in Deg\n",
+      "Z3r  =  100;#  in  ohm\n",
+      "Z3a = -90; # in Deg\n",
+      "\n",
+      "#calculation\n",
+      "Z1  =  Z1r*math.cos(Z1a*math.pi/180)  +  1j*Z1r*math.sin(Z1a*math.pi/180)\n",
+      "Z2  =  Z2r*math.cos(Z2a*math.pi/180)  +  1j*Z2r*math.sin(Z2a*math.pi/180)\n",
+      "Z3  =  Z3r*math.cos(Z3a*math.pi/180)  +  1j*Z3r*math.sin(Z3a*math.pi/180)\n",
+      "\n",
+      "ZA = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z2\n",
+      "ZB = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z3\n",
+      "ZC = (Z1*Z2 + Z2*Z3 + Z3*Z1)/Z1\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  Resistances of equivalent Delta connection are, ZA =\",round(ZA.real,0),\"  +  (\",round(ZA.imag,0),\")i  ohmn, ZB =\",round(ZB.real,0),\"  +  (\",round(ZB.imag,0),\")i  Ohm and ZC =\",round(ZC.real,0),\"  - (\",round(ZC.imag,0),\")i  Ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  Resistances of equivalent Delta connection are, ZA = 50.0   +  ( -250.0 )i  ohmn, ZB = 140.0   +  ( 80.0 )i  Ohm and ZC = 80.0   - ( -140.0 )i  Ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint.ipynb
new file mode 100755
index 00000000..7462b3a9
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint.ipynb
@@ -0,0 +1,637 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 35: Maximum power transfer theorems and impedance matching</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of R for maximum power to be transferred from the source to the load,\n",
+      "#and (b) the value of the maximum power delivered to R\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "Z  =  15  +  1j*20;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  occurs  when  R  =  mod(Z)\n",
+      "R  =  abs(Z)\n",
+      " #the  total  circuit  impedance\n",
+      "ZT  =  Z  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  delivered\n",
+      "P  =  R*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",abs(P),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   25.0   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   180.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine\n",
+      "#(a) the value of Z that results in maximum power transfer, and\n",
+      "#(b) the value of the maximum power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "Z  =  15  +  1j*20;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  occurs  when  X  =  -1*imag(Z)  and  R  =  real(Z)\n",
+      "z  =  Z.real  -  1j*Z.imag\n",
+      " #Total  circuit  impedance  at  maximum  power  transfer  condition,\n",
+      "ZT  =  Z  +  z\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  delivered\n",
+      "P  =  Z.real*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  Z  is  \",z.real,\"  +  (\",  z.imag,\")i  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",abs(P),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  Z  is   15.0   +  ( -20.0 )i  ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   240.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine\n",
+      "#(a) the value of the load resistance R required for maximum power transfer, and\n",
+      "#(b) the value of the maximum power transferred.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  200;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  100;#  in  ohm\n",
+      "C  =  1E-6;#  in  farad\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Hence  source  impedance,\n",
+      "z  =  R1*(1j*Xc)/(R1  +  1j*Xc)\n",
+      " #maximum  power  transfer  is  achieved  when  R  =  mod(z)\n",
+      "R  =  abs(z)\n",
+      " #Total  circuit  impedance  at  maximum  power  transfer  condition,\n",
+      "ZT  =  z  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",round(R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   84.67   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   127.9   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 621</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of R for which the power transferred to the load is a maximum, and \n",
+      "#(b) the value of the maximum power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  60;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "XL  =  10;#  in  ohm\n",
+      "Xc  =  7;#  in  ohm\n",
+      "R2  =  XL*1j;#  in  ohm\n",
+      "R3  =  -1j*Xc;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  is  achieved  when\n",
+      "R  =  (R1**2  +  (XL  -  Xc)**2)**0.5\n",
+      " #Hence  source  impedance,\n",
+      "ZT  =  R1  +  R2  +  R3  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   5.0   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   200.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 622</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the load resistance R\n",
+      "#calculate the value of this power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  20;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  15;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #R  is  removed  from  the  network  as  shown  in  Figure  35.6\n",
+      " #P.d.  across  AB,  E\n",
+      "E  =  (R2/(R1  +  R2))*V\n",
+      " #Impedance  \u2018looking-in\u2019  at  terminals  AB  with  the  source  removed  is  given  by\n",
+      "r  =  R1*R2/(R1  +  R2)\n",
+      " #The  equivalent  Th\u00b4evenin  circuit  supplying  terminals  AB  is  shown  in  Figure  35.7.  \n",
+      "    #From  condition  (2),  for  maximum  power  transfer\n",
+      "R  =  r\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  E/(R  +  r)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   15.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 622</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the values of R and X that will result in maximum power being transferred across terminals AB, and \n",
+      "#(b) the value of the maximum power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  10j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Resistance  R  and  reactance  X  are  removed  from  the  network  as  shown  in  Figure  35.9\n",
+      " #P.d.  across  AB,\n",
+      "E  =  ((R2  +  R3)/(R1  +  R2  +  R3))*V\n",
+      " #With  the  source  removed  the  impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by:\n",
+      "z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)\n",
+      " #The  equivalent  Th\u00b4evenin  circuit  is  shown  in  Figure  35.10.  From  condition  3, \n",
+      "    #maximum  power  transfer  is  achieved  when  X  =  -1*imag(z)  and  R  =  real(z)\n",
+      "X  =  -1*z.imag\n",
+      "R  =  z.real\n",
+      "Z  =  R  +  1j*X\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  E/(z  +  Z)\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm  and  X  is  \",  X,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm  and  X  is   -1.25   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   416.67   W"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 624</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the optimum value of load resistance for maximum power transfer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ro  =  448;#  in  ohm\n",
+      "tr  =  8;#  turn  ratio  N1/N2\n",
+      "\n",
+      " #calculation:  \n",
+      " #The  equivalent  input  resistance  r  of  the  transformer  must  be  Ro  for  maximum  power  transfer.\n",
+      "r  =  Ro\n",
+      "RL  =  r*(1/tr)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  value  of  load  resistance  is  \",RL,\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  value  of  load  resistance  is   7.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 624</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the turns ratio of an ideal transformer necessary to match the generator to a load of (40 + j19) ohm \n",
+      "#for maximum transfer of power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  450  +  1j*60;#  in  ohm\n",
+      "ZL  =  40  +  1j*19;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #transformer  turns  ratio  tr  =  (N1/N2)\n",
+      "Zomag  =  abs(Zo)\n",
+      "ZLmag  =  abs(ZL)\n",
+      "tr  =  (Zomag/ZLmag)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  turns  ratio  is  \",round(tr,2),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  turns  ratio  is   3.2 "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 625</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the primary current flowing, and \n",
+      "#(b) the power dissipated in the load resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  240;#  in  volts\n",
+      "V2  =  1920;#  in  volts\n",
+      "R1  =  5;#  in  ohms\n",
+      "R2  =  1600;#  in  ohms\n",
+      "\n",
+      "#calculation:  \n",
+      " #The  network  is  shown  in  Figure  35.12.\n",
+      " #turn  ratio  N1/N2  =  V1/V2\n",
+      "tr  =  V1/V2\n",
+      " #Equivalent  input  resistance  of  the  transformer,\n",
+      "RL  =  R2\n",
+      "r  =  RL*tr**2\n",
+      " #Total  input  resistance,\n",
+      "Rin  =  R1  +  r\n",
+      " #primary  current,  I1\n",
+      "I1  =  V1/Rin\n",
+      " #For  an  ideal  transformer  V1/V2  =  I2/I1\n",
+      "I2  =  I1*(V1/V2)\n",
+      " #Power  dissipated  in  the  load  resistance\n",
+      "P  =  RL*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  primary  current  flowing  is  \",I1,\"  A\"\n",
+      "print  \"\\n  (b)  Power  dissipated  in  the  load  resistance  is  \",P,\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  primary  current  flowing  is   8.0   A\n",
+        "\n",
+        "  (b)  Power  dissipated  in  the  load  resistance  is   1600.0 W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 625</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for maximum power transfer (a) the value of the load resistance,\n",
+      "# and (b) the power dissipated in the load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  30;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "r  =  20000;#  in  ohms\n",
+      "tr  =  20;#  turn  ratio\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)  \n",
+      " #The  network  diagram  is  shown  in  Figure  35.13.\n",
+      " #For  maximum  power  transfer,  r1  must  be  equal  to\n",
+      "r1  =  r\n",
+      " #load  resistance  RL\n",
+      "RL  =  r1/tr**2\n",
+      " #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is\n",
+      "RT  =  r  +  r1\n",
+      " #Primary  current\n",
+      "I1  =  V/RT\n",
+      " #N1/N2  =  I2/I1\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistance  RL  is  given  by\n",
+      "P  =  RL*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  the  load  resistance  is  \",RL,\"  ohm\"\n",
+      "print  \"\\n  (b)  Power  dissipated  in  the  load  resistance  is  \",abs(P*1000),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  the  load  resistance  is   50.0   ohm\n",
+        "\n",
+        "  (b)  Power  dissipated  in  the  load  resistance  is   11.25 mW"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_1.ipynb
new file mode 100755
index 00000000..7462b3a9
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_1.ipynb
@@ -0,0 +1,637 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 35: Maximum power transfer theorems and impedance matching</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of R for maximum power to be transferred from the source to the load,\n",
+      "#and (b) the value of the maximum power delivered to R\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "Z  =  15  +  1j*20;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  occurs  when  R  =  mod(Z)\n",
+      "R  =  abs(Z)\n",
+      " #the  total  circuit  impedance\n",
+      "ZT  =  Z  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  delivered\n",
+      "P  =  R*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",abs(P),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   25.0   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   180.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine\n",
+      "#(a) the value of Z that results in maximum power transfer, and\n",
+      "#(b) the value of the maximum power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "Z  =  15  +  1j*20;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  occurs  when  X  =  -1*imag(Z)  and  R  =  real(Z)\n",
+      "z  =  Z.real  -  1j*Z.imag\n",
+      " #Total  circuit  impedance  at  maximum  power  transfer  condition,\n",
+      "ZT  =  Z  +  z\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  delivered\n",
+      "P  =  Z.real*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  Z  is  \",z.real,\"  +  (\",  z.imag,\")i  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",abs(P),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  Z  is   15.0   +  ( -20.0 )i  ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   240.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine\n",
+      "#(a) the value of the load resistance R required for maximum power transfer, and\n",
+      "#(b) the value of the maximum power transferred.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  200;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  100;#  in  ohm\n",
+      "C  =  1E-6;#  in  farad\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Hence  source  impedance,\n",
+      "z  =  R1*(1j*Xc)/(R1  +  1j*Xc)\n",
+      " #maximum  power  transfer  is  achieved  when  R  =  mod(z)\n",
+      "R  =  abs(z)\n",
+      " #Total  circuit  impedance  at  maximum  power  transfer  condition,\n",
+      "ZT  =  z  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",round(R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   84.67   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   127.9   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 621</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of R for which the power transferred to the load is a maximum, and \n",
+      "#(b) the value of the maximum power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  60;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "XL  =  10;#  in  ohm\n",
+      "Xc  =  7;#  in  ohm\n",
+      "R2  =  XL*1j;#  in  ohm\n",
+      "R3  =  -1j*Xc;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  is  achieved  when\n",
+      "R  =  (R1**2  +  (XL  -  Xc)**2)**0.5\n",
+      " #Hence  source  impedance,\n",
+      "ZT  =  R1  +  R2  +  R3  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   5.0   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   200.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 622</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the load resistance R\n",
+      "#calculate the value of this power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  20;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  15;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #R  is  removed  from  the  network  as  shown  in  Figure  35.6\n",
+      " #P.d.  across  AB,  E\n",
+      "E  =  (R2/(R1  +  R2))*V\n",
+      " #Impedance  \u2018looking-in\u2019  at  terminals  AB  with  the  source  removed  is  given  by\n",
+      "r  =  R1*R2/(R1  +  R2)\n",
+      " #The  equivalent  Th\u00b4evenin  circuit  supplying  terminals  AB  is  shown  in  Figure  35.7.  \n",
+      "    #From  condition  (2),  for  maximum  power  transfer\n",
+      "R  =  r\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  E/(R  +  r)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   15.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 622</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the values of R and X that will result in maximum power being transferred across terminals AB, and \n",
+      "#(b) the value of the maximum power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  10j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Resistance  R  and  reactance  X  are  removed  from  the  network  as  shown  in  Figure  35.9\n",
+      " #P.d.  across  AB,\n",
+      "E  =  ((R2  +  R3)/(R1  +  R2  +  R3))*V\n",
+      " #With  the  source  removed  the  impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by:\n",
+      "z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)\n",
+      " #The  equivalent  Th\u00b4evenin  circuit  is  shown  in  Figure  35.10.  From  condition  3, \n",
+      "    #maximum  power  transfer  is  achieved  when  X  =  -1*imag(z)  and  R  =  real(z)\n",
+      "X  =  -1*z.imag\n",
+      "R  =  z.real\n",
+      "Z  =  R  +  1j*X\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  E/(z  +  Z)\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm  and  X  is  \",  X,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm  and  X  is   -1.25   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   416.67   W"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 624</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the optimum value of load resistance for maximum power transfer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ro  =  448;#  in  ohm\n",
+      "tr  =  8;#  turn  ratio  N1/N2\n",
+      "\n",
+      " #calculation:  \n",
+      " #The  equivalent  input  resistance  r  of  the  transformer  must  be  Ro  for  maximum  power  transfer.\n",
+      "r  =  Ro\n",
+      "RL  =  r*(1/tr)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  value  of  load  resistance  is  \",RL,\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  value  of  load  resistance  is   7.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 624</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the turns ratio of an ideal transformer necessary to match the generator to a load of (40 + j19) ohm \n",
+      "#for maximum transfer of power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  450  +  1j*60;#  in  ohm\n",
+      "ZL  =  40  +  1j*19;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #transformer  turns  ratio  tr  =  (N1/N2)\n",
+      "Zomag  =  abs(Zo)\n",
+      "ZLmag  =  abs(ZL)\n",
+      "tr  =  (Zomag/ZLmag)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  turns  ratio  is  \",round(tr,2),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  turns  ratio  is   3.2 "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 625</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the primary current flowing, and \n",
+      "#(b) the power dissipated in the load resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  240;#  in  volts\n",
+      "V2  =  1920;#  in  volts\n",
+      "R1  =  5;#  in  ohms\n",
+      "R2  =  1600;#  in  ohms\n",
+      "\n",
+      "#calculation:  \n",
+      " #The  network  is  shown  in  Figure  35.12.\n",
+      " #turn  ratio  N1/N2  =  V1/V2\n",
+      "tr  =  V1/V2\n",
+      " #Equivalent  input  resistance  of  the  transformer,\n",
+      "RL  =  R2\n",
+      "r  =  RL*tr**2\n",
+      " #Total  input  resistance,\n",
+      "Rin  =  R1  +  r\n",
+      " #primary  current,  I1\n",
+      "I1  =  V1/Rin\n",
+      " #For  an  ideal  transformer  V1/V2  =  I2/I1\n",
+      "I2  =  I1*(V1/V2)\n",
+      " #Power  dissipated  in  the  load  resistance\n",
+      "P  =  RL*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  primary  current  flowing  is  \",I1,\"  A\"\n",
+      "print  \"\\n  (b)  Power  dissipated  in  the  load  resistance  is  \",P,\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  primary  current  flowing  is   8.0   A\n",
+        "\n",
+        "  (b)  Power  dissipated  in  the  load  resistance  is   1600.0 W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 625</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for maximum power transfer (a) the value of the load resistance,\n",
+      "# and (b) the power dissipated in the load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  30;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "r  =  20000;#  in  ohms\n",
+      "tr  =  20;#  turn  ratio\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)  \n",
+      " #The  network  diagram  is  shown  in  Figure  35.13.\n",
+      " #For  maximum  power  transfer,  r1  must  be  equal  to\n",
+      "r1  =  r\n",
+      " #load  resistance  RL\n",
+      "RL  =  r1/tr**2\n",
+      " #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is\n",
+      "RT  =  r  +  r1\n",
+      " #Primary  current\n",
+      "I1  =  V/RT\n",
+      " #N1/N2  =  I2/I1\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistance  RL  is  given  by\n",
+      "P  =  RL*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  the  load  resistance  is  \",RL,\"  ohm\"\n",
+      "print  \"\\n  (b)  Power  dissipated  in  the  load  resistance  is  \",abs(P*1000),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  the  load  resistance  is   50.0   ohm\n",
+        "\n",
+        "  (b)  Power  dissipated  in  the  load  resistance  is   11.25 mW"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_2.ipynb
new file mode 100755
index 00000000..7462b3a9
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_2.ipynb
@@ -0,0 +1,637 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 35: Maximum power transfer theorems and impedance matching</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of R for maximum power to be transferred from the source to the load,\n",
+      "#and (b) the value of the maximum power delivered to R\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "Z  =  15  +  1j*20;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  occurs  when  R  =  mod(Z)\n",
+      "R  =  abs(Z)\n",
+      " #the  total  circuit  impedance\n",
+      "ZT  =  Z  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  delivered\n",
+      "P  =  R*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",abs(P),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   25.0   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   180.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine\n",
+      "#(a) the value of Z that results in maximum power transfer, and\n",
+      "#(b) the value of the maximum power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "Z  =  15  +  1j*20;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  occurs  when  X  =  -1*imag(Z)  and  R  =  real(Z)\n",
+      "z  =  Z.real  -  1j*Z.imag\n",
+      " #Total  circuit  impedance  at  maximum  power  transfer  condition,\n",
+      "ZT  =  Z  +  z\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  delivered\n",
+      "P  =  Z.real*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  Z  is  \",z.real,\"  +  (\",  z.imag,\")i  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",abs(P),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  Z  is   15.0   +  ( -20.0 )i  ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   240.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine\n",
+      "#(a) the value of the load resistance R required for maximum power transfer, and\n",
+      "#(b) the value of the maximum power transferred.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  200;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  100;#  in  ohm\n",
+      "C  =  1E-6;#  in  farad\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Hence  source  impedance,\n",
+      "z  =  R1*(1j*Xc)/(R1  +  1j*Xc)\n",
+      " #maximum  power  transfer  is  achieved  when  R  =  mod(z)\n",
+      "R  =  abs(z)\n",
+      " #Total  circuit  impedance  at  maximum  power  transfer  condition,\n",
+      "ZT  =  z  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",round(R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   84.67   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   127.9   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 621</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of R for which the power transferred to the load is a maximum, and \n",
+      "#(b) the value of the maximum power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  60;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "XL  =  10;#  in  ohm\n",
+      "Xc  =  7;#  in  ohm\n",
+      "R2  =  XL*1j;#  in  ohm\n",
+      "R3  =  -1j*Xc;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  is  achieved  when\n",
+      "R  =  (R1**2  +  (XL  -  Xc)**2)**0.5\n",
+      " #Hence  source  impedance,\n",
+      "ZT  =  R1  +  R2  +  R3  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   5.0   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   200.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 622</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the load resistance R\n",
+      "#calculate the value of this power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  20;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  15;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #R  is  removed  from  the  network  as  shown  in  Figure  35.6\n",
+      " #P.d.  across  AB,  E\n",
+      "E  =  (R2/(R1  +  R2))*V\n",
+      " #Impedance  \u2018looking-in\u2019  at  terminals  AB  with  the  source  removed  is  given  by\n",
+      "r  =  R1*R2/(R1  +  R2)\n",
+      " #The  equivalent  Th\u00b4evenin  circuit  supplying  terminals  AB  is  shown  in  Figure  35.7.  \n",
+      "    #From  condition  (2),  for  maximum  power  transfer\n",
+      "R  =  r\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  E/(R  +  r)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   15.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 622</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the values of R and X that will result in maximum power being transferred across terminals AB, and \n",
+      "#(b) the value of the maximum power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  10j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Resistance  R  and  reactance  X  are  removed  from  the  network  as  shown  in  Figure  35.9\n",
+      " #P.d.  across  AB,\n",
+      "E  =  ((R2  +  R3)/(R1  +  R2  +  R3))*V\n",
+      " #With  the  source  removed  the  impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by:\n",
+      "z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)\n",
+      " #The  equivalent  Th\u00b4evenin  circuit  is  shown  in  Figure  35.10.  From  condition  3, \n",
+      "    #maximum  power  transfer  is  achieved  when  X  =  -1*imag(z)  and  R  =  real(z)\n",
+      "X  =  -1*z.imag\n",
+      "R  =  z.real\n",
+      "Z  =  R  +  1j*X\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  E/(z  +  Z)\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm  and  X  is  \",  X,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm  and  X  is   -1.25   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   416.67   W"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 624</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the optimum value of load resistance for maximum power transfer\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ro  =  448;#  in  ohm\n",
+      "tr  =  8;#  turn  ratio  N1/N2\n",
+      "\n",
+      " #calculation:  \n",
+      " #The  equivalent  input  resistance  r  of  the  transformer  must  be  Ro  for  maximum  power  transfer.\n",
+      "r  =  Ro\n",
+      "RL  =  r*(1/tr)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  value  of  load  resistance  is  \",RL,\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  value  of  load  resistance  is   7.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 624</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the turns ratio of an ideal transformer necessary to match the generator to a load of (40 + j19) ohm \n",
+      "#for maximum transfer of power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  450  +  1j*60;#  in  ohm\n",
+      "ZL  =  40  +  1j*19;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #transformer  turns  ratio  tr  =  (N1/N2)\n",
+      "Zomag  =  abs(Zo)\n",
+      "ZLmag  =  abs(ZL)\n",
+      "tr  =  (Zomag/ZLmag)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  turns  ratio  is  \",round(tr,2),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  turns  ratio  is   3.2 "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 625</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the primary current flowing, and \n",
+      "#(b) the power dissipated in the load resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  240;#  in  volts\n",
+      "V2  =  1920;#  in  volts\n",
+      "R1  =  5;#  in  ohms\n",
+      "R2  =  1600;#  in  ohms\n",
+      "\n",
+      "#calculation:  \n",
+      " #The  network  is  shown  in  Figure  35.12.\n",
+      " #turn  ratio  N1/N2  =  V1/V2\n",
+      "tr  =  V1/V2\n",
+      " #Equivalent  input  resistance  of  the  transformer,\n",
+      "RL  =  R2\n",
+      "r  =  RL*tr**2\n",
+      " #Total  input  resistance,\n",
+      "Rin  =  R1  +  r\n",
+      " #primary  current,  I1\n",
+      "I1  =  V1/Rin\n",
+      " #For  an  ideal  transformer  V1/V2  =  I2/I1\n",
+      "I2  =  I1*(V1/V2)\n",
+      " #Power  dissipated  in  the  load  resistance\n",
+      "P  =  RL*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  primary  current  flowing  is  \",I1,\"  A\"\n",
+      "print  \"\\n  (b)  Power  dissipated  in  the  load  resistance  is  \",P,\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  primary  current  flowing  is   8.0   A\n",
+        "\n",
+        "  (b)  Power  dissipated  in  the  load  resistance  is   1600.0 W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 625</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for maximum power transfer (a) the value of the load resistance,\n",
+      "# and (b) the power dissipated in the load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  30;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "r  =  20000;#  in  ohms\n",
+      "tr  =  20;#  turn  ratio\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)  \n",
+      " #The  network  diagram  is  shown  in  Figure  35.13.\n",
+      " #For  maximum  power  transfer,  r1  must  be  equal  to\n",
+      "r1  =  r\n",
+      " #load  resistance  RL\n",
+      "RL  =  r1/tr**2\n",
+      " #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is\n",
+      "RT  =  r  +  r1\n",
+      " #Primary  current\n",
+      "I1  =  V/RT\n",
+      " #N1/N2  =  I2/I1\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistance  RL  is  given  by\n",
+      "P  =  RL*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  the  load  resistance  is  \",RL,\"  ohm\"\n",
+      "print  \"\\n  (b)  Power  dissipated  in  the  load  resistance  is  \",abs(P*1000),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  the  load  resistance  is   50.0   ohm\n",
+        "\n",
+        "  (b)  Power  dissipated  in  the  load  resistance  is   11.25 mW"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_3.ipynb
new file mode 100755
index 00000000..d4ffa9bf
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_35-checkpoint_3.ipynb
@@ -0,0 +1,627 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a02e331c8801853e98b04884a3697abba7050e1171bfe09d4dc7ba891e3a3f21"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 35: Maximum power transfer theorems and impedance matching</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "Z  =  15  +  1j*20;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  occurs  when  R  =  mod(Z)\n",
+      "R  =  abs(Z)\n",
+      " #the  total  circuit  impedance\n",
+      "ZT  =  Z  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  delivered\n",
+      "P  =  R*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",abs(P),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   25.0   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   180.0   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  120;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "Z  =  15  +  1j*20;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  occurs  when  X  =  -1*imag(Z)  and  R  =  real(Z)\n",
+      "z  =  Z.real  -  1j*Z.imag\n",
+      " #Total  circuit  impedance  at  maximum  power  transfer  condition,\n",
+      "ZT  =  Z  +  z\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  delivered\n",
+      "P  =  Z.real*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  Z  is  \",z.real,\"  +  (\",  z.imag,\")i  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",abs(P),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  Z  is   15.0   +  ( -20.0 )i  ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   240.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 620</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  200;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  100;#  in  ohm\n",
+      "C  =  1E-6;#  in  farad\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Capacitive  reactance,  Xc\n",
+      "Xc  =  1/(2*math.pi*f*C)\n",
+      " #Hence  source  impedance,\n",
+      "z  =  R1*(1j*Xc)/(R1  +  1j*Xc)\n",
+      " #maximum  power  transfer  is  achieved  when  R  =  mod(z)\n",
+      "R  =  abs(z)\n",
+      " #Total  circuit  impedance  at  maximum  power  transfer  condition,\n",
+      "ZT  =  z  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",round(R,2),\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   84.67   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   127.9   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 621</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  60;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "R1  =  4;#  in  ohm\n",
+      "XL  =  10;#  in  ohm\n",
+      "Xc  =  7;#  in  ohm\n",
+      "R2  =  XL*1j;#  in  ohm\n",
+      "R3  =  -1j*Xc;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #maximum  power  transfer  is  achieved  when\n",
+      "R  =  (R1**2  +  (XL  -  Xc)**2)**0.5\n",
+      " #Hence  source  impedance,\n",
+      "ZT  =  R1  +  R2  +  R3  +  R\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  V/ZT\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   5.0   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   200.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 622</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  20;#  in  volts\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  15;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #R  is  removed  from  the  network  as  shown  in  Figure  35.6\n",
+      " #P.d.  across  AB,  E\n",
+      "E  =  (R2/(R1  +  R2))*V\n",
+      " #Impedance  \u2018looking-in\u2019  at  terminals  AB  with  the  source  removed  is  given  by\n",
+      "r  =  R1*R2/(R1  +  R2)\n",
+      " #The  equivalent  Th\u00b4evenin  circuit  supplying  terminals  AB  is  shown  in  Figure  35.7.  \n",
+      "    #From  condition  (2),  for  maximum  power  transfer\n",
+      "R  =  r\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  E/(R  +  r)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*I**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   15.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 622</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  100;#  in  volts\n",
+      "thetav  =  30;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "R2  =  5;#  in  ohm\n",
+      "R3  =  10j;#  in  ohm\n",
+      "\n",
+      "#calculation:  \n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
+      " #Resistance  R  and  reactance  X  are  removed  from  the  network  as  shown  in  Figure  35.9\n",
+      " #P.d.  across  AB,\n",
+      "E  =  ((R2  +  R3)/(R1  +  R2  +  R3))*V\n",
+      " #With  the  source  removed  the  impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by:\n",
+      "z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)\n",
+      " #The  equivalent  Th\u00b4evenin  circuit  is  shown  in  Figure  35.10.  From  condition  3, \n",
+      "    #maximum  power  transfer  is  achieved  when  X  =  -1*imag(z)  and  R  =  real(z)\n",
+      "X  =  -1*z.imag\n",
+      "R  =  z.real\n",
+      "Z  =  R  +  1j*X\n",
+      " #Current  I  flowing  in  the  load  is  given  by\n",
+      "I  =  E/(z  +  Z)\n",
+      "Imag  =  abs(I)\n",
+      " #maximum  power  transferred,\n",
+      "P  =  R*Imag**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)maximum  power  transfer  occurs  when  R  is  \",R,\"  ohm  and  X  is  \",  X,\"  ohm\"\n",
+      "print  \"\\n  (b)  maximum  power  delivered  is  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)maximum  power  transfer  occurs  when  R  is   3.75   ohm  and  X  is   -1.25   ohm\n",
+        "\n",
+        "  (b)  maximum  power  delivered  is   416.67   W"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 624</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ro  =  448;#  in  ohm\n",
+      "tr  =  8;#  turn  ratio  N1/N2\n",
+      "\n",
+      " #calculation:  \n",
+      " #The  equivalent  input  resistance  r  of  the  transformer  must  be  Ro  for  maximum  power  transfer.\n",
+      "r  =  Ro\n",
+      "RL  =  r*(1/tr)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  optimum  value  of  load  resistance  is  \",RL,\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  optimum  value  of  load  resistance  is   7.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 624</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  450  +  1j*60;#  in  ohm\n",
+      "ZL  =  40  +  1j*19;#  in  ohm\n",
+      "\n",
+      " #calculation:  \n",
+      " #transformer  turns  ratio  tr  =  (N1/N2)\n",
+      "Zomag  =  abs(Zo)\n",
+      "ZLmag  =  abs(ZL)\n",
+      "tr  =  (Zomag/ZLmag)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  transformer  turns  ratio  is  \",round(tr,2),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  transformer  turns  ratio  is   3.2 "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 625</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  240;#  in  volts\n",
+      "V2  =  1920;#  in  volts\n",
+      "R1  =  5;#  in  ohms\n",
+      "R2  =  1600;#  in  ohms\n",
+      "\n",
+      "#calculation:  \n",
+      " #The  network  is  shown  in  Figure  35.12.\n",
+      " #turn  ratio  N1/N2  =  V1/V2\n",
+      "tr  =  V1/V2\n",
+      " #Equivalent  input  resistance  of  the  transformer,\n",
+      "RL  =  R2\n",
+      "r  =  RL*tr**2\n",
+      " #Total  input  resistance,\n",
+      "Rin  =  R1  +  r\n",
+      " #primary  current,  I1\n",
+      "I1  =  V1/Rin\n",
+      " #For  an  ideal  transformer  V1/V2  =  I2/I1\n",
+      "I2  =  I1*(V1/V2)\n",
+      " #Power  dissipated  in  the  load  resistance\n",
+      "P  =  RL*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)  primary  current  flowing  is  \",I1,\"  A\"\n",
+      "print  \"\\n  (b)  Power  dissipated  in  the  load  resistance  is  \",P,\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)  primary  current  flowing  is   8.0   A\n",
+        "\n",
+        "  (b)  Power  dissipated  in  the  load  resistance  is   1600.0 W"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 625</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rv  =  30;#  in  volts\n",
+      "thetav  =  0;#  in  degrees\n",
+      "r  =  20000;#  in  ohms\n",
+      "tr  =  20;#  turn  ratio\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V  =  rv*math.cos(thetav*math.pi/180)  + 1j*rv*math.sin(thetav*math.pi/180)  \n",
+      " #The  network  diagram  is  shown  in  Figure  35.13.\n",
+      " #For  maximum  power  transfer,  r1  must  be  equal  to\n",
+      "r1  =  r\n",
+      " #load  resistance  RL\n",
+      "RL  =  r1/tr**2\n",
+      " #The  total  input  resistance  when  the  source  is  connected  to  the  matching  transformer  is\n",
+      "RT  =  r  +  r1\n",
+      " #Primary  current\n",
+      "I1  =  V/RT\n",
+      " #N1/N2  =  I2/I1\n",
+      "I2  =  I1*tr\n",
+      " #Power  dissipated  in  load  resistance  RL  is  given  by\n",
+      "P  =  RL*I2**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  value  of  the  load  resistance  is  \",RL,\"  ohm\"\n",
+      "print  \"\\n  (b)  Power  dissipated  in  the  load  resistance  is  \",abs(P*1000),\"mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  value  of  the  load  resistance  is   50.0   ohm\n",
+        "\n",
+        "  (b)  Power  dissipated  in  the  load  resistance  is   11.25 mW"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint.ipynb
new file mode 100755
index 00000000..39a16efe
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint.ipynb
@@ -0,0 +1,1017 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 36: Complex Waveforms</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 643</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Write down an expression to represent voltage v.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V = 240; # in Volts\n",
+      "f = 50; # in Hz\n",
+      "x = 0.2;\n",
+      "phi3 = 3*math.pi/4; # in Rad\n",
+      "\n",
+      " #calculation:\n",
+      "Vamp = V*2**0.5\n",
+      "w = 2*math.pi*f\n",
+      "T = 1/f\n",
+      "V3 = Vamp*x\n",
+      "f3 = 3*f\n",
+      "w3 = 3*w\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage, V =\",round(Vamp,1),\"sin(\",round(w,1),\"t) + \",round(V3,1),\"sin(\", round(w3,1),\"t - \", round(phi3,1),\") volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage, V = 339.4 sin( 314.2 t) +  67.9 sin( 942.5 t -  2.4 ) volts"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 648</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the rms value of the current waveform\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  0.100;#  in  amperes\n",
+      "A3  =  0.020;#  in  amperes\n",
+      "A5  =  0.010;#  in  amperes\n",
+      "\n",
+      " #calculation:\n",
+      " #the  rms  value  of  current  is  given  by\n",
+      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  rms  value  of  current  is  \",round(Irms*1000,2),\" mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  rms  value  of  current  is   72.46  mA"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 649</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the voltage, (a) the rms value, (b) the mean value and (c) the form factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  10;#  in  volts\n",
+      "A3  =  3;#  in  volts\n",
+      "A5  =  2;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #the  rms  value  of  voltage  is  given  by\n",
+      "Vrms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      " #the  mean  value  of  voltage  is  given  by\n",
+      " #x  =  wt\n",
+      "Vav  =  (1/math.pi)*((10 + 1 + 2/5)-(-10 - 1 - 2/5))\n",
+      " #form  factor  is  given  by\n",
+      "ff  =  Vrms/Vav\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
+      "print  \"\\n  (b)the  mean  value  of  voltage  is  \",round(Vav,2),\"  V\"\n",
+      "print  \"\\n  (c)form  factor  is  \",round(ff,3),\"  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  rms  value  of  voltage  is   7.52   V\n",
+        "\n",
+        "  (b)the  mean  value  of  voltage  is   7.26   V\n",
+        "\n",
+        "  (c)form  factor  is   1.036   "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 649</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the rms value of the fundamental and each harmonic. \n",
+      "#(b) Write down an expression to represent the complex voltage waveform if the frequency of the fundamental is 31.83 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  volts\n",
+      "x  =  0.3;#  for  third  harmonic\n",
+      "y  =  0.1;#  for  fifth  harmonic\n",
+      "f  =  31.83;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #V3  =  x*V1\n",
+      " #V5  =  y*V1\n",
+      " #the  rms  value  of  the  fundamental,\n",
+      "V1  =  ((V**2)/(1  +  x**2  +  y**2))**0.5\n",
+      " #Rms  value  of  the  third  harmonic\n",
+      "V3  =  x*V1\n",
+      " #the  rms  value  of  the  fifth  harmonic,\n",
+      "V5  =  y*V1\n",
+      " #Maximum  value  of  the  fundamental,\n",
+      "V1m  =  V1*2**0.5\n",
+      " #Maximum  value  of  the  third  harmonic,\n",
+      "V3m  =  V3*2**0.5\n",
+      " #Maximum  value  of  the  fifth  harmonic,\n",
+      "V5m  =  V5*2**0.5\n",
+      "w  =  2*math.pi*f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"v  =  \",round(V1m,2),\"sin\",round(w,2),\"t  +  \",round(V3m,2),\"sin\",round((3*w),2),\"t  +  \",round(V5m,2),\"sin\",round((5*w),2),\"t  Volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "v  =   323.62 sin 199.99 t  +   97.08 sin 599.98 t  +   32.36 sin 999.97 t  Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 652</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the average power in a 20 ohm resistance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  12;#  in  amperes\n",
+      "A3  =  5;#  in  amperes\n",
+      "A5  =  2;#  in  amperes\n",
+      "R  =  20;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #rms  current\n",
+      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      " #average  power\n",
+      "P  =  R*Irms**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  average  power  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  average  power   1730.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 652</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the total active power supplied to the circuit, and (b) the overall power factor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  2;#  in  amperes\n",
+      "Ia3  =  0.3;#  in  amperes\n",
+      "Ia5  =  0.1;#  in  amperes\n",
+      "Va1  =  60;#  in  volts\n",
+      "Va3  =  15;#  in  volts\n",
+      "Va5  =  10;#  in  volts\n",
+      "Phii1  =  -1*math.pi/6;#  in  radians\n",
+      "Phii3  =  -1*math.pi/12;#  in  radians\n",
+      "Phii5  =  -8*math.pi/9;#  in  radians\n",
+      "Phiv1  =  0;#  in  radians\n",
+      "Phiv3  =  math.pi/4;#  in  radians\n",
+      "Phiv5  =  -1*math.pi/2;#  in  radians\n",
+      "\n",
+      "\n",
+      " #calculation:\n",
+      " #rms  values;\n",
+      "I1  =  Ia1/(2**0.5);#  in  amperes\n",
+      "I3  =  Ia3/(2**0.5);#  in  amperes\n",
+      "I5  =  Ia5/(2**0.5);#  in  amperes\n",
+      "V1  =  Va1/(2**0.5);#  in  volts\n",
+      "V3  =  Va3/(2**0.5);#  in  volts\n",
+      "V5  =  Va5/(2**0.5);#  in  volts\n",
+      " #total  power  supplied,\n",
+      "P  =  V1*I1*math.cos(Phiv1  -  Phii1)  +  V3*I3*math.cos(Phiv3  -  Phii3)  +  V5*I5*math.cos(Phiv5  -  Phii5)\n",
+      " #rms  current\n",
+      "Irms  =  ((I1**2  +  I3**2  +  I5**2))**0.5\n",
+      " #rms  voltage\n",
+      "Vrms  =  ((V1**2  +  V3**2  +  V5**2))**0.5\n",
+      " #overall  power  factor\n",
+      "pf  =  P/(Vrms*Irms)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  total  active  power  supplied  to  the  circuit  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(b)overall  power  factor  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  total  active  power  supplied  to  the  circuit   53.26   W\n",
+        "\n",
+        "(b)overall  power  factor   0.84"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 655</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for each case an expression for the current flowing if the fundamental frequency is 1 kHz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  40;#  in  ohm\n",
+      "L  =  7.96E-3;#  in  Henry\n",
+      "C = 25E-6; # in Farad\n",
+      "f = 1000; # in Hx\n",
+      "\n",
+      "#calculation:\n",
+      "wL = 2*math.pi*1000*L\n",
+      "wC = 2*math.pi*1000*C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)i  =  \",round(100/R1,2),\"sin(wt)  +\",round(30/R1,2),\"sin(3wt - pi/3)  +\",round(10/R1,2),\"sin(5wt - pi/6) A\"\n",
+      "print  \"(b)i  =  \",round(100/wL,2),\"sin(wt - pi/2)  +\",round(30/(3*wL),2),\"sin(3wt - pi/6)  +\",round(10/(5*wL),2),\"sin(5wt - 2pi/3) A\"\n",
+      "print  \"(c)i  =  \",round(100*wC,2),\"sin(wt + pi/2)  +\",round(30*3*wC,2),\"sin(3wt + 5pi/6)  +\",round(10*5*wC,2),\"sin(5wt + pi/3) A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)i  =   2.5 sin(wt)  + 0.75 sin(3wt - pi/3)  + 0.25 sin(5wt - pi/6) A\n",
+        "(b)i  =   2.0 sin(wt - pi/2)  + 0.2 sin(3wt - pi/6)  + 0.04 sin(5wt - 2pi/3) A\n",
+        "(c)i  =   15.71 sin(wt + pi/2)  + 14.14 sin(3wt + 5pi/6)  + 7.85 sin(5wt + pi/3) A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 656</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) an expression to represent the instantaneous value of the current, \n",
+      "#(b) the rms voltage, (c) the rms current, (d) the power dissipated, and (e) the overall power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  240;#  in  volts\n",
+      "V3m  =  40;#  in  volts\n",
+      "V5m  =  30;#  in  volts\n",
+      "w1  =  314;#  fundamental\n",
+      "R  =  12;#  in  ohm\n",
+      "L  =  0.00955;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #fundamental  or  first  harmonic\n",
+      " #inductive  reactance,\n",
+      "XL1  =  w1*L\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*XL1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1m/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Third  harmonic\n",
+      "XL3  =  3*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*XL3\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3m/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #fifth  harmonic\n",
+      "XL5  =  5*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z5  =  R  +  1j*XL5\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I5m  =  V5m/Z5\n",
+      "I5mag  =  abs(I5m)\n",
+      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
+      " #rms  voltage\n",
+      "Vrms  =  ((V1m**2  +  V3m**2  +  V5m**2)/2)**0.5\n",
+      " #rms  current\n",
+      "Irms  =  ((I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
+      " #power  dissipated\n",
+      "P  =  R*Irms**2\n",
+      " #overall  power  factor\n",
+      "pf  =  P/(Vrms*Irms)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)i  =  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
+      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
+      "print  \"\\n(c)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
+      "print  \"\\n(d)the  total  power  dissipated  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(e)overall  power  factor  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)i  =   19.4 sin( 314.0 t  +  ( -0.24 ))  +   2.67 sin( 942.0 t  +  ( -0.64 ))  +   1.56 sin( 1570.0 t  +  ( -0.9 ))  A\n",
+        "\n",
+        "(b)the  rms  value  of  current  is   13.89   A\n",
+        "\n",
+        "(c)the  rms  value  of  voltage  is   173.35   V\n",
+        "\n",
+        "(d)the  total  power  dissipated   2316.26   W\n",
+        "\n",
+        "(e)overall  power  factor   0.96"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 658</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Obtain an expression for the current flowing and hence determine the rms value of current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  50;#  in  volts\n",
+      "V1m  =  200;#  in  volts\n",
+      "V2m  =  40;#  in  volts\n",
+      "V4m  =  5;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  50;#  in  ohm\n",
+      "C  =  100E-6;#  in  farad\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv2  =  -1*math.pi/2;#  in  rad\n",
+      "phiv4  =  math.pi/4;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V2  =  V2m*math.cos(phiv2)  +  1j*V2m*math.sin(phiv2)\n",
+      "V4  =  V4m*math.cos(phiv4)  +  1j*V4m*math.sin(phiv4)\n",
+      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
+      "Iom  =  0\n",
+      " #fundamental  or  first  harmonic\n",
+      "w1  =  2*math.pi*f\n",
+      " #inductive  reactance,\n",
+      "Xc1  =  1/(w1*C)\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*Xc1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #second  harmonic\n",
+      "Xc2  =  Xc1/2\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z2  =  R  +  1j*Xc2\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I2m  =  V2/Z2\n",
+      "I2mag  =  abs(I2m)\n",
+      "phii2  =  cmath.phase(complex(I2m.real,I2m.imag))\n",
+      " #fourth  harmonic\n",
+      "Xc4  =  Xc1/4\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z4  =  R  +  1j*Xc4\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I4m  =  V4/Z4\n",
+      "I4mag  =  abs(I4m)\n",
+      "phii4  =  cmath.phase(complex(I4m.real,I4m.imag))\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1mag**2  +  I2mag**2  +  I4mag**2)/2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I2mag,2),\"sin(\",round((w1*2),2),\"t  +  (\",round(phii2,2),\"))  +  \",round(I4mag,2),\"sin(\",round((w1*4),2),\"t  +  (\",round(phii4,2),\"))  A\"\n",
+      "print  \"(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)i =  0.0  +  3.37 sin( 314.16 t  +  ( -0.57 ))  +   0.76 sin( 628.32 t  +  ( -1.88 ))  +   0.1 sin( 1256.64 t  +  ( 0.63 ))  A\n",
+        "(b)the  rms  value  of  current  is   2.45   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 659</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) an expression to represent the current flowing in the circuit, \n",
+      "#(b) the rms value of current, correct to two decimal places, and \n",
+      "#(c) the power dissipated in the circuit, correct to three significant figures.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  25;#  in  volts\n",
+      "V1m  =  100;#  in  volts\n",
+      "V3m  =  40;#  in  volts\n",
+      "V5m  =  20;#  in  volts\n",
+      "w1  =  10000;#  fundamental\n",
+      "R  =  5;#  in  ohm\n",
+      "L  =  500E-6;#  in  Henry\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv3  =  math.pi/6;#  in  rad\n",
+      "phiv5  =  math.pi/12;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V3  =  V3m*math.cos(phiv3)  +  1j*V3m*math.sin(phiv3)\n",
+      "V5  =  V5m*math.cos(phiv5)  +  1j*V5m*math.sin(phiv5)\n",
+      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
+      "Iom  =  Vom/R\n",
+      " #fundamental  or  first  harmonic\n",
+      " #inductive  reactance,\n",
+      "XL1  =  w1*L\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*XL1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      "#Third  harmonic\n",
+      "XL3  =  3*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*XL3\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #fifth  harmonic\n",
+      "XL5  =  5*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z5  =  R  +  1j*XL5\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I5m  =  V5/Z5\n",
+      "I5mag  =  abs(I5m)\n",
+      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
+      " #power  dissipated\n",
+      "P  =  R*Irms**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)i  =  \",round(Iom,2),\"  +  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
+      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
+      "print  \"\\n(c)the  total  power  dissipated  \",round(P,3),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)i  =   5.0   +   14.14 sin( 10000.0 t  +  ( -0.79 ))  +   2.53 sin( 30000.0 t  +  ( -0.73 ))  +   0.78 sin( 50000.0 t  +  ( -1.11 ))  A\n",
+        "\n",
+        "(b)the  rms  value  of  current  is   11.34   A\n",
+        "\n",
+        "(c)the  total  power  dissipated   642.538   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 661</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the average power supplied, (b) the type of components present, and (c) the values of the components.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  30;#  in  volts\n",
+      "V1m  =  40;#  in  volts\n",
+      "V2m  =  25;#  in  volts\n",
+      "V4m  =  15;#  in  volts\n",
+      "Iom  =  0;#  in  amperes\n",
+      "I1m  =  0.743;#  in  Amperes\n",
+      "I2m  =  0.781;#  in  Amperes\n",
+      "I4m  =  0.636;#  in  Amperes\n",
+      "phii1  =  1.190;#  in  rad\n",
+      "phii2  =  0.896;#  in  rad\n",
+      "phii4  =  0.559;#  in  rad\n",
+      "w  =  1000;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #the  average  power  P  is  given  by\n",
+      "P  =  Vom*Iom+(0.707*V1m)*(0.707*I1m)*math.cos(phii1)+(0.707*V2m)*(0.707*I2m)*math.cos(phii2) + (0.707*V4m)*(0.707*I4m)*math.cos(phii4)\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1m**2  +  I2m**2  +  I4m**2)/2)**0.5\n",
+      " #resistance  R\n",
+      "R  =  P/(Irms**2)\n",
+      " #impedance\n",
+      "Z1  =  V1m/I1m\n",
+      " #Xc1\n",
+      "Xc1  =  (Z1**2  -  R**2)**0.5\n",
+      " #capacitance\n",
+      "C  =  1/(w*Xc1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  average  power  P  is  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(c)the  resistance  R  \",round(R,2),\"  ohm  and  capacitance  \",round(C*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  average  power  P  is   15.66   W\n",
+        "\n",
+        "(c)the  resistance  R   19.99   ohm  and  capacitance   20.01 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 662</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) an expression for the supply current, i, (b) the percentage harmonic content of the supply current, (c) the total power dissipated, (d) an expression for the p.d. shown as \t1, and (e) an expression for current ic.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  300;#  in  volts\n",
+      "V3m  =  120;#  in  volts\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv2  =  0.698;#  in  rad\n",
+      "w1  =  314;#  in  rad\n",
+      "C  =  2.123E-6;#  in  farads\n",
+      "R1  =  560;#  in  ohms\n",
+      "R2  =  2000;#  in  Ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V3  =  V3m*math.cos(phiv2)  +  1j*V3m*math.sin(phiv2)\n",
+      " #capacitive  reactance,\n",
+      "Xc1  =  1/(w1*C)\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R1  -  1j*Xc1*R2/(R2  -  1j*Xc1)\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Third  harmonic\n",
+      "Xc3  =  Xc1/3\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R1  -  1j*Xc3*R2/(R2  -  1j*Xc3)\n",
+      "I1m = V1m/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #Percentage  harmonic  content  of  the  supply  current  is  given  by\n",
+      "percent  =  I3mag*100/I1mag\n",
+      " #total  active  power\n",
+      "P  =  (0.707*V1m)*(0.707*I1mag)*math.cos(phiv1  -  phii1)  +  (0.707*V3m)*(0.707*I3m)*math.cos(phiv2  -  phii3)\n",
+      "\n",
+      "I1 = I1m*R2/(R2 - 1j*Xc1)\n",
+      "I3 = I3m*R2/(R2 - 1j*Xc3)\n",
+      "\n",
+      "I1nmag  =  abs(I1)\n",
+      "phini1  =  cmath.phase(complex(I1.real,I1.imag))\n",
+      "I3nmag  =  abs(I3)\n",
+      "phini3  =  cmath.phase(complex(I3.real,I3.imag))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)supply current, i=\", round(I1mag,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
+      "print  \"\\n(b)Percentage  harmonic  content  of  the  supply  current  is  \",round(percent,2),\"  percent\"\n",
+      "print  \"\\n(c)total  active  power  is  \",round(abs(P),2),\"  W\"\n",
+      "print  \"\\n(d)Voltage, v1 =\", round(I1mag*R1,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag*R1,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
+      "print  \"\\n(e)current, ic =\", round(I1nmag,3),\"sin(\", w1,\"t +\",round(phini1,3),\") + \",round(I3nmag,3),\"sin(\", 3*w1,\"t +\",round(phini3,3),\") A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)supply current, i= 0.187 sin( 314 t + 0.643 ) +  0.145 sin( 942 t + 1.305 ) A\n",
+        "\n",
+        "(b)Percentage  harmonic  content  of  the  supply  current  is   77.57   percent\n",
+        "\n",
+        "(c)total  active  power  is   25.34   W\n",
+        "\n",
+        "(d)Voltage, v1 = 104.996 sin( 314 t + 0.643 ) +  81.45 sin( 942 t + 1.305 ) A\n",
+        "\n",
+        "(e)current, ic = 0.15 sin( 314 t + 1.287 ) +  0.141 sin( 942 t + 1.55 ) A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 664</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the fundamental frequency for resonance with the third harmonic, and \n",
+      "#(b) the maximum value of the fundamental and third harmonic components of current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  400;#  in  volts\n",
+      "V3m  =  10;#  in  volts\n",
+      "C  =  0.2E-6;#  in  farads\n",
+      "R  =  2;#  in  ohms\n",
+      "L  =  0.5;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #Resonance  with  the  third  harmonic  means  that\n",
+      "w  =  (1/(9*L*C))**0.5\n",
+      " #fundamental  frequency,  f\n",
+      "f  =  w/(2*math.pi)\n",
+      " #At  the  fundamental  frequency,\n",
+      " #impedance  Z1\n",
+      "Z1  =  R  +  1j*(w*L  -  1/(w*C))\n",
+      "Z1mag  =  abs(Z1)\n",
+      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
+      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
+      "I1m  =  V1m/Z1mag\n",
+      " #At  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*(3*w*L  -  1/(3*w*C))\n",
+      "Z3mag  =  abs(Z3)\n",
+      "phiZ3  =  cmath.phase(complex(Z3.real,Z3.imag))\n",
+      " #Maximum  value  of  current  at  the  third  harmonic  frequency,\n",
+      "I3m  =  V3m/Z3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is  \",round(f,2),\"  Hz\"\n",
+      "print  \"(b)Maximum value of current at fundamental freq. is\",round(abs(I1m),3),\"A \"\n",
+      "print   \"and  at  the  third  harmonic  frequency  \",  abs(I3m),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is   167.76   Hz\n",
+        "(b)Maximum value of current at fundamental freq. is 0.095 A \n",
+        "and  at  the  third  harmonic  frequency   5.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 665</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of n, (b) the maximum value of current at the nth harmonic, \n",
+      "#(c) the p.d. across the capacitor at the nth harmonic and\n",
+      "#(d) the maximum value of the fundamental current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  800;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "x  =  0.015;\n",
+      "C  =  0.122E-6;#  in  farads\n",
+      "R  =  5;#  in  ohms\n",
+      "L  =  0.369;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage  at  nth  harmonic\n",
+      "Vnm  =  x*V1m\n",
+      "w  =  2*math.pi*f\n",
+      " #For  resonance  at  the  nth  harmonic  nwL  =  1/nwC\n",
+      "n  =  1/(w*(L*C)**0.5)\n",
+      " #At  resonance,  impedance\n",
+      "Zn  =  R\n",
+      " #the  maximum  value  of  current  at  the  nth  harmonic\n",
+      "Inm  =  Vnm/Zn\n",
+      " #capacitive  reactance,  at  nth  harmonic\n",
+      "Xcn  =  1/(n*w*C)\n",
+      " #the  p.d.  across  the  capacitor  at  the  nth  harmonic\n",
+      "Vcn  =  Inm*Xcn\n",
+      " #At  the  fundamental  frequency,  inductive  reactance,\n",
+      "XL1  =  w*L\n",
+      " #capacitive  reactance\n",
+      "Xc1  =  1/(w*C)\n",
+      " #Impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*(XL1  -  Xc1)\n",
+      "Z1mag  =  abs(Z1)\n",
+      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
+      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
+      "I1m  =  V1m/Z1mag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)n  =  \",round(n,2),\"\"\n",
+      "print  \"\\n(b)the  maximum  value  of  current  at  the  nth  harmonic  \",round(Inm,2),\"  A\"\n",
+      "print  \"\\n(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is  \",round(Vcn,2),\"\"\n",
+      "print  \"\\n(d)the  maximum  value  of  the  fundamental  current.  \",round(I1m,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)n  =   15.0 \n",
+        "\n",
+        "(b)the  maximum  value  of  current  at  the  nth  harmonic   2.4   A\n",
+        "\n",
+        "(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is   4173.92 \n",
+        "\n",
+        "(d)the  maximum  value  of  the  fundamental  current.   0.03   A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_1.ipynb
new file mode 100755
index 00000000..39a16efe
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_1.ipynb
@@ -0,0 +1,1017 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 36: Complex Waveforms</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 643</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Write down an expression to represent voltage v.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V = 240; # in Volts\n",
+      "f = 50; # in Hz\n",
+      "x = 0.2;\n",
+      "phi3 = 3*math.pi/4; # in Rad\n",
+      "\n",
+      " #calculation:\n",
+      "Vamp = V*2**0.5\n",
+      "w = 2*math.pi*f\n",
+      "T = 1/f\n",
+      "V3 = Vamp*x\n",
+      "f3 = 3*f\n",
+      "w3 = 3*w\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage, V =\",round(Vamp,1),\"sin(\",round(w,1),\"t) + \",round(V3,1),\"sin(\", round(w3,1),\"t - \", round(phi3,1),\") volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage, V = 339.4 sin( 314.2 t) +  67.9 sin( 942.5 t -  2.4 ) volts"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 648</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the rms value of the current waveform\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  0.100;#  in  amperes\n",
+      "A3  =  0.020;#  in  amperes\n",
+      "A5  =  0.010;#  in  amperes\n",
+      "\n",
+      " #calculation:\n",
+      " #the  rms  value  of  current  is  given  by\n",
+      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  rms  value  of  current  is  \",round(Irms*1000,2),\" mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  rms  value  of  current  is   72.46  mA"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 649</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the voltage, (a) the rms value, (b) the mean value and (c) the form factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  10;#  in  volts\n",
+      "A3  =  3;#  in  volts\n",
+      "A5  =  2;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #the  rms  value  of  voltage  is  given  by\n",
+      "Vrms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      " #the  mean  value  of  voltage  is  given  by\n",
+      " #x  =  wt\n",
+      "Vav  =  (1/math.pi)*((10 + 1 + 2/5)-(-10 - 1 - 2/5))\n",
+      " #form  factor  is  given  by\n",
+      "ff  =  Vrms/Vav\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
+      "print  \"\\n  (b)the  mean  value  of  voltage  is  \",round(Vav,2),\"  V\"\n",
+      "print  \"\\n  (c)form  factor  is  \",round(ff,3),\"  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  rms  value  of  voltage  is   7.52   V\n",
+        "\n",
+        "  (b)the  mean  value  of  voltage  is   7.26   V\n",
+        "\n",
+        "  (c)form  factor  is   1.036   "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 649</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the rms value of the fundamental and each harmonic. \n",
+      "#(b) Write down an expression to represent the complex voltage waveform if the frequency of the fundamental is 31.83 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  volts\n",
+      "x  =  0.3;#  for  third  harmonic\n",
+      "y  =  0.1;#  for  fifth  harmonic\n",
+      "f  =  31.83;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #V3  =  x*V1\n",
+      " #V5  =  y*V1\n",
+      " #the  rms  value  of  the  fundamental,\n",
+      "V1  =  ((V**2)/(1  +  x**2  +  y**2))**0.5\n",
+      " #Rms  value  of  the  third  harmonic\n",
+      "V3  =  x*V1\n",
+      " #the  rms  value  of  the  fifth  harmonic,\n",
+      "V5  =  y*V1\n",
+      " #Maximum  value  of  the  fundamental,\n",
+      "V1m  =  V1*2**0.5\n",
+      " #Maximum  value  of  the  third  harmonic,\n",
+      "V3m  =  V3*2**0.5\n",
+      " #Maximum  value  of  the  fifth  harmonic,\n",
+      "V5m  =  V5*2**0.5\n",
+      "w  =  2*math.pi*f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"v  =  \",round(V1m,2),\"sin\",round(w,2),\"t  +  \",round(V3m,2),\"sin\",round((3*w),2),\"t  +  \",round(V5m,2),\"sin\",round((5*w),2),\"t  Volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "v  =   323.62 sin 199.99 t  +   97.08 sin 599.98 t  +   32.36 sin 999.97 t  Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 652</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the average power in a 20 ohm resistance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  12;#  in  amperes\n",
+      "A3  =  5;#  in  amperes\n",
+      "A5  =  2;#  in  amperes\n",
+      "R  =  20;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #rms  current\n",
+      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      " #average  power\n",
+      "P  =  R*Irms**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  average  power  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  average  power   1730.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 652</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the total active power supplied to the circuit, and (b) the overall power factor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  2;#  in  amperes\n",
+      "Ia3  =  0.3;#  in  amperes\n",
+      "Ia5  =  0.1;#  in  amperes\n",
+      "Va1  =  60;#  in  volts\n",
+      "Va3  =  15;#  in  volts\n",
+      "Va5  =  10;#  in  volts\n",
+      "Phii1  =  -1*math.pi/6;#  in  radians\n",
+      "Phii3  =  -1*math.pi/12;#  in  radians\n",
+      "Phii5  =  -8*math.pi/9;#  in  radians\n",
+      "Phiv1  =  0;#  in  radians\n",
+      "Phiv3  =  math.pi/4;#  in  radians\n",
+      "Phiv5  =  -1*math.pi/2;#  in  radians\n",
+      "\n",
+      "\n",
+      " #calculation:\n",
+      " #rms  values;\n",
+      "I1  =  Ia1/(2**0.5);#  in  amperes\n",
+      "I3  =  Ia3/(2**0.5);#  in  amperes\n",
+      "I5  =  Ia5/(2**0.5);#  in  amperes\n",
+      "V1  =  Va1/(2**0.5);#  in  volts\n",
+      "V3  =  Va3/(2**0.5);#  in  volts\n",
+      "V5  =  Va5/(2**0.5);#  in  volts\n",
+      " #total  power  supplied,\n",
+      "P  =  V1*I1*math.cos(Phiv1  -  Phii1)  +  V3*I3*math.cos(Phiv3  -  Phii3)  +  V5*I5*math.cos(Phiv5  -  Phii5)\n",
+      " #rms  current\n",
+      "Irms  =  ((I1**2  +  I3**2  +  I5**2))**0.5\n",
+      " #rms  voltage\n",
+      "Vrms  =  ((V1**2  +  V3**2  +  V5**2))**0.5\n",
+      " #overall  power  factor\n",
+      "pf  =  P/(Vrms*Irms)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  total  active  power  supplied  to  the  circuit  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(b)overall  power  factor  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  total  active  power  supplied  to  the  circuit   53.26   W\n",
+        "\n",
+        "(b)overall  power  factor   0.84"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 655</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for each case an expression for the current flowing if the fundamental frequency is 1 kHz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  40;#  in  ohm\n",
+      "L  =  7.96E-3;#  in  Henry\n",
+      "C = 25E-6; # in Farad\n",
+      "f = 1000; # in Hx\n",
+      "\n",
+      "#calculation:\n",
+      "wL = 2*math.pi*1000*L\n",
+      "wC = 2*math.pi*1000*C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)i  =  \",round(100/R1,2),\"sin(wt)  +\",round(30/R1,2),\"sin(3wt - pi/3)  +\",round(10/R1,2),\"sin(5wt - pi/6) A\"\n",
+      "print  \"(b)i  =  \",round(100/wL,2),\"sin(wt - pi/2)  +\",round(30/(3*wL),2),\"sin(3wt - pi/6)  +\",round(10/(5*wL),2),\"sin(5wt - 2pi/3) A\"\n",
+      "print  \"(c)i  =  \",round(100*wC,2),\"sin(wt + pi/2)  +\",round(30*3*wC,2),\"sin(3wt + 5pi/6)  +\",round(10*5*wC,2),\"sin(5wt + pi/3) A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)i  =   2.5 sin(wt)  + 0.75 sin(3wt - pi/3)  + 0.25 sin(5wt - pi/6) A\n",
+        "(b)i  =   2.0 sin(wt - pi/2)  + 0.2 sin(3wt - pi/6)  + 0.04 sin(5wt - 2pi/3) A\n",
+        "(c)i  =   15.71 sin(wt + pi/2)  + 14.14 sin(3wt + 5pi/6)  + 7.85 sin(5wt + pi/3) A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 656</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) an expression to represent the instantaneous value of the current, \n",
+      "#(b) the rms voltage, (c) the rms current, (d) the power dissipated, and (e) the overall power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  240;#  in  volts\n",
+      "V3m  =  40;#  in  volts\n",
+      "V5m  =  30;#  in  volts\n",
+      "w1  =  314;#  fundamental\n",
+      "R  =  12;#  in  ohm\n",
+      "L  =  0.00955;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #fundamental  or  first  harmonic\n",
+      " #inductive  reactance,\n",
+      "XL1  =  w1*L\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*XL1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1m/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Third  harmonic\n",
+      "XL3  =  3*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*XL3\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3m/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #fifth  harmonic\n",
+      "XL5  =  5*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z5  =  R  +  1j*XL5\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I5m  =  V5m/Z5\n",
+      "I5mag  =  abs(I5m)\n",
+      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
+      " #rms  voltage\n",
+      "Vrms  =  ((V1m**2  +  V3m**2  +  V5m**2)/2)**0.5\n",
+      " #rms  current\n",
+      "Irms  =  ((I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
+      " #power  dissipated\n",
+      "P  =  R*Irms**2\n",
+      " #overall  power  factor\n",
+      "pf  =  P/(Vrms*Irms)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)i  =  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
+      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
+      "print  \"\\n(c)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
+      "print  \"\\n(d)the  total  power  dissipated  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(e)overall  power  factor  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)i  =   19.4 sin( 314.0 t  +  ( -0.24 ))  +   2.67 sin( 942.0 t  +  ( -0.64 ))  +   1.56 sin( 1570.0 t  +  ( -0.9 ))  A\n",
+        "\n",
+        "(b)the  rms  value  of  current  is   13.89   A\n",
+        "\n",
+        "(c)the  rms  value  of  voltage  is   173.35   V\n",
+        "\n",
+        "(d)the  total  power  dissipated   2316.26   W\n",
+        "\n",
+        "(e)overall  power  factor   0.96"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 658</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Obtain an expression for the current flowing and hence determine the rms value of current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  50;#  in  volts\n",
+      "V1m  =  200;#  in  volts\n",
+      "V2m  =  40;#  in  volts\n",
+      "V4m  =  5;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  50;#  in  ohm\n",
+      "C  =  100E-6;#  in  farad\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv2  =  -1*math.pi/2;#  in  rad\n",
+      "phiv4  =  math.pi/4;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V2  =  V2m*math.cos(phiv2)  +  1j*V2m*math.sin(phiv2)\n",
+      "V4  =  V4m*math.cos(phiv4)  +  1j*V4m*math.sin(phiv4)\n",
+      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
+      "Iom  =  0\n",
+      " #fundamental  or  first  harmonic\n",
+      "w1  =  2*math.pi*f\n",
+      " #inductive  reactance,\n",
+      "Xc1  =  1/(w1*C)\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*Xc1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #second  harmonic\n",
+      "Xc2  =  Xc1/2\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z2  =  R  +  1j*Xc2\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I2m  =  V2/Z2\n",
+      "I2mag  =  abs(I2m)\n",
+      "phii2  =  cmath.phase(complex(I2m.real,I2m.imag))\n",
+      " #fourth  harmonic\n",
+      "Xc4  =  Xc1/4\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z4  =  R  +  1j*Xc4\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I4m  =  V4/Z4\n",
+      "I4mag  =  abs(I4m)\n",
+      "phii4  =  cmath.phase(complex(I4m.real,I4m.imag))\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1mag**2  +  I2mag**2  +  I4mag**2)/2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I2mag,2),\"sin(\",round((w1*2),2),\"t  +  (\",round(phii2,2),\"))  +  \",round(I4mag,2),\"sin(\",round((w1*4),2),\"t  +  (\",round(phii4,2),\"))  A\"\n",
+      "print  \"(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)i =  0.0  +  3.37 sin( 314.16 t  +  ( -0.57 ))  +   0.76 sin( 628.32 t  +  ( -1.88 ))  +   0.1 sin( 1256.64 t  +  ( 0.63 ))  A\n",
+        "(b)the  rms  value  of  current  is   2.45   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 659</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) an expression to represent the current flowing in the circuit, \n",
+      "#(b) the rms value of current, correct to two decimal places, and \n",
+      "#(c) the power dissipated in the circuit, correct to three significant figures.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  25;#  in  volts\n",
+      "V1m  =  100;#  in  volts\n",
+      "V3m  =  40;#  in  volts\n",
+      "V5m  =  20;#  in  volts\n",
+      "w1  =  10000;#  fundamental\n",
+      "R  =  5;#  in  ohm\n",
+      "L  =  500E-6;#  in  Henry\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv3  =  math.pi/6;#  in  rad\n",
+      "phiv5  =  math.pi/12;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V3  =  V3m*math.cos(phiv3)  +  1j*V3m*math.sin(phiv3)\n",
+      "V5  =  V5m*math.cos(phiv5)  +  1j*V5m*math.sin(phiv5)\n",
+      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
+      "Iom  =  Vom/R\n",
+      " #fundamental  or  first  harmonic\n",
+      " #inductive  reactance,\n",
+      "XL1  =  w1*L\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*XL1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      "#Third  harmonic\n",
+      "XL3  =  3*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*XL3\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #fifth  harmonic\n",
+      "XL5  =  5*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z5  =  R  +  1j*XL5\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I5m  =  V5/Z5\n",
+      "I5mag  =  abs(I5m)\n",
+      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
+      " #power  dissipated\n",
+      "P  =  R*Irms**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)i  =  \",round(Iom,2),\"  +  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
+      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
+      "print  \"\\n(c)the  total  power  dissipated  \",round(P,3),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)i  =   5.0   +   14.14 sin( 10000.0 t  +  ( -0.79 ))  +   2.53 sin( 30000.0 t  +  ( -0.73 ))  +   0.78 sin( 50000.0 t  +  ( -1.11 ))  A\n",
+        "\n",
+        "(b)the  rms  value  of  current  is   11.34   A\n",
+        "\n",
+        "(c)the  total  power  dissipated   642.538   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 661</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the average power supplied, (b) the type of components present, and (c) the values of the components.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  30;#  in  volts\n",
+      "V1m  =  40;#  in  volts\n",
+      "V2m  =  25;#  in  volts\n",
+      "V4m  =  15;#  in  volts\n",
+      "Iom  =  0;#  in  amperes\n",
+      "I1m  =  0.743;#  in  Amperes\n",
+      "I2m  =  0.781;#  in  Amperes\n",
+      "I4m  =  0.636;#  in  Amperes\n",
+      "phii1  =  1.190;#  in  rad\n",
+      "phii2  =  0.896;#  in  rad\n",
+      "phii4  =  0.559;#  in  rad\n",
+      "w  =  1000;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #the  average  power  P  is  given  by\n",
+      "P  =  Vom*Iom+(0.707*V1m)*(0.707*I1m)*math.cos(phii1)+(0.707*V2m)*(0.707*I2m)*math.cos(phii2) + (0.707*V4m)*(0.707*I4m)*math.cos(phii4)\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1m**2  +  I2m**2  +  I4m**2)/2)**0.5\n",
+      " #resistance  R\n",
+      "R  =  P/(Irms**2)\n",
+      " #impedance\n",
+      "Z1  =  V1m/I1m\n",
+      " #Xc1\n",
+      "Xc1  =  (Z1**2  -  R**2)**0.5\n",
+      " #capacitance\n",
+      "C  =  1/(w*Xc1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  average  power  P  is  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(c)the  resistance  R  \",round(R,2),\"  ohm  and  capacitance  \",round(C*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  average  power  P  is   15.66   W\n",
+        "\n",
+        "(c)the  resistance  R   19.99   ohm  and  capacitance   20.01 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 662</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) an expression for the supply current, i, (b) the percentage harmonic content of the supply current, (c) the total power dissipated, (d) an expression for the p.d. shown as \t1, and (e) an expression for current ic.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  300;#  in  volts\n",
+      "V3m  =  120;#  in  volts\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv2  =  0.698;#  in  rad\n",
+      "w1  =  314;#  in  rad\n",
+      "C  =  2.123E-6;#  in  farads\n",
+      "R1  =  560;#  in  ohms\n",
+      "R2  =  2000;#  in  Ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V3  =  V3m*math.cos(phiv2)  +  1j*V3m*math.sin(phiv2)\n",
+      " #capacitive  reactance,\n",
+      "Xc1  =  1/(w1*C)\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R1  -  1j*Xc1*R2/(R2  -  1j*Xc1)\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Third  harmonic\n",
+      "Xc3  =  Xc1/3\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R1  -  1j*Xc3*R2/(R2  -  1j*Xc3)\n",
+      "I1m = V1m/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #Percentage  harmonic  content  of  the  supply  current  is  given  by\n",
+      "percent  =  I3mag*100/I1mag\n",
+      " #total  active  power\n",
+      "P  =  (0.707*V1m)*(0.707*I1mag)*math.cos(phiv1  -  phii1)  +  (0.707*V3m)*(0.707*I3m)*math.cos(phiv2  -  phii3)\n",
+      "\n",
+      "I1 = I1m*R2/(R2 - 1j*Xc1)\n",
+      "I3 = I3m*R2/(R2 - 1j*Xc3)\n",
+      "\n",
+      "I1nmag  =  abs(I1)\n",
+      "phini1  =  cmath.phase(complex(I1.real,I1.imag))\n",
+      "I3nmag  =  abs(I3)\n",
+      "phini3  =  cmath.phase(complex(I3.real,I3.imag))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)supply current, i=\", round(I1mag,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
+      "print  \"\\n(b)Percentage  harmonic  content  of  the  supply  current  is  \",round(percent,2),\"  percent\"\n",
+      "print  \"\\n(c)total  active  power  is  \",round(abs(P),2),\"  W\"\n",
+      "print  \"\\n(d)Voltage, v1 =\", round(I1mag*R1,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag*R1,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
+      "print  \"\\n(e)current, ic =\", round(I1nmag,3),\"sin(\", w1,\"t +\",round(phini1,3),\") + \",round(I3nmag,3),\"sin(\", 3*w1,\"t +\",round(phini3,3),\") A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)supply current, i= 0.187 sin( 314 t + 0.643 ) +  0.145 sin( 942 t + 1.305 ) A\n",
+        "\n",
+        "(b)Percentage  harmonic  content  of  the  supply  current  is   77.57   percent\n",
+        "\n",
+        "(c)total  active  power  is   25.34   W\n",
+        "\n",
+        "(d)Voltage, v1 = 104.996 sin( 314 t + 0.643 ) +  81.45 sin( 942 t + 1.305 ) A\n",
+        "\n",
+        "(e)current, ic = 0.15 sin( 314 t + 1.287 ) +  0.141 sin( 942 t + 1.55 ) A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 664</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the fundamental frequency for resonance with the third harmonic, and \n",
+      "#(b) the maximum value of the fundamental and third harmonic components of current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  400;#  in  volts\n",
+      "V3m  =  10;#  in  volts\n",
+      "C  =  0.2E-6;#  in  farads\n",
+      "R  =  2;#  in  ohms\n",
+      "L  =  0.5;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #Resonance  with  the  third  harmonic  means  that\n",
+      "w  =  (1/(9*L*C))**0.5\n",
+      " #fundamental  frequency,  f\n",
+      "f  =  w/(2*math.pi)\n",
+      " #At  the  fundamental  frequency,\n",
+      " #impedance  Z1\n",
+      "Z1  =  R  +  1j*(w*L  -  1/(w*C))\n",
+      "Z1mag  =  abs(Z1)\n",
+      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
+      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
+      "I1m  =  V1m/Z1mag\n",
+      " #At  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*(3*w*L  -  1/(3*w*C))\n",
+      "Z3mag  =  abs(Z3)\n",
+      "phiZ3  =  cmath.phase(complex(Z3.real,Z3.imag))\n",
+      " #Maximum  value  of  current  at  the  third  harmonic  frequency,\n",
+      "I3m  =  V3m/Z3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is  \",round(f,2),\"  Hz\"\n",
+      "print  \"(b)Maximum value of current at fundamental freq. is\",round(abs(I1m),3),\"A \"\n",
+      "print   \"and  at  the  third  harmonic  frequency  \",  abs(I3m),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is   167.76   Hz\n",
+        "(b)Maximum value of current at fundamental freq. is 0.095 A \n",
+        "and  at  the  third  harmonic  frequency   5.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 665</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of n, (b) the maximum value of current at the nth harmonic, \n",
+      "#(c) the p.d. across the capacitor at the nth harmonic and\n",
+      "#(d) the maximum value of the fundamental current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  800;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "x  =  0.015;\n",
+      "C  =  0.122E-6;#  in  farads\n",
+      "R  =  5;#  in  ohms\n",
+      "L  =  0.369;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage  at  nth  harmonic\n",
+      "Vnm  =  x*V1m\n",
+      "w  =  2*math.pi*f\n",
+      " #For  resonance  at  the  nth  harmonic  nwL  =  1/nwC\n",
+      "n  =  1/(w*(L*C)**0.5)\n",
+      " #At  resonance,  impedance\n",
+      "Zn  =  R\n",
+      " #the  maximum  value  of  current  at  the  nth  harmonic\n",
+      "Inm  =  Vnm/Zn\n",
+      " #capacitive  reactance,  at  nth  harmonic\n",
+      "Xcn  =  1/(n*w*C)\n",
+      " #the  p.d.  across  the  capacitor  at  the  nth  harmonic\n",
+      "Vcn  =  Inm*Xcn\n",
+      " #At  the  fundamental  frequency,  inductive  reactance,\n",
+      "XL1  =  w*L\n",
+      " #capacitive  reactance\n",
+      "Xc1  =  1/(w*C)\n",
+      " #Impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*(XL1  -  Xc1)\n",
+      "Z1mag  =  abs(Z1)\n",
+      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
+      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
+      "I1m  =  V1m/Z1mag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)n  =  \",round(n,2),\"\"\n",
+      "print  \"\\n(b)the  maximum  value  of  current  at  the  nth  harmonic  \",round(Inm,2),\"  A\"\n",
+      "print  \"\\n(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is  \",round(Vcn,2),\"\"\n",
+      "print  \"\\n(d)the  maximum  value  of  the  fundamental  current.  \",round(I1m,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)n  =   15.0 \n",
+        "\n",
+        "(b)the  maximum  value  of  current  at  the  nth  harmonic   2.4   A\n",
+        "\n",
+        "(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is   4173.92 \n",
+        "\n",
+        "(d)the  maximum  value  of  the  fundamental  current.   0.03   A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_2.ipynb
new file mode 100755
index 00000000..39a16efe
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_2.ipynb
@@ -0,0 +1,1017 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 36: Complex Waveforms</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 643</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Write down an expression to represent voltage v.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V = 240; # in Volts\n",
+      "f = 50; # in Hz\n",
+      "x = 0.2;\n",
+      "phi3 = 3*math.pi/4; # in Rad\n",
+      "\n",
+      " #calculation:\n",
+      "Vamp = V*2**0.5\n",
+      "w = 2*math.pi*f\n",
+      "T = 1/f\n",
+      "V3 = Vamp*x\n",
+      "f3 = 3*f\n",
+      "w3 = 3*w\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage, V =\",round(Vamp,1),\"sin(\",round(w,1),\"t) + \",round(V3,1),\"sin(\", round(w3,1),\"t - \", round(phi3,1),\") volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage, V = 339.4 sin( 314.2 t) +  67.9 sin( 942.5 t -  2.4 ) volts"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 648</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the rms value of the current waveform\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  0.100;#  in  amperes\n",
+      "A3  =  0.020;#  in  amperes\n",
+      "A5  =  0.010;#  in  amperes\n",
+      "\n",
+      " #calculation:\n",
+      " #the  rms  value  of  current  is  given  by\n",
+      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  rms  value  of  current  is  \",round(Irms*1000,2),\" mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  rms  value  of  current  is   72.46  mA"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 649</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the voltage, (a) the rms value, (b) the mean value and (c) the form factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  10;#  in  volts\n",
+      "A3  =  3;#  in  volts\n",
+      "A5  =  2;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #the  rms  value  of  voltage  is  given  by\n",
+      "Vrms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      " #the  mean  value  of  voltage  is  given  by\n",
+      " #x  =  wt\n",
+      "Vav  =  (1/math.pi)*((10 + 1 + 2/5)-(-10 - 1 - 2/5))\n",
+      " #form  factor  is  given  by\n",
+      "ff  =  Vrms/Vav\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
+      "print  \"\\n  (b)the  mean  value  of  voltage  is  \",round(Vav,2),\"  V\"\n",
+      "print  \"\\n  (c)form  factor  is  \",round(ff,3),\"  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  rms  value  of  voltage  is   7.52   V\n",
+        "\n",
+        "  (b)the  mean  value  of  voltage  is   7.26   V\n",
+        "\n",
+        "  (c)form  factor  is   1.036   "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 649</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the rms value of the fundamental and each harmonic. \n",
+      "#(b) Write down an expression to represent the complex voltage waveform if the frequency of the fundamental is 31.83 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  volts\n",
+      "x  =  0.3;#  for  third  harmonic\n",
+      "y  =  0.1;#  for  fifth  harmonic\n",
+      "f  =  31.83;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #V3  =  x*V1\n",
+      " #V5  =  y*V1\n",
+      " #the  rms  value  of  the  fundamental,\n",
+      "V1  =  ((V**2)/(1  +  x**2  +  y**2))**0.5\n",
+      " #Rms  value  of  the  third  harmonic\n",
+      "V3  =  x*V1\n",
+      " #the  rms  value  of  the  fifth  harmonic,\n",
+      "V5  =  y*V1\n",
+      " #Maximum  value  of  the  fundamental,\n",
+      "V1m  =  V1*2**0.5\n",
+      " #Maximum  value  of  the  third  harmonic,\n",
+      "V3m  =  V3*2**0.5\n",
+      " #Maximum  value  of  the  fifth  harmonic,\n",
+      "V5m  =  V5*2**0.5\n",
+      "w  =  2*math.pi*f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"v  =  \",round(V1m,2),\"sin\",round(w,2),\"t  +  \",round(V3m,2),\"sin\",round((3*w),2),\"t  +  \",round(V5m,2),\"sin\",round((5*w),2),\"t  Volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "v  =   323.62 sin 199.99 t  +   97.08 sin 599.98 t  +   32.36 sin 999.97 t  Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 652</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the average power in a 20 ohm resistance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  12;#  in  amperes\n",
+      "A3  =  5;#  in  amperes\n",
+      "A5  =  2;#  in  amperes\n",
+      "R  =  20;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #rms  current\n",
+      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      " #average  power\n",
+      "P  =  R*Irms**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  average  power  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  average  power   1730.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 652</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the total active power supplied to the circuit, and (b) the overall power factor\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  2;#  in  amperes\n",
+      "Ia3  =  0.3;#  in  amperes\n",
+      "Ia5  =  0.1;#  in  amperes\n",
+      "Va1  =  60;#  in  volts\n",
+      "Va3  =  15;#  in  volts\n",
+      "Va5  =  10;#  in  volts\n",
+      "Phii1  =  -1*math.pi/6;#  in  radians\n",
+      "Phii3  =  -1*math.pi/12;#  in  radians\n",
+      "Phii5  =  -8*math.pi/9;#  in  radians\n",
+      "Phiv1  =  0;#  in  radians\n",
+      "Phiv3  =  math.pi/4;#  in  radians\n",
+      "Phiv5  =  -1*math.pi/2;#  in  radians\n",
+      "\n",
+      "\n",
+      " #calculation:\n",
+      " #rms  values;\n",
+      "I1  =  Ia1/(2**0.5);#  in  amperes\n",
+      "I3  =  Ia3/(2**0.5);#  in  amperes\n",
+      "I5  =  Ia5/(2**0.5);#  in  amperes\n",
+      "V1  =  Va1/(2**0.5);#  in  volts\n",
+      "V3  =  Va3/(2**0.5);#  in  volts\n",
+      "V5  =  Va5/(2**0.5);#  in  volts\n",
+      " #total  power  supplied,\n",
+      "P  =  V1*I1*math.cos(Phiv1  -  Phii1)  +  V3*I3*math.cos(Phiv3  -  Phii3)  +  V5*I5*math.cos(Phiv5  -  Phii5)\n",
+      " #rms  current\n",
+      "Irms  =  ((I1**2  +  I3**2  +  I5**2))**0.5\n",
+      " #rms  voltage\n",
+      "Vrms  =  ((V1**2  +  V3**2  +  V5**2))**0.5\n",
+      " #overall  power  factor\n",
+      "pf  =  P/(Vrms*Irms)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  total  active  power  supplied  to  the  circuit  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(b)overall  power  factor  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  total  active  power  supplied  to  the  circuit   53.26   W\n",
+        "\n",
+        "(b)overall  power  factor   0.84"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 655</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for each case an expression for the current flowing if the fundamental frequency is 1 kHz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  40;#  in  ohm\n",
+      "L  =  7.96E-3;#  in  Henry\n",
+      "C = 25E-6; # in Farad\n",
+      "f = 1000; # in Hx\n",
+      "\n",
+      "#calculation:\n",
+      "wL = 2*math.pi*1000*L\n",
+      "wC = 2*math.pi*1000*C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)i  =  \",round(100/R1,2),\"sin(wt)  +\",round(30/R1,2),\"sin(3wt - pi/3)  +\",round(10/R1,2),\"sin(5wt - pi/6) A\"\n",
+      "print  \"(b)i  =  \",round(100/wL,2),\"sin(wt - pi/2)  +\",round(30/(3*wL),2),\"sin(3wt - pi/6)  +\",round(10/(5*wL),2),\"sin(5wt - 2pi/3) A\"\n",
+      "print  \"(c)i  =  \",round(100*wC,2),\"sin(wt + pi/2)  +\",round(30*3*wC,2),\"sin(3wt + 5pi/6)  +\",round(10*5*wC,2),\"sin(5wt + pi/3) A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)i  =   2.5 sin(wt)  + 0.75 sin(3wt - pi/3)  + 0.25 sin(5wt - pi/6) A\n",
+        "(b)i  =   2.0 sin(wt - pi/2)  + 0.2 sin(3wt - pi/6)  + 0.04 sin(5wt - 2pi/3) A\n",
+        "(c)i  =   15.71 sin(wt + pi/2)  + 14.14 sin(3wt + 5pi/6)  + 7.85 sin(5wt + pi/3) A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 656</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) an expression to represent the instantaneous value of the current, \n",
+      "#(b) the rms voltage, (c) the rms current, (d) the power dissipated, and (e) the overall power factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  240;#  in  volts\n",
+      "V3m  =  40;#  in  volts\n",
+      "V5m  =  30;#  in  volts\n",
+      "w1  =  314;#  fundamental\n",
+      "R  =  12;#  in  ohm\n",
+      "L  =  0.00955;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #fundamental  or  first  harmonic\n",
+      " #inductive  reactance,\n",
+      "XL1  =  w1*L\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*XL1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1m/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Third  harmonic\n",
+      "XL3  =  3*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*XL3\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3m/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #fifth  harmonic\n",
+      "XL5  =  5*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z5  =  R  +  1j*XL5\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I5m  =  V5m/Z5\n",
+      "I5mag  =  abs(I5m)\n",
+      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
+      " #rms  voltage\n",
+      "Vrms  =  ((V1m**2  +  V3m**2  +  V5m**2)/2)**0.5\n",
+      " #rms  current\n",
+      "Irms  =  ((I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
+      " #power  dissipated\n",
+      "P  =  R*Irms**2\n",
+      " #overall  power  factor\n",
+      "pf  =  P/(Vrms*Irms)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)i  =  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
+      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
+      "print  \"\\n(c)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
+      "print  \"\\n(d)the  total  power  dissipated  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(e)overall  power  factor  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)i  =   19.4 sin( 314.0 t  +  ( -0.24 ))  +   2.67 sin( 942.0 t  +  ( -0.64 ))  +   1.56 sin( 1570.0 t  +  ( -0.9 ))  A\n",
+        "\n",
+        "(b)the  rms  value  of  current  is   13.89   A\n",
+        "\n",
+        "(c)the  rms  value  of  voltage  is   173.35   V\n",
+        "\n",
+        "(d)the  total  power  dissipated   2316.26   W\n",
+        "\n",
+        "(e)overall  power  factor   0.96"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 658</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Obtain an expression for the current flowing and hence determine the rms value of current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  50;#  in  volts\n",
+      "V1m  =  200;#  in  volts\n",
+      "V2m  =  40;#  in  volts\n",
+      "V4m  =  5;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  50;#  in  ohm\n",
+      "C  =  100E-6;#  in  farad\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv2  =  -1*math.pi/2;#  in  rad\n",
+      "phiv4  =  math.pi/4;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V2  =  V2m*math.cos(phiv2)  +  1j*V2m*math.sin(phiv2)\n",
+      "V4  =  V4m*math.cos(phiv4)  +  1j*V4m*math.sin(phiv4)\n",
+      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
+      "Iom  =  0\n",
+      " #fundamental  or  first  harmonic\n",
+      "w1  =  2*math.pi*f\n",
+      " #inductive  reactance,\n",
+      "Xc1  =  1/(w1*C)\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*Xc1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #second  harmonic\n",
+      "Xc2  =  Xc1/2\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z2  =  R  +  1j*Xc2\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I2m  =  V2/Z2\n",
+      "I2mag  =  abs(I2m)\n",
+      "phii2  =  cmath.phase(complex(I2m.real,I2m.imag))\n",
+      " #fourth  harmonic\n",
+      "Xc4  =  Xc1/4\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z4  =  R  +  1j*Xc4\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I4m  =  V4/Z4\n",
+      "I4mag  =  abs(I4m)\n",
+      "phii4  =  cmath.phase(complex(I4m.real,I4m.imag))\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1mag**2  +  I2mag**2  +  I4mag**2)/2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I2mag,2),\"sin(\",round((w1*2),2),\"t  +  (\",round(phii2,2),\"))  +  \",round(I4mag,2),\"sin(\",round((w1*4),2),\"t  +  (\",round(phii4,2),\"))  A\"\n",
+      "print  \"(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)i =  0.0  +  3.37 sin( 314.16 t  +  ( -0.57 ))  +   0.76 sin( 628.32 t  +  ( -1.88 ))  +   0.1 sin( 1256.64 t  +  ( 0.63 ))  A\n",
+        "(b)the  rms  value  of  current  is   2.45   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 659</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) an expression to represent the current flowing in the circuit, \n",
+      "#(b) the rms value of current, correct to two decimal places, and \n",
+      "#(c) the power dissipated in the circuit, correct to three significant figures.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  25;#  in  volts\n",
+      "V1m  =  100;#  in  volts\n",
+      "V3m  =  40;#  in  volts\n",
+      "V5m  =  20;#  in  volts\n",
+      "w1  =  10000;#  fundamental\n",
+      "R  =  5;#  in  ohm\n",
+      "L  =  500E-6;#  in  Henry\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv3  =  math.pi/6;#  in  rad\n",
+      "phiv5  =  math.pi/12;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V3  =  V3m*math.cos(phiv3)  +  1j*V3m*math.sin(phiv3)\n",
+      "V5  =  V5m*math.cos(phiv5)  +  1j*V5m*math.sin(phiv5)\n",
+      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
+      "Iom  =  Vom/R\n",
+      " #fundamental  or  first  harmonic\n",
+      " #inductive  reactance,\n",
+      "XL1  =  w1*L\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*XL1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      "#Third  harmonic\n",
+      "XL3  =  3*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*XL3\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #fifth  harmonic\n",
+      "XL5  =  5*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z5  =  R  +  1j*XL5\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I5m  =  V5/Z5\n",
+      "I5mag  =  abs(I5m)\n",
+      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
+      " #power  dissipated\n",
+      "P  =  R*Irms**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)i  =  \",round(Iom,2),\"  +  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
+      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
+      "print  \"\\n(c)the  total  power  dissipated  \",round(P,3),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)i  =   5.0   +   14.14 sin( 10000.0 t  +  ( -0.79 ))  +   2.53 sin( 30000.0 t  +  ( -0.73 ))  +   0.78 sin( 50000.0 t  +  ( -1.11 ))  A\n",
+        "\n",
+        "(b)the  rms  value  of  current  is   11.34   A\n",
+        "\n",
+        "(c)the  total  power  dissipated   642.538   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 661</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the average power supplied, (b) the type of components present, and (c) the values of the components.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  30;#  in  volts\n",
+      "V1m  =  40;#  in  volts\n",
+      "V2m  =  25;#  in  volts\n",
+      "V4m  =  15;#  in  volts\n",
+      "Iom  =  0;#  in  amperes\n",
+      "I1m  =  0.743;#  in  Amperes\n",
+      "I2m  =  0.781;#  in  Amperes\n",
+      "I4m  =  0.636;#  in  Amperes\n",
+      "phii1  =  1.190;#  in  rad\n",
+      "phii2  =  0.896;#  in  rad\n",
+      "phii4  =  0.559;#  in  rad\n",
+      "w  =  1000;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #the  average  power  P  is  given  by\n",
+      "P  =  Vom*Iom+(0.707*V1m)*(0.707*I1m)*math.cos(phii1)+(0.707*V2m)*(0.707*I2m)*math.cos(phii2) + (0.707*V4m)*(0.707*I4m)*math.cos(phii4)\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1m**2  +  I2m**2  +  I4m**2)/2)**0.5\n",
+      " #resistance  R\n",
+      "R  =  P/(Irms**2)\n",
+      " #impedance\n",
+      "Z1  =  V1m/I1m\n",
+      " #Xc1\n",
+      "Xc1  =  (Z1**2  -  R**2)**0.5\n",
+      " #capacitance\n",
+      "C  =  1/(w*Xc1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  average  power  P  is  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(c)the  resistance  R  \",round(R,2),\"  ohm  and  capacitance  \",round(C*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  average  power  P  is   15.66   W\n",
+        "\n",
+        "(c)the  resistance  R   19.99   ohm  and  capacitance   20.01 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 662</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) an expression for the supply current, i, (b) the percentage harmonic content of the supply current, (c) the total power dissipated, (d) an expression for the p.d. shown as \t1, and (e) an expression for current ic.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  300;#  in  volts\n",
+      "V3m  =  120;#  in  volts\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv2  =  0.698;#  in  rad\n",
+      "w1  =  314;#  in  rad\n",
+      "C  =  2.123E-6;#  in  farads\n",
+      "R1  =  560;#  in  ohms\n",
+      "R2  =  2000;#  in  Ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V3  =  V3m*math.cos(phiv2)  +  1j*V3m*math.sin(phiv2)\n",
+      " #capacitive  reactance,\n",
+      "Xc1  =  1/(w1*C)\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R1  -  1j*Xc1*R2/(R2  -  1j*Xc1)\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Third  harmonic\n",
+      "Xc3  =  Xc1/3\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R1  -  1j*Xc3*R2/(R2  -  1j*Xc3)\n",
+      "I1m = V1m/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #Percentage  harmonic  content  of  the  supply  current  is  given  by\n",
+      "percent  =  I3mag*100/I1mag\n",
+      " #total  active  power\n",
+      "P  =  (0.707*V1m)*(0.707*I1mag)*math.cos(phiv1  -  phii1)  +  (0.707*V3m)*(0.707*I3m)*math.cos(phiv2  -  phii3)\n",
+      "\n",
+      "I1 = I1m*R2/(R2 - 1j*Xc1)\n",
+      "I3 = I3m*R2/(R2 - 1j*Xc3)\n",
+      "\n",
+      "I1nmag  =  abs(I1)\n",
+      "phini1  =  cmath.phase(complex(I1.real,I1.imag))\n",
+      "I3nmag  =  abs(I3)\n",
+      "phini3  =  cmath.phase(complex(I3.real,I3.imag))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)supply current, i=\", round(I1mag,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
+      "print  \"\\n(b)Percentage  harmonic  content  of  the  supply  current  is  \",round(percent,2),\"  percent\"\n",
+      "print  \"\\n(c)total  active  power  is  \",round(abs(P),2),\"  W\"\n",
+      "print  \"\\n(d)Voltage, v1 =\", round(I1mag*R1,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag*R1,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
+      "print  \"\\n(e)current, ic =\", round(I1nmag,3),\"sin(\", w1,\"t +\",round(phini1,3),\") + \",round(I3nmag,3),\"sin(\", 3*w1,\"t +\",round(phini3,3),\") A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)supply current, i= 0.187 sin( 314 t + 0.643 ) +  0.145 sin( 942 t + 1.305 ) A\n",
+        "\n",
+        "(b)Percentage  harmonic  content  of  the  supply  current  is   77.57   percent\n",
+        "\n",
+        "(c)total  active  power  is   25.34   W\n",
+        "\n",
+        "(d)Voltage, v1 = 104.996 sin( 314 t + 0.643 ) +  81.45 sin( 942 t + 1.305 ) A\n",
+        "\n",
+        "(e)current, ic = 0.15 sin( 314 t + 1.287 ) +  0.141 sin( 942 t + 1.55 ) A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 664</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the fundamental frequency for resonance with the third harmonic, and \n",
+      "#(b) the maximum value of the fundamental and third harmonic components of current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  400;#  in  volts\n",
+      "V3m  =  10;#  in  volts\n",
+      "C  =  0.2E-6;#  in  farads\n",
+      "R  =  2;#  in  ohms\n",
+      "L  =  0.5;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #Resonance  with  the  third  harmonic  means  that\n",
+      "w  =  (1/(9*L*C))**0.5\n",
+      " #fundamental  frequency,  f\n",
+      "f  =  w/(2*math.pi)\n",
+      " #At  the  fundamental  frequency,\n",
+      " #impedance  Z1\n",
+      "Z1  =  R  +  1j*(w*L  -  1/(w*C))\n",
+      "Z1mag  =  abs(Z1)\n",
+      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
+      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
+      "I1m  =  V1m/Z1mag\n",
+      " #At  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*(3*w*L  -  1/(3*w*C))\n",
+      "Z3mag  =  abs(Z3)\n",
+      "phiZ3  =  cmath.phase(complex(Z3.real,Z3.imag))\n",
+      " #Maximum  value  of  current  at  the  third  harmonic  frequency,\n",
+      "I3m  =  V3m/Z3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is  \",round(f,2),\"  Hz\"\n",
+      "print  \"(b)Maximum value of current at fundamental freq. is\",round(abs(I1m),3),\"A \"\n",
+      "print   \"and  at  the  third  harmonic  frequency  \",  abs(I3m),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is   167.76   Hz\n",
+        "(b)Maximum value of current at fundamental freq. is 0.095 A \n",
+        "and  at  the  third  harmonic  frequency   5.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 665</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of n, (b) the maximum value of current at the nth harmonic, \n",
+      "#(c) the p.d. across the capacitor at the nth harmonic and\n",
+      "#(d) the maximum value of the fundamental current.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  800;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "x  =  0.015;\n",
+      "C  =  0.122E-6;#  in  farads\n",
+      "R  =  5;#  in  ohms\n",
+      "L  =  0.369;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage  at  nth  harmonic\n",
+      "Vnm  =  x*V1m\n",
+      "w  =  2*math.pi*f\n",
+      " #For  resonance  at  the  nth  harmonic  nwL  =  1/nwC\n",
+      "n  =  1/(w*(L*C)**0.5)\n",
+      " #At  resonance,  impedance\n",
+      "Zn  =  R\n",
+      " #the  maximum  value  of  current  at  the  nth  harmonic\n",
+      "Inm  =  Vnm/Zn\n",
+      " #capacitive  reactance,  at  nth  harmonic\n",
+      "Xcn  =  1/(n*w*C)\n",
+      " #the  p.d.  across  the  capacitor  at  the  nth  harmonic\n",
+      "Vcn  =  Inm*Xcn\n",
+      " #At  the  fundamental  frequency,  inductive  reactance,\n",
+      "XL1  =  w*L\n",
+      " #capacitive  reactance\n",
+      "Xc1  =  1/(w*C)\n",
+      " #Impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*(XL1  -  Xc1)\n",
+      "Z1mag  =  abs(Z1)\n",
+      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
+      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
+      "I1m  =  V1m/Z1mag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)n  =  \",round(n,2),\"\"\n",
+      "print  \"\\n(b)the  maximum  value  of  current  at  the  nth  harmonic  \",round(Inm,2),\"  A\"\n",
+      "print  \"\\n(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is  \",round(Vcn,2),\"\"\n",
+      "print  \"\\n(d)the  maximum  value  of  the  fundamental  current.  \",round(I1m,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)n  =   15.0 \n",
+        "\n",
+        "(b)the  maximum  value  of  current  at  the  nth  harmonic   2.4   A\n",
+        "\n",
+        "(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is   4173.92 \n",
+        "\n",
+        "(d)the  maximum  value  of  the  fundamental  current.   0.03   A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_3.ipynb
new file mode 100755
index 00000000..3cc6ab7f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint_3.ipynb
@@ -0,0 +1,1011 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:fe5e092c877ff6677e6a5e3ccdd3dedece58eaf02a85044a49674acc32664be2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 36: Complex Waveforms</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 643</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V = 240; # in Volts\n",
+      "f = 50; # in Hz\n",
+      "x = 0.2;\n",
+      "phi3 = 3*math.pi/4; # in Rad\n",
+      "\n",
+      " #calculation:\n",
+      "Vamp = V*2**0.5\n",
+      "w = 2*math.pi*f\n",
+      "T = 1/f\n",
+      "V3 = Vamp*x\n",
+      "f3 = 3*f\n",
+      "w3 = 3*w\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage, V =\",round(Vamp,1),\"sin(\",round(w,1),\"t) + \",round(V3,1),\"sin(\", round(w3,1),\"t - \", round(phi3,1),\") volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage, V = 339.4 sin( 314.2 t) +  67.9 sin( 942.5 t -  2.4 ) volts"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 648</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  0.100;#  in  amperes\n",
+      "A3  =  0.020;#  in  amperes\n",
+      "A5  =  0.010;#  in  amperes\n",
+      "\n",
+      " #calculation:\n",
+      " #the  rms  value  of  current  is  given  by\n",
+      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  rms  value  of  current  is  \",round(Irms*1000,2),\" mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  rms  value  of  current  is   72.46  mA"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 649</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  10;#  in  volts\n",
+      "A3  =  3;#  in  volts\n",
+      "A5  =  2;#  in  volts\n",
+      "\n",
+      "#calculation:\n",
+      " #the  rms  value  of  voltage  is  given  by\n",
+      "Vrms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      " #the  mean  value  of  voltage  is  given  by\n",
+      " #x  =  wt\n",
+      "Vav  =  (1/math.pi)*((10 + 1 + 2/5)-(-10 - 1 - 2/5))\n",
+      " #form  factor  is  given  by\n",
+      "ff  =  Vrms/Vav\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
+      "print  \"\\n  (b)the  mean  value  of  voltage  is  \",round(Vav,2),\"  V\"\n",
+      "print  \"\\n  (c)form  factor  is  \",round(ff,3),\"  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  rms  value  of  voltage  is   7.52   V\n",
+        "\n",
+        "  (b)the  mean  value  of  voltage  is   7.26   V\n",
+        "\n",
+        "  (c)form  factor  is   1.036   "
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 649</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  volts\n",
+      "x  =  0.3;#  for  third  harmonic\n",
+      "y  =  0.1;#  for  fifth  harmonic\n",
+      "f  =  31.83;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #V3  =  x*V1\n",
+      " #V5  =  y*V1\n",
+      " #the  rms  value  of  the  fundamental,\n",
+      "V1  =  ((V**2)/(1  +  x**2  +  y**2))**0.5\n",
+      " #Rms  value  of  the  third  harmonic\n",
+      "V3  =  x*V1\n",
+      " #the  rms  value  of  the  fifth  harmonic,\n",
+      "V5  =  y*V1\n",
+      " #Maximum  value  of  the  fundamental,\n",
+      "V1m  =  V1*2**0.5\n",
+      " #Maximum  value  of  the  third  harmonic,\n",
+      "V3m  =  V3*2**0.5\n",
+      " #Maximum  value  of  the  fifth  harmonic,\n",
+      "V5m  =  V5*2**0.5\n",
+      "w  =  2*math.pi*f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"v  =  \",round(V1m,2),\"sin\",round(w,2),\"t  +  \",round(V3m,2),\"sin\",round((3*w),2),\"t  +  \",round(V5m,2),\"sin\",round((5*w),2),\"t  Volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "v  =   323.62 sin 199.99 t  +   97.08 sin 599.98 t  +   32.36 sin 999.97 t  Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 652</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A1  =  12;#  in  amperes\n",
+      "A3  =  5;#  in  amperes\n",
+      "A5  =  2;#  in  amperes\n",
+      "R  =  20;#  in  ohms\n",
+      "\n",
+      "#calculation:\n",
+      " #rms  current\n",
+      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
+      " #average  power\n",
+      "P  =  R*Irms**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  average  power  \",P,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  average  power   1730.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 652</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ia1  =  2;#  in  amperes\n",
+      "Ia3  =  0.3;#  in  amperes\n",
+      "Ia5  =  0.1;#  in  amperes\n",
+      "Va1  =  60;#  in  volts\n",
+      "Va3  =  15;#  in  volts\n",
+      "Va5  =  10;#  in  volts\n",
+      "Phii1  =  -1*math.pi/6;#  in  radians\n",
+      "Phii3  =  -1*math.pi/12;#  in  radians\n",
+      "Phii5  =  -8*math.pi/9;#  in  radians\n",
+      "Phiv1  =  0;#  in  radians\n",
+      "Phiv3  =  math.pi/4;#  in  radians\n",
+      "Phiv5  =  -1*math.pi/2;#  in  radians\n",
+      "\n",
+      "\n",
+      " #calculation:\n",
+      " #rms  values;\n",
+      "I1  =  Ia1/(2**0.5);#  in  amperes\n",
+      "I3  =  Ia3/(2**0.5);#  in  amperes\n",
+      "I5  =  Ia5/(2**0.5);#  in  amperes\n",
+      "V1  =  Va1/(2**0.5);#  in  volts\n",
+      "V3  =  Va3/(2**0.5);#  in  volts\n",
+      "V5  =  Va5/(2**0.5);#  in  volts\n",
+      " #total  power  supplied,\n",
+      "P  =  V1*I1*math.cos(Phiv1  -  Phii1)  +  V3*I3*math.cos(Phiv3  -  Phii3)  +  V5*I5*math.cos(Phiv5  -  Phii5)\n",
+      " #rms  current\n",
+      "Irms  =  ((I1**2  +  I3**2  +  I5**2))**0.5\n",
+      " #rms  voltage\n",
+      "Vrms  =  ((V1**2  +  V3**2  +  V5**2))**0.5\n",
+      " #overall  power  factor\n",
+      "pf  =  P/(Vrms*Irms)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  total  active  power  supplied  to  the  circuit  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(b)overall  power  factor  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  total  active  power  supplied  to  the  circuit   53.26   W\n",
+        "\n",
+        "(b)overall  power  factor   0.84"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 655</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  40;#  in  ohm\n",
+      "L  =  7.96E-3;#  in  Henry\n",
+      "C = 25E-6; # in Farad\n",
+      "f = 1000; # in Hx\n",
+      "\n",
+      "#calculation:\n",
+      "wL = 2*math.pi*1000*L\n",
+      "wC = 2*math.pi*1000*C\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)i  =  \",round(100/R1,2),\"sin(wt)  +\",round(30/R1,2),\"sin(3wt - pi/3)  +\",round(10/R1,2),\"sin(5wt - pi/6) A\"\n",
+      "print  \"(b)i  =  \",round(100/wL,2),\"sin(wt - pi/2)  +\",round(30/(3*wL),2),\"sin(3wt - pi/6)  +\",round(10/(5*wL),2),\"sin(5wt - 2pi/3) A\"\n",
+      "print  \"(c)i  =  \",round(100*wC,2),\"sin(wt + pi/2)  +\",round(30*3*wC,2),\"sin(3wt + 5pi/6)  +\",round(10*5*wC,2),\"sin(5wt + pi/3) A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)i  =   2.5 sin(wt)  + 0.75 sin(3wt - pi/3)  + 0.25 sin(5wt - pi/6) A\n",
+        "(b)i  =   2.0 sin(wt - pi/2)  + 0.2 sin(3wt - pi/6)  + 0.04 sin(5wt - 2pi/3) A\n",
+        "(c)i  =   15.71 sin(wt + pi/2)  + 14.14 sin(3wt + 5pi/6)  + 7.85 sin(5wt + pi/3) A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 656</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  240;#  in  volts\n",
+      "V3m  =  40;#  in  volts\n",
+      "V5m  =  30;#  in  volts\n",
+      "w1  =  314;#  fundamental\n",
+      "R  =  12;#  in  ohm\n",
+      "L  =  0.00955;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #fundamental  or  first  harmonic\n",
+      " #inductive  reactance,\n",
+      "XL1  =  w1*L\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*XL1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1m/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Third  harmonic\n",
+      "XL3  =  3*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*XL3\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3m/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #fifth  harmonic\n",
+      "XL5  =  5*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z5  =  R  +  1j*XL5\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I5m  =  V5m/Z5\n",
+      "I5mag  =  abs(I5m)\n",
+      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
+      " #rms  voltage\n",
+      "Vrms  =  ((V1m**2  +  V3m**2  +  V5m**2)/2)**0.5\n",
+      " #rms  current\n",
+      "Irms  =  ((I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
+      " #power  dissipated\n",
+      "P  =  R*Irms**2\n",
+      " #overall  power  factor\n",
+      "pf  =  P/(Vrms*Irms)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)i  =  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
+      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
+      "print  \"\\n(c)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
+      "print  \"\\n(d)the  total  power  dissipated  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(e)overall  power  factor  \",round(pf,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)i  =   19.4 sin( 314.0 t  +  ( -0.24 ))  +   2.67 sin( 942.0 t  +  ( -0.64 ))  +   1.56 sin( 1570.0 t  +  ( -0.9 ))  A\n",
+        "\n",
+        "(b)the  rms  value  of  current  is   13.89   A\n",
+        "\n",
+        "(c)the  rms  value  of  voltage  is   173.35   V\n",
+        "\n",
+        "(d)the  total  power  dissipated   2316.26   W\n",
+        "\n",
+        "(e)overall  power  factor   0.96"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 658</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  50;#  in  volts\n",
+      "V1m  =  200;#  in  volts\n",
+      "V2m  =  40;#  in  volts\n",
+      "V4m  =  5;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "R  =  50;#  in  ohm\n",
+      "C  =  100E-6;#  in  farad\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv2  =  -1*math.pi/2;#  in  rad\n",
+      "phiv4  =  math.pi/4;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V2  =  V2m*math.cos(phiv2)  +  1j*V2m*math.sin(phiv2)\n",
+      "V4  =  V4m*math.cos(phiv4)  +  1j*V4m*math.sin(phiv4)\n",
+      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
+      "Iom  =  0\n",
+      " #fundamental  or  first  harmonic\n",
+      "w1  =  2*math.pi*f\n",
+      " #inductive  reactance,\n",
+      "Xc1  =  1/(w1*C)\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*Xc1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #second  harmonic\n",
+      "Xc2  =  Xc1/2\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z2  =  R  +  1j*Xc2\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I2m  =  V2/Z2\n",
+      "I2mag  =  abs(I2m)\n",
+      "phii2  =  cmath.phase(complex(I2m.real,I2m.imag))\n",
+      " #fourth  harmonic\n",
+      "Xc4  =  Xc1/4\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z4  =  R  +  1j*Xc4\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I4m  =  V4/Z4\n",
+      "I4mag  =  abs(I4m)\n",
+      "phii4  =  cmath.phase(complex(I4m.real,I4m.imag))\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1mag**2  +  I2mag**2  +  I4mag**2)/2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I2mag,2),\"sin(\",round((w1*2),2),\"t  +  (\",round(phii2,2),\"))  +  \",round(I4mag,2),\"sin(\",round((w1*4),2),\"t  +  (\",round(phii4,2),\"))  A\"\n",
+      "print  \"(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)i =  0.0  +  3.37 sin( 314.16 t  +  ( -0.57 ))  +   0.76 sin( 628.32 t  +  ( -1.88 ))  +   0.1 sin( 1256.64 t  +  ( 0.63 ))  A\n",
+        "(b)the  rms  value  of  current  is   2.45   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 659</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  25;#  in  volts\n",
+      "V1m  =  100;#  in  volts\n",
+      "V3m  =  40;#  in  volts\n",
+      "V5m  =  20;#  in  volts\n",
+      "w1  =  10000;#  fundamental\n",
+      "R  =  5;#  in  ohm\n",
+      "L  =  500E-6;#  in  Henry\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv3  =  math.pi/6;#  in  rad\n",
+      "phiv5  =  math.pi/12;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V3  =  V3m*math.cos(phiv3)  +  1j*V3m*math.sin(phiv3)\n",
+      "V5  =  V5m*math.cos(phiv5)  +  1j*V5m*math.sin(phiv5)\n",
+      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
+      "Iom  =  Vom/R\n",
+      " #fundamental  or  first  harmonic\n",
+      " #inductive  reactance,\n",
+      "XL1  =  w1*L\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*XL1\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      "#Third  harmonic\n",
+      "XL3  =  3*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*XL3\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #fifth  harmonic\n",
+      "XL5  =  5*XL1\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z5  =  R  +  1j*XL5\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I5m  =  V5/Z5\n",
+      "I5mag  =  abs(I5m)\n",
+      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
+      " #power  dissipated\n",
+      "P  =  R*Irms**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)i  =  \",round(Iom,2),\"  +  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
+      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
+      "print  \"\\n(c)the  total  power  dissipated  \",round(P,3),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)i  =   5.0   +   14.14 sin( 10000.0 t  +  ( -0.79 ))  +   2.53 sin( 30000.0 t  +  ( -0.73 ))  +   0.78 sin( 50000.0 t  +  ( -1.11 ))  A\n",
+        "\n",
+        "(b)the  rms  value  of  current  is   11.34   A\n",
+        "\n",
+        "(c)the  total  power  dissipated   642.538   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 661</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vom  =  30;#  in  volts\n",
+      "V1m  =  40;#  in  volts\n",
+      "V2m  =  25;#  in  volts\n",
+      "V4m  =  15;#  in  volts\n",
+      "Iom  =  0;#  in  amperes\n",
+      "I1m  =  0.743;#  in  Amperes\n",
+      "I2m  =  0.781;#  in  Amperes\n",
+      "I4m  =  0.636;#  in  Amperes\n",
+      "phii1  =  1.190;#  in  rad\n",
+      "phii2  =  0.896;#  in  rad\n",
+      "phii4  =  0.559;#  in  rad\n",
+      "w  =  1000;#  in  rad\n",
+      "\n",
+      " #calculation:\n",
+      " #the  average  power  P  is  given  by\n",
+      "P  =  Vom*Iom+(0.707*V1m)*(0.707*I1m)*math.cos(phii1)+(0.707*V2m)*(0.707*I2m)*math.cos(phii2) + (0.707*V4m)*(0.707*I4m)*math.cos(phii4)\n",
+      " #rms  current\n",
+      "Irms  =  (Iom**2  +  (I1m**2  +  I2m**2  +  I4m**2)/2)**0.5\n",
+      " #resistance  R\n",
+      "R  =  P/(Irms**2)\n",
+      " #impedance\n",
+      "Z1  =  V1m/I1m\n",
+      " #Xc1\n",
+      "Xc1  =  (Z1**2  -  R**2)**0.5\n",
+      " #capacitance\n",
+      "C  =  1/(w*Xc1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)the  average  power  P  is  \",round(P,2),\"  W\"\n",
+      "print  \"\\n(c)the  resistance  R  \",round(R,2),\"  ohm  and  capacitance  \",round(C*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)the  average  power  P  is   15.66   W\n",
+        "\n",
+        "(c)the  resistance  R   19.99   ohm  and  capacitance   20.01 uF\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 662</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  300;#  in  volts\n",
+      "V3m  =  120;#  in  volts\n",
+      "phiv1  =  0;#  in  rad\n",
+      "phiv2  =  0.698;#  in  rad\n",
+      "w1  =  314;#  in  rad\n",
+      "C  =  2.123E-6;#  in  farads\n",
+      "R1  =  560;#  in  ohms\n",
+      "R2  =  2000;#  in  Ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
+      "V3  =  V3m*math.cos(phiv2)  +  1j*V3m*math.sin(phiv2)\n",
+      " #capacitive  reactance,\n",
+      "Xc1  =  1/(w1*C)\n",
+      " #impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R1  -  1j*Xc1*R2/(R2  -  1j*Xc1)\n",
+      " #Maximum  current  at  fundamental  frequency\n",
+      "I1m  =  V1/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Third  harmonic\n",
+      "Xc3  =  Xc1/3\n",
+      " #impedance  at  the  third  harmonic  frequency,\n",
+      "Z3  =  R1  -  1j*Xc3*R2/(R2  -  1j*Xc3)\n",
+      "I1m = V1m/Z1\n",
+      "I1mag  =  abs(I1m)\n",
+      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
+      " #Maximum  current  at  third  harmonic  frequency\n",
+      "I3m  =  V3/Z3\n",
+      "I3mag  =  abs(I3m)\n",
+      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
+      " #Percentage  harmonic  content  of  the  supply  current  is  given  by\n",
+      "percent  =  I3mag*100/I1mag\n",
+      " #total  active  power\n",
+      "P  =  (0.707*V1m)*(0.707*I1mag)*math.cos(phiv1  -  phii1)  +  (0.707*V3m)*(0.707*I3m)*math.cos(phiv2  -  phii3)\n",
+      "\n",
+      "I1 = I1m*R2/(R2 - 1j*Xc1)\n",
+      "I3 = I3m*R2/(R2 - 1j*Xc3)\n",
+      "\n",
+      "I1nmag  =  abs(I1)\n",
+      "phini1  =  cmath.phase(complex(I1.real,I1.imag))\n",
+      "I3nmag  =  abs(I3)\n",
+      "phini3  =  cmath.phase(complex(I3.real,I3.imag))\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)supply current, i=\", round(I1mag,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
+      "print  \"\\n(b)Percentage  harmonic  content  of  the  supply  current  is  \",round(percent,2),\"  percent\"\n",
+      "print  \"\\n(c)total  active  power  is  \",round(abs(P),2),\"  W\"\n",
+      "print  \"\\n(d)Voltage, v1 =\", round(I1mag*R1,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag*R1,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
+      "print  \"\\n(e)current, ic =\", round(I1nmag,3),\"sin(\", w1,\"t +\",round(phini1,3),\") + \",round(I3nmag,3),\"sin(\", 3*w1,\"t +\",round(phini3,3),\") A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)supply current, i= 0.187 sin( 314 t + 0.643 ) +  0.145 sin( 942 t + 1.305 ) A\n",
+        "\n",
+        "(b)Percentage  harmonic  content  of  the  supply  current  is   77.57   percent\n",
+        "\n",
+        "(c)total  active  power  is   25.34   W\n",
+        "\n",
+        "(d)Voltage, v1 = 104.996 sin( 314 t + 0.643 ) +  81.45 sin( 942 t + 1.305 ) A\n",
+        "\n",
+        "(e)current, ic = 0.15 sin( 314 t + 1.287 ) +  0.141 sin( 942 t + 1.55 ) A"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 664</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  400;#  in  volts\n",
+      "V3m  =  10;#  in  volts\n",
+      "C  =  0.2E-6;#  in  farads\n",
+      "R  =  2;#  in  ohms\n",
+      "L  =  0.5;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #Resonance  with  the  third  harmonic  means  that\n",
+      "w  =  (1/(9*L*C))**0.5\n",
+      " #fundamental  frequency,  f\n",
+      "f  =  w/(2*math.pi)\n",
+      " #At  the  fundamental  frequency,\n",
+      " #impedance  Z1\n",
+      "Z1  =  R  +  1j*(w*L  -  1/(w*C))\n",
+      "Z1mag  =  abs(Z1)\n",
+      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
+      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
+      "I1m  =  V1m/Z1mag\n",
+      " #At  the  third  harmonic  frequency,\n",
+      "Z3  =  R  +  1j*(3*w*L  -  1/(3*w*C))\n",
+      "Z3mag  =  abs(Z3)\n",
+      "phiZ3  =  cmath.phase(complex(Z3.real,Z3.imag))\n",
+      " #Maximum  value  of  current  at  the  third  harmonic  frequency,\n",
+      "I3m  =  V3m/Z3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is  \",round(f,2),\"  Hz\"\n",
+      "print  \"(b)Maximum value of current at fundamental freq. is\",round(abs(I1m),3),\"A \"\n",
+      "print   \"and  at  the  third  harmonic  frequency  \",  abs(I3m),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is   167.76   Hz\n",
+        "(b)Maximum value of current at fundamental freq. is 0.095 A \n",
+        "and  at  the  third  harmonic  frequency   5.0   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 665</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1m  =  800;#  in  volts\n",
+      "f  =  50;#  in  Hz\n",
+      "x  =  0.015;\n",
+      "C  =  0.122E-6;#  in  farads\n",
+      "R  =  5;#  in  ohms\n",
+      "L  =  0.369;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage  at  nth  harmonic\n",
+      "Vnm  =  x*V1m\n",
+      "w  =  2*math.pi*f\n",
+      " #For  resonance  at  the  nth  harmonic  nwL  =  1/nwC\n",
+      "n  =  1/(w*(L*C)**0.5)\n",
+      " #At  resonance,  impedance\n",
+      "Zn  =  R\n",
+      " #the  maximum  value  of  current  at  the  nth  harmonic\n",
+      "Inm  =  Vnm/Zn\n",
+      " #capacitive  reactance,  at  nth  harmonic\n",
+      "Xcn  =  1/(n*w*C)\n",
+      " #the  p.d.  across  the  capacitor  at  the  nth  harmonic\n",
+      "Vcn  =  Inm*Xcn\n",
+      " #At  the  fundamental  frequency,  inductive  reactance,\n",
+      "XL1  =  w*L\n",
+      " #capacitive  reactance\n",
+      "Xc1  =  1/(w*C)\n",
+      " #Impedance  at  the  fundamental  frequency,\n",
+      "Z1  =  R  +  1j*(XL1  -  Xc1)\n",
+      "Z1mag  =  abs(Z1)\n",
+      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
+      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
+      "I1m  =  V1m/Z1mag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)n  =  \",round(n,2),\"\"\n",
+      "print  \"\\n(b)the  maximum  value  of  current  at  the  nth  harmonic  \",round(Inm,2),\"  A\"\n",
+      "print  \"\\n(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is  \",round(Vcn,2),\"\"\n",
+      "print  \"\\n(d)the  maximum  value  of  the  fundamental  current.  \",round(I1m,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)n  =   15.0 \n",
+        "\n",
+        "(b)the  maximum  value  of  current  at  the  nth  harmonic   2.4   A\n",
+        "\n",
+        "(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is   4173.92 \n",
+        "\n",
+        "(d)the  maximum  value  of  the  fundamental  current.   0.03   A"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint.ipynb
new file mode 100755
index 00000000..c4a1d2a6
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint.ipynb
@@ -0,0 +1,631 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 38: Magnetic materials</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 694</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the hysteresis loss per m3 per cycle, and (b) the hysteresis loss per m3 at a frequency of 50 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A  =  12.5;#  in  cm2\n",
+      "x  =  500;#  horizontal  axis  1  cm  =  500  A/m\n",
+      "y  =  0.2;#  vertical  axis  1  cm  =  0.2  T\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  per  cycle\n",
+      "HL  =  A*x*y\n",
+      " #At  50  Hz  frequency,  hysteresis  loss\n",
+      "HLf  =  HL*f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)hysteresis  loss  per  cycle  is  =  \",HL,\"  J/m3\"\n",
+      "print  \"\\n(b)At  50  Hz  frequency,  hysteresis  loss  \",HLf,\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)hysteresis  loss  per  cycle  is  =   1250.0   J/m3\n",
+        "\n",
+        "(b)At  50  Hz  frequency,  hysteresis  loss   62500.0   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 695</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the hysteresis loss per m3 for a maximum flux density of 1.1 T and frequency of 25 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "n  =  1.6;#  the  Steinmetz  index\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  25;#  in  Hz\n",
+      "Bm1  =  1.5;#  in  Tesla\n",
+      "Bm2  =  1.1;#  in  Tesla\n",
+      "Ph1  =  62500;#  in  W/m3\n",
+      "v  =  1;\n",
+      "\n",
+      "#calculation:  \n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n\n",
+      "kh  =  Ph1/(v*f1*(Bm1)**n)\n",
+      " #When  f  =  25  Hz  and  Bm  =  1.1  T,\n",
+      "Ph2  =  kh*v*f2*(Bm2)**n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  hysteresis  loss  When  f  =  25  Hz  and  Bm  =  1.1  T,  is  =  \",round(Ph2,2),\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  hysteresis  loss  When  f  =  25  Hz  and  Bm  =  1.1  T,  is  =   19025.33   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 695</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the hysteresis loss at a frequency of 80 Hz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "csa  =  0.002;#  in  m2\n",
+      "l  =  1;#  in  m\n",
+      "a  =  400/0.01;#  10  mm  =  400  A/m  \n",
+      "b  =  0.1/0.01;#  10  mm  =  0.1  T  \n",
+      "A  =  0.01;#  in  m2\n",
+      "f  =  80;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #hysteresis  loss  per  cycle\n",
+      "HL  =  A*a*b\n",
+      " #At  a  frequency  of  80  Hz,\n",
+      " #hysteresis  loss\n",
+      "HLf  =  HL*f\n",
+      " #Volume  of  ring\n",
+      "v  =  csa*l\n",
+      " #hysteresis  loss\n",
+      "Ph  =  HLf*v\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  hysteresis  loss  at  a  frequency  of  80  Hz  is  \",Ph,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  hysteresis  loss  at  a  frequency  of  80  Hz  is   640.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 696</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the hysteresis loss when the maximum core flux is 8 mWb and the frequency is 50 Hz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Phi1  =  0.01;#  in  Wb\n",
+      "Phi2  =  0.008;#  in  Wb\n",
+      "csa  =  0.008;#  in  m2\n",
+      "v  =  0.005;#  in  m3\n",
+      "f  =  50;#  in  Hz\n",
+      "n  =  1.7;#  the  Steinmetz  constant\n",
+      "Ph1  =  100;#  in  Watt\n",
+      "\n",
+      " #calculation:  \n",
+      " #maximum  flux  density\n",
+      "Bm1  =  Phi1/csa\n",
+      " #hysteresis  loss  Ph1  =  kh*v*f*(Bm1)**n\n",
+      "kh  =  Ph1/(v*f*(Bm1)**n)\n",
+      " #When  the  maximum  core  flux  is  8  mWb,\n",
+      "Bm2  =  Phi2/csa\n",
+      " #hysteresis  loss,  Ph2\n",
+      "Ph2  =  kh*v*f*(Bm2)**n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"value  of hysteresis  loss  when  maximum  core  flux  is  8  mWb  and  the  frequency  is  50  Hz  is  \",round(Ph2,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "value  of hysteresis  loss  when  maximum  core  flux  is  8  mWb  and  the  frequency  is  50  Hz  is   68.43   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 699</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the new value of eddy current loss per cubic metre.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pe1  =  10;#  in  W/m3\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  30;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #When  the  eddy  current  loss  is  10  W/m3,  frequency  f  is  50  Hz.\n",
+      " #constant  k\n",
+      "k  =  Pe1/(f1**2)\n",
+      " #When  the  frequency  is  30  Hz,  eddy  current  loss,\n",
+      "Pe2  =  k*(f2**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\neddy  current  loss  per  cubic  metre  is  \",Pe2,\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "eddy  current  loss  per  cubic  metre  is   3.6   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 699</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the necessary new thickness of the laminations.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pe  =  100;#  in  W/m3\n",
+      "f1  =  50;#  in  Hz\n",
+      "t1  =  0.0005;#  in  m\n",
+      "x  =  1/3;\n",
+      "f2  =  250;#  in  Hz\n",
+      "Bm1 = 1;\n",
+      " #calculation:  \n",
+      " #Pe  =  ke*(Bm1*f1*t1)**2\n",
+      " #Hence,  at  50  Hz  frequency\n",
+      "ke  =  Pe/(Bm1*f1*t1)**2\n",
+      " #At  250  Hz  frequency\n",
+      "Bm2  =  x*Bm1\n",
+      "t2  =  ((Pe/ke)**0.5)/(Bm2*f2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nlamination  thickness  is  \",t2,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "lamination  thickness  is   0.0003 m"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 700</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the values of the losses if the frequency is increased to 60 Hz. \n",
+      "#(b) What will be the total core loss if the frequency is 50 Hz and the lamination are made one-half of their original thickness?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ph1  =  40;#  in  W\n",
+      "Pe1  =  20;#  in  W\n",
+      "f1  =  50;#  in  Hz\n",
+      "x  =  1/2;\n",
+      "f2  =  60;#  in  Hz\n",
+      "t1  =  1;\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n  =  k1*f\n",
+      " #Thus  when  the  hysteresis  is  40  W  and  the  frequency  50  Hz,\n",
+      "k1  =  Ph1/f1\n",
+      " #If  the  frequency  is  increased  to  60  Hz,\n",
+      "Ph2  =  k1*f2\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*f**2\n",
+      " #since  the  flux  density  and  lamination  thickness  are  constant.\n",
+      " #When  the  eddy  current  loss  is  20  W  the  frequency  is  50  Hz.  Thus\n",
+      "k2  =  Pe1/(f1**2)\n",
+      " #If  the  frequency  is  increased  to  60  Hz,\n",
+      "Pe2  =  k2*(f2**2)\n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n,  is  independent  of  the  thickness  of  the  laminations.  \n",
+      "    #Thus,  if  the  thickness  of  the  laminations  is  halved,  the  hysteresis  loss  remains  at  \n",
+      "Phb2  =  Ph1\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*t**3\n",
+      "k3  =  Pe1/(t1**3)\n",
+      "t2  =  0.5*t1\n",
+      "Peb2  =  k3*t2**3\n",
+      " #total  core  loss  when  the  thickness  of  the  laminations  is  halved  is  given  by\n",
+      "TL  =  Phb2  +  Peb2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)If  the  frequency  is  increased  to  60  Hz,hysteresis  loss  is  \",Ph2,\"  W  and  eddy  current  loss  \",  Pe2,\"  W\"\n",
+      "print  \"\\n(b)the  total  core  loss  when  the  thickness  of  the  laminations  is  halved  \",TL,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)If  the  frequency  is  increased  to  60  Hz,hysteresis  loss  is   48.0   W  and  eddy  current  loss   28.8   W\n",
+        "\n",
+        "(b)the  total  core  loss  when  the  thickness  of  the  laminations  is  halved   42.5   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 701</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the new total core loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  500;#  in  Volts\n",
+      "V2  =  1000;#  in  Volts\n",
+      "Ph1  =  400;#  in  W\n",
+      "Pe1  =  150;#  in  W\n",
+      "f1  =  50;#  in  Hz\n",
+      "n  =  1.6;#  Steinmetz  index\n",
+      "f2  =  100;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  Ph  =  k1*f*(E/f)**n\n",
+      " #At  500  V  and  50  Hz\n",
+      "k1  =  Ph1/(f1*(V1/f1)**1.6)\n",
+      " #At  1000  V  and  100  Hz,\n",
+      "Ph2  =  k1*f2*(V2/f2)**1.6\n",
+      " #eddy  current  loss,  Pe  =  k2*E**2\n",
+      " #At  500  V,\n",
+      "k2  =  Pe1/(V1**2)\n",
+      " #At  1000  V,\n",
+      "Pe2  =  k2*(V2**2)\n",
+      " #the  new  total  core  loss\n",
+      "TL  =  Ph2  +  Pe2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  new  total  core  loss  \",TL,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  new  total  core  loss   1400.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 702</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the separate values of the hysteresis and eddy current losses at frequencies of (a) 50 Hz and (b) 60 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  60;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "k1  =  0.5\n",
+      "k2  =  0.032\n",
+      "f = [30, 50, 70,90]\n",
+      "Pcf = [1.5, 2.1, 2.7, 3.4]\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(f,Pcf,'-')\n",
+      "xlabel('f')\n",
+      "ylabel('Pc/f')\n",
+      "show()\n",
+      "HL1 = k1*f1\n",
+      "ECL1 = k2*f1**2\n",
+      "\n",
+      "HL2 = k1*f2\n",
+      "ECL2 = k2*f2**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a) at Frequency = 50 Hz, hysteresis loss is\", HL1,\" W and eddy current loss is\", ECL1,\" W\"\n",
+      "print  \"\\n  (b) at Frequency = 60 Hz, hysteresis loss is\", HL2,\" W and eddy current loss is\", ECL2,\" W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'."
+       ]
+      },
+      {
+       "output_type": "display_data",
+       "png": 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+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a) at Frequency = 50 Hz, hysteresis loss is 25.0  W and eddy current loss is 80.0  W\n",
+        "\n",
+        "  (b) at Frequency = 60 Hz, hysteresis loss is 30.0  W and eddy current loss is 115.2  W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 703</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the hysteresis and eddy current losses at 50 Hz (b) determine the hysteresis and eddy current losses at 50 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "TL1  =  400;#  in  Watt\n",
+      "TL2  =  498;#  in  Watt\n",
+      "x  =  0.25;\n",
+      "y  =  0.4;\n",
+      "f1  =  50;#  in  Hz\n",
+      "n  =  1.7;#  Steinmetz  index\n",
+      "f2  =  60;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #if  volume  v  and  the  maximum  flux  density  are  constant\n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n  =  k1*f\n",
+      " #(if  the  maximum  flux  density  and  the  lamination  thickness  are  constant)\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*f**2\n",
+      " #At  50  Hz  frequency,  TL1  =  k1*f1  +  k2*f1**2\n",
+      " #At  60  Hz  frequency,  TL2  =  k1*f2  +  k2*f2**2\n",
+      " #Solving  equations  gives  the  values  of  k1  and  k2.\n",
+      "k2  =  (5*TL2  -  6*TL1)/(5*(f2**2)  -  6*(f1**2))\n",
+      "k1  =  (TL1  -  k2*f1**2)/f1\n",
+      " #hysteresis  loss  Ph  =  k1*f\n",
+      "Ph1  =  k1*f1\n",
+      " #eddy  current  loss\n",
+      "Pe1  =  k2*f1**2\n",
+      " #Since  at  50  Hz  the  flux  density  is  increased  by  25%,  the  new  hysteresis  loss  is\n",
+      "Ph2  =  Ph1*(1  +  x)**1.7\n",
+      " #Since  at  50  Hz  the  flux  density  is  increased  by  25%,  and  the  lamination  thickness  is  increased  by  40%, \n",
+      "    #the  new  eddy  current  loss  is\n",
+      "Pe2  =  Pe1*((1  +  x)**2)*(1  +  y)**3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  hysteresis  and  eddy  current  losses  at  50  Hz  are  \",round(Ph1,2),\"  W  and  \",round(  Pe1,2),\"  W  resp.\"\n",
+      "print  \"\\n  (b)the  hysteresis  and  eddy  current  losses  at  50  Hz  after  increement  are  \",round(Ph2,2),\"  W  and  \",round(  Pe2,2),\"  W  resp.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  hysteresis  and  eddy  current  losses  at  50  Hz  are   325.0   W  and   75.0   W  resp.\n",
+        "\n",
+        "  (b)the  hysteresis  and  eddy  current  losses  at  50  Hz  after  increement  are   474.93   W  and   321.56   W  resp.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_1.ipynb
new file mode 100755
index 00000000..c4a1d2a6
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_1.ipynb
@@ -0,0 +1,631 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 38: Magnetic materials</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 694</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the hysteresis loss per m3 per cycle, and (b) the hysteresis loss per m3 at a frequency of 50 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A  =  12.5;#  in  cm2\n",
+      "x  =  500;#  horizontal  axis  1  cm  =  500  A/m\n",
+      "y  =  0.2;#  vertical  axis  1  cm  =  0.2  T\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  per  cycle\n",
+      "HL  =  A*x*y\n",
+      " #At  50  Hz  frequency,  hysteresis  loss\n",
+      "HLf  =  HL*f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)hysteresis  loss  per  cycle  is  =  \",HL,\"  J/m3\"\n",
+      "print  \"\\n(b)At  50  Hz  frequency,  hysteresis  loss  \",HLf,\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)hysteresis  loss  per  cycle  is  =   1250.0   J/m3\n",
+        "\n",
+        "(b)At  50  Hz  frequency,  hysteresis  loss   62500.0   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 695</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the hysteresis loss per m3 for a maximum flux density of 1.1 T and frequency of 25 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "n  =  1.6;#  the  Steinmetz  index\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  25;#  in  Hz\n",
+      "Bm1  =  1.5;#  in  Tesla\n",
+      "Bm2  =  1.1;#  in  Tesla\n",
+      "Ph1  =  62500;#  in  W/m3\n",
+      "v  =  1;\n",
+      "\n",
+      "#calculation:  \n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n\n",
+      "kh  =  Ph1/(v*f1*(Bm1)**n)\n",
+      " #When  f  =  25  Hz  and  Bm  =  1.1  T,\n",
+      "Ph2  =  kh*v*f2*(Bm2)**n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  hysteresis  loss  When  f  =  25  Hz  and  Bm  =  1.1  T,  is  =  \",round(Ph2,2),\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  hysteresis  loss  When  f  =  25  Hz  and  Bm  =  1.1  T,  is  =   19025.33   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 695</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the hysteresis loss at a frequency of 80 Hz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "csa  =  0.002;#  in  m2\n",
+      "l  =  1;#  in  m\n",
+      "a  =  400/0.01;#  10  mm  =  400  A/m  \n",
+      "b  =  0.1/0.01;#  10  mm  =  0.1  T  \n",
+      "A  =  0.01;#  in  m2\n",
+      "f  =  80;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #hysteresis  loss  per  cycle\n",
+      "HL  =  A*a*b\n",
+      " #At  a  frequency  of  80  Hz,\n",
+      " #hysteresis  loss\n",
+      "HLf  =  HL*f\n",
+      " #Volume  of  ring\n",
+      "v  =  csa*l\n",
+      " #hysteresis  loss\n",
+      "Ph  =  HLf*v\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  hysteresis  loss  at  a  frequency  of  80  Hz  is  \",Ph,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  hysteresis  loss  at  a  frequency  of  80  Hz  is   640.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 696</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the hysteresis loss when the maximum core flux is 8 mWb and the frequency is 50 Hz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Phi1  =  0.01;#  in  Wb\n",
+      "Phi2  =  0.008;#  in  Wb\n",
+      "csa  =  0.008;#  in  m2\n",
+      "v  =  0.005;#  in  m3\n",
+      "f  =  50;#  in  Hz\n",
+      "n  =  1.7;#  the  Steinmetz  constant\n",
+      "Ph1  =  100;#  in  Watt\n",
+      "\n",
+      " #calculation:  \n",
+      " #maximum  flux  density\n",
+      "Bm1  =  Phi1/csa\n",
+      " #hysteresis  loss  Ph1  =  kh*v*f*(Bm1)**n\n",
+      "kh  =  Ph1/(v*f*(Bm1)**n)\n",
+      " #When  the  maximum  core  flux  is  8  mWb,\n",
+      "Bm2  =  Phi2/csa\n",
+      " #hysteresis  loss,  Ph2\n",
+      "Ph2  =  kh*v*f*(Bm2)**n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"value  of hysteresis  loss  when  maximum  core  flux  is  8  mWb  and  the  frequency  is  50  Hz  is  \",round(Ph2,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "value  of hysteresis  loss  when  maximum  core  flux  is  8  mWb  and  the  frequency  is  50  Hz  is   68.43   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 699</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the new value of eddy current loss per cubic metre.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pe1  =  10;#  in  W/m3\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  30;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #When  the  eddy  current  loss  is  10  W/m3,  frequency  f  is  50  Hz.\n",
+      " #constant  k\n",
+      "k  =  Pe1/(f1**2)\n",
+      " #When  the  frequency  is  30  Hz,  eddy  current  loss,\n",
+      "Pe2  =  k*(f2**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\neddy  current  loss  per  cubic  metre  is  \",Pe2,\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "eddy  current  loss  per  cubic  metre  is   3.6   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 699</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the necessary new thickness of the laminations.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pe  =  100;#  in  W/m3\n",
+      "f1  =  50;#  in  Hz\n",
+      "t1  =  0.0005;#  in  m\n",
+      "x  =  1/3;\n",
+      "f2  =  250;#  in  Hz\n",
+      "Bm1 = 1;\n",
+      " #calculation:  \n",
+      " #Pe  =  ke*(Bm1*f1*t1)**2\n",
+      " #Hence,  at  50  Hz  frequency\n",
+      "ke  =  Pe/(Bm1*f1*t1)**2\n",
+      " #At  250  Hz  frequency\n",
+      "Bm2  =  x*Bm1\n",
+      "t2  =  ((Pe/ke)**0.5)/(Bm2*f2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nlamination  thickness  is  \",t2,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "lamination  thickness  is   0.0003 m"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 700</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the values of the losses if the frequency is increased to 60 Hz. \n",
+      "#(b) What will be the total core loss if the frequency is 50 Hz and the lamination are made one-half of their original thickness?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ph1  =  40;#  in  W\n",
+      "Pe1  =  20;#  in  W\n",
+      "f1  =  50;#  in  Hz\n",
+      "x  =  1/2;\n",
+      "f2  =  60;#  in  Hz\n",
+      "t1  =  1;\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n  =  k1*f\n",
+      " #Thus  when  the  hysteresis  is  40  W  and  the  frequency  50  Hz,\n",
+      "k1  =  Ph1/f1\n",
+      " #If  the  frequency  is  increased  to  60  Hz,\n",
+      "Ph2  =  k1*f2\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*f**2\n",
+      " #since  the  flux  density  and  lamination  thickness  are  constant.\n",
+      " #When  the  eddy  current  loss  is  20  W  the  frequency  is  50  Hz.  Thus\n",
+      "k2  =  Pe1/(f1**2)\n",
+      " #If  the  frequency  is  increased  to  60  Hz,\n",
+      "Pe2  =  k2*(f2**2)\n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n,  is  independent  of  the  thickness  of  the  laminations.  \n",
+      "    #Thus,  if  the  thickness  of  the  laminations  is  halved,  the  hysteresis  loss  remains  at  \n",
+      "Phb2  =  Ph1\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*t**3\n",
+      "k3  =  Pe1/(t1**3)\n",
+      "t2  =  0.5*t1\n",
+      "Peb2  =  k3*t2**3\n",
+      " #total  core  loss  when  the  thickness  of  the  laminations  is  halved  is  given  by\n",
+      "TL  =  Phb2  +  Peb2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)If  the  frequency  is  increased  to  60  Hz,hysteresis  loss  is  \",Ph2,\"  W  and  eddy  current  loss  \",  Pe2,\"  W\"\n",
+      "print  \"\\n(b)the  total  core  loss  when  the  thickness  of  the  laminations  is  halved  \",TL,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)If  the  frequency  is  increased  to  60  Hz,hysteresis  loss  is   48.0   W  and  eddy  current  loss   28.8   W\n",
+        "\n",
+        "(b)the  total  core  loss  when  the  thickness  of  the  laminations  is  halved   42.5   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 701</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the new total core loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  500;#  in  Volts\n",
+      "V2  =  1000;#  in  Volts\n",
+      "Ph1  =  400;#  in  W\n",
+      "Pe1  =  150;#  in  W\n",
+      "f1  =  50;#  in  Hz\n",
+      "n  =  1.6;#  Steinmetz  index\n",
+      "f2  =  100;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  Ph  =  k1*f*(E/f)**n\n",
+      " #At  500  V  and  50  Hz\n",
+      "k1  =  Ph1/(f1*(V1/f1)**1.6)\n",
+      " #At  1000  V  and  100  Hz,\n",
+      "Ph2  =  k1*f2*(V2/f2)**1.6\n",
+      " #eddy  current  loss,  Pe  =  k2*E**2\n",
+      " #At  500  V,\n",
+      "k2  =  Pe1/(V1**2)\n",
+      " #At  1000  V,\n",
+      "Pe2  =  k2*(V2**2)\n",
+      " #the  new  total  core  loss\n",
+      "TL  =  Ph2  +  Pe2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  new  total  core  loss  \",TL,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  new  total  core  loss   1400.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 702</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the separate values of the hysteresis and eddy current losses at frequencies of (a) 50 Hz and (b) 60 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  60;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "k1  =  0.5\n",
+      "k2  =  0.032\n",
+      "f = [30, 50, 70,90]\n",
+      "Pcf = [1.5, 2.1, 2.7, 3.4]\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(f,Pcf,'-')\n",
+      "xlabel('f')\n",
+      "ylabel('Pc/f')\n",
+      "show()\n",
+      "HL1 = k1*f1\n",
+      "ECL1 = k2*f1**2\n",
+      "\n",
+      "HL2 = k1*f2\n",
+      "ECL2 = k2*f2**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a) at Frequency = 50 Hz, hysteresis loss is\", HL1,\" W and eddy current loss is\", ECL1,\" W\"\n",
+      "print  \"\\n  (b) at Frequency = 60 Hz, hysteresis loss is\", HL2,\" W and eddy current loss is\", ECL2,\" W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'."
+       ]
+      },
+      {
+       "output_type": "display_data",
+       "png": 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+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a) at Frequency = 50 Hz, hysteresis loss is 25.0  W and eddy current loss is 80.0  W\n",
+        "\n",
+        "  (b) at Frequency = 60 Hz, hysteresis loss is 30.0  W and eddy current loss is 115.2  W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 703</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the hysteresis and eddy current losses at 50 Hz (b) determine the hysteresis and eddy current losses at 50 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "TL1  =  400;#  in  Watt\n",
+      "TL2  =  498;#  in  Watt\n",
+      "x  =  0.25;\n",
+      "y  =  0.4;\n",
+      "f1  =  50;#  in  Hz\n",
+      "n  =  1.7;#  Steinmetz  index\n",
+      "f2  =  60;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #if  volume  v  and  the  maximum  flux  density  are  constant\n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n  =  k1*f\n",
+      " #(if  the  maximum  flux  density  and  the  lamination  thickness  are  constant)\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*f**2\n",
+      " #At  50  Hz  frequency,  TL1  =  k1*f1  +  k2*f1**2\n",
+      " #At  60  Hz  frequency,  TL2  =  k1*f2  +  k2*f2**2\n",
+      " #Solving  equations  gives  the  values  of  k1  and  k2.\n",
+      "k2  =  (5*TL2  -  6*TL1)/(5*(f2**2)  -  6*(f1**2))\n",
+      "k1  =  (TL1  -  k2*f1**2)/f1\n",
+      " #hysteresis  loss  Ph  =  k1*f\n",
+      "Ph1  =  k1*f1\n",
+      " #eddy  current  loss\n",
+      "Pe1  =  k2*f1**2\n",
+      " #Since  at  50  Hz  the  flux  density  is  increased  by  25%,  the  new  hysteresis  loss  is\n",
+      "Ph2  =  Ph1*(1  +  x)**1.7\n",
+      " #Since  at  50  Hz  the  flux  density  is  increased  by  25%,  and  the  lamination  thickness  is  increased  by  40%, \n",
+      "    #the  new  eddy  current  loss  is\n",
+      "Pe2  =  Pe1*((1  +  x)**2)*(1  +  y)**3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  hysteresis  and  eddy  current  losses  at  50  Hz  are  \",round(Ph1,2),\"  W  and  \",round(  Pe1,2),\"  W  resp.\"\n",
+      "print  \"\\n  (b)the  hysteresis  and  eddy  current  losses  at  50  Hz  after  increement  are  \",round(Ph2,2),\"  W  and  \",round(  Pe2,2),\"  W  resp.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  hysteresis  and  eddy  current  losses  at  50  Hz  are   325.0   W  and   75.0   W  resp.\n",
+        "\n",
+        "  (b)the  hysteresis  and  eddy  current  losses  at  50  Hz  after  increement  are   474.93   W  and   321.56   W  resp.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_2.ipynb
new file mode 100755
index 00000000..c4a1d2a6
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_2.ipynb
@@ -0,0 +1,631 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 38: Magnetic materials</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 694</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the hysteresis loss per m3 per cycle, and (b) the hysteresis loss per m3 at a frequency of 50 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A  =  12.5;#  in  cm2\n",
+      "x  =  500;#  horizontal  axis  1  cm  =  500  A/m\n",
+      "y  =  0.2;#  vertical  axis  1  cm  =  0.2  T\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  per  cycle\n",
+      "HL  =  A*x*y\n",
+      " #At  50  Hz  frequency,  hysteresis  loss\n",
+      "HLf  =  HL*f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)hysteresis  loss  per  cycle  is  =  \",HL,\"  J/m3\"\n",
+      "print  \"\\n(b)At  50  Hz  frequency,  hysteresis  loss  \",HLf,\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)hysteresis  loss  per  cycle  is  =   1250.0   J/m3\n",
+        "\n",
+        "(b)At  50  Hz  frequency,  hysteresis  loss   62500.0   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 695</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the hysteresis loss per m3 for a maximum flux density of 1.1 T and frequency of 25 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "n  =  1.6;#  the  Steinmetz  index\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  25;#  in  Hz\n",
+      "Bm1  =  1.5;#  in  Tesla\n",
+      "Bm2  =  1.1;#  in  Tesla\n",
+      "Ph1  =  62500;#  in  W/m3\n",
+      "v  =  1;\n",
+      "\n",
+      "#calculation:  \n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n\n",
+      "kh  =  Ph1/(v*f1*(Bm1)**n)\n",
+      " #When  f  =  25  Hz  and  Bm  =  1.1  T,\n",
+      "Ph2  =  kh*v*f2*(Bm2)**n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  hysteresis  loss  When  f  =  25  Hz  and  Bm  =  1.1  T,  is  =  \",round(Ph2,2),\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  hysteresis  loss  When  f  =  25  Hz  and  Bm  =  1.1  T,  is  =   19025.33   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 695</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the hysteresis loss at a frequency of 80 Hz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "csa  =  0.002;#  in  m2\n",
+      "l  =  1;#  in  m\n",
+      "a  =  400/0.01;#  10  mm  =  400  A/m  \n",
+      "b  =  0.1/0.01;#  10  mm  =  0.1  T  \n",
+      "A  =  0.01;#  in  m2\n",
+      "f  =  80;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #hysteresis  loss  per  cycle\n",
+      "HL  =  A*a*b\n",
+      " #At  a  frequency  of  80  Hz,\n",
+      " #hysteresis  loss\n",
+      "HLf  =  HL*f\n",
+      " #Volume  of  ring\n",
+      "v  =  csa*l\n",
+      " #hysteresis  loss\n",
+      "Ph  =  HLf*v\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  hysteresis  loss  at  a  frequency  of  80  Hz  is  \",Ph,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  hysteresis  loss  at  a  frequency  of  80  Hz  is   640.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 696</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the value of the hysteresis loss when the maximum core flux is 8 mWb and the frequency is 50 Hz\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Phi1  =  0.01;#  in  Wb\n",
+      "Phi2  =  0.008;#  in  Wb\n",
+      "csa  =  0.008;#  in  m2\n",
+      "v  =  0.005;#  in  m3\n",
+      "f  =  50;#  in  Hz\n",
+      "n  =  1.7;#  the  Steinmetz  constant\n",
+      "Ph1  =  100;#  in  Watt\n",
+      "\n",
+      " #calculation:  \n",
+      " #maximum  flux  density\n",
+      "Bm1  =  Phi1/csa\n",
+      " #hysteresis  loss  Ph1  =  kh*v*f*(Bm1)**n\n",
+      "kh  =  Ph1/(v*f*(Bm1)**n)\n",
+      " #When  the  maximum  core  flux  is  8  mWb,\n",
+      "Bm2  =  Phi2/csa\n",
+      " #hysteresis  loss,  Ph2\n",
+      "Ph2  =  kh*v*f*(Bm2)**n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"value  of hysteresis  loss  when  maximum  core  flux  is  8  mWb  and  the  frequency  is  50  Hz  is  \",round(Ph2,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "value  of hysteresis  loss  when  maximum  core  flux  is  8  mWb  and  the  frequency  is  50  Hz  is   68.43   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 699</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the new value of eddy current loss per cubic metre.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pe1  =  10;#  in  W/m3\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  30;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #When  the  eddy  current  loss  is  10  W/m3,  frequency  f  is  50  Hz.\n",
+      " #constant  k\n",
+      "k  =  Pe1/(f1**2)\n",
+      " #When  the  frequency  is  30  Hz,  eddy  current  loss,\n",
+      "Pe2  =  k*(f2**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\neddy  current  loss  per  cubic  metre  is  \",Pe2,\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "eddy  current  loss  per  cubic  metre  is   3.6   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 699</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the necessary new thickness of the laminations.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pe  =  100;#  in  W/m3\n",
+      "f1  =  50;#  in  Hz\n",
+      "t1  =  0.0005;#  in  m\n",
+      "x  =  1/3;\n",
+      "f2  =  250;#  in  Hz\n",
+      "Bm1 = 1;\n",
+      " #calculation:  \n",
+      " #Pe  =  ke*(Bm1*f1*t1)**2\n",
+      " #Hence,  at  50  Hz  frequency\n",
+      "ke  =  Pe/(Bm1*f1*t1)**2\n",
+      " #At  250  Hz  frequency\n",
+      "Bm2  =  x*Bm1\n",
+      "t2  =  ((Pe/ke)**0.5)/(Bm2*f2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nlamination  thickness  is  \",t2,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "lamination  thickness  is   0.0003 m"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 700</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the values of the losses if the frequency is increased to 60 Hz. \n",
+      "#(b) What will be the total core loss if the frequency is 50 Hz and the lamination are made one-half of their original thickness?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ph1  =  40;#  in  W\n",
+      "Pe1  =  20;#  in  W\n",
+      "f1  =  50;#  in  Hz\n",
+      "x  =  1/2;\n",
+      "f2  =  60;#  in  Hz\n",
+      "t1  =  1;\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n  =  k1*f\n",
+      " #Thus  when  the  hysteresis  is  40  W  and  the  frequency  50  Hz,\n",
+      "k1  =  Ph1/f1\n",
+      " #If  the  frequency  is  increased  to  60  Hz,\n",
+      "Ph2  =  k1*f2\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*f**2\n",
+      " #since  the  flux  density  and  lamination  thickness  are  constant.\n",
+      " #When  the  eddy  current  loss  is  20  W  the  frequency  is  50  Hz.  Thus\n",
+      "k2  =  Pe1/(f1**2)\n",
+      " #If  the  frequency  is  increased  to  60  Hz,\n",
+      "Pe2  =  k2*(f2**2)\n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n,  is  independent  of  the  thickness  of  the  laminations.  \n",
+      "    #Thus,  if  the  thickness  of  the  laminations  is  halved,  the  hysteresis  loss  remains  at  \n",
+      "Phb2  =  Ph1\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*t**3\n",
+      "k3  =  Pe1/(t1**3)\n",
+      "t2  =  0.5*t1\n",
+      "Peb2  =  k3*t2**3\n",
+      " #total  core  loss  when  the  thickness  of  the  laminations  is  halved  is  given  by\n",
+      "TL  =  Phb2  +  Peb2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)If  the  frequency  is  increased  to  60  Hz,hysteresis  loss  is  \",Ph2,\"  W  and  eddy  current  loss  \",  Pe2,\"  W\"\n",
+      "print  \"\\n(b)the  total  core  loss  when  the  thickness  of  the  laminations  is  halved  \",TL,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)If  the  frequency  is  increased  to  60  Hz,hysteresis  loss  is   48.0   W  and  eddy  current  loss   28.8   W\n",
+        "\n",
+        "(b)the  total  core  loss  when  the  thickness  of  the  laminations  is  halved   42.5   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 701</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the new total core loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  500;#  in  Volts\n",
+      "V2  =  1000;#  in  Volts\n",
+      "Ph1  =  400;#  in  W\n",
+      "Pe1  =  150;#  in  W\n",
+      "f1  =  50;#  in  Hz\n",
+      "n  =  1.6;#  Steinmetz  index\n",
+      "f2  =  100;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  Ph  =  k1*f*(E/f)**n\n",
+      " #At  500  V  and  50  Hz\n",
+      "k1  =  Ph1/(f1*(V1/f1)**1.6)\n",
+      " #At  1000  V  and  100  Hz,\n",
+      "Ph2  =  k1*f2*(V2/f2)**1.6\n",
+      " #eddy  current  loss,  Pe  =  k2*E**2\n",
+      " #At  500  V,\n",
+      "k2  =  Pe1/(V1**2)\n",
+      " #At  1000  V,\n",
+      "Pe2  =  k2*(V2**2)\n",
+      " #the  new  total  core  loss\n",
+      "TL  =  Ph2  +  Pe2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  new  total  core  loss  \",TL,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  new  total  core  loss   1400.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 702</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the separate values of the hysteresis and eddy current losses at frequencies of (a) 50 Hz and (b) 60 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  60;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "k1  =  0.5\n",
+      "k2  =  0.032\n",
+      "f = [30, 50, 70,90]\n",
+      "Pcf = [1.5, 2.1, 2.7, 3.4]\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(f,Pcf,'-')\n",
+      "xlabel('f')\n",
+      "ylabel('Pc/f')\n",
+      "show()\n",
+      "HL1 = k1*f1\n",
+      "ECL1 = k2*f1**2\n",
+      "\n",
+      "HL2 = k1*f2\n",
+      "ECL2 = k2*f2**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a) at Frequency = 50 Hz, hysteresis loss is\", HL1,\" W and eddy current loss is\", ECL1,\" W\"\n",
+      "print  \"\\n  (b) at Frequency = 60 Hz, hysteresis loss is\", HL2,\" W and eddy current loss is\", ECL2,\" W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'."
+       ]
+      },
+      {
+       "output_type": "display_data",
+       "png": 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+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a) at Frequency = 50 Hz, hysteresis loss is 25.0  W and eddy current loss is 80.0  W\n",
+        "\n",
+        "  (b) at Frequency = 60 Hz, hysteresis loss is 30.0  W and eddy current loss is 115.2  W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 703</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine the hysteresis and eddy current losses at 50 Hz (b) determine the hysteresis and eddy current losses at 50 Hz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "TL1  =  400;#  in  Watt\n",
+      "TL2  =  498;#  in  Watt\n",
+      "x  =  0.25;\n",
+      "y  =  0.4;\n",
+      "f1  =  50;#  in  Hz\n",
+      "n  =  1.7;#  Steinmetz  index\n",
+      "f2  =  60;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #if  volume  v  and  the  maximum  flux  density  are  constant\n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n  =  k1*f\n",
+      " #(if  the  maximum  flux  density  and  the  lamination  thickness  are  constant)\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*f**2\n",
+      " #At  50  Hz  frequency,  TL1  =  k1*f1  +  k2*f1**2\n",
+      " #At  60  Hz  frequency,  TL2  =  k1*f2  +  k2*f2**2\n",
+      " #Solving  equations  gives  the  values  of  k1  and  k2.\n",
+      "k2  =  (5*TL2  -  6*TL1)/(5*(f2**2)  -  6*(f1**2))\n",
+      "k1  =  (TL1  -  k2*f1**2)/f1\n",
+      " #hysteresis  loss  Ph  =  k1*f\n",
+      "Ph1  =  k1*f1\n",
+      " #eddy  current  loss\n",
+      "Pe1  =  k2*f1**2\n",
+      " #Since  at  50  Hz  the  flux  density  is  increased  by  25%,  the  new  hysteresis  loss  is\n",
+      "Ph2  =  Ph1*(1  +  x)**1.7\n",
+      " #Since  at  50  Hz  the  flux  density  is  increased  by  25%,  and  the  lamination  thickness  is  increased  by  40%, \n",
+      "    #the  new  eddy  current  loss  is\n",
+      "Pe2  =  Pe1*((1  +  x)**2)*(1  +  y)**3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  hysteresis  and  eddy  current  losses  at  50  Hz  are  \",round(Ph1,2),\"  W  and  \",round(  Pe1,2),\"  W  resp.\"\n",
+      "print  \"\\n  (b)the  hysteresis  and  eddy  current  losses  at  50  Hz  after  increement  are  \",round(Ph2,2),\"  W  and  \",round(  Pe2,2),\"  W  resp.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  hysteresis  and  eddy  current  losses  at  50  Hz  are   325.0   W  and   75.0   W  resp.\n",
+        "\n",
+        "  (b)the  hysteresis  and  eddy  current  losses  at  50  Hz  after  increement  are   474.93   W  and   321.56   W  resp.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_3.ipynb
new file mode 100755
index 00000000..68af6e5d
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_38-checkpoint_3.ipynb
@@ -0,0 +1,631 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7389552750eb30891c831f10215e616f84c75b4bc1c44e309d45236439691034"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 38: Magnetic materials</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 694</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "A  =  12.5;#  in  cm2\n",
+      "x  =  500;#  horizontal  axis  1  cm  =  500  A/m\n",
+      "y  =  0.2;#  vertical  axis  1  cm  =  0.2  T\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  per  cycle\n",
+      "HL  =  A*x*y\n",
+      " #At  50  Hz  frequency,  hysteresis  loss\n",
+      "HLf  =  HL*f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)hysteresis  loss  per  cycle  is  =  \",HL,\"  J/m3\"\n",
+      "print  \"\\n(b)At  50  Hz  frequency,  hysteresis  loss  \",HLf,\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)hysteresis  loss  per  cycle  is  =   1250.0   J/m3\n",
+        "\n",
+        "(b)At  50  Hz  frequency,  hysteresis  loss   62500.0   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 695</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "n  =  1.6;#  the  Steinmetz  index\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  25;#  in  Hz\n",
+      "Bm1  =  1.5;#  in  Tesla\n",
+      "Bm2  =  1.1;#  in  Tesla\n",
+      "Ph1  =  62500;#  in  W/m3\n",
+      "v  =  1;\n",
+      "\n",
+      "#calculation:  \n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n\n",
+      "kh  =  Ph1/(v*f1*(Bm1)**n)\n",
+      " #When  f  =  25  Hz  and  Bm  =  1.1  T,\n",
+      "Ph2  =  kh*v*f2*(Bm2)**n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  hysteresis  loss  When  f  =  25  Hz  and  Bm  =  1.1  T,  is  =  \",round(Ph2,2),\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  hysteresis  loss  When  f  =  25  Hz  and  Bm  =  1.1  T,  is  =   19025.33   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 695</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "csa  =  0.002;#  in  m2\n",
+      "l  =  1;#  in  m\n",
+      "a  =  400/0.01;#  10  mm  =  400  A/m  \n",
+      "b  =  0.1/0.01;#  10  mm  =  0.1  T  \n",
+      "A  =  0.01;#  in  m2\n",
+      "f  =  80;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #hysteresis  loss  per  cycle\n",
+      "HL  =  A*a*b\n",
+      " #At  a  frequency  of  80  Hz,\n",
+      " #hysteresis  loss\n",
+      "HLf  =  HL*f\n",
+      " #Volume  of  ring\n",
+      "v  =  csa*l\n",
+      " #hysteresis  loss\n",
+      "Ph  =  HLf*v\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  hysteresis  loss  at  a  frequency  of  80  Hz  is  \",Ph,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  hysteresis  loss  at  a  frequency  of  80  Hz  is   640.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 696</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Phi1  =  0.01;#  in  Wb\n",
+      "Phi2  =  0.008;#  in  Wb\n",
+      "csa  =  0.008;#  in  m2\n",
+      "v  =  0.005;#  in  m3\n",
+      "f  =  50;#  in  Hz\n",
+      "n  =  1.7;#  the  Steinmetz  constant\n",
+      "Ph1  =  100;#  in  Watt\n",
+      "\n",
+      " #calculation:  \n",
+      " #maximum  flux  density\n",
+      "Bm1  =  Phi1/csa\n",
+      " #hysteresis  loss  Ph1  =  kh*v*f*(Bm1)**n\n",
+      "kh  =  Ph1/(v*f*(Bm1)**n)\n",
+      " #When  the  maximum  core  flux  is  8  mWb,\n",
+      "Bm2  =  Phi2/csa\n",
+      " #hysteresis  loss,  Ph2\n",
+      "Ph2  =  kh*v*f*(Bm2)**n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"value  of hysteresis  loss  when  maximum  core  flux  is  8  mWb  and  the  frequency  is  50  Hz  is  \",round(Ph2,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "value  of hysteresis  loss  when  maximum  core  flux  is  8  mWb  and  the  frequency  is  50  Hz  is   68.43   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 699</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pe1  =  10;#  in  W/m3\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  30;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #When  the  eddy  current  loss  is  10  W/m3,  frequency  f  is  50  Hz.\n",
+      " #constant  k\n",
+      "k  =  Pe1/(f1**2)\n",
+      " #When  the  frequency  is  30  Hz,  eddy  current  loss,\n",
+      "Pe2  =  k*(f2**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\neddy  current  loss  per  cubic  metre  is  \",Pe2,\"  W/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "eddy  current  loss  per  cubic  metre  is   3.6   W/m3"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 699</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Pe  =  100;#  in  W/m3\n",
+      "f1  =  50;#  in  Hz\n",
+      "t1  =  0.0005;#  in  m\n",
+      "x  =  1/3;\n",
+      "f2  =  250;#  in  Hz\n",
+      "Bm1 = 1;\n",
+      " #calculation:  \n",
+      " #Pe  =  ke*(Bm1*f1*t1)**2\n",
+      " #Hence,  at  50  Hz  frequency\n",
+      "ke  =  Pe/(Bm1*f1*t1)**2\n",
+      " #At  250  Hz  frequency\n",
+      "Bm2  =  x*Bm1\n",
+      "t2  =  ((Pe/ke)**0.5)/(Bm2*f2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nlamination  thickness  is  \",t2,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "lamination  thickness  is   0.0003 m"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 700</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ph1  =  40;#  in  W\n",
+      "Pe1  =  20;#  in  W\n",
+      "f1  =  50;#  in  Hz\n",
+      "x  =  1/2;\n",
+      "f2  =  60;#  in  Hz\n",
+      "t1  =  1;\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n  =  k1*f\n",
+      " #Thus  when  the  hysteresis  is  40  W  and  the  frequency  50  Hz,\n",
+      "k1  =  Ph1/f1\n",
+      " #If  the  frequency  is  increased  to  60  Hz,\n",
+      "Ph2  =  k1*f2\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*f**2\n",
+      " #since  the  flux  density  and  lamination  thickness  are  constant.\n",
+      " #When  the  eddy  current  loss  is  20  W  the  frequency  is  50  Hz.  Thus\n",
+      "k2  =  Pe1/(f1**2)\n",
+      " #If  the  frequency  is  increased  to  60  Hz,\n",
+      "Pe2  =  k2*(f2**2)\n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n,  is  independent  of  the  thickness  of  the  laminations.  \n",
+      "    #Thus,  if  the  thickness  of  the  laminations  is  halved,  the  hysteresis  loss  remains  at  \n",
+      "Phb2  =  Ph1\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*t**3\n",
+      "k3  =  Pe1/(t1**3)\n",
+      "t2  =  0.5*t1\n",
+      "Peb2  =  k3*t2**3\n",
+      " #total  core  loss  when  the  thickness  of  the  laminations  is  halved  is  given  by\n",
+      "TL  =  Phb2  +  Peb2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a)If  the  frequency  is  increased  to  60  Hz,hysteresis  loss  is  \",Ph2,\"  W  and  eddy  current  loss  \",  Pe2,\"  W\"\n",
+      "print  \"\\n(b)the  total  core  loss  when  the  thickness  of  the  laminations  is  halved  \",TL,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a)If  the  frequency  is  increased  to  60  Hz,hysteresis  loss  is   48.0   W  and  eddy  current  loss   28.8   W\n",
+        "\n",
+        "(b)the  total  core  loss  when  the  thickness  of  the  laminations  is  halved   42.5   W\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 701</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V1  =  500;#  in  Volts\n",
+      "V2  =  1000;#  in  Volts\n",
+      "Ph1  =  400;#  in  W\n",
+      "Pe1  =  150;#  in  W\n",
+      "f1  =  50;#  in  Hz\n",
+      "n  =  1.6;#  Steinmetz  index\n",
+      "f2  =  100;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #hysteresis  loss  Ph  =  k1*f*(E/f)**n\n",
+      " #At  500  V  and  50  Hz\n",
+      "k1  =  Ph1/(f1*(V1/f1)**1.6)\n",
+      " #At  1000  V  and  100  Hz,\n",
+      "Ph2  =  k1*f2*(V2/f2)**1.6\n",
+      " #eddy  current  loss,  Pe  =  k2*E**2\n",
+      " #At  500  V,\n",
+      "k2  =  Pe1/(V1**2)\n",
+      " #At  1000  V,\n",
+      "Pe2  =  k2*(V2**2)\n",
+      " #the  new  total  core  loss\n",
+      "TL  =  Ph2  +  Pe2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  new  total  core  loss  \",TL,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  new  total  core  loss   1400.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 702</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#from pylab import *\n",
+      "%pylab inline\n",
+      "#initializing  the  variables:\n",
+      "f1  =  50;#  in  Hz\n",
+      "f2  =  60;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "k1  =  0.5\n",
+      "k2  =  0.032\n",
+      "f = [30, 50, 70,90]\n",
+      "Pcf = [1.5, 2.1, 2.7, 3.4]\n",
+      "fig  = plt.figure()\n",
+      "ax = fig.add_subplot(1, 1, 1)\n",
+      "ax.plot(f,Pcf,'-')\n",
+      "xlabel('f')\n",
+      "ylabel('Pc/f')\n",
+      "show()\n",
+      "HL1 = k1*f1\n",
+      "ECL1 = k2*f1**2\n",
+      "\n",
+      "HL2 = k1*f2\n",
+      "ECL2 = k2*f2**2\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a) at Frequency = 50 Hz, hysteresis loss is\", HL1,\" W and eddy current loss is\", ECL1,\" W\"\n",
+      "print  \"\\n  (b) at Frequency = 60 Hz, hysteresis loss is\", HL2,\" W and eddy current loss is\", ECL2,\" W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n",
+        "For more information, type 'help(pylab)'."
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a) at Frequency = 50 Hz, hysteresis loss is 25.0  W and eddy current loss is 80.0  W\n",
+        "\n",
+        "  (b) at Frequency = 60 Hz, hysteresis loss is 30.0  W and eddy current loss is 115.2  W"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 703</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "TL1  =  400;#  in  Watt\n",
+      "TL2  =  498;#  in  Watt\n",
+      "x  =  0.25;\n",
+      "y  =  0.4;\n",
+      "f1  =  50;#  in  Hz\n",
+      "n  =  1.7;#  Steinmetz  index\n",
+      "f2  =  60;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #if  volume  v  and  the  maximum  flux  density  are  constant\n",
+      " #hysteresis  loss  Ph  =  kh*v*f*(Bm)**n  =  k1*f\n",
+      " #(if  the  maximum  flux  density  and  the  lamination  thickness  are  constant)\n",
+      " #eddy  current  loss,  Pe  =  ke*(Bm1*f1*t1)**2  =  k2*f**2\n",
+      " #At  50  Hz  frequency,  TL1  =  k1*f1  +  k2*f1**2\n",
+      " #At  60  Hz  frequency,  TL2  =  k1*f2  +  k2*f2**2\n",
+      " #Solving  equations  gives  the  values  of  k1  and  k2.\n",
+      "k2  =  (5*TL2  -  6*TL1)/(5*(f2**2)  -  6*(f1**2))\n",
+      "k1  =  (TL1  -  k2*f1**2)/f1\n",
+      " #hysteresis  loss  Ph  =  k1*f\n",
+      "Ph1  =  k1*f1\n",
+      " #eddy  current  loss\n",
+      "Pe1  =  k2*f1**2\n",
+      " #Since  at  50  Hz  the  flux  density  is  increased  by  25%,  the  new  hysteresis  loss  is\n",
+      "Ph2  =  Ph1*(1  +  x)**1.7\n",
+      " #Since  at  50  Hz  the  flux  density  is  increased  by  25%,  and  the  lamination  thickness  is  increased  by  40%, \n",
+      "    #the  new  eddy  current  loss  is\n",
+      "Pe2  =  Pe1*((1  +  x)**2)*(1  +  y)**3\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)the  hysteresis  and  eddy  current  losses  at  50  Hz  are  \",round(Ph1,2),\"  W  and  \",round(  Pe1,2),\"  W  resp.\"\n",
+      "print  \"\\n  (b)the  hysteresis  and  eddy  current  losses  at  50  Hz  after  increement  are  \",round(Ph2,2),\"  W  and  \",round(  Pe2,2),\"  W  resp.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)the  hysteresis  and  eddy  current  losses  at  50  Hz  are   325.0   W  and   75.0   W  resp.\n",
+        "\n",
+        "  (b)the  hysteresis  and  eddy  current  losses  at  50  Hz  after  increement  are   474.93   W  and   321.56   W  resp.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint.ipynb
new file mode 100755
index 00000000..33240359
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint.ipynb
@@ -0,0 +1,200 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 39: Dielectrics and dielectric loss</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 717</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the capacitor, at a frequency of 8 MHz, \n",
+      "#(a) the loss angle, (b) the power factor, (c) the Q-factor, and (d) the dissipation factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rs  =  1.5;#  in  ohms\n",
+      "Cs  =  400E-12;#  in  Farads\n",
+      "f  =  8E6;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #for  a  series  equivalent  circuit,\n",
+      " #tan(del)  =  Rs*w*Cs\n",
+      " #loss  angle,\n",
+      "de  =  math.atan(Rs*Cs*(2*math.pi*f))\n",
+      " #power  factor\n",
+      "pf  =  math.cos(de)\n",
+      " #the  Q-factor\n",
+      "Q  =  1/math.tan(de)\n",
+      " #dissipation  factor,\n",
+      "D  =  1/Q\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)loss  angle  \",round(de,2),\"  rad.\"\n",
+      "print  \"\\n  (b)power  factor  \",round(de,2),\"  rad.\"\n",
+      "print  \"\\n  (c)Q-factor  is  \",round(Q,2)\n",
+      "print  \"\\n  (d)dissipation  factor  \",round(D,2),\"  rad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)loss  angle   0.03   rad.\n",
+        "\n",
+        "  (b)power  factor   0.03   rad.\n",
+        "\n",
+        "  (c)Q-factor  is   33.16\n",
+        "\n",
+        "  (d)dissipation  factor   0.03   rad."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 718</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the component values of the equivalent parallel circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "de  =  0.025;#  in  rad.\n",
+      "V  =  5000;#  in  Volts\n",
+      "PL  =  20;#  power  loss\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #power  loss  =  w*C*V**2*tan(del)\n",
+      "Cp  =  PL/(2*math.pi*f*V*V*math.tan(de))\n",
+      " #for  a  parallel  equivalent  circuit,\n",
+      " #tan(del)  =  1/(Rp*w*Cp)\n",
+      "Rp  =  1/(2*math.pi*f*Cp*math.tan(de))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  C  \",round(Cp*1E6,2),\"uF  and  parallel  resistance  \",round(Rp,2),\"ohm.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  C   0.1 uF  and  parallel  resistance   1250000.0 ohm."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 718</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the loss angle, (b) the equivalent series loss resistance, and (c) the equivalent parallel loss resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "P  =  500E-6;#  in  Watt\n",
+      "C  =  2000E-12;#  in  Farads\n",
+      "V  =  20;#  in  Volts\n",
+      "f  =  10000;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #power  loss  =  w*C*V**2*tan(del)\n",
+      " #loss  angle\n",
+      "de =  math.atan(P/(2*math.pi*f*V*V*C))\n",
+      " #for  an  equivalent  series  circuit,\n",
+      " #tan(del)  =  (Rs*w*Cs)\n",
+      "Cs  =  C\n",
+      "Rs  =  (math.tan(de))/(2*math.pi*f*Cs)\n",
+      " #for  an  equivalent  parallel  circuit\n",
+      " #tan(del)  =  1/(Rp*w*Cp)\n",
+      "Cp  =  C\n",
+      "Rp  =  1/(2*math.pi*f*Cp*math.tan(de))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)loss  angle  \",round(de*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  (b)series  resistance  \",round(Rs,2),\"  ohm.\"\n",
+      "print  \"\\n  (c)parallel  resistance  \",round(Rp/1000,2),\"Kohm.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)loss  angle   0.57 deg\n",
+        "\n",
+        "  (b)series  resistance   79.16   ohm.\n",
+        "\n",
+        "  (c)parallel  resistance   800.0 Kohm."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_1.ipynb
new file mode 100755
index 00000000..33240359
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_1.ipynb
@@ -0,0 +1,200 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 39: Dielectrics and dielectric loss</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 717</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the capacitor, at a frequency of 8 MHz, \n",
+      "#(a) the loss angle, (b) the power factor, (c) the Q-factor, and (d) the dissipation factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rs  =  1.5;#  in  ohms\n",
+      "Cs  =  400E-12;#  in  Farads\n",
+      "f  =  8E6;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #for  a  series  equivalent  circuit,\n",
+      " #tan(del)  =  Rs*w*Cs\n",
+      " #loss  angle,\n",
+      "de  =  math.atan(Rs*Cs*(2*math.pi*f))\n",
+      " #power  factor\n",
+      "pf  =  math.cos(de)\n",
+      " #the  Q-factor\n",
+      "Q  =  1/math.tan(de)\n",
+      " #dissipation  factor,\n",
+      "D  =  1/Q\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)loss  angle  \",round(de,2),\"  rad.\"\n",
+      "print  \"\\n  (b)power  factor  \",round(de,2),\"  rad.\"\n",
+      "print  \"\\n  (c)Q-factor  is  \",round(Q,2)\n",
+      "print  \"\\n  (d)dissipation  factor  \",round(D,2),\"  rad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)loss  angle   0.03   rad.\n",
+        "\n",
+        "  (b)power  factor   0.03   rad.\n",
+        "\n",
+        "  (c)Q-factor  is   33.16\n",
+        "\n",
+        "  (d)dissipation  factor   0.03   rad."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 718</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the component values of the equivalent parallel circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "de  =  0.025;#  in  rad.\n",
+      "V  =  5000;#  in  Volts\n",
+      "PL  =  20;#  power  loss\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #power  loss  =  w*C*V**2*tan(del)\n",
+      "Cp  =  PL/(2*math.pi*f*V*V*math.tan(de))\n",
+      " #for  a  parallel  equivalent  circuit,\n",
+      " #tan(del)  =  1/(Rp*w*Cp)\n",
+      "Rp  =  1/(2*math.pi*f*Cp*math.tan(de))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  C  \",round(Cp*1E6,2),\"uF  and  parallel  resistance  \",round(Rp,2),\"ohm.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  C   0.1 uF  and  parallel  resistance   1250000.0 ohm."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 718</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the loss angle, (b) the equivalent series loss resistance, and (c) the equivalent parallel loss resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "P  =  500E-6;#  in  Watt\n",
+      "C  =  2000E-12;#  in  Farads\n",
+      "V  =  20;#  in  Volts\n",
+      "f  =  10000;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #power  loss  =  w*C*V**2*tan(del)\n",
+      " #loss  angle\n",
+      "de =  math.atan(P/(2*math.pi*f*V*V*C))\n",
+      " #for  an  equivalent  series  circuit,\n",
+      " #tan(del)  =  (Rs*w*Cs)\n",
+      "Cs  =  C\n",
+      "Rs  =  (math.tan(de))/(2*math.pi*f*Cs)\n",
+      " #for  an  equivalent  parallel  circuit\n",
+      " #tan(del)  =  1/(Rp*w*Cp)\n",
+      "Cp  =  C\n",
+      "Rp  =  1/(2*math.pi*f*Cp*math.tan(de))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)loss  angle  \",round(de*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  (b)series  resistance  \",round(Rs,2),\"  ohm.\"\n",
+      "print  \"\\n  (c)parallel  resistance  \",round(Rp/1000,2),\"Kohm.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)loss  angle   0.57 deg\n",
+        "\n",
+        "  (b)series  resistance   79.16   ohm.\n",
+        "\n",
+        "  (c)parallel  resistance   800.0 Kohm."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_2.ipynb
new file mode 100755
index 00000000..33240359
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_2.ipynb
@@ -0,0 +1,200 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 39: Dielectrics and dielectric loss</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 717</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the capacitor, at a frequency of 8 MHz, \n",
+      "#(a) the loss angle, (b) the power factor, (c) the Q-factor, and (d) the dissipation factor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rs  =  1.5;#  in  ohms\n",
+      "Cs  =  400E-12;#  in  Farads\n",
+      "f  =  8E6;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #for  a  series  equivalent  circuit,\n",
+      " #tan(del)  =  Rs*w*Cs\n",
+      " #loss  angle,\n",
+      "de  =  math.atan(Rs*Cs*(2*math.pi*f))\n",
+      " #power  factor\n",
+      "pf  =  math.cos(de)\n",
+      " #the  Q-factor\n",
+      "Q  =  1/math.tan(de)\n",
+      " #dissipation  factor,\n",
+      "D  =  1/Q\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)loss  angle  \",round(de,2),\"  rad.\"\n",
+      "print  \"\\n  (b)power  factor  \",round(de,2),\"  rad.\"\n",
+      "print  \"\\n  (c)Q-factor  is  \",round(Q,2)\n",
+      "print  \"\\n  (d)dissipation  factor  \",round(D,2),\"  rad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)loss  angle   0.03   rad.\n",
+        "\n",
+        "  (b)power  factor   0.03   rad.\n",
+        "\n",
+        "  (c)Q-factor  is   33.16\n",
+        "\n",
+        "  (d)dissipation  factor   0.03   rad."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 718</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the component values of the equivalent parallel circuit.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "de  =  0.025;#  in  rad.\n",
+      "V  =  5000;#  in  Volts\n",
+      "PL  =  20;#  power  loss\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #power  loss  =  w*C*V**2*tan(del)\n",
+      "Cp  =  PL/(2*math.pi*f*V*V*math.tan(de))\n",
+      " #for  a  parallel  equivalent  circuit,\n",
+      " #tan(del)  =  1/(Rp*w*Cp)\n",
+      "Rp  =  1/(2*math.pi*f*Cp*math.tan(de))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  C  \",round(Cp*1E6,2),\"uF  and  parallel  resistance  \",round(Rp,2),\"ohm.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  C   0.1 uF  and  parallel  resistance   1250000.0 ohm."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 718</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the loss angle, (b) the equivalent series loss resistance, and (c) the equivalent parallel loss resistance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "P  =  500E-6;#  in  Watt\n",
+      "C  =  2000E-12;#  in  Farads\n",
+      "V  =  20;#  in  Volts\n",
+      "f  =  10000;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #power  loss  =  w*C*V**2*tan(del)\n",
+      " #loss  angle\n",
+      "de =  math.atan(P/(2*math.pi*f*V*V*C))\n",
+      " #for  an  equivalent  series  circuit,\n",
+      " #tan(del)  =  (Rs*w*Cs)\n",
+      "Cs  =  C\n",
+      "Rs  =  (math.tan(de))/(2*math.pi*f*Cs)\n",
+      " #for  an  equivalent  parallel  circuit\n",
+      " #tan(del)  =  1/(Rp*w*Cp)\n",
+      "Cp  =  C\n",
+      "Rp  =  1/(2*math.pi*f*Cp*math.tan(de))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)loss  angle  \",round(de*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  (b)series  resistance  \",round(Rs,2),\"  ohm.\"\n",
+      "print  \"\\n  (c)parallel  resistance  \",round(Rp/1000,2),\"Kohm.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)loss  angle   0.57 deg\n",
+        "\n",
+        "  (b)series  resistance   79.16   ohm.\n",
+        "\n",
+        "  (c)parallel  resistance   800.0 Kohm."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_3.ipynb
new file mode 100755
index 00000000..6edb5773
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_39-checkpoint_3.ipynb
@@ -0,0 +1,200 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1ec371cb843067028dd1e055f435f5ad6cdc4ce9c5ed963c08c525b60770abca"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 39: Dielectrics and dielectric loss</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 717</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "Rs  =  1.5;#  in  ohms\n",
+      "Cs  =  400E-12;#  in  Farads\n",
+      "f  =  8E6;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #for  a  series  equivalent  circuit,\n",
+      " #tan(del)  =  Rs*w*Cs\n",
+      " #loss  angle,\n",
+      "de  =  math.atan(Rs*Cs*(2*math.pi*f))\n",
+      " #power  factor\n",
+      "pf  =  math.cos(de)\n",
+      " #the  Q-factor\n",
+      "Q  =  1/math.tan(de)\n",
+      " #dissipation  factor,\n",
+      "D  =  1/Q\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)loss  angle  \",round(de,2),\"  rad.\"\n",
+      "print  \"\\n  (b)power  factor  \",round(de,2),\"  rad.\"\n",
+      "print  \"\\n  (c)Q-factor  is  \",round(Q,2)\n",
+      "print  \"\\n  (d)dissipation  factor  \",round(D,2),\"  rad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)loss  angle   0.03   rad.\n",
+        "\n",
+        "  (b)power  factor   0.03   rad.\n",
+        "\n",
+        "  (c)Q-factor  is   33.16\n",
+        "\n",
+        "  (d)dissipation  factor   0.03   rad."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 718</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "de  =  0.025;#  in  rad.\n",
+      "V  =  5000;#  in  Volts\n",
+      "PL  =  20;#  power  loss\n",
+      "f  =  50;#  in  Hz\n",
+      "\n",
+      "#calculation:  \n",
+      " #power  loss  =  w*C*V**2*tan(del)\n",
+      "Cp  =  PL/(2*math.pi*f*V*V*math.tan(de))\n",
+      " #for  a  parallel  equivalent  circuit,\n",
+      " #tan(del)  =  1/(Rp*w*Cp)\n",
+      "Rp  =  1/(2*math.pi*f*Cp*math.tan(de))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  C  \",round(Cp*1E6,2),\"uF  and  parallel  resistance  \",round(Rp,2),\"ohm.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  C   0.1 uF  and  parallel  resistance   1250000.0 ohm."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 718</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "P  =  500E-6;#  in  Watt\n",
+      "C  =  2000E-12;#  in  Farads\n",
+      "V  =  20;#  in  Volts\n",
+      "f  =  10000;#  in  Hz\n",
+      "\n",
+      " #calculation:  \n",
+      " #power  loss  =  w*C*V**2*tan(del)\n",
+      " #loss  angle\n",
+      "de =  math.atan(P/(2*math.pi*f*V*V*C))\n",
+      " #for  an  equivalent  series  circuit,\n",
+      " #tan(del)  =  (Rs*w*Cs)\n",
+      "Cs  =  C\n",
+      "Rs  =  (math.tan(de))/(2*math.pi*f*Cs)\n",
+      " #for  an  equivalent  parallel  circuit\n",
+      " #tan(del)  =  1/(Rp*w*Cp)\n",
+      "Cp  =  C\n",
+      "Rp  =  1/(2*math.pi*f*Cp*math.tan(de))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)loss  angle  \",round(de*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  (b)series  resistance  \",round(Rs,2),\"  ohm.\"\n",
+      "print  \"\\n  (c)parallel  resistance  \",round(Rp/1000,2),\"Kohm.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)loss  angle   0.57 deg\n",
+        "\n",
+        "  (b)series  resistance   79.16   ohm.\n",
+        "\n",
+        "  (c)parallel  resistance   800.0 Kohm."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb
new file mode 100755
index 00000000..3f3b710b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb
@@ -0,0 +1,1328 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 40: Field theory</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 725</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitance per metre length of the system\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.8;\n",
+      "l  =  1;#  in  m\n",
+      "\n",
+      "#calculation:  \n",
+      " #From  Figure  40.9\n",
+      "m  =  16;#  number  of  parallel  squares  measured  along  each  equipotential\n",
+      "n  =  6;#  the  number  of  series  squares  measured  along  each  line  of  force\n",
+      "C  =  e0*er*l*m/n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   66.08 pFarad."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 725</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitance of a 100 m length of the cable.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.4;\n",
+      "l  =  100;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #From  Figure  40.10\n",
+      "m  =  13;#  number  of  parallel  squares  measured  along  each  equipotential\n",
+      "n  =  4;#  the  number  of  series  squares  measured  along  each  line  of  force\n",
+      "C  =  e0*er*l*m/n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E9,2),\"nFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   9.78 nFarad."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 726</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitance per metre length of the cable\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.7;\n",
+      "ri  =  0.0005;#  in  m\n",
+      "ro  =  0.006;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(ro/ri))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   60.42 pFarad."
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 727</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the internal diameter of the sheath.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  80E-12;#  in  Farads\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "d0  =  0.008;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #internal  diameter\n",
+      "di  =  d0*(math.e**(2*math.pi*e0*er/C))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  internal  diameter  is  \",round(di,2),\"  m.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  internal  diameter  is   0.09   m."
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 728</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the capacitance per kilometre length of the cable, \n",
+      "#(b) the dielectric stress at a radius of 30 mm, and\n",
+      "#(c) the maximum and minimum values of dielectric stress.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "di  =  0.08;#  in  m\n",
+      "d0  =  0.032;#  in  m\n",
+      "r  =  0.03;#  in  m\n",
+      "V  =  40000;#  in  Volts\n",
+      "\n",
+      "#calculation:  \n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(di/d0))\n",
+      " #dielectric  stress  at  radius  r,\n",
+      "E  =  V/(r*math.log(di/d0))\n",
+      " #maximum  dielectric  stress,\n",
+      "Emax  =  V/((d0/2)*(math.log((di/d0))))\n",
+      " #minimum  dielectric  stress,\n",
+      "Emin  =  V/((di/2)*(math.log((di/d0))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pF/km\"\n",
+      "print  \"\\n  dielectric  stress  at  radius  r  is  \",round(E,2),\"V/m\"\n",
+      "print  \"\\n  maximum  dielectric  stress,  is  \",round(Emax,2),\"V/m  minimum  dielectric  stress  \",round(  Emin,2),\"V/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   212.4 pF/km\n",
+        "\n",
+        "  dielectric  stress  at  radius  r  is   1455142.22 V/m\n",
+        "\n",
+        "  maximum  dielectric  stress,  is   2728391.67 V/m  minimum  dielectric  stress   1091356.67 V/m"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 729</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the core and inner sheath radii for the most economical cable,\n",
+      "#(b) the capacitance per metre length, and (c) the charging current per kilometre run.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "V  =  60000;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "Em  =  10E6;#  in  V/m\n",
+      "\n",
+      "\n",
+      "#calculation:  \n",
+      " #core  radius,  a\n",
+      "a  =  V/Em\n",
+      " #internal  sheath  radius,\n",
+      "b  =  a*math.e**1\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er/(math.log(b/a))\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*C\n",
+      " #charging  current  per  kilometre\n",
+      "Ipkm  =  I*1000\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  core  radius  is  \",round(a*1000,2),\"mm  and  internal  sheath  radius  \",round(b*1000,1),\"mm\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,0),\"pF/m\"\n",
+      "print  \"\\n  the  charging  current  per  kilometre  \",round(Ipkm,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  core  radius  is   6.0 mm  and  internal  sheath  radius   16.3 mm\n",
+        "\n",
+        "  capacitance  is   195.0 pF/m\n",
+        "\n",
+        "  the  charging  current  per  kilometre   3.67   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 730</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for a 1 km length of the cable (a) the capacitance, (b) the charging current and (c) the power loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.5;\n",
+      "di  =  0.08;#  in  m\n",
+      "d0  =  0.025;#  in  m\n",
+      "r  =  1000;#  in  m\n",
+      "V  =  132000;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "de  =  3.5E-3;#  rad.\n",
+      "\n",
+      " #calculation:\n",
+      " #core  radius,  a\n",
+      "a  =  d0/2\n",
+      " #internal  sheath  radius,\n",
+      "b  =  di/2\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er*1E3/(math.log(b/a))\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*C\n",
+      " #power  loss\n",
+      "P  =  (2*math.pi*f*C*math.tan(de))*V**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance  for  a  1  km  length  is  \",round(C*1E6,2),\"uF\"\n",
+      "print  \"\\n  (b)the  charging  current  \",round(I,2),\"A/km\"\n",
+      "print  \"\\n  (c)power  loss  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance  for  a  1  km  length  is   0.12 uF\n",
+        "\n",
+        "  (b)the  charging  current   4.96 A/km\n",
+        "\n",
+        "  (c)power  loss   2289.78   W"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 732</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitance of the cable per metre length by the method of curvilinear squares\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.2;\n",
+      "di  =  0.06;#  in  m\n",
+      "d0  =  0.020;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #core  radius,  a\n",
+      "a  =  d0/2\n",
+      " #internal  sheath  radius,\n",
+      "b  =  di/2\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er/(math.log(b/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  per  m  of  length  is  \",round(C*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  per  m  of  length  is   0.16 nF"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 736</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitance of the line if the total length is 200 m.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "D  =  0.05;#  in  m\n",
+      "d  =  0.005;#  in  m\n",
+      "l  =  200;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  math.pi*e0*er/(math.log(D/(d/2)))\n",
+      " #capacitance  of  a  200  m  length\n",
+      "C200  =  C*l\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  of  a  200  m  length  is  \",round(C200*1E6,5),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  of  a  200  m  length  is   0.00186 uF"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 736</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for a 1 km length of line, (a) the capacitance of the  conductors, \n",
+      "#(b) the value of charge carried by each conductor, and (c) the charging current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "D  =  1.2;#  in  m\n",
+      "r  =  0.004;#  in  m\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  15000;#  in  Volts\n",
+      "l  =  1000;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  math.pi*e0*er/(math.log(D/r))\n",
+      " #capacitance  of  a  1  km  length\n",
+      "Cpkm  =  C*l\n",
+      " #Charge  Q\n",
+      "Q  =  Cpkm*V\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*Cpkm\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  per  1km  length  is  \",round(Cpkm*1E9,2),\"nF\"\n",
+      "print  \"\\n  Charge  Q  is  \",round(Q*1E6,2),\"uC\"\n",
+      "print  \"\\n  Charging  current  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  per  1km  length  is   4.87 nF\n",
+        "\n",
+        "  Charge  Q  is   73.12 uC\n",
+        "\n",
+        "  Charging  current  is   0.02   A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 737</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the maximum value required for the capacitance per metre length,\n",
+      "#and (b) the maximum diameter of each conductor if their distance between centres is 1.25 m.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "I  =  0.015;#  in  Amperes\n",
+      "d  =  1.25;#  in  m\n",
+      "r  =  800;#  in  m\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  10000;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  I/(2*math.pi*f*V)\n",
+      " #required  maximum  value  of  capacitance\n",
+      "Cmax  =  C/r\n",
+      " #maximum  diameter  of  each  conductor\n",
+      "D  =  2*d/(math.e**(math.pi*e0*er/Cmax))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  required  maximum  value  of  capacitance  is  \",round(Cmax*1E12,2),\"pF/m\"\n",
+      "print  \"\\nthe  maximum  diameter  of  each  conductor  is  \",round(D,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  required  maximum  value  of  capacitance  is   5.97 pF/m\n",
+        "\n",
+        "the  maximum  diameter  of  each  conductor  is   0.02   m"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 739</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and \n",
+      "#the average power developed if this energy is dissipated in 10 \u03bcs\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "C  =  10E-9;#  in  Farad\n",
+      "V  =  1000;#  in  Volts\n",
+      "t  =  10E-6;#  in  sec\n",
+      "\n",
+      " #calculation:\n",
+      " #energy  stored,Wf\n",
+      "Wf  =  C*V*V/2\n",
+      " #average  power  developed\n",
+      "Pav  =  Wf/t\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  energy  stored  is  \",Wf,\"J\"\n",
+      "print  \"\\nthe  average  power  developed  is  \",Pav,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  energy  stored  is   0.005 J\n",
+        "\n",
+        "the  average  power  developed  is   500.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 739</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the voltage across the plates and (b) the capacitance of the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "Q  =  5E-3;#  in  Coulomb\n",
+      "W  =  0.625;#  in  Joules\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage  across  the  plates\n",
+      "V  =  2*W/Q\n",
+      " #Capacitance  C\n",
+      "C  =  Q/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  across  the  plates  is  \",V,\"  V\"\n",
+      "print  \"\\n  Capacitance  C  is  \",C*1E6,\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  across  the  plates  is   250.0   V\n",
+        "\n",
+        "  Capacitance  C  is   20.0 uF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 740</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the required thickness of the ceramic dielectric, \n",
+      "#(b) the area of plate required if the relative permittivity of the ceramic is 10, and \n",
+      "#(c) the maximum energy stored by the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  10;\n",
+      "C  =  0.01E-6;#  in  Farad\n",
+      "E  =  10E6;#  in  V/m\n",
+      "V  =  2500;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #thickness  of  ceramic  dielectric,\n",
+      "d  =  V/E\n",
+      " #cross-sectional  area  of  plate\n",
+      "A  =  C*d/(e0*er)\n",
+      " #Maximum  energy  stored,\n",
+      "W  =  C*V*V/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  thickness  of  ceramic  dielectric  is  \",d*1000,\"mm\"\n",
+      "print  \"\\n  cross-sectional  area  of  plate,  is  \",round(A,2),\"m2\"\n",
+      "print  \"\\n  Maximum  energy  stored  is  \",round(W,3),\"  J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  thickness  of  ceramic  dielectric  is   0.25 mm\n",
+        "\n",
+        "  cross-sectional  area  of  plate,  is   0.03 m2\n",
+        "\n",
+        "  Maximum  energy  stored  is   0.031   J"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 740</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the energy stored per cubic metre of the dielectric.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.3;\n",
+      "A  =  0.02;#  in  m2\n",
+      "C  =  400E-12;#  in  Farad\n",
+      "V  =  100;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #energy  stored  per  unit  volume  of  dielectric,\n",
+      "W  =  ((C*V)**2)/(2*e0*er*A**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  energy  stored  per  unit  volume  of  dielectric  is  \",round(W,2),\"  J/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  energy  stored  per  unit  volume  of  dielectric  is   0.1   J/m3"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 744</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the inductance of the cable per metre length.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "a  =  0.001;#  in  m\n",
+      "b  =  0.004;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  L\n",
+      "L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(b/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  L  is  \",round(L*1E6,2),\"uH/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  L  is   0.33 uH/m"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 744</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the diameter of the sheath.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "da  =  0.010;#  in  m\n",
+      "L  =  4E-7;#  in  H/m\n",
+      "\n",
+      " #calculation:\n",
+      " #diameter  of  the  sheath\n",
+      "db  =  da*(math.e**(L/(u0*ur/(2*math.pi))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  diameter  of  the  sheath  is  \",round(db,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  diameter  of  the  sheath  is   0.07   m"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 745</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the cable (a) the inductance, assuming nonmagnetic materials, and \n",
+      "#(b) the capacitance, assuming a dielectric of relative permittivity 3.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  3;\n",
+      "da  =  0.010;#  in  m\n",
+      "db  =  0.025;#  in  m\n",
+      "l  =  7500;#  in  m\n",
+      "\n",
+      "#calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(db/da))\n",
+      " #Since  the  cable  is  7500  m  long,\n",
+      "L7500  =  L*7500\n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(db/da))\n",
+      " #//Since  the  cable  is  7500  m  long,\n",
+      "C7500  =  C*7500\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L7500*1000,2),\" mH\"\n",
+      "print  \"\\ncapCItance  is  \",round(C7500*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   1.75  mH\n",
+        "\n",
+        "capCItance  is   1.37 uF"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 748</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the inductance of the line per metre length ignoring internal linkages\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  3;\n",
+      "D  =  1.2;#  in  m\n",
+      "a  =  0.008;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(math.pi))*(math.log(D/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L*1E6,2),\"uH/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   2.0 uH/m"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 748</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the loop inductance, and \n",
+      "#(b) the capacitance of a 1 km length of single-phase twin line having conductors of diameter 10 mm and spaced 800 mm apart in air.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  1;\n",
+      "l  =  1000;#  in  m\n",
+      "D  =  0.8;#  in  m\n",
+      "a  =  0.01/2;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(math.pi))*(0.25  +  math.log(D/a))\n",
+      " #Since  the  cable  is  1000  m  long,\n",
+      "L1k  =  L*l\n",
+      " #capacitance  C\n",
+      "C  =  math.pi*e0*er/(math.log(D/a))\n",
+      " #//Since  the  cable  is  1000  m  long,\n",
+      "C1k  =  C*l\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L1k*1000,2),\" mH\"\n",
+      "print  \"\\ncapcitance  is  \",round(C1k*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   2.13  mH\n",
+        "\n",
+        "capcitance  is   5.48 nF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 749</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the distance between their centres.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  2.185E-6;#  in  H/m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "a  =  0.012/2;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #distance  D\n",
+      "D  =  a*math.e**((L*math.pi)/(u0*ur)  -  0.25)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ndistance  D  is  \",round(D,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "distance  D  is   1.1   m"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What value of current would double the energy stored?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.2;#  in  H\n",
+      "I  =  0.05;#  in  Amperes\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored  in  inductor\n",
+      "W  =  L*I*I/2\n",
+      " #current  I\n",
+      "I  =  (2*2*W/L)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nenergy  stored  in  inductor  is  \",round(W*1000,2),\"mJ\"\n",
+      "print  \"\\ncurrent  I  is  \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "energy  stored  in  inductor  is   0.25 mJ\n",
+        "\n",
+        "current  I  is   0.07 A"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the total energy stored in the magnetic field of the airgap.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.05;#  in  Tesla\n",
+      "A  =  500E-6;#  in  m2\n",
+      "l  =  0.002;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored\n",
+      "W  =  (B**2)/(2*u0)\n",
+      " #Volume  of  airgap\n",
+      "v  =  A*l\n",
+      " #energy  stored  in  airgap\n",
+      "W  =  W*v\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nenergy  stored  in  the  airgap  is  \",round(W*1E6,2),\"uJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "energy  stored  in  the  airgap  is   994.72 uJ"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the strength of a uniform electric fi\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.8;#  in  Tesla\n",
+      "A  =  500E-6;#  in  m2\n",
+      "l  =  0.002;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  1;\n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored  in  mag.  field\n",
+      "W  =  (B**2)/(2*u0)\n",
+      " #electric  field\n",
+      "E  =  (2*W/(e0*er))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nelectric  field  strength  is  \",round(E/1E6,2),\"MV/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "electric  field  strength  is   239.89 MV/m"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_1.ipynb
new file mode 100755
index 00000000..3f3b710b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_1.ipynb
@@ -0,0 +1,1328 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 40: Field theory</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 725</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitance per metre length of the system\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.8;\n",
+      "l  =  1;#  in  m\n",
+      "\n",
+      "#calculation:  \n",
+      " #From  Figure  40.9\n",
+      "m  =  16;#  number  of  parallel  squares  measured  along  each  equipotential\n",
+      "n  =  6;#  the  number  of  series  squares  measured  along  each  line  of  force\n",
+      "C  =  e0*er*l*m/n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   66.08 pFarad."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 725</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitance of a 100 m length of the cable.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.4;\n",
+      "l  =  100;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #From  Figure  40.10\n",
+      "m  =  13;#  number  of  parallel  squares  measured  along  each  equipotential\n",
+      "n  =  4;#  the  number  of  series  squares  measured  along  each  line  of  force\n",
+      "C  =  e0*er*l*m/n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E9,2),\"nFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   9.78 nFarad."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 726</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitance per metre length of the cable\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.7;\n",
+      "ri  =  0.0005;#  in  m\n",
+      "ro  =  0.006;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(ro/ri))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   60.42 pFarad."
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 727</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the internal diameter of the sheath.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  80E-12;#  in  Farads\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "d0  =  0.008;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #internal  diameter\n",
+      "di  =  d0*(math.e**(2*math.pi*e0*er/C))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  internal  diameter  is  \",round(di,2),\"  m.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  internal  diameter  is   0.09   m."
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 728</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the capacitance per kilometre length of the cable, \n",
+      "#(b) the dielectric stress at a radius of 30 mm, and\n",
+      "#(c) the maximum and minimum values of dielectric stress.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "di  =  0.08;#  in  m\n",
+      "d0  =  0.032;#  in  m\n",
+      "r  =  0.03;#  in  m\n",
+      "V  =  40000;#  in  Volts\n",
+      "\n",
+      "#calculation:  \n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(di/d0))\n",
+      " #dielectric  stress  at  radius  r,\n",
+      "E  =  V/(r*math.log(di/d0))\n",
+      " #maximum  dielectric  stress,\n",
+      "Emax  =  V/((d0/2)*(math.log((di/d0))))\n",
+      " #minimum  dielectric  stress,\n",
+      "Emin  =  V/((di/2)*(math.log((di/d0))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pF/km\"\n",
+      "print  \"\\n  dielectric  stress  at  radius  r  is  \",round(E,2),\"V/m\"\n",
+      "print  \"\\n  maximum  dielectric  stress,  is  \",round(Emax,2),\"V/m  minimum  dielectric  stress  \",round(  Emin,2),\"V/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   212.4 pF/km\n",
+        "\n",
+        "  dielectric  stress  at  radius  r  is   1455142.22 V/m\n",
+        "\n",
+        "  maximum  dielectric  stress,  is   2728391.67 V/m  minimum  dielectric  stress   1091356.67 V/m"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 729</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the core and inner sheath radii for the most economical cable,\n",
+      "#(b) the capacitance per metre length, and (c) the charging current per kilometre run.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "V  =  60000;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "Em  =  10E6;#  in  V/m\n",
+      "\n",
+      "\n",
+      "#calculation:  \n",
+      " #core  radius,  a\n",
+      "a  =  V/Em\n",
+      " #internal  sheath  radius,\n",
+      "b  =  a*math.e**1\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er/(math.log(b/a))\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*C\n",
+      " #charging  current  per  kilometre\n",
+      "Ipkm  =  I*1000\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  core  radius  is  \",round(a*1000,2),\"mm  and  internal  sheath  radius  \",round(b*1000,1),\"mm\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,0),\"pF/m\"\n",
+      "print  \"\\n  the  charging  current  per  kilometre  \",round(Ipkm,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  core  radius  is   6.0 mm  and  internal  sheath  radius   16.3 mm\n",
+        "\n",
+        "  capacitance  is   195.0 pF/m\n",
+        "\n",
+        "  the  charging  current  per  kilometre   3.67   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 730</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for a 1 km length of the cable (a) the capacitance, (b) the charging current and (c) the power loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.5;\n",
+      "di  =  0.08;#  in  m\n",
+      "d0  =  0.025;#  in  m\n",
+      "r  =  1000;#  in  m\n",
+      "V  =  132000;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "de  =  3.5E-3;#  rad.\n",
+      "\n",
+      " #calculation:\n",
+      " #core  radius,  a\n",
+      "a  =  d0/2\n",
+      " #internal  sheath  radius,\n",
+      "b  =  di/2\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er*1E3/(math.log(b/a))\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*C\n",
+      " #power  loss\n",
+      "P  =  (2*math.pi*f*C*math.tan(de))*V**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance  for  a  1  km  length  is  \",round(C*1E6,2),\"uF\"\n",
+      "print  \"\\n  (b)the  charging  current  \",round(I,2),\"A/km\"\n",
+      "print  \"\\n  (c)power  loss  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance  for  a  1  km  length  is   0.12 uF\n",
+        "\n",
+        "  (b)the  charging  current   4.96 A/km\n",
+        "\n",
+        "  (c)power  loss   2289.78   W"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 732</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitance of the cable per metre length by the method of curvilinear squares\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.2;\n",
+      "di  =  0.06;#  in  m\n",
+      "d0  =  0.020;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #core  radius,  a\n",
+      "a  =  d0/2\n",
+      " #internal  sheath  radius,\n",
+      "b  =  di/2\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er/(math.log(b/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  per  m  of  length  is  \",round(C*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  per  m  of  length  is   0.16 nF"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 736</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitance of the line if the total length is 200 m.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "D  =  0.05;#  in  m\n",
+      "d  =  0.005;#  in  m\n",
+      "l  =  200;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  math.pi*e0*er/(math.log(D/(d/2)))\n",
+      " #capacitance  of  a  200  m  length\n",
+      "C200  =  C*l\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  of  a  200  m  length  is  \",round(C200*1E6,5),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  of  a  200  m  length  is   0.00186 uF"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 736</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for a 1 km length of line, (a) the capacitance of the  conductors, \n",
+      "#(b) the value of charge carried by each conductor, and (c) the charging current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "D  =  1.2;#  in  m\n",
+      "r  =  0.004;#  in  m\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  15000;#  in  Volts\n",
+      "l  =  1000;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  math.pi*e0*er/(math.log(D/r))\n",
+      " #capacitance  of  a  1  km  length\n",
+      "Cpkm  =  C*l\n",
+      " #Charge  Q\n",
+      "Q  =  Cpkm*V\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*Cpkm\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  per  1km  length  is  \",round(Cpkm*1E9,2),\"nF\"\n",
+      "print  \"\\n  Charge  Q  is  \",round(Q*1E6,2),\"uC\"\n",
+      "print  \"\\n  Charging  current  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  per  1km  length  is   4.87 nF\n",
+        "\n",
+        "  Charge  Q  is   73.12 uC\n",
+        "\n",
+        "  Charging  current  is   0.02   A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 737</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the maximum value required for the capacitance per metre length,\n",
+      "#and (b) the maximum diameter of each conductor if their distance between centres is 1.25 m.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "I  =  0.015;#  in  Amperes\n",
+      "d  =  1.25;#  in  m\n",
+      "r  =  800;#  in  m\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  10000;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  I/(2*math.pi*f*V)\n",
+      " #required  maximum  value  of  capacitance\n",
+      "Cmax  =  C/r\n",
+      " #maximum  diameter  of  each  conductor\n",
+      "D  =  2*d/(math.e**(math.pi*e0*er/Cmax))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  required  maximum  value  of  capacitance  is  \",round(Cmax*1E12,2),\"pF/m\"\n",
+      "print  \"\\nthe  maximum  diameter  of  each  conductor  is  \",round(D,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  required  maximum  value  of  capacitance  is   5.97 pF/m\n",
+        "\n",
+        "the  maximum  diameter  of  each  conductor  is   0.02   m"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 739</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and \n",
+      "#the average power developed if this energy is dissipated in 10 \u03bcs\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "C  =  10E-9;#  in  Farad\n",
+      "V  =  1000;#  in  Volts\n",
+      "t  =  10E-6;#  in  sec\n",
+      "\n",
+      " #calculation:\n",
+      " #energy  stored,Wf\n",
+      "Wf  =  C*V*V/2\n",
+      " #average  power  developed\n",
+      "Pav  =  Wf/t\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  energy  stored  is  \",Wf,\"J\"\n",
+      "print  \"\\nthe  average  power  developed  is  \",Pav,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  energy  stored  is   0.005 J\n",
+        "\n",
+        "the  average  power  developed  is   500.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 739</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the voltage across the plates and (b) the capacitance of the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "Q  =  5E-3;#  in  Coulomb\n",
+      "W  =  0.625;#  in  Joules\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage  across  the  plates\n",
+      "V  =  2*W/Q\n",
+      " #Capacitance  C\n",
+      "C  =  Q/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  across  the  plates  is  \",V,\"  V\"\n",
+      "print  \"\\n  Capacitance  C  is  \",C*1E6,\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  across  the  plates  is   250.0   V\n",
+        "\n",
+        "  Capacitance  C  is   20.0 uF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 740</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the required thickness of the ceramic dielectric, \n",
+      "#(b) the area of plate required if the relative permittivity of the ceramic is 10, and \n",
+      "#(c) the maximum energy stored by the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  10;\n",
+      "C  =  0.01E-6;#  in  Farad\n",
+      "E  =  10E6;#  in  V/m\n",
+      "V  =  2500;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #thickness  of  ceramic  dielectric,\n",
+      "d  =  V/E\n",
+      " #cross-sectional  area  of  plate\n",
+      "A  =  C*d/(e0*er)\n",
+      " #Maximum  energy  stored,\n",
+      "W  =  C*V*V/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  thickness  of  ceramic  dielectric  is  \",d*1000,\"mm\"\n",
+      "print  \"\\n  cross-sectional  area  of  plate,  is  \",round(A,2),\"m2\"\n",
+      "print  \"\\n  Maximum  energy  stored  is  \",round(W,3),\"  J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  thickness  of  ceramic  dielectric  is   0.25 mm\n",
+        "\n",
+        "  cross-sectional  area  of  plate,  is   0.03 m2\n",
+        "\n",
+        "  Maximum  energy  stored  is   0.031   J"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 740</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the energy stored per cubic metre of the dielectric.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.3;\n",
+      "A  =  0.02;#  in  m2\n",
+      "C  =  400E-12;#  in  Farad\n",
+      "V  =  100;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #energy  stored  per  unit  volume  of  dielectric,\n",
+      "W  =  ((C*V)**2)/(2*e0*er*A**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  energy  stored  per  unit  volume  of  dielectric  is  \",round(W,2),\"  J/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  energy  stored  per  unit  volume  of  dielectric  is   0.1   J/m3"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 744</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the inductance of the cable per metre length.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "a  =  0.001;#  in  m\n",
+      "b  =  0.004;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  L\n",
+      "L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(b/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  L  is  \",round(L*1E6,2),\"uH/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  L  is   0.33 uH/m"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 744</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the diameter of the sheath.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "da  =  0.010;#  in  m\n",
+      "L  =  4E-7;#  in  H/m\n",
+      "\n",
+      " #calculation:\n",
+      " #diameter  of  the  sheath\n",
+      "db  =  da*(math.e**(L/(u0*ur/(2*math.pi))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  diameter  of  the  sheath  is  \",round(db,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  diameter  of  the  sheath  is   0.07   m"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 745</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the cable (a) the inductance, assuming nonmagnetic materials, and \n",
+      "#(b) the capacitance, assuming a dielectric of relative permittivity 3.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  3;\n",
+      "da  =  0.010;#  in  m\n",
+      "db  =  0.025;#  in  m\n",
+      "l  =  7500;#  in  m\n",
+      "\n",
+      "#calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(db/da))\n",
+      " #Since  the  cable  is  7500  m  long,\n",
+      "L7500  =  L*7500\n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(db/da))\n",
+      " #//Since  the  cable  is  7500  m  long,\n",
+      "C7500  =  C*7500\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L7500*1000,2),\" mH\"\n",
+      "print  \"\\ncapCItance  is  \",round(C7500*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   1.75  mH\n",
+        "\n",
+        "capCItance  is   1.37 uF"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 748</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the inductance of the line per metre length ignoring internal linkages\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  3;\n",
+      "D  =  1.2;#  in  m\n",
+      "a  =  0.008;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(math.pi))*(math.log(D/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L*1E6,2),\"uH/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   2.0 uH/m"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 748</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the loop inductance, and \n",
+      "#(b) the capacitance of a 1 km length of single-phase twin line having conductors of diameter 10 mm and spaced 800 mm apart in air.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  1;\n",
+      "l  =  1000;#  in  m\n",
+      "D  =  0.8;#  in  m\n",
+      "a  =  0.01/2;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(math.pi))*(0.25  +  math.log(D/a))\n",
+      " #Since  the  cable  is  1000  m  long,\n",
+      "L1k  =  L*l\n",
+      " #capacitance  C\n",
+      "C  =  math.pi*e0*er/(math.log(D/a))\n",
+      " #//Since  the  cable  is  1000  m  long,\n",
+      "C1k  =  C*l\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L1k*1000,2),\" mH\"\n",
+      "print  \"\\ncapcitance  is  \",round(C1k*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   2.13  mH\n",
+        "\n",
+        "capcitance  is   5.48 nF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 749</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the distance between their centres.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  2.185E-6;#  in  H/m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "a  =  0.012/2;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #distance  D\n",
+      "D  =  a*math.e**((L*math.pi)/(u0*ur)  -  0.25)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ndistance  D  is  \",round(D,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "distance  D  is   1.1   m"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What value of current would double the energy stored?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.2;#  in  H\n",
+      "I  =  0.05;#  in  Amperes\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored  in  inductor\n",
+      "W  =  L*I*I/2\n",
+      " #current  I\n",
+      "I  =  (2*2*W/L)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nenergy  stored  in  inductor  is  \",round(W*1000,2),\"mJ\"\n",
+      "print  \"\\ncurrent  I  is  \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "energy  stored  in  inductor  is   0.25 mJ\n",
+        "\n",
+        "current  I  is   0.07 A"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the total energy stored in the magnetic field of the airgap.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.05;#  in  Tesla\n",
+      "A  =  500E-6;#  in  m2\n",
+      "l  =  0.002;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored\n",
+      "W  =  (B**2)/(2*u0)\n",
+      " #Volume  of  airgap\n",
+      "v  =  A*l\n",
+      " #energy  stored  in  airgap\n",
+      "W  =  W*v\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nenergy  stored  in  the  airgap  is  \",round(W*1E6,2),\"uJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "energy  stored  in  the  airgap  is   994.72 uJ"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the strength of a uniform electric fi\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.8;#  in  Tesla\n",
+      "A  =  500E-6;#  in  m2\n",
+      "l  =  0.002;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  1;\n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored  in  mag.  field\n",
+      "W  =  (B**2)/(2*u0)\n",
+      " #electric  field\n",
+      "E  =  (2*W/(e0*er))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nelectric  field  strength  is  \",round(E/1E6,2),\"MV/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "electric  field  strength  is   239.89 MV/m"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_2.ipynb
new file mode 100755
index 00000000..3f3b710b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_2.ipynb
@@ -0,0 +1,1328 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 40: Field theory</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 725</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitance per metre length of the system\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.8;\n",
+      "l  =  1;#  in  m\n",
+      "\n",
+      "#calculation:  \n",
+      " #From  Figure  40.9\n",
+      "m  =  16;#  number  of  parallel  squares  measured  along  each  equipotential\n",
+      "n  =  6;#  the  number  of  series  squares  measured  along  each  line  of  force\n",
+      "C  =  e0*er*l*m/n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   66.08 pFarad."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 725</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitance of a 100 m length of the cable.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.4;\n",
+      "l  =  100;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #From  Figure  40.10\n",
+      "m  =  13;#  number  of  parallel  squares  measured  along  each  equipotential\n",
+      "n  =  4;#  the  number  of  series  squares  measured  along  each  line  of  force\n",
+      "C  =  e0*er*l*m/n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E9,2),\"nFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   9.78 nFarad."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 726</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitance per metre length of the cable\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.7;\n",
+      "ri  =  0.0005;#  in  m\n",
+      "ro  =  0.006;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(ro/ri))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   60.42 pFarad."
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 727</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the internal diameter of the sheath.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  80E-12;#  in  Farads\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "d0  =  0.008;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #internal  diameter\n",
+      "di  =  d0*(math.e**(2*math.pi*e0*er/C))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  internal  diameter  is  \",round(di,2),\"  m.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  internal  diameter  is   0.09   m."
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 728</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the capacitance per kilometre length of the cable, \n",
+      "#(b) the dielectric stress at a radius of 30 mm, and\n",
+      "#(c) the maximum and minimum values of dielectric stress.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "di  =  0.08;#  in  m\n",
+      "d0  =  0.032;#  in  m\n",
+      "r  =  0.03;#  in  m\n",
+      "V  =  40000;#  in  Volts\n",
+      "\n",
+      "#calculation:  \n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(di/d0))\n",
+      " #dielectric  stress  at  radius  r,\n",
+      "E  =  V/(r*math.log(di/d0))\n",
+      " #maximum  dielectric  stress,\n",
+      "Emax  =  V/((d0/2)*(math.log((di/d0))))\n",
+      " #minimum  dielectric  stress,\n",
+      "Emin  =  V/((di/2)*(math.log((di/d0))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pF/km\"\n",
+      "print  \"\\n  dielectric  stress  at  radius  r  is  \",round(E,2),\"V/m\"\n",
+      "print  \"\\n  maximum  dielectric  stress,  is  \",round(Emax,2),\"V/m  minimum  dielectric  stress  \",round(  Emin,2),\"V/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   212.4 pF/km\n",
+        "\n",
+        "  dielectric  stress  at  radius  r  is   1455142.22 V/m\n",
+        "\n",
+        "  maximum  dielectric  stress,  is   2728391.67 V/m  minimum  dielectric  stress   1091356.67 V/m"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 729</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the core and inner sheath radii for the most economical cable,\n",
+      "#(b) the capacitance per metre length, and (c) the charging current per kilometre run.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "V  =  60000;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "Em  =  10E6;#  in  V/m\n",
+      "\n",
+      "\n",
+      "#calculation:  \n",
+      " #core  radius,  a\n",
+      "a  =  V/Em\n",
+      " #internal  sheath  radius,\n",
+      "b  =  a*math.e**1\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er/(math.log(b/a))\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*C\n",
+      " #charging  current  per  kilometre\n",
+      "Ipkm  =  I*1000\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  core  radius  is  \",round(a*1000,2),\"mm  and  internal  sheath  radius  \",round(b*1000,1),\"mm\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,0),\"pF/m\"\n",
+      "print  \"\\n  the  charging  current  per  kilometre  \",round(Ipkm,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  core  radius  is   6.0 mm  and  internal  sheath  radius   16.3 mm\n",
+        "\n",
+        "  capacitance  is   195.0 pF/m\n",
+        "\n",
+        "  the  charging  current  per  kilometre   3.67   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 730</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for a 1 km length of the cable (a) the capacitance, (b) the charging current and (c) the power loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.5;\n",
+      "di  =  0.08;#  in  m\n",
+      "d0  =  0.025;#  in  m\n",
+      "r  =  1000;#  in  m\n",
+      "V  =  132000;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "de  =  3.5E-3;#  rad.\n",
+      "\n",
+      " #calculation:\n",
+      " #core  radius,  a\n",
+      "a  =  d0/2\n",
+      " #internal  sheath  radius,\n",
+      "b  =  di/2\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er*1E3/(math.log(b/a))\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*C\n",
+      " #power  loss\n",
+      "P  =  (2*math.pi*f*C*math.tan(de))*V**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance  for  a  1  km  length  is  \",round(C*1E6,2),\"uF\"\n",
+      "print  \"\\n  (b)the  charging  current  \",round(I,2),\"A/km\"\n",
+      "print  \"\\n  (c)power  loss  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance  for  a  1  km  length  is   0.12 uF\n",
+        "\n",
+        "  (b)the  charging  current   4.96 A/km\n",
+        "\n",
+        "  (c)power  loss   2289.78   W"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 732</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the capacitance of the cable per metre length by the method of curvilinear squares\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.2;\n",
+      "di  =  0.06;#  in  m\n",
+      "d0  =  0.020;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #core  radius,  a\n",
+      "a  =  d0/2\n",
+      " #internal  sheath  radius,\n",
+      "b  =  di/2\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er/(math.log(b/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  per  m  of  length  is  \",round(C*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  per  m  of  length  is   0.16 nF"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 736</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the capacitance of the line if the total length is 200 m.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "D  =  0.05;#  in  m\n",
+      "d  =  0.005;#  in  m\n",
+      "l  =  200;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  math.pi*e0*er/(math.log(D/(d/2)))\n",
+      " #capacitance  of  a  200  m  length\n",
+      "C200  =  C*l\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  of  a  200  m  length  is  \",round(C200*1E6,5),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  of  a  200  m  length  is   0.00186 uF"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 736</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for a 1 km length of line, (a) the capacitance of the  conductors, \n",
+      "#(b) the value of charge carried by each conductor, and (c) the charging current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "D  =  1.2;#  in  m\n",
+      "r  =  0.004;#  in  m\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  15000;#  in  Volts\n",
+      "l  =  1000;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  math.pi*e0*er/(math.log(D/r))\n",
+      " #capacitance  of  a  1  km  length\n",
+      "Cpkm  =  C*l\n",
+      " #Charge  Q\n",
+      "Q  =  Cpkm*V\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*Cpkm\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  per  1km  length  is  \",round(Cpkm*1E9,2),\"nF\"\n",
+      "print  \"\\n  Charge  Q  is  \",round(Q*1E6,2),\"uC\"\n",
+      "print  \"\\n  Charging  current  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  per  1km  length  is   4.87 nF\n",
+        "\n",
+        "  Charge  Q  is   73.12 uC\n",
+        "\n",
+        "  Charging  current  is   0.02   A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 737</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the maximum value required for the capacitance per metre length,\n",
+      "#and (b) the maximum diameter of each conductor if their distance between centres is 1.25 m.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "I  =  0.015;#  in  Amperes\n",
+      "d  =  1.25;#  in  m\n",
+      "r  =  800;#  in  m\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  10000;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  I/(2*math.pi*f*V)\n",
+      " #required  maximum  value  of  capacitance\n",
+      "Cmax  =  C/r\n",
+      " #maximum  diameter  of  each  conductor\n",
+      "D  =  2*d/(math.e**(math.pi*e0*er/Cmax))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  required  maximum  value  of  capacitance  is  \",round(Cmax*1E12,2),\"pF/m\"\n",
+      "print  \"\\nthe  maximum  diameter  of  each  conductor  is  \",round(D,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  required  maximum  value  of  capacitance  is   5.97 pF/m\n",
+        "\n",
+        "the  maximum  diameter  of  each  conductor  is   0.02   m"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 739</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and \n",
+      "#the average power developed if this energy is dissipated in 10 \u03bcs\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "C  =  10E-9;#  in  Farad\n",
+      "V  =  1000;#  in  Volts\n",
+      "t  =  10E-6;#  in  sec\n",
+      "\n",
+      " #calculation:\n",
+      " #energy  stored,Wf\n",
+      "Wf  =  C*V*V/2\n",
+      " #average  power  developed\n",
+      "Pav  =  Wf/t\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  energy  stored  is  \",Wf,\"J\"\n",
+      "print  \"\\nthe  average  power  developed  is  \",Pav,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  energy  stored  is   0.005 J\n",
+        "\n",
+        "the  average  power  developed  is   500.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 739</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the voltage across the plates and (b) the capacitance of the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "Q  =  5E-3;#  in  Coulomb\n",
+      "W  =  0.625;#  in  Joules\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage  across  the  plates\n",
+      "V  =  2*W/Q\n",
+      " #Capacitance  C\n",
+      "C  =  Q/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  across  the  plates  is  \",V,\"  V\"\n",
+      "print  \"\\n  Capacitance  C  is  \",C*1E6,\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  across  the  plates  is   250.0   V\n",
+        "\n",
+        "  Capacitance  C  is   20.0 uF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 740</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the required thickness of the ceramic dielectric, \n",
+      "#(b) the area of plate required if the relative permittivity of the ceramic is 10, and \n",
+      "#(c) the maximum energy stored by the capacitor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  10;\n",
+      "C  =  0.01E-6;#  in  Farad\n",
+      "E  =  10E6;#  in  V/m\n",
+      "V  =  2500;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #thickness  of  ceramic  dielectric,\n",
+      "d  =  V/E\n",
+      " #cross-sectional  area  of  plate\n",
+      "A  =  C*d/(e0*er)\n",
+      " #Maximum  energy  stored,\n",
+      "W  =  C*V*V/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  thickness  of  ceramic  dielectric  is  \",d*1000,\"mm\"\n",
+      "print  \"\\n  cross-sectional  area  of  plate,  is  \",round(A,2),\"m2\"\n",
+      "print  \"\\n  Maximum  energy  stored  is  \",round(W,3),\"  J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  thickness  of  ceramic  dielectric  is   0.25 mm\n",
+        "\n",
+        "  cross-sectional  area  of  plate,  is   0.03 m2\n",
+        "\n",
+        "  Maximum  energy  stored  is   0.031   J"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 740</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the energy stored per cubic metre of the dielectric.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.3;\n",
+      "A  =  0.02;#  in  m2\n",
+      "C  =  400E-12;#  in  Farad\n",
+      "V  =  100;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #energy  stored  per  unit  volume  of  dielectric,\n",
+      "W  =  ((C*V)**2)/(2*e0*er*A**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  energy  stored  per  unit  volume  of  dielectric  is  \",round(W,2),\"  J/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  energy  stored  per  unit  volume  of  dielectric  is   0.1   J/m3"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 744</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the inductance of the cable per metre length.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "a  =  0.001;#  in  m\n",
+      "b  =  0.004;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  L\n",
+      "L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(b/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  L  is  \",round(L*1E6,2),\"uH/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  L  is   0.33 uH/m"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 744</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the diameter of the sheath.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "da  =  0.010;#  in  m\n",
+      "L  =  4E-7;#  in  H/m\n",
+      "\n",
+      " #calculation:\n",
+      " #diameter  of  the  sheath\n",
+      "db  =  da*(math.e**(L/(u0*ur/(2*math.pi))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  diameter  of  the  sheath  is  \",round(db,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  diameter  of  the  sheath  is   0.07   m"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 745</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the cable (a) the inductance, assuming nonmagnetic materials, and \n",
+      "#(b) the capacitance, assuming a dielectric of relative permittivity 3.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  3;\n",
+      "da  =  0.010;#  in  m\n",
+      "db  =  0.025;#  in  m\n",
+      "l  =  7500;#  in  m\n",
+      "\n",
+      "#calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(db/da))\n",
+      " #Since  the  cable  is  7500  m  long,\n",
+      "L7500  =  L*7500\n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(db/da))\n",
+      " #//Since  the  cable  is  7500  m  long,\n",
+      "C7500  =  C*7500\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L7500*1000,2),\" mH\"\n",
+      "print  \"\\ncapCItance  is  \",round(C7500*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   1.75  mH\n",
+        "\n",
+        "capCItance  is   1.37 uF"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 748</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the inductance of the line per metre length ignoring internal linkages\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  3;\n",
+      "D  =  1.2;#  in  m\n",
+      "a  =  0.008;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(math.pi))*(math.log(D/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L*1E6,2),\"uH/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   2.0 uH/m"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 748</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the loop inductance, and \n",
+      "#(b) the capacitance of a 1 km length of single-phase twin line having conductors of diameter 10 mm and spaced 800 mm apart in air.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  1;\n",
+      "l  =  1000;#  in  m\n",
+      "D  =  0.8;#  in  m\n",
+      "a  =  0.01/2;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(math.pi))*(0.25  +  math.log(D/a))\n",
+      " #Since  the  cable  is  1000  m  long,\n",
+      "L1k  =  L*l\n",
+      " #capacitance  C\n",
+      "C  =  math.pi*e0*er/(math.log(D/a))\n",
+      " #//Since  the  cable  is  1000  m  long,\n",
+      "C1k  =  C*l\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L1k*1000,2),\" mH\"\n",
+      "print  \"\\ncapcitance  is  \",round(C1k*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   2.13  mH\n",
+        "\n",
+        "capcitance  is   5.48 nF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 749</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the distance between their centres.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  2.185E-6;#  in  H/m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "a  =  0.012/2;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #distance  D\n",
+      "D  =  a*math.e**((L*math.pi)/(u0*ur)  -  0.25)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ndistance  D  is  \",round(D,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "distance  D  is   1.1   m"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#What value of current would double the energy stored?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.2;#  in  H\n",
+      "I  =  0.05;#  in  Amperes\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored  in  inductor\n",
+      "W  =  L*I*I/2\n",
+      " #current  I\n",
+      "I  =  (2*2*W/L)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nenergy  stored  in  inductor  is  \",round(W*1000,2),\"mJ\"\n",
+      "print  \"\\ncurrent  I  is  \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "energy  stored  in  inductor  is   0.25 mJ\n",
+        "\n",
+        "current  I  is   0.07 A"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the total energy stored in the magnetic field of the airgap.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.05;#  in  Tesla\n",
+      "A  =  500E-6;#  in  m2\n",
+      "l  =  0.002;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored\n",
+      "W  =  (B**2)/(2*u0)\n",
+      " #Volume  of  airgap\n",
+      "v  =  A*l\n",
+      " #energy  stored  in  airgap\n",
+      "W  =  W*v\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nenergy  stored  in  the  airgap  is  \",round(W*1E6,2),\"uJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "energy  stored  in  the  airgap  is   994.72 uJ"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the strength of a uniform electric fi\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.8;#  in  Tesla\n",
+      "A  =  500E-6;#  in  m2\n",
+      "l  =  0.002;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  1;\n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored  in  mag.  field\n",
+      "W  =  (B**2)/(2*u0)\n",
+      " #electric  field\n",
+      "E  =  (2*W/(e0*er))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nelectric  field  strength  is  \",round(E/1E6,2),\"MV/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "electric  field  strength  is   239.89 MV/m"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb
new file mode 100755
index 00000000..181b437b
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb
@@ -0,0 +1,1319 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b1ccfc3b90d3e80d897264262435b8c5eeae5b8f515ec94674405489033f24fc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 40: Field theory</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 725</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.8;\n",
+      "l  =  1;#  in  m\n",
+      "\n",
+      "#calculation:  \n",
+      " #From  Figure  40.9\n",
+      "m  =  16;#  number  of  parallel  squares  measured  along  each  equipotential\n",
+      "n  =  6;#  the  number  of  series  squares  measured  along  each  line  of  force\n",
+      "C  =  e0*er*l*m/n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   66.08 pFarad."
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 725</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.4;\n",
+      "l  =  100;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #From  Figure  40.10\n",
+      "m  =  13;#  number  of  parallel  squares  measured  along  each  equipotential\n",
+      "n  =  4;#  the  number  of  series  squares  measured  along  each  line  of  force\n",
+      "C  =  e0*er*l*m/n\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E9,2),\"nFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   9.78 nFarad."
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 726</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.7;\n",
+      "ri  =  0.0005;#  in  m\n",
+      "ro  =  0.006;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(ro/ri))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pFarad.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   60.42 pFarad."
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 727</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "C  =  80E-12;#  in  Farads\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "d0  =  0.008;#  in  m\n",
+      "\n",
+      " #calculation:  \n",
+      " #internal  diameter\n",
+      "di  =  d0*(math.e**(2*math.pi*e0*er/C))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  internal  diameter  is  \",round(di,2),\"  m.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  internal  diameter  is   0.09   m."
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 728</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "di  =  0.08;#  in  m\n",
+      "d0  =  0.032;#  in  m\n",
+      "r  =  0.03;#  in  m\n",
+      "V  =  40000;#  in  Volts\n",
+      "\n",
+      "#calculation:  \n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(di/d0))\n",
+      " #dielectric  stress  at  radius  r,\n",
+      "E  =  V/(r*math.log(di/d0))\n",
+      " #maximum  dielectric  stress,\n",
+      "Emax  =  V/((d0/2)*(math.log((di/d0))))\n",
+      " #minimum  dielectric  stress,\n",
+      "Emin  =  V/((di/2)*(math.log((di/d0))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,2),\"pF/km\"\n",
+      "print  \"\\n  dielectric  stress  at  radius  r  is  \",round(E,2),\"V/m\"\n",
+      "print  \"\\n  maximum  dielectric  stress,  is  \",round(Emax,2),\"V/m  minimum  dielectric  stress  \",round(  Emin,2),\"V/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  is   212.4 pF/km\n",
+        "\n",
+        "  dielectric  stress  at  radius  r  is   1455142.22 V/m\n",
+        "\n",
+        "  maximum  dielectric  stress,  is   2728391.67 V/m  minimum  dielectric  stress   1091356.67 V/m"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 729</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.5;\n",
+      "V  =  60000;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "Em  =  10E6;#  in  V/m\n",
+      "\n",
+      "\n",
+      "#calculation:  \n",
+      " #core  radius,  a\n",
+      "a  =  V/Em\n",
+      " #internal  sheath  radius,\n",
+      "b  =  a*math.e**1\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er/(math.log(b/a))\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*C\n",
+      " #charging  current  per  kilometre\n",
+      "Ipkm  =  I*1000\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  core  radius  is  \",round(a*1000,2),\"mm  and  internal  sheath  radius  \",round(b*1000,1),\"mm\"\n",
+      "print  \"\\n  capacitance  is  \",round(C*1E12,0),\"pF/m\"\n",
+      "print  \"\\n  the  charging  current  per  kilometre  \",round(Ipkm,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  core  radius  is   6.0 mm  and  internal  sheath  radius   16.3 mm\n",
+        "\n",
+        "  capacitance  is   195.0 pF/m\n",
+        "\n",
+        "  the  charging  current  per  kilometre   3.67   A"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 730</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.5;\n",
+      "di  =  0.08;#  in  m\n",
+      "d0  =  0.025;#  in  m\n",
+      "r  =  1000;#  in  m\n",
+      "V  =  132000;#  in  Volts\n",
+      "f  =  50;#  in  Hz\n",
+      "de  =  3.5E-3;#  rad.\n",
+      "\n",
+      " #calculation:\n",
+      " #core  radius,  a\n",
+      "a  =  d0/2\n",
+      " #internal  sheath  radius,\n",
+      "b  =  di/2\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er*1E3/(math.log(b/a))\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*C\n",
+      " #power  loss\n",
+      "P  =  (2*math.pi*f*C*math.tan(de))*V**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)capacitance  for  a  1  km  length  is  \",round(C*1E6,2),\"uF\"\n",
+      "print  \"\\n  (b)the  charging  current  \",round(I,2),\"A/km\"\n",
+      "print  \"\\n  (c)power  loss  \",round(P,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)capacitance  for  a  1  km  length  is   0.12 uF\n",
+        "\n",
+        "  (b)the  charging  current   4.96 A/km\n",
+        "\n",
+        "  (c)power  loss   2289.78   W"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 732</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  3.2;\n",
+      "di  =  0.06;#  in  m\n",
+      "d0  =  0.020;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #core  radius,  a\n",
+      "a  =  d0/2\n",
+      " #internal  sheath  radius,\n",
+      "b  =  di/2\n",
+      " #capacitance\n",
+      "C  =  2*math.pi*e0*er/(math.log(b/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  per  m  of  length  is  \",round(C*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  per  m  of  length  is   0.16 nF"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 736</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "D  =  0.05;#  in  m\n",
+      "d  =  0.005;#  in  m\n",
+      "l  =  200;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  math.pi*e0*er/(math.log(D/(d/2)))\n",
+      " #capacitance  of  a  200  m  length\n",
+      "C200  =  C*l\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  of  a  200  m  length  is  \",round(C200*1E6,5),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  of  a  200  m  length  is   0.00186 uF"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 736</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "D  =  1.2;#  in  m\n",
+      "r  =  0.004;#  in  m\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  15000;#  in  Volts\n",
+      "l  =  1000;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  math.pi*e0*er/(math.log(D/r))\n",
+      " #capacitance  of  a  1  km  length\n",
+      "Cpkm  =  C*l\n",
+      " #Charge  Q\n",
+      "Q  =  Cpkm*V\n",
+      " #Charging  current\n",
+      "I  =  V*2*math.pi*f*Cpkm\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  capacitance  per  1km  length  is  \",round(Cpkm*1E9,2),\"nF\"\n",
+      "print  \"\\n  Charge  Q  is  \",round(Q*1E6,2),\"uC\"\n",
+      "print  \"\\n  Charging  current  is  \",round(I,2),\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  capacitance  per  1km  length  is   4.87 nF\n",
+        "\n",
+        "  Charge  Q  is   73.12 uC\n",
+        "\n",
+        "  Charging  current  is   0.02   A"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 737</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "I  =  0.015;#  in  Amperes\n",
+      "d  =  1.25;#  in  m\n",
+      "r  =  800;#  in  m\n",
+      "f  =  50;#  in  Hz\n",
+      "V  =  10000;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  I/(2*math.pi*f*V)\n",
+      " #required  maximum  value  of  capacitance\n",
+      "Cmax  =  C/r\n",
+      " #maximum  diameter  of  each  conductor\n",
+      "D  =  2*d/(math.e**(math.pi*e0*er/Cmax))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  required  maximum  value  of  capacitance  is  \",round(Cmax*1E12,2),\"pF/m\"\n",
+      "print  \"\\nthe  maximum  diameter  of  each  conductor  is  \",round(D,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  required  maximum  value  of  capacitance  is   5.97 pF/m\n",
+        "\n",
+        "the  maximum  diameter  of  each  conductor  is   0.02   m"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 739</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "C  =  10E-9;#  in  Farad\n",
+      "V  =  1000;#  in  Volts\n",
+      "t  =  10E-6;#  in  sec\n",
+      "\n",
+      " #calculation:\n",
+      " #energy  stored,Wf\n",
+      "Wf  =  C*V*V/2\n",
+      " #average  power  developed\n",
+      "Pav  =  Wf/t\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  energy  stored  is  \",Wf,\"J\"\n",
+      "print  \"\\nthe  average  power  developed  is  \",Pav,\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  energy  stored  is   0.005 J\n",
+        "\n",
+        "the  average  power  developed  is   500.0   W"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 739</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  1;\n",
+      "Q  =  5E-3;#  in  Coulomb\n",
+      "W  =  0.625;#  in  Joules\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage  across  the  plates\n",
+      "V  =  2*W/Q\n",
+      " #Capacitance  C\n",
+      "C  =  Q/V\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  voltage  across  the  plates  is  \",V,\"  V\"\n",
+      "print  \"\\n  Capacitance  C  is  \",C*1E6,\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  voltage  across  the  plates  is   250.0   V\n",
+        "\n",
+        "  Capacitance  C  is   20.0 uF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 740</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  10;\n",
+      "C  =  0.01E-6;#  in  Farad\n",
+      "E  =  10E6;#  in  V/m\n",
+      "V  =  2500;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #thickness  of  ceramic  dielectric,\n",
+      "d  =  V/E\n",
+      " #cross-sectional  area  of  plate\n",
+      "A  =  C*d/(e0*er)\n",
+      " #Maximum  energy  stored,\n",
+      "W  =  C*V*V/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  thickness  of  ceramic  dielectric  is  \",d*1000,\"mm\"\n",
+      "print  \"\\n  cross-sectional  area  of  plate,  is  \",round(A,2),\"m2\"\n",
+      "print  \"\\n  Maximum  energy  stored  is  \",round(W,3),\"  J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  thickness  of  ceramic  dielectric  is   0.25 mm\n",
+        "\n",
+        "  cross-sectional  area  of  plate,  is   0.03 m2\n",
+        "\n",
+        "  Maximum  energy  stored  is   0.031   J"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 740</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "e0  =  8.85E-12;  \n",
+      "er  =  2.3;\n",
+      "A  =  0.02;#  in  m2\n",
+      "C  =  400E-12;#  in  Farad\n",
+      "V  =  100;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #energy  stored  per  unit  volume  of  dielectric,\n",
+      "W  =  ((C*V)**2)/(2*e0*er*A**2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  energy  stored  per  unit  volume  of  dielectric  is  \",round(W,2),\"  J/m3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  energy  stored  per  unit  volume  of  dielectric  is   0.1   J/m3"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 744</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "a  =  0.001;#  in  m\n",
+      "b  =  0.004;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  L\n",
+      "L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(b/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  L  is  \",round(L*1E6,2),\"uH/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  L  is   0.33 uH/m"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 744</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "da  =  0.010;#  in  m\n",
+      "L  =  4E-7;#  in  H/m\n",
+      "\n",
+      " #calculation:\n",
+      " #diameter  of  the  sheath\n",
+      "db  =  da*(math.e**(L/(u0*ur/(2*math.pi))))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  diameter  of  the  sheath  is  \",round(db,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  diameter  of  the  sheath  is   0.07   m"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 745</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  3;\n",
+      "da  =  0.010;#  in  m\n",
+      "db  =  0.025;#  in  m\n",
+      "l  =  7500;#  in  m\n",
+      "\n",
+      "#calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(2*math.pi))*(0.25  +  math.log(db/da))\n",
+      " #Since  the  cable  is  7500  m  long,\n",
+      "L7500  =  L*7500\n",
+      " #capacitance  C\n",
+      "C  =  2*math.pi*e0*er/(math.log(db/da))\n",
+      " #//Since  the  cable  is  7500  m  long,\n",
+      "C7500  =  C*7500\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L7500*1000,2),\" mH\"\n",
+      "print  \"\\ncapCItance  is  \",round(C7500*1E6,2),\"uF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   1.75  mH\n",
+        "\n",
+        "capCItance  is   1.37 uF"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 19, page no. 748</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  3;\n",
+      "D  =  1.2;#  in  m\n",
+      "a  =  0.008;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(math.pi))*(math.log(D/a))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L*1E6,2),\"uH/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   2.0 uH/m"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 20, page no. 748</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  1;\n",
+      "l  =  1000;#  in  m\n",
+      "D  =  0.8;#  in  m\n",
+      "a  =  0.01/2;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #inductance  per  metre  length\n",
+      "L  =  (u0*ur/(math.pi))*(0.25  +  math.log(D/a))\n",
+      " #Since  the  cable  is  1000  m  long,\n",
+      "L1k  =  L*l\n",
+      " #capacitance  C\n",
+      "C  =  math.pi*e0*er/(math.log(D/a))\n",
+      " #//Since  the  cable  is  1000  m  long,\n",
+      "C1k  =  C*l\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ninductance  is  \",round(L1k*1000,2),\" mH\"\n",
+      "print  \"\\ncapcitance  is  \",round(C1k*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "inductance  is   2.13  mH\n",
+        "\n",
+        "capcitance  is   5.48 nF\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 21, page no. 749</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  2.185E-6;#  in  H/m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "a  =  0.012/2;#  in  m\n",
+      "\n",
+      " #calculation:\n",
+      " #distance  D\n",
+      "D  =  a*math.e**((L*math.pi)/(u0*ur)  -  0.25)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ndistance  D  is  \",round(D,2),\"  m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "distance  D  is   1.1   m"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 22, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.2;#  in  H\n",
+      "I  =  0.05;#  in  Amperes\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored  in  inductor\n",
+      "W  =  L*I*I/2\n",
+      " #current  I\n",
+      "I  =  (2*2*W/L)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nenergy  stored  in  inductor  is  \",round(W*1000,2),\"mJ\"\n",
+      "print  \"\\ncurrent  I  is  \",round(I,2),\"A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "energy  stored  in  inductor  is   0.25 mJ\n",
+        "\n",
+        "current  I  is   0.07 A"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 23, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.05;#  in  Tesla\n",
+      "A  =  500E-6;#  in  m2\n",
+      "l  =  0.002;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored\n",
+      "W  =  (B**2)/(2*u0)\n",
+      " #Volume  of  airgap\n",
+      "v  =  A*l\n",
+      " #energy  stored  in  airgap\n",
+      "W  =  W*v\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nenergy  stored  in  the  airgap  is  \",round(W*1E6,2),\"uJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "energy  stored  in  the  airgap  is   994.72 uJ"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 24, page no. 752</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "B  =  0.8;#  in  Tesla\n",
+      "A  =  500E-6;#  in  m2\n",
+      "l  =  0.002;#  in  m\n",
+      "u0  =  4*math.pi*1E-7;  \n",
+      "ur  =  1;\n",
+      "e0  =  8.85E-12;\n",
+      "er  =  1;\n",
+      "\n",
+      "#calculation:\n",
+      " #energy  stored  in  mag.  field\n",
+      "W  =  (B**2)/(2*u0)\n",
+      " #electric  field\n",
+      "E  =  (2*W/(e0*er))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nelectric  field  strength  is  \",round(E/1E6,2),\"MV/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "electric  field  strength  is   239.89 MV/m"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb
new file mode 100755
index 00000000..788f6f5e
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint.ipynb
@@ -0,0 +1,1005 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 41: Attenuators</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 763</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power ratio in each case (i) in decibels and (ii) in nepers.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      " #ratio  of  output  power  to  input  power\n",
+      "rp1  =  2;\n",
+      "rp2  =  25;  \n",
+      "rp3  =  1000;\n",
+      "rp4  =  0.01;\n",
+      "\n",
+      "#calculation:\n",
+      " #power  ratio  in  decibels\n",
+      "rpd1  =  10*(1/2.303)*math.log(rp1)\n",
+      "rpd2  =  10*(1/2.303)*math.log(rp2)\n",
+      "rpd3  =  10*(1/2.303)*math.log(rp3)\n",
+      "rpd4  =  10*(1/2.303)*math.log(rp4)\n",
+      " #power  ratio  in  nepers\n",
+      "rpn1  =  (math.log(rp1))/2\n",
+      "rpn2  =  (math.log(rp2))/2\n",
+      "rpn3  =  (math.log(rp3))/2\n",
+      "rpn4  =  (math.log(rp4))/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  ratio  in  decibels  are  (a)\",round(rpd1,2),\"  dB  (b)\",round(rpd2,2),\"  dB \"\n",
+      "print  \"(c)  \",round(rpd3,2),\"  dB  and  (d)  \",round(rpd4,2),\"  dB\"\n",
+      "print  \"\\n  power  ratio  in  nepers  are  (a)\",round(rpn1,2),\"  Np  (b)\",round(rpn2,2),\"  Np\"\n",
+      "print   \"(c)  \",round(rpn3,2),\"  Np  and  (d)  \",round(rpn4,2),\"  Np\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  ratio  in  decibels  are  (a) 3.01   dB  (b) 13.98   dB \n",
+        "(c)   29.99   dB  and  (d)   -20.0   dB\n",
+        "\n",
+        "  power  ratio  in  nepers  are  (a) 0.35   Np  (b) 1.61   Np\n",
+        "(c)   3.45   Np  and  (d)   -2.3   Np\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 763</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the attenuation in decibels\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rp  =  0.05;#  power  ratio  P2/P1\n",
+      "\n",
+      " #calculation:\n",
+      " #power  ratio  in  decibels\n",
+      "rpd  =  10*(1/2.303)*math.log(rp)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nthe  attenuation  is  \",round(abs(rpd),2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "the  attenuation  is   13.01   dB"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 764</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the output power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "gain  =  1.5;#  in  dB\n",
+      "Pi  =  0.012;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      " #output  power\n",
+      "Po  =  Pi*10**gain\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\noutput  power  is  \",round(Po,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "output  power  is   0.38   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 764</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current input and (b) the current ratio expressed in decibels.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "I2  =  0.05;#  in  Amperes\n",
+      "rin  =  1.32;#  in  Np\n",
+      "\n",
+      " #calculation:\n",
+      " #current  input,  I1\n",
+      "I1  =  I2*math.e**(rin)\n",
+      " #current  ratio  in  decibels\n",
+      "rid  =  20*(1/2.303)*math.log(I2/I1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncurrent  input,  I1  is  \",round(I1,2),\"  A\"\n",
+      "print  \"\\ncurrent  ratio  in  decibels  is  \",round(rid,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "current  input,  I1  is   0.19   A\n",
+        "\n",
+        "current  ratio  in  decibels  is   -11.46   dB"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 769</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the characteristic impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ra1 = 8; # in ohms\n",
+      "Ra2 = 8; # in ohms\n",
+      "Ra3 = 21; # in ohms\n",
+      "Rb1 = 10; # in ohms\n",
+      "Rb2 = 10; # in ohms\n",
+      "Rb3 = 15; # in ohms\n",
+      "Rc1 = 200; # in ohms\n",
+      "Rc2 = 200; # in ohms\n",
+      "Rc3 = 56.25; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R01 = (Ra1**2 + 2*Ra2*Ra3)**0.5\n",
+      "R02 = (Rb1**2 + 2*Rb2*Rb3)**0.5\n",
+      "R03 = (Rc1**2 + 2*Rc2*Rc3)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a) the characteristic impedance, R0  is  \",R01,\"  ohm\"\n",
+      "print  \"\\n(b) the characteristic impedance, R0  is  \",R02,\"  ohm\"\n",
+      "print  \"\\n(c) the characteristic impedance, R0  is  \",R03,\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a) the characteristic impedance, R0  is   20.0   ohm\n",
+        "\n",
+        "(b) the characteristic impedance, R0  is   20.0   ohm\n",
+        "\n",
+        "(c) the characteristic impedance, R0  is   250.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 769</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the characteristic impedance, and (b) the attenuation (in dB) produced by the pad\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  500;#  in  ohm\n",
+      "R2  =  1000;#  in  ohm\n",
+      "I1  =  1;#  in  ampere  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #  for  symmetrical  pi-attenuator  section\n",
+      " #characteristic  impedance,  R0\n",
+      "R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5\n",
+      " #current  Ix\n",
+      "Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R0))*Ix\n",
+      "ri  =  I1/I2;#  retio  of  currents\n",
+      " #attenuation\n",
+      "attn  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  characteristic  impedance  is  \",round(R0,2),\"  ohm\"\n",
+      "print  \"\\n  attenuation  is  \",round(attn,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  characteristic  impedance  is   447.21   ohm\n",
+        "\n",
+        "  attenuation  is   8.36   dB"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 770</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the input resistance when the output port is open-circuited, \n",
+      "#(b) the input resistance when the output port is short-circuited, and (c) the characteristic impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ra1  =  15;#  in  ohm\n",
+      "Ra2  =  15;#  in  ohm\n",
+      "Ra3  =  10;#  in  ohm\n",
+      "Rb1  =  15;#  in  ohm\n",
+      "Rb2  =  5;#  in  ohm\n",
+      "Rb3  =  5;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Roc1 = Ra1 + Ra3\n",
+      "Rsc1 = Ra1 + Ra2*Ra3/(Ra2+Ra3)\n",
+      "R01 = (Roc1*Rsc1)**0.5\n",
+      "\n",
+      "Roc2 = Rb2* (Rb1 + Rb3)/(Rb1 + Rb2 + Rb3)\n",
+      "Rsc2 = Rb2*Rb1/(Rb2+Rb1)\n",
+      "R02 = (Roc2*Rsc2)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a) the input resistance when the output port is open-circuited  is\", Roc1,\" ohm for T-Network\"\n",
+      "print   \"and \",Roc2,\" ohm for pi-Network \"\n",
+      "print  \"\\n  (b) the input resistance when the output port is short-circuited  is,\", Rsc1,\" ohm for T-Network\"\n",
+      "print   \"and \",Rsc2,\" ohm for pi-Network \"\n",
+      "print  \"\\n  (c) the characteristic impedance.  is,\",round(R01,1),\" ohm for T-Network and ,\",round(R02,2),\" ohm for pi-Network \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a) the input resistance when the output port is open-circuited  is 25  ohm for T-Network\n",
+        "and  4.0  ohm for pi-Network \n",
+        "\n",
+        "  (b) the input resistance when the output port is short-circuited  is, 21.0  ohm for T-Network\n",
+        "and  3.75  ohm for pi-Network \n",
+        "\n",
+        "  (c) the characteristic impedance.  is, 22.9  ohm for T-Network and , 3.87  ohm for pi-Network \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 770</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design a T-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
+      "#having a characteristic impedance of 600 ohm.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vat = 20; # in db\n",
+      "R0  =  600;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "N = math.e**(Vat*2.3/20)\n",
+      "R1 = R0*(N-1)/(N+1)\n",
+      "R2 = R0*2*N/(N**2 - 1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,0),\" ohm and Resistance R2 is\",round(R2,1),\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " For a T-section symmetrical attenuator pad, Resistance R1 is 491.0  ohm and Resistance R2 is 121.5  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 771</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design a pi-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
+      "#having a characteristic impedance of 600 ohm.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vat = 20; # in db\n",
+      "R0  =  600;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "N = math.e**(Vat*2.303/20)\n",
+      "R1 = R0*(N**2 - 1)/(2*N)\n",
+      "R2 = R0*(N+1)/(N-1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n For a pi-section symmetrical attenuator pad, Resistance R1 is\",round(R1/1000,2),\" Kohm\"\n",
+      "print   \"and Resistance R2 is\",round(R2,0),\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " For a pi-section symmetrical attenuator pad, Resistance R1 is 2.97  Kohm\n",
+        "and Resistance R2 is 733.0  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 772</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the characteristic impedance R0, and (b) the insertion loss in decibels\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  300;#  in  ohm\n",
+      "R2  =  450;#  in  ohm\n",
+      "I1  =  1;#  in  ampere  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #the  characteristic  impedance  of  a  symmetric  T-pad  attenuator  is  given  by\n",
+      "R0  =  (R1**2  +  2*R1*R2)**0.5\n",
+      " #By  current  division\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R1+  R0))*I1\n",
+      "ri  =  I1/I2;#  ratio  of  currents\n",
+      " #insertion  loss\n",
+      "il  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  characteristic  impedance  is  \",R0,\"  ohm\"\n",
+      "print  \"\\n  insertion  loss  is  \",round(il,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  characteristic  impedance  is   600.0   ohm\n",
+        "\n",
+        "  insertion  loss  is   9.54   dB"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 773</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the insertion loss at a tapping of (a) 2 kohm, (b) 1 kohm.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  500;#  in  ohm\n",
+      "Rhm  =  3000;#  in  ohm\n",
+      "RL  =  2000;#  in  ohm\n",
+      "r1  =  2000;#  in  ohm\n",
+      "r2  =  1000;#  in  ohm\n",
+      "E  =  1;#  in  volts  (lets  say)\n",
+      "\n",
+      " #calculation:\n",
+      " #Without  the  rheostat  in  the  circuit  the  voltage  across  the  2  kohm\u0018  load,  VL\n",
+      "VL  =  (RL/(RL  +  r))*E\n",
+      " #voltage  V2  with  2kohm  tapping\n",
+      "V2  =  ((RL*r1/(r1  +  RL))/((RL*r1/(r1  +  RL))  +  Rhm  -  r1  +  r))*E\n",
+      "rv1  =  VL/V2;#  ratio  of  currents\n",
+      " #insertion  loss  \n",
+      "il1  =  20*(1/2.303)*math.log(rv1)\n",
+      " #voltage  V1  with  1kohm  tapping\n",
+      "V1  =  ((RL*r2/(r2  +  RL))/((RL*r2/(r2  +  RL))  +  Rhm  -  r2  +  r))*E\n",
+      "rv2  =  VL/V1;#  ratio  of  currents\n",
+      " #insertion  loss  \n",
+      "il2  =  20*(1/2.303)*math.log(rv2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  insertion  loss  for  2kohm  tap  is  \",round(il1,2),\"  dB\"\n",
+      "print  \"\\n  insertion  loss  for  1kohm  tap  is  \",round(il2,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  insertion  loss  for  2kohm  tap  is   6.02   dB\n",
+        "\n",
+        "  insertion  loss  for  1kohm  tap  is   11.59   dB"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 774</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) its characteristic impedance, and (b) the insertion loss (in decibels) when feeding a matched load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  500;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #characteristic  impedance  of  a  symmetrical  attenuator\n",
+      "R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5\n",
+      " #current  Ix\n",
+      "Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R0))*Ix\n",
+      "ri  =  I1/I2;#  retio  of  currents\n",
+      " #insertion  loss  \n",
+      "il  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  characteristic  impedance  is  \",round(R0,2),\"  ohm\"\n",
+      "print  \"\\n  insertion  loss  is  \",round(il,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  characteristic  impedance  is   353.55   ohm\n",
+        "\n",
+        "  insertion  loss  is   15.31   dB"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 776</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  100;#  in  ohm\n",
+      "R2  =  200;#  in  ohm\n",
+      "R3  =  300;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #image  impedance  Roa\n",
+      "Roa  =  ((R1  +  R2)*(R2  +  (R1*R3/(R1  +  R3))))**0.5\n",
+      " #image  impedance  Rob\n",
+      "Rob  =  ((R1  +  R3)*(R3  +  (R1*R2/(R1  +  R2))))**0.5\n",
+      " #The  iterative  impedance  at  port  1\n",
+      "Ri1  =  (-1*R1  +  (R1**2  -  (-1*4*((R2*(R1  +  R3))  +  (R3*R1))))**0.5)/2\n",
+      " #The  iterative  impedance  at  port  2\n",
+      "Ri2  =  (R1  +  (R1**2  -  (-1*4*((R3*(R1  +  R2))  +  (R2*R1))))**0.5)/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  image  impedance  are  \",round(Roa,2),\"  ohm  and  \",round(Rob,2),\"  ohm  \"\n",
+      "print  \"\\n  iterative  impedances  are  \",round(Ri1,2),\"  ohm  and  \",round(Ri2,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  image  impedance  are   287.23   ohm  and   382.97   ohm  \n",
+        "\n",
+        "  iterative  impedances  are   285.41   ohm  and   385.41   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 777</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  2000;#  in  ohm\n",
+      "R3  =  3000;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #image  impedance  Roa\n",
+      "Roa  =  (((R3  +  R2)*R1/(R1  +  R2  +  R3))*(R1*R3/(R1  +  R3)))**0.5\n",
+      " #image  impedance  Rob\n",
+      "Rob  =  (((R3  +  R1)*R2/(R1  +  R2  +  R3))*(R2*R3/(R2  +  R3)))**0.5\n",
+      " #The  iterative  impedance  at  port  1\n",
+      "Ri1  =  (-1*R1  +  ((R1**2)  -  (-1*4*2*R2*R1))**0.5)/(2*2)\n",
+      " #The  iterative  impedance  at  port  2\n",
+      "Ri2  =  (R1  +  ((-1*R1)**2  -  (-1*4*2*R2*R1))**0.5)/(2*2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  image  impedance  are  \",round(Roa,2),\"  ohm  and  \",round(Rob,2),\"  ohm  \"\n",
+      "print  \"\\n  iterative  impedances  are  \",round(Ri1,2),\"  ohm  and  \",round(Ri2,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  image  impedance  are   790.57   ohm  and   1264.91   ohm  \n",
+        "\n",
+        "  iterative  impedances  are   780.78   ohm  and   1280.78   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 780</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the values of resistance R1 and R2, (b) the attenuation of the pad in decibels, and (c) its insertion loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  500;#  in  ohm\n",
+      "RL  =  100;#  in  ohm\n",
+      "E  =  1;#  in  volts  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #res.\n",
+      "R1  =  (r*(r  -  RL))**0.5\n",
+      "R2  =  (r*RL**2/(r  -  RL))**0.5\n",
+      " #current  I1\n",
+      "I1  =  E/(r  +  R1  +  R2*RL/(RL  +  R2))\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  RL))*I1\n",
+      " #input  power\n",
+      "P1  =  r*I1**2\n",
+      " #output  power\n",
+      "P2  =  RL*I2**2\n",
+      " #attenuation\n",
+      "attn  =  10*(1/2.303)*math.log(P1/P2)\n",
+      " #Load  current,  IL\n",
+      "IL  =  E/(r  +  RL)\n",
+      " #voltage,  VL\n",
+      "VL  =  IL*RL\n",
+      " #voltage,  V1\n",
+      "V1  =  E  -  I1*r\n",
+      " #voltage,  V2\n",
+      "V2  =  V1  -  I1*R1\n",
+      " #insertion  loss\n",
+      "il  =  20*(1/2.303)*math.log(VL/V2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  R1  =  \",round(R1,2),\"  ohm  and  R2  =  \",round(R2,2),\"  ohm  \"\n",
+      "print  \"\\n  attenuation  is  \",round(attn,2),\"  dB  \"\n",
+      "print  \"\\n  In  decibels,  the  insertion  loss  is  \",round(il,2),\"  dB  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  R1  =   447.21   ohm  and  R2  =   111.8   ohm  \n",
+        "\n",
+        "  attenuation  is   12.54   dB  \n",
+        "\n",
+        "  In  decibels,  the  insertion  loss  is   9.98   dB  "
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 783</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the attenuation of each individual attenuation section, \n",
+      "#(b) the voltage output of the final stage, and (c) the voltage output of the third stage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "attnO  =  70;#  in  dB\n",
+      "n  =  5;#  numbers  of  identical  atteneurs\n",
+      "V1  =  0.02;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #attenuation  of  each  section\n",
+      "attn  =  attnO/n\n",
+      " #output  of  the  final  stage\n",
+      "Vo  =  V1/(10**(attnO/20))\n",
+      " #voltage  output  of  the  third  stage\n",
+      "V3  =  V1/(10**(3*attn/20))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  attenuation  of  each  section  =  \",round(attn,2),\"  dB  \"\n",
+      "print  \"\\n  output  of  the  final  stage  is  \",round(Vo*1E6,2),\"uV  \"\n",
+      "print  \"\\n  voltage  output  of  the  third  stage  is  \",round(V3*1E3,3),\"mV  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  attenuation  of  each  section  =   14.0   dB  \n",
+        "\n",
+        "  output  of  the  final  stage  is   6.32 uV  \n",
+        "\n",
+        "  voltage  output  of  the  third  stage  is   0.159 mV  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 784</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Design a T-network attenuator pad\n",
+      "#(b)determine the fraction of the initial current \n",
+      "#(c) Determine the attenuation in decibels given by four such sections\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r = 450; # in ohm\n",
+      "R0 = 450; # in ohms\n",
+      "x = 1/8\n",
+      "\n",
+      "#calculation:\n",
+      "N = 1/x\n",
+      "R1 = R0*(N-1)/(N+1)\n",
+      "R2 = R0*2*N/(N**2 - 1)\n",
+      "\n",
+      "Io = x*x\n",
+      "\n",
+      "attn  =  20*math.log10(N)\n",
+      "attnO = 4*attn\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,2),\" ohm \"\n",
+      "print   \"and Resistance R2 is\",round(R2,0),\" ohm\"\n",
+      "print  \"\\n  (b)current flows in the load =  \",round(Io,2),\"of the original current.\"\n",
+      "print  \"\\n  (c)overall attenuation is  \",round(attnO,2),\"dB  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For a T-section symmetrical attenuator pad, Resistance R1 is 350.0  ohm \n",
+        "and Resistance R2 is 114.0  ohm\n",
+        "\n",
+        "  (b)current flows in the load =   0.02 of the original current.\n",
+        "\n",
+        "  (c)overall attenuation is   72.25 dB  \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_1.ipynb
new file mode 100755
index 00000000..788f6f5e
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_1.ipynb
@@ -0,0 +1,1005 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 41: Attenuators</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 763</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power ratio in each case (i) in decibels and (ii) in nepers.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      " #ratio  of  output  power  to  input  power\n",
+      "rp1  =  2;\n",
+      "rp2  =  25;  \n",
+      "rp3  =  1000;\n",
+      "rp4  =  0.01;\n",
+      "\n",
+      "#calculation:\n",
+      " #power  ratio  in  decibels\n",
+      "rpd1  =  10*(1/2.303)*math.log(rp1)\n",
+      "rpd2  =  10*(1/2.303)*math.log(rp2)\n",
+      "rpd3  =  10*(1/2.303)*math.log(rp3)\n",
+      "rpd4  =  10*(1/2.303)*math.log(rp4)\n",
+      " #power  ratio  in  nepers\n",
+      "rpn1  =  (math.log(rp1))/2\n",
+      "rpn2  =  (math.log(rp2))/2\n",
+      "rpn3  =  (math.log(rp3))/2\n",
+      "rpn4  =  (math.log(rp4))/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  ratio  in  decibels  are  (a)\",round(rpd1,2),\"  dB  (b)\",round(rpd2,2),\"  dB \"\n",
+      "print  \"(c)  \",round(rpd3,2),\"  dB  and  (d)  \",round(rpd4,2),\"  dB\"\n",
+      "print  \"\\n  power  ratio  in  nepers  are  (a)\",round(rpn1,2),\"  Np  (b)\",round(rpn2,2),\"  Np\"\n",
+      "print   \"(c)  \",round(rpn3,2),\"  Np  and  (d)  \",round(rpn4,2),\"  Np\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  ratio  in  decibels  are  (a) 3.01   dB  (b) 13.98   dB \n",
+        "(c)   29.99   dB  and  (d)   -20.0   dB\n",
+        "\n",
+        "  power  ratio  in  nepers  are  (a) 0.35   Np  (b) 1.61   Np\n",
+        "(c)   3.45   Np  and  (d)   -2.3   Np\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 763</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the attenuation in decibels\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rp  =  0.05;#  power  ratio  P2/P1\n",
+      "\n",
+      " #calculation:\n",
+      " #power  ratio  in  decibels\n",
+      "rpd  =  10*(1/2.303)*math.log(rp)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nthe  attenuation  is  \",round(abs(rpd),2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "the  attenuation  is   13.01   dB"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 764</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the output power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "gain  =  1.5;#  in  dB\n",
+      "Pi  =  0.012;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      " #output  power\n",
+      "Po  =  Pi*10**gain\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\noutput  power  is  \",round(Po,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "output  power  is   0.38   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 764</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current input and (b) the current ratio expressed in decibels.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "I2  =  0.05;#  in  Amperes\n",
+      "rin  =  1.32;#  in  Np\n",
+      "\n",
+      " #calculation:\n",
+      " #current  input,  I1\n",
+      "I1  =  I2*math.e**(rin)\n",
+      " #current  ratio  in  decibels\n",
+      "rid  =  20*(1/2.303)*math.log(I2/I1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncurrent  input,  I1  is  \",round(I1,2),\"  A\"\n",
+      "print  \"\\ncurrent  ratio  in  decibels  is  \",round(rid,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "current  input,  I1  is   0.19   A\n",
+        "\n",
+        "current  ratio  in  decibels  is   -11.46   dB"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 769</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the characteristic impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ra1 = 8; # in ohms\n",
+      "Ra2 = 8; # in ohms\n",
+      "Ra3 = 21; # in ohms\n",
+      "Rb1 = 10; # in ohms\n",
+      "Rb2 = 10; # in ohms\n",
+      "Rb3 = 15; # in ohms\n",
+      "Rc1 = 200; # in ohms\n",
+      "Rc2 = 200; # in ohms\n",
+      "Rc3 = 56.25; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R01 = (Ra1**2 + 2*Ra2*Ra3)**0.5\n",
+      "R02 = (Rb1**2 + 2*Rb2*Rb3)**0.5\n",
+      "R03 = (Rc1**2 + 2*Rc2*Rc3)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a) the characteristic impedance, R0  is  \",R01,\"  ohm\"\n",
+      "print  \"\\n(b) the characteristic impedance, R0  is  \",R02,\"  ohm\"\n",
+      "print  \"\\n(c) the characteristic impedance, R0  is  \",R03,\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a) the characteristic impedance, R0  is   20.0   ohm\n",
+        "\n",
+        "(b) the characteristic impedance, R0  is   20.0   ohm\n",
+        "\n",
+        "(c) the characteristic impedance, R0  is   250.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 769</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the characteristic impedance, and (b) the attenuation (in dB) produced by the pad\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  500;#  in  ohm\n",
+      "R2  =  1000;#  in  ohm\n",
+      "I1  =  1;#  in  ampere  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #  for  symmetrical  pi-attenuator  section\n",
+      " #characteristic  impedance,  R0\n",
+      "R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5\n",
+      " #current  Ix\n",
+      "Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R0))*Ix\n",
+      "ri  =  I1/I2;#  retio  of  currents\n",
+      " #attenuation\n",
+      "attn  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  characteristic  impedance  is  \",round(R0,2),\"  ohm\"\n",
+      "print  \"\\n  attenuation  is  \",round(attn,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  characteristic  impedance  is   447.21   ohm\n",
+        "\n",
+        "  attenuation  is   8.36   dB"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 770</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the input resistance when the output port is open-circuited, \n",
+      "#(b) the input resistance when the output port is short-circuited, and (c) the characteristic impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ra1  =  15;#  in  ohm\n",
+      "Ra2  =  15;#  in  ohm\n",
+      "Ra3  =  10;#  in  ohm\n",
+      "Rb1  =  15;#  in  ohm\n",
+      "Rb2  =  5;#  in  ohm\n",
+      "Rb3  =  5;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Roc1 = Ra1 + Ra3\n",
+      "Rsc1 = Ra1 + Ra2*Ra3/(Ra2+Ra3)\n",
+      "R01 = (Roc1*Rsc1)**0.5\n",
+      "\n",
+      "Roc2 = Rb2* (Rb1 + Rb3)/(Rb1 + Rb2 + Rb3)\n",
+      "Rsc2 = Rb2*Rb1/(Rb2+Rb1)\n",
+      "R02 = (Roc2*Rsc2)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a) the input resistance when the output port is open-circuited  is\", Roc1,\" ohm for T-Network\"\n",
+      "print   \"and \",Roc2,\" ohm for pi-Network \"\n",
+      "print  \"\\n  (b) the input resistance when the output port is short-circuited  is,\", Rsc1,\" ohm for T-Network\"\n",
+      "print   \"and \",Rsc2,\" ohm for pi-Network \"\n",
+      "print  \"\\n  (c) the characteristic impedance.  is,\",round(R01,1),\" ohm for T-Network and ,\",round(R02,2),\" ohm for pi-Network \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a) the input resistance when the output port is open-circuited  is 25  ohm for T-Network\n",
+        "and  4.0  ohm for pi-Network \n",
+        "\n",
+        "  (b) the input resistance when the output port is short-circuited  is, 21.0  ohm for T-Network\n",
+        "and  3.75  ohm for pi-Network \n",
+        "\n",
+        "  (c) the characteristic impedance.  is, 22.9  ohm for T-Network and , 3.87  ohm for pi-Network \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 770</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design a T-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
+      "#having a characteristic impedance of 600 ohm.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vat = 20; # in db\n",
+      "R0  =  600;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "N = math.e**(Vat*2.3/20)\n",
+      "R1 = R0*(N-1)/(N+1)\n",
+      "R2 = R0*2*N/(N**2 - 1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,0),\" ohm and Resistance R2 is\",round(R2,1),\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " For a T-section symmetrical attenuator pad, Resistance R1 is 491.0  ohm and Resistance R2 is 121.5  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 771</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design a pi-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
+      "#having a characteristic impedance of 600 ohm.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vat = 20; # in db\n",
+      "R0  =  600;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "N = math.e**(Vat*2.303/20)\n",
+      "R1 = R0*(N**2 - 1)/(2*N)\n",
+      "R2 = R0*(N+1)/(N-1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n For a pi-section symmetrical attenuator pad, Resistance R1 is\",round(R1/1000,2),\" Kohm\"\n",
+      "print   \"and Resistance R2 is\",round(R2,0),\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " For a pi-section symmetrical attenuator pad, Resistance R1 is 2.97  Kohm\n",
+        "and Resistance R2 is 733.0  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 772</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the characteristic impedance R0, and (b) the insertion loss in decibels\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  300;#  in  ohm\n",
+      "R2  =  450;#  in  ohm\n",
+      "I1  =  1;#  in  ampere  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #the  characteristic  impedance  of  a  symmetric  T-pad  attenuator  is  given  by\n",
+      "R0  =  (R1**2  +  2*R1*R2)**0.5\n",
+      " #By  current  division\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R1+  R0))*I1\n",
+      "ri  =  I1/I2;#  ratio  of  currents\n",
+      " #insertion  loss\n",
+      "il  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  characteristic  impedance  is  \",R0,\"  ohm\"\n",
+      "print  \"\\n  insertion  loss  is  \",round(il,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  characteristic  impedance  is   600.0   ohm\n",
+        "\n",
+        "  insertion  loss  is   9.54   dB"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 773</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the insertion loss at a tapping of (a) 2 kohm, (b) 1 kohm.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  500;#  in  ohm\n",
+      "Rhm  =  3000;#  in  ohm\n",
+      "RL  =  2000;#  in  ohm\n",
+      "r1  =  2000;#  in  ohm\n",
+      "r2  =  1000;#  in  ohm\n",
+      "E  =  1;#  in  volts  (lets  say)\n",
+      "\n",
+      " #calculation:\n",
+      " #Without  the  rheostat  in  the  circuit  the  voltage  across  the  2  kohm\u0018  load,  VL\n",
+      "VL  =  (RL/(RL  +  r))*E\n",
+      " #voltage  V2  with  2kohm  tapping\n",
+      "V2  =  ((RL*r1/(r1  +  RL))/((RL*r1/(r1  +  RL))  +  Rhm  -  r1  +  r))*E\n",
+      "rv1  =  VL/V2;#  ratio  of  currents\n",
+      " #insertion  loss  \n",
+      "il1  =  20*(1/2.303)*math.log(rv1)\n",
+      " #voltage  V1  with  1kohm  tapping\n",
+      "V1  =  ((RL*r2/(r2  +  RL))/((RL*r2/(r2  +  RL))  +  Rhm  -  r2  +  r))*E\n",
+      "rv2  =  VL/V1;#  ratio  of  currents\n",
+      " #insertion  loss  \n",
+      "il2  =  20*(1/2.303)*math.log(rv2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  insertion  loss  for  2kohm  tap  is  \",round(il1,2),\"  dB\"\n",
+      "print  \"\\n  insertion  loss  for  1kohm  tap  is  \",round(il2,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  insertion  loss  for  2kohm  tap  is   6.02   dB\n",
+        "\n",
+        "  insertion  loss  for  1kohm  tap  is   11.59   dB"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 774</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) its characteristic impedance, and (b) the insertion loss (in decibels) when feeding a matched load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  500;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #characteristic  impedance  of  a  symmetrical  attenuator\n",
+      "R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5\n",
+      " #current  Ix\n",
+      "Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R0))*Ix\n",
+      "ri  =  I1/I2;#  retio  of  currents\n",
+      " #insertion  loss  \n",
+      "il  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  characteristic  impedance  is  \",round(R0,2),\"  ohm\"\n",
+      "print  \"\\n  insertion  loss  is  \",round(il,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  characteristic  impedance  is   353.55   ohm\n",
+        "\n",
+        "  insertion  loss  is   15.31   dB"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 776</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  100;#  in  ohm\n",
+      "R2  =  200;#  in  ohm\n",
+      "R3  =  300;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #image  impedance  Roa\n",
+      "Roa  =  ((R1  +  R2)*(R2  +  (R1*R3/(R1  +  R3))))**0.5\n",
+      " #image  impedance  Rob\n",
+      "Rob  =  ((R1  +  R3)*(R3  +  (R1*R2/(R1  +  R2))))**0.5\n",
+      " #The  iterative  impedance  at  port  1\n",
+      "Ri1  =  (-1*R1  +  (R1**2  -  (-1*4*((R2*(R1  +  R3))  +  (R3*R1))))**0.5)/2\n",
+      " #The  iterative  impedance  at  port  2\n",
+      "Ri2  =  (R1  +  (R1**2  -  (-1*4*((R3*(R1  +  R2))  +  (R2*R1))))**0.5)/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  image  impedance  are  \",round(Roa,2),\"  ohm  and  \",round(Rob,2),\"  ohm  \"\n",
+      "print  \"\\n  iterative  impedances  are  \",round(Ri1,2),\"  ohm  and  \",round(Ri2,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  image  impedance  are   287.23   ohm  and   382.97   ohm  \n",
+        "\n",
+        "  iterative  impedances  are   285.41   ohm  and   385.41   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 777</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  2000;#  in  ohm\n",
+      "R3  =  3000;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #image  impedance  Roa\n",
+      "Roa  =  (((R3  +  R2)*R1/(R1  +  R2  +  R3))*(R1*R3/(R1  +  R3)))**0.5\n",
+      " #image  impedance  Rob\n",
+      "Rob  =  (((R3  +  R1)*R2/(R1  +  R2  +  R3))*(R2*R3/(R2  +  R3)))**0.5\n",
+      " #The  iterative  impedance  at  port  1\n",
+      "Ri1  =  (-1*R1  +  ((R1**2)  -  (-1*4*2*R2*R1))**0.5)/(2*2)\n",
+      " #The  iterative  impedance  at  port  2\n",
+      "Ri2  =  (R1  +  ((-1*R1)**2  -  (-1*4*2*R2*R1))**0.5)/(2*2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  image  impedance  are  \",round(Roa,2),\"  ohm  and  \",round(Rob,2),\"  ohm  \"\n",
+      "print  \"\\n  iterative  impedances  are  \",round(Ri1,2),\"  ohm  and  \",round(Ri2,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  image  impedance  are   790.57   ohm  and   1264.91   ohm  \n",
+        "\n",
+        "  iterative  impedances  are   780.78   ohm  and   1280.78   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 780</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the values of resistance R1 and R2, (b) the attenuation of the pad in decibels, and (c) its insertion loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  500;#  in  ohm\n",
+      "RL  =  100;#  in  ohm\n",
+      "E  =  1;#  in  volts  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #res.\n",
+      "R1  =  (r*(r  -  RL))**0.5\n",
+      "R2  =  (r*RL**2/(r  -  RL))**0.5\n",
+      " #current  I1\n",
+      "I1  =  E/(r  +  R1  +  R2*RL/(RL  +  R2))\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  RL))*I1\n",
+      " #input  power\n",
+      "P1  =  r*I1**2\n",
+      " #output  power\n",
+      "P2  =  RL*I2**2\n",
+      " #attenuation\n",
+      "attn  =  10*(1/2.303)*math.log(P1/P2)\n",
+      " #Load  current,  IL\n",
+      "IL  =  E/(r  +  RL)\n",
+      " #voltage,  VL\n",
+      "VL  =  IL*RL\n",
+      " #voltage,  V1\n",
+      "V1  =  E  -  I1*r\n",
+      " #voltage,  V2\n",
+      "V2  =  V1  -  I1*R1\n",
+      " #insertion  loss\n",
+      "il  =  20*(1/2.303)*math.log(VL/V2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  R1  =  \",round(R1,2),\"  ohm  and  R2  =  \",round(R2,2),\"  ohm  \"\n",
+      "print  \"\\n  attenuation  is  \",round(attn,2),\"  dB  \"\n",
+      "print  \"\\n  In  decibels,  the  insertion  loss  is  \",round(il,2),\"  dB  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  R1  =   447.21   ohm  and  R2  =   111.8   ohm  \n",
+        "\n",
+        "  attenuation  is   12.54   dB  \n",
+        "\n",
+        "  In  decibels,  the  insertion  loss  is   9.98   dB  "
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 783</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the attenuation of each individual attenuation section, \n",
+      "#(b) the voltage output of the final stage, and (c) the voltage output of the third stage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "attnO  =  70;#  in  dB\n",
+      "n  =  5;#  numbers  of  identical  atteneurs\n",
+      "V1  =  0.02;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #attenuation  of  each  section\n",
+      "attn  =  attnO/n\n",
+      " #output  of  the  final  stage\n",
+      "Vo  =  V1/(10**(attnO/20))\n",
+      " #voltage  output  of  the  third  stage\n",
+      "V3  =  V1/(10**(3*attn/20))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  attenuation  of  each  section  =  \",round(attn,2),\"  dB  \"\n",
+      "print  \"\\n  output  of  the  final  stage  is  \",round(Vo*1E6,2),\"uV  \"\n",
+      "print  \"\\n  voltage  output  of  the  third  stage  is  \",round(V3*1E3,3),\"mV  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  attenuation  of  each  section  =   14.0   dB  \n",
+        "\n",
+        "  output  of  the  final  stage  is   6.32 uV  \n",
+        "\n",
+        "  voltage  output  of  the  third  stage  is   0.159 mV  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 784</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Design a T-network attenuator pad\n",
+      "#(b)determine the fraction of the initial current \n",
+      "#(c) Determine the attenuation in decibels given by four such sections\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r = 450; # in ohm\n",
+      "R0 = 450; # in ohms\n",
+      "x = 1/8\n",
+      "\n",
+      "#calculation:\n",
+      "N = 1/x\n",
+      "R1 = R0*(N-1)/(N+1)\n",
+      "R2 = R0*2*N/(N**2 - 1)\n",
+      "\n",
+      "Io = x*x\n",
+      "\n",
+      "attn  =  20*math.log10(N)\n",
+      "attnO = 4*attn\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,2),\" ohm \"\n",
+      "print   \"and Resistance R2 is\",round(R2,0),\" ohm\"\n",
+      "print  \"\\n  (b)current flows in the load =  \",round(Io,2),\"of the original current.\"\n",
+      "print  \"\\n  (c)overall attenuation is  \",round(attnO,2),\"dB  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For a T-section symmetrical attenuator pad, Resistance R1 is 350.0  ohm \n",
+        "and Resistance R2 is 114.0  ohm\n",
+        "\n",
+        "  (b)current flows in the load =   0.02 of the original current.\n",
+        "\n",
+        "  (c)overall attenuation is   72.25 dB  \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_2.ipynb
new file mode 100755
index 00000000..788f6f5e
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_2.ipynb
@@ -0,0 +1,1005 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 41: Attenuators</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 763</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the power ratio in each case (i) in decibels and (ii) in nepers.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      " #ratio  of  output  power  to  input  power\n",
+      "rp1  =  2;\n",
+      "rp2  =  25;  \n",
+      "rp3  =  1000;\n",
+      "rp4  =  0.01;\n",
+      "\n",
+      "#calculation:\n",
+      " #power  ratio  in  decibels\n",
+      "rpd1  =  10*(1/2.303)*math.log(rp1)\n",
+      "rpd2  =  10*(1/2.303)*math.log(rp2)\n",
+      "rpd3  =  10*(1/2.303)*math.log(rp3)\n",
+      "rpd4  =  10*(1/2.303)*math.log(rp4)\n",
+      " #power  ratio  in  nepers\n",
+      "rpn1  =  (math.log(rp1))/2\n",
+      "rpn2  =  (math.log(rp2))/2\n",
+      "rpn3  =  (math.log(rp3))/2\n",
+      "rpn4  =  (math.log(rp4))/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  ratio  in  decibels  are  (a)\",round(rpd1,2),\"  dB  (b)\",round(rpd2,2),\"  dB \"\n",
+      "print  \"(c)  \",round(rpd3,2),\"  dB  and  (d)  \",round(rpd4,2),\"  dB\"\n",
+      "print  \"\\n  power  ratio  in  nepers  are  (a)\",round(rpn1,2),\"  Np  (b)\",round(rpn2,2),\"  Np\"\n",
+      "print   \"(c)  \",round(rpn3,2),\"  Np  and  (d)  \",round(rpn4,2),\"  Np\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  ratio  in  decibels  are  (a) 3.01   dB  (b) 13.98   dB \n",
+        "(c)   29.99   dB  and  (d)   -20.0   dB\n",
+        "\n",
+        "  power  ratio  in  nepers  are  (a) 0.35   Np  (b) 1.61   Np\n",
+        "(c)   3.45   Np  and  (d)   -2.3   Np\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 763</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the attenuation in decibels\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rp  =  0.05;#  power  ratio  P2/P1\n",
+      "\n",
+      " #calculation:\n",
+      " #power  ratio  in  decibels\n",
+      "rpd  =  10*(1/2.303)*math.log(rp)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nthe  attenuation  is  \",round(abs(rpd),2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "the  attenuation  is   13.01   dB"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 764</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the output power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "gain  =  1.5;#  in  dB\n",
+      "Pi  =  0.012;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      " #output  power\n",
+      "Po  =  Pi*10**gain\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\noutput  power  is  \",round(Po,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "output  power  is   0.38   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 764</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the current input and (b) the current ratio expressed in decibels.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "I2  =  0.05;#  in  Amperes\n",
+      "rin  =  1.32;#  in  Np\n",
+      "\n",
+      " #calculation:\n",
+      " #current  input,  I1\n",
+      "I1  =  I2*math.e**(rin)\n",
+      " #current  ratio  in  decibels\n",
+      "rid  =  20*(1/2.303)*math.log(I2/I1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncurrent  input,  I1  is  \",round(I1,2),\"  A\"\n",
+      "print  \"\\ncurrent  ratio  in  decibels  is  \",round(rid,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "current  input,  I1  is   0.19   A\n",
+        "\n",
+        "current  ratio  in  decibels  is   -11.46   dB"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 769</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the characteristic impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ra1 = 8; # in ohms\n",
+      "Ra2 = 8; # in ohms\n",
+      "Ra3 = 21; # in ohms\n",
+      "Rb1 = 10; # in ohms\n",
+      "Rb2 = 10; # in ohms\n",
+      "Rb3 = 15; # in ohms\n",
+      "Rc1 = 200; # in ohms\n",
+      "Rc2 = 200; # in ohms\n",
+      "Rc3 = 56.25; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R01 = (Ra1**2 + 2*Ra2*Ra3)**0.5\n",
+      "R02 = (Rb1**2 + 2*Rb2*Rb3)**0.5\n",
+      "R03 = (Rc1**2 + 2*Rc2*Rc3)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a) the characteristic impedance, R0  is  \",R01,\"  ohm\"\n",
+      "print  \"\\n(b) the characteristic impedance, R0  is  \",R02,\"  ohm\"\n",
+      "print  \"\\n(c) the characteristic impedance, R0  is  \",R03,\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a) the characteristic impedance, R0  is   20.0   ohm\n",
+        "\n",
+        "(b) the characteristic impedance, R0  is   20.0   ohm\n",
+        "\n",
+        "(c) the characteristic impedance, R0  is   250.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 769</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the characteristic impedance, and (b) the attenuation (in dB) produced by the pad\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  500;#  in  ohm\n",
+      "R2  =  1000;#  in  ohm\n",
+      "I1  =  1;#  in  ampere  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #  for  symmetrical  pi-attenuator  section\n",
+      " #characteristic  impedance,  R0\n",
+      "R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5\n",
+      " #current  Ix\n",
+      "Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R0))*Ix\n",
+      "ri  =  I1/I2;#  retio  of  currents\n",
+      " #attenuation\n",
+      "attn  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  characteristic  impedance  is  \",round(R0,2),\"  ohm\"\n",
+      "print  \"\\n  attenuation  is  \",round(attn,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  characteristic  impedance  is   447.21   ohm\n",
+        "\n",
+        "  attenuation  is   8.36   dB"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 770</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the input resistance when the output port is open-circuited, \n",
+      "#(b) the input resistance when the output port is short-circuited, and (c) the characteristic impedance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ra1  =  15;#  in  ohm\n",
+      "Ra2  =  15;#  in  ohm\n",
+      "Ra3  =  10;#  in  ohm\n",
+      "Rb1  =  15;#  in  ohm\n",
+      "Rb2  =  5;#  in  ohm\n",
+      "Rb3  =  5;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Roc1 = Ra1 + Ra3\n",
+      "Rsc1 = Ra1 + Ra2*Ra3/(Ra2+Ra3)\n",
+      "R01 = (Roc1*Rsc1)**0.5\n",
+      "\n",
+      "Roc2 = Rb2* (Rb1 + Rb3)/(Rb1 + Rb2 + Rb3)\n",
+      "Rsc2 = Rb2*Rb1/(Rb2+Rb1)\n",
+      "R02 = (Roc2*Rsc2)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a) the input resistance when the output port is open-circuited  is\", Roc1,\" ohm for T-Network\"\n",
+      "print   \"and \",Roc2,\" ohm for pi-Network \"\n",
+      "print  \"\\n  (b) the input resistance when the output port is short-circuited  is,\", Rsc1,\" ohm for T-Network\"\n",
+      "print   \"and \",Rsc2,\" ohm for pi-Network \"\n",
+      "print  \"\\n  (c) the characteristic impedance.  is,\",round(R01,1),\" ohm for T-Network and ,\",round(R02,2),\" ohm for pi-Network \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a) the input resistance when the output port is open-circuited  is 25  ohm for T-Network\n",
+        "and  4.0  ohm for pi-Network \n",
+        "\n",
+        "  (b) the input resistance when the output port is short-circuited  is, 21.0  ohm for T-Network\n",
+        "and  3.75  ohm for pi-Network \n",
+        "\n",
+        "  (c) the characteristic impedance.  is, 22.9  ohm for T-Network and , 3.87  ohm for pi-Network \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 770</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design a T-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
+      "#having a characteristic impedance of 600 ohm.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vat = 20; # in db\n",
+      "R0  =  600;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "N = math.e**(Vat*2.3/20)\n",
+      "R1 = R0*(N-1)/(N+1)\n",
+      "R2 = R0*2*N/(N**2 - 1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,0),\" ohm and Resistance R2 is\",round(R2,1),\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " For a T-section symmetrical attenuator pad, Resistance R1 is 491.0  ohm and Resistance R2 is 121.5  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 771</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design a pi-section symmetrical attenuator pad to provide a voltage attenuation of 20 dB and \n",
+      "#having a characteristic impedance of 600 ohm.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vat = 20; # in db\n",
+      "R0  =  600;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "N = math.e**(Vat*2.303/20)\n",
+      "R1 = R0*(N**2 - 1)/(2*N)\n",
+      "R2 = R0*(N+1)/(N-1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n For a pi-section symmetrical attenuator pad, Resistance R1 is\",round(R1/1000,2),\" Kohm\"\n",
+      "print   \"and Resistance R2 is\",round(R2,0),\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " For a pi-section symmetrical attenuator pad, Resistance R1 is 2.97  Kohm\n",
+        "and Resistance R2 is 733.0  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 772</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the characteristic impedance R0, and (b) the insertion loss in decibels\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  300;#  in  ohm\n",
+      "R2  =  450;#  in  ohm\n",
+      "I1  =  1;#  in  ampere  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #the  characteristic  impedance  of  a  symmetric  T-pad  attenuator  is  given  by\n",
+      "R0  =  (R1**2  +  2*R1*R2)**0.5\n",
+      " #By  current  division\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R1+  R0))*I1\n",
+      "ri  =  I1/I2;#  ratio  of  currents\n",
+      " #insertion  loss\n",
+      "il  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  characteristic  impedance  is  \",R0,\"  ohm\"\n",
+      "print  \"\\n  insertion  loss  is  \",round(il,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  characteristic  impedance  is   600.0   ohm\n",
+        "\n",
+        "  insertion  loss  is   9.54   dB"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 773</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the insertion loss at a tapping of (a) 2 kohm, (b) 1 kohm.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  500;#  in  ohm\n",
+      "Rhm  =  3000;#  in  ohm\n",
+      "RL  =  2000;#  in  ohm\n",
+      "r1  =  2000;#  in  ohm\n",
+      "r2  =  1000;#  in  ohm\n",
+      "E  =  1;#  in  volts  (lets  say)\n",
+      "\n",
+      " #calculation:\n",
+      " #Without  the  rheostat  in  the  circuit  the  voltage  across  the  2  kohm\u0018  load,  VL\n",
+      "VL  =  (RL/(RL  +  r))*E\n",
+      " #voltage  V2  with  2kohm  tapping\n",
+      "V2  =  ((RL*r1/(r1  +  RL))/((RL*r1/(r1  +  RL))  +  Rhm  -  r1  +  r))*E\n",
+      "rv1  =  VL/V2;#  ratio  of  currents\n",
+      " #insertion  loss  \n",
+      "il1  =  20*(1/2.303)*math.log(rv1)\n",
+      " #voltage  V1  with  1kohm  tapping\n",
+      "V1  =  ((RL*r2/(r2  +  RL))/((RL*r2/(r2  +  RL))  +  Rhm  -  r2  +  r))*E\n",
+      "rv2  =  VL/V1;#  ratio  of  currents\n",
+      " #insertion  loss  \n",
+      "il2  =  20*(1/2.303)*math.log(rv2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  insertion  loss  for  2kohm  tap  is  \",round(il1,2),\"  dB\"\n",
+      "print  \"\\n  insertion  loss  for  1kohm  tap  is  \",round(il2,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  insertion  loss  for  2kohm  tap  is   6.02   dB\n",
+        "\n",
+        "  insertion  loss  for  1kohm  tap  is   11.59   dB"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 774</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) its characteristic impedance, and (b) the insertion loss (in decibels) when feeding a matched load.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  500;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #characteristic  impedance  of  a  symmetrical  attenuator\n",
+      "R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5\n",
+      " #current  Ix\n",
+      "Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R0))*Ix\n",
+      "ri  =  I1/I2;#  retio  of  currents\n",
+      " #insertion  loss  \n",
+      "il  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  characteristic  impedance  is  \",round(R0,2),\"  ohm\"\n",
+      "print  \"\\n  insertion  loss  is  \",round(il,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  characteristic  impedance  is   353.55   ohm\n",
+        "\n",
+        "  insertion  loss  is   15.31   dB"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 776</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  100;#  in  ohm\n",
+      "R2  =  200;#  in  ohm\n",
+      "R3  =  300;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #image  impedance  Roa\n",
+      "Roa  =  ((R1  +  R2)*(R2  +  (R1*R3/(R1  +  R3))))**0.5\n",
+      " #image  impedance  Rob\n",
+      "Rob  =  ((R1  +  R3)*(R3  +  (R1*R2/(R1  +  R2))))**0.5\n",
+      " #The  iterative  impedance  at  port  1\n",
+      "Ri1  =  (-1*R1  +  (R1**2  -  (-1*4*((R2*(R1  +  R3))  +  (R3*R1))))**0.5)/2\n",
+      " #The  iterative  impedance  at  port  2\n",
+      "Ri2  =  (R1  +  (R1**2  -  (-1*4*((R3*(R1  +  R2))  +  (R2*R1))))**0.5)/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  image  impedance  are  \",round(Roa,2),\"  ohm  and  \",round(Rob,2),\"  ohm  \"\n",
+      "print  \"\\n  iterative  impedances  are  \",round(Ri1,2),\"  ohm  and  \",round(Ri2,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  image  impedance  are   287.23   ohm  and   382.97   ohm  \n",
+        "\n",
+        "  iterative  impedances  are   285.41   ohm  and   385.41   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 777</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the section (a) the image impedances, and (b) the iterative impedances\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  2000;#  in  ohm\n",
+      "R3  =  3000;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #image  impedance  Roa\n",
+      "Roa  =  (((R3  +  R2)*R1/(R1  +  R2  +  R3))*(R1*R3/(R1  +  R3)))**0.5\n",
+      " #image  impedance  Rob\n",
+      "Rob  =  (((R3  +  R1)*R2/(R1  +  R2  +  R3))*(R2*R3/(R2  +  R3)))**0.5\n",
+      " #The  iterative  impedance  at  port  1\n",
+      "Ri1  =  (-1*R1  +  ((R1**2)  -  (-1*4*2*R2*R1))**0.5)/(2*2)\n",
+      " #The  iterative  impedance  at  port  2\n",
+      "Ri2  =  (R1  +  ((-1*R1)**2  -  (-1*4*2*R2*R1))**0.5)/(2*2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  image  impedance  are  \",round(Roa,2),\"  ohm  and  \",round(Rob,2),\"  ohm  \"\n",
+      "print  \"\\n  iterative  impedances  are  \",round(Ri1,2),\"  ohm  and  \",round(Ri2,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  image  impedance  are   790.57   ohm  and   1264.91   ohm  \n",
+        "\n",
+        "  iterative  impedances  are   780.78   ohm  and   1280.78   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 780</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the values of resistance R1 and R2, (b) the attenuation of the pad in decibels, and (c) its insertion loss.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  500;#  in  ohm\n",
+      "RL  =  100;#  in  ohm\n",
+      "E  =  1;#  in  volts  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #res.\n",
+      "R1  =  (r*(r  -  RL))**0.5\n",
+      "R2  =  (r*RL**2/(r  -  RL))**0.5\n",
+      " #current  I1\n",
+      "I1  =  E/(r  +  R1  +  R2*RL/(RL  +  R2))\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  RL))*I1\n",
+      " #input  power\n",
+      "P1  =  r*I1**2\n",
+      " #output  power\n",
+      "P2  =  RL*I2**2\n",
+      " #attenuation\n",
+      "attn  =  10*(1/2.303)*math.log(P1/P2)\n",
+      " #Load  current,  IL\n",
+      "IL  =  E/(r  +  RL)\n",
+      " #voltage,  VL\n",
+      "VL  =  IL*RL\n",
+      " #voltage,  V1\n",
+      "V1  =  E  -  I1*r\n",
+      " #voltage,  V2\n",
+      "V2  =  V1  -  I1*R1\n",
+      " #insertion  loss\n",
+      "il  =  20*(1/2.303)*math.log(VL/V2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  R1  =  \",round(R1,2),\"  ohm  and  R2  =  \",round(R2,2),\"  ohm  \"\n",
+      "print  \"\\n  attenuation  is  \",round(attn,2),\"  dB  \"\n",
+      "print  \"\\n  In  decibels,  the  insertion  loss  is  \",round(il,2),\"  dB  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  R1  =   447.21   ohm  and  R2  =   111.8   ohm  \n",
+        "\n",
+        "  attenuation  is   12.54   dB  \n",
+        "\n",
+        "  In  decibels,  the  insertion  loss  is   9.98   dB  "
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 783</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the attenuation of each individual attenuation section, \n",
+      "#(b) the voltage output of the final stage, and (c) the voltage output of the third stage.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "attnO  =  70;#  in  dB\n",
+      "n  =  5;#  numbers  of  identical  atteneurs\n",
+      "V1  =  0.02;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #attenuation  of  each  section\n",
+      "attn  =  attnO/n\n",
+      " #output  of  the  final  stage\n",
+      "Vo  =  V1/(10**(attnO/20))\n",
+      " #voltage  output  of  the  third  stage\n",
+      "V3  =  V1/(10**(3*attn/20))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  attenuation  of  each  section  =  \",round(attn,2),\"  dB  \"\n",
+      "print  \"\\n  output  of  the  final  stage  is  \",round(Vo*1E6,2),\"uV  \"\n",
+      "print  \"\\n  voltage  output  of  the  third  stage  is  \",round(V3*1E3,3),\"mV  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  attenuation  of  each  section  =   14.0   dB  \n",
+        "\n",
+        "  output  of  the  final  stage  is   6.32 uV  \n",
+        "\n",
+        "  voltage  output  of  the  third  stage  is   0.159 mV  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 784</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a)Design a T-network attenuator pad\n",
+      "#(b)determine the fraction of the initial current \n",
+      "#(c) Determine the attenuation in decibels given by four such sections\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r = 450; # in ohm\n",
+      "R0 = 450; # in ohms\n",
+      "x = 1/8\n",
+      "\n",
+      "#calculation:\n",
+      "N = 1/x\n",
+      "R1 = R0*(N-1)/(N+1)\n",
+      "R2 = R0*2*N/(N**2 - 1)\n",
+      "\n",
+      "Io = x*x\n",
+      "\n",
+      "attn  =  20*math.log10(N)\n",
+      "attnO = 4*attn\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,2),\" ohm \"\n",
+      "print   \"and Resistance R2 is\",round(R2,0),\" ohm\"\n",
+      "print  \"\\n  (b)current flows in the load =  \",round(Io,2),\"of the original current.\"\n",
+      "print  \"\\n  (c)overall attenuation is  \",round(attnO,2),\"dB  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For a T-section symmetrical attenuator pad, Resistance R1 is 350.0  ohm \n",
+        "and Resistance R2 is 114.0  ohm\n",
+        "\n",
+        "  (b)current flows in the load =   0.02 of the original current.\n",
+        "\n",
+        "  (c)overall attenuation is   72.25 dB  \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_3.ipynb
new file mode 100755
index 00000000..4bdaf71c
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_41-checkpoint_3.ipynb
@@ -0,0 +1,1000 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c5300e3088f971e53c15dd8fc5e0190bde02f3350126276135402bbe82ffef31"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 41: Attenuators</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 763</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      " #ratio  of  output  power  to  input  power\n",
+      "rp1  =  2;\n",
+      "rp2  =  25;  \n",
+      "rp3  =  1000;\n",
+      "rp4  =  0.01;\n",
+      "\n",
+      "#calculation:\n",
+      " #power  ratio  in  decibels\n",
+      "rpd1  =  10*(1/2.303)*math.log(rp1)\n",
+      "rpd2  =  10*(1/2.303)*math.log(rp2)\n",
+      "rpd3  =  10*(1/2.303)*math.log(rp3)\n",
+      "rpd4  =  10*(1/2.303)*math.log(rp4)\n",
+      " #power  ratio  in  nepers\n",
+      "rpn1  =  (math.log(rp1))/2\n",
+      "rpn2  =  (math.log(rp2))/2\n",
+      "rpn3  =  (math.log(rp3))/2\n",
+      "rpn4  =  (math.log(rp4))/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  power  ratio  in  decibels  are  (a)\",round(rpd1,2),\"  dB  (b)\",round(rpd2,2),\"  dB \"\n",
+      "print  \"(c)  \",round(rpd3,2),\"  dB  and  (d)  \",round(rpd4,2),\"  dB\"\n",
+      "print  \"\\n  power  ratio  in  nepers  are  (a)\",round(rpn1,2),\"  Np  (b)\",round(rpn2,2),\"  Np\"\n",
+      "print   \"(c)  \",round(rpn3,2),\"  Np  and  (d)  \",round(rpn4,2),\"  Np\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  power  ratio  in  decibels  are  (a) 3.01   dB  (b) 13.98   dB \n",
+        "(c)   29.99   dB  and  (d)   -20.0   dB\n",
+        "\n",
+        "  power  ratio  in  nepers  are  (a) 0.35   Np  (b) 1.61   Np\n",
+        "(c)   3.45   Np  and  (d)   -2.3   Np\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 763</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rp  =  0.05;#  power  ratio  P2/P1\n",
+      "\n",
+      " #calculation:\n",
+      " #power  ratio  in  decibels\n",
+      "rpd  =  10*(1/2.303)*math.log(rp)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nthe  attenuation  is  \",round(abs(rpd),2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "the  attenuation  is   13.01   dB"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 764</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "gain  =  1.5;#  in  dB\n",
+      "Pi  =  0.012;#  in  Watt\n",
+      "\n",
+      "#calculation:\n",
+      " #output  power\n",
+      "Po  =  Pi*10**gain\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\noutput  power  is  \",round(Po,2),\"  W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "output  power  is   0.38   W"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 764</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "I2  =  0.05;#  in  Amperes\n",
+      "rin  =  1.32;#  in  Np\n",
+      "\n",
+      " #calculation:\n",
+      " #current  input,  I1\n",
+      "I1  =  I2*math.e**(rin)\n",
+      " #current  ratio  in  decibels\n",
+      "rid  =  20*(1/2.303)*math.log(I2/I1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncurrent  input,  I1  is  \",round(I1,2),\"  A\"\n",
+      "print  \"\\ncurrent  ratio  in  decibels  is  \",round(rid,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "current  input,  I1  is   0.19   A\n",
+        "\n",
+        "current  ratio  in  decibels  is   -11.46   dB"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 769</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ra1 = 8; # in ohms\n",
+      "Ra2 = 8; # in ohms\n",
+      "Ra3 = 21; # in ohms\n",
+      "Rb1 = 10; # in ohms\n",
+      "Rb2 = 10; # in ohms\n",
+      "Rb3 = 15; # in ohms\n",
+      "Rc1 = 200; # in ohms\n",
+      "Rc2 = 200; # in ohms\n",
+      "Rc3 = 56.25; # in ohms\n",
+      "\n",
+      "#calculation:\n",
+      "R01 = (Ra1**2 + 2*Ra2*Ra3)**0.5\n",
+      "R02 = (Rb1**2 + 2*Rb2*Rb3)**0.5\n",
+      "R03 = (Rc1**2 + 2*Rc2*Rc3)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n(a) the characteristic impedance, R0  is  \",R01,\"  ohm\"\n",
+      "print  \"\\n(b) the characteristic impedance, R0  is  \",R02,\"  ohm\"\n",
+      "print  \"\\n(c) the characteristic impedance, R0  is  \",R03,\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "(a) the characteristic impedance, R0  is   20.0   ohm\n",
+        "\n",
+        "(b) the characteristic impedance, R0  is   20.0   ohm\n",
+        "\n",
+        "(c) the characteristic impedance, R0  is   250.0   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 769</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  500;#  in  ohm\n",
+      "R2  =  1000;#  in  ohm\n",
+      "I1  =  1;#  in  ampere  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #  for  symmetrical  pi-attenuator  section\n",
+      " #characteristic  impedance,  R0\n",
+      "R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5\n",
+      " #current  Ix\n",
+      "Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R0))*Ix\n",
+      "ri  =  I1/I2;#  retio  of  currents\n",
+      " #attenuation\n",
+      "attn  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  characteristic  impedance  is  \",round(R0,2),\"  ohm\"\n",
+      "print  \"\\n  attenuation  is  \",round(attn,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  characteristic  impedance  is   447.21   ohm\n",
+        "\n",
+        "  attenuation  is   8.36   dB"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 770</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Ra1  =  15;#  in  ohm\n",
+      "Ra2  =  15;#  in  ohm\n",
+      "Ra3  =  10;#  in  ohm\n",
+      "Rb1  =  15;#  in  ohm\n",
+      "Rb2  =  5;#  in  ohm\n",
+      "Rb3  =  5;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "Roc1 = Ra1 + Ra3\n",
+      "Rsc1 = Ra1 + Ra2*Ra3/(Ra2+Ra3)\n",
+      "R01 = (Roc1*Rsc1)**0.5\n",
+      "\n",
+      "Roc2 = Rb2* (Rb1 + Rb3)/(Rb1 + Rb2 + Rb3)\n",
+      "Rsc2 = Rb2*Rb1/(Rb2+Rb1)\n",
+      "R02 = (Roc2*Rsc2)**0.5\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a) the input resistance when the output port is open-circuited  is\", Roc1,\" ohm for T-Network\"\n",
+      "print   \"and \",Roc2,\" ohm for pi-Network \"\n",
+      "print  \"\\n  (b) the input resistance when the output port is short-circuited  is,\", Rsc1,\" ohm for T-Network\"\n",
+      "print   \"and \",Rsc2,\" ohm for pi-Network \"\n",
+      "print  \"\\n  (c) the characteristic impedance.  is,\",round(R01,1),\" ohm for T-Network and ,\",round(R02,2),\" ohm for pi-Network \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a) the input resistance when the output port is open-circuited  is 25  ohm for T-Network\n",
+        "and  4.0  ohm for pi-Network \n",
+        "\n",
+        "  (b) the input resistance when the output port is short-circuited  is, 21.0  ohm for T-Network\n",
+        "and  3.75  ohm for pi-Network \n",
+        "\n",
+        "  (c) the characteristic impedance.  is, 22.9  ohm for T-Network and , 3.87  ohm for pi-Network \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 770</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vat = 20; # in db\n",
+      "R0  =  600;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "N = math.e**(Vat*2.3/20)\n",
+      "R1 = R0*(N-1)/(N+1)\n",
+      "R2 = R0*2*N/(N**2 - 1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,0),\" ohm and Resistance R2 is\",round(R2,1),\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " For a T-section symmetrical attenuator pad, Resistance R1 is 491.0  ohm and Resistance R2 is 121.5  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 771</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vat = 20; # in db\n",
+      "R0  =  600;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      "N = math.e**(Vat*2.303/20)\n",
+      "R1 = R0*(N**2 - 1)/(2*N)\n",
+      "R2 = R0*(N+1)/(N-1)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n For a pi-section symmetrical attenuator pad, Resistance R1 is\",round(R1/1000,2),\" Kohm\"\n",
+      "print   \"and Resistance R2 is\",round(R2,0),\" ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        " For a pi-section symmetrical attenuator pad, Resistance R1 is 2.97  Kohm\n",
+        "and Resistance R2 is 733.0  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 772</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "R1  =  300;#  in  ohm\n",
+      "R2  =  450;#  in  ohm\n",
+      "I1  =  1;#  in  ampere  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #the  characteristic  impedance  of  a  symmetric  T-pad  attenuator  is  given  by\n",
+      "R0  =  (R1**2  +  2*R1*R2)**0.5\n",
+      " #By  current  division\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R1+  R0))*I1\n",
+      "ri  =  I1/I2;#  ratio  of  currents\n",
+      " #insertion  loss\n",
+      "il  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  characteristic  impedance  is  \",R0,\"  ohm\"\n",
+      "print  \"\\n  insertion  loss  is  \",round(il,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  characteristic  impedance  is   600.0   ohm\n",
+        "\n",
+        "  insertion  loss  is   9.54   dB"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 773</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  500;#  in  ohm\n",
+      "Rhm  =  3000;#  in  ohm\n",
+      "RL  =  2000;#  in  ohm\n",
+      "r1  =  2000;#  in  ohm\n",
+      "r2  =  1000;#  in  ohm\n",
+      "E  =  1;#  in  volts  (lets  say)\n",
+      "\n",
+      " #calculation:\n",
+      " #Without  the  rheostat  in  the  circuit  the  voltage  across  the  2  kohm\u0018  load,  VL\n",
+      "VL  =  (RL/(RL  +  r))*E\n",
+      " #voltage  V2  with  2kohm  tapping\n",
+      "V2  =  ((RL*r1/(r1  +  RL))/((RL*r1/(r1  +  RL))  +  Rhm  -  r1  +  r))*E\n",
+      "rv1  =  VL/V2;#  ratio  of  currents\n",
+      " #insertion  loss  \n",
+      "il1  =  20*(1/2.303)*math.log(rv1)\n",
+      " #voltage  V1  with  1kohm  tapping\n",
+      "V1  =  ((RL*r2/(r2  +  RL))/((RL*r2/(r2  +  RL))  +  Rhm  -  r2  +  r))*E\n",
+      "rv2  =  VL/V1;#  ratio  of  currents\n",
+      " #insertion  loss  \n",
+      "il2  =  20*(1/2.303)*math.log(rv2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  insertion  loss  for  2kohm  tap  is  \",round(il1,2),\"  dB\"\n",
+      "print  \"\\n  insertion  loss  for  1kohm  tap  is  \",round(il2,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  insertion  loss  for  2kohm  tap  is   6.02   dB\n",
+        "\n",
+        "  insertion  loss  for  1kohm  tap  is   11.59   dB"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 774</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  500;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #characteristic  impedance  of  a  symmetrical  attenuator\n",
+      "R0  =  (R1*(R2**2)/(R1  +  2*R2))**0.5\n",
+      " #current  Ix\n",
+      "Ix  =  (R2/(R2  +  R1  +  (R2*R0/(R2  +  R0))))*I1\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  R0))*Ix\n",
+      "ri  =  I1/I2;#  retio  of  currents\n",
+      " #insertion  loss  \n",
+      "il  =  20*(1/2.303)*math.log(ri)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  characteristic  impedance  is  \",round(R0,2),\"  ohm\"\n",
+      "print  \"\\n  insertion  loss  is  \",round(il,2),\"  dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  characteristic  impedance  is   353.55   ohm\n",
+        "\n",
+        "  insertion  loss  is   15.31   dB"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 776</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  100;#  in  ohm\n",
+      "R2  =  200;#  in  ohm\n",
+      "R3  =  300;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #image  impedance  Roa\n",
+      "Roa  =  ((R1  +  R2)*(R2  +  (R1*R3/(R1  +  R3))))**0.5\n",
+      " #image  impedance  Rob\n",
+      "Rob  =  ((R1  +  R3)*(R3  +  (R1*R2/(R1  +  R2))))**0.5\n",
+      " #The  iterative  impedance  at  port  1\n",
+      "Ri1  =  (-1*R1  +  (R1**2  -  (-1*4*((R2*(R1  +  R3))  +  (R3*R1))))**0.5)/2\n",
+      " #The  iterative  impedance  at  port  2\n",
+      "Ri2  =  (R1  +  (R1**2  -  (-1*4*((R3*(R1  +  R2))  +  (R2*R1))))**0.5)/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  image  impedance  are  \",round(Roa,2),\"  ohm  and  \",round(Rob,2),\"  ohm  \"\n",
+      "print  \"\\n  iterative  impedances  are  \",round(Ri1,2),\"  ohm  and  \",round(Ri2,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  image  impedance  are   287.23   ohm  and   382.97   ohm  \n",
+        "\n",
+        "  iterative  impedances  are   285.41   ohm  and   385.41   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 777</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R1  =  1000;#  in  ohm\n",
+      "R2  =  2000;#  in  ohm\n",
+      "R3  =  3000;#  in  ohm\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #image  impedance  Roa\n",
+      "Roa  =  (((R3  +  R2)*R1/(R1  +  R2  +  R3))*(R1*R3/(R1  +  R3)))**0.5\n",
+      " #image  impedance  Rob\n",
+      "Rob  =  (((R3  +  R1)*R2/(R1  +  R2  +  R3))*(R2*R3/(R2  +  R3)))**0.5\n",
+      " #The  iterative  impedance  at  port  1\n",
+      "Ri1  =  (-1*R1  +  ((R1**2)  -  (-1*4*2*R2*R1))**0.5)/(2*2)\n",
+      " #The  iterative  impedance  at  port  2\n",
+      "Ri2  =  (R1  +  ((-1*R1)**2  -  (-1*4*2*R2*R1))**0.5)/(2*2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  image  impedance  are  \",round(Roa,2),\"  ohm  and  \",round(Rob,2),\"  ohm  \"\n",
+      "print  \"\\n  iterative  impedances  are  \",round(Ri1,2),\"  ohm  and  \",round(Ri2,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  image  impedance  are   790.57   ohm  and   1264.91   ohm  \n",
+        "\n",
+        "  iterative  impedances  are   780.78   ohm  and   1280.78   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 780</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r  =  500;#  in  ohm\n",
+      "RL  =  100;#  in  ohm\n",
+      "E  =  1;#  in  volts  (lets  say)\n",
+      "\n",
+      "#calculation:\n",
+      " #res.\n",
+      "R1  =  (r*(r  -  RL))**0.5\n",
+      "R2  =  (r*RL**2/(r  -  RL))**0.5\n",
+      " #current  I1\n",
+      "I1  =  E/(r  +  R1  +  R2*RL/(RL  +  R2))\n",
+      " #current  I2\n",
+      "I2  =  (R2/(R2  +  RL))*I1\n",
+      " #input  power\n",
+      "P1  =  r*I1**2\n",
+      " #output  power\n",
+      "P2  =  RL*I2**2\n",
+      " #attenuation\n",
+      "attn  =  10*(1/2.303)*math.log(P1/P2)\n",
+      " #Load  current,  IL\n",
+      "IL  =  E/(r  +  RL)\n",
+      " #voltage,  VL\n",
+      "VL  =  IL*RL\n",
+      " #voltage,  V1\n",
+      "V1  =  E  -  I1*r\n",
+      " #voltage,  V2\n",
+      "V2  =  V1  -  I1*R1\n",
+      " #insertion  loss\n",
+      "il  =  20*(1/2.303)*math.log(VL/V2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  R1  =  \",round(R1,2),\"  ohm  and  R2  =  \",round(R2,2),\"  ohm  \"\n",
+      "print  \"\\n  attenuation  is  \",round(attn,2),\"  dB  \"\n",
+      "print  \"\\n  In  decibels,  the  insertion  loss  is  \",round(il,2),\"  dB  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  R1  =   447.21   ohm  and  R2  =   111.8   ohm  \n",
+        "\n",
+        "  attenuation  is   12.54   dB  \n",
+        "\n",
+        "  In  decibels,  the  insertion  loss  is   9.98   dB  "
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 783</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "attnO  =  70;#  in  dB\n",
+      "n  =  5;#  numbers  of  identical  atteneurs\n",
+      "V1  =  0.02;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #attenuation  of  each  section\n",
+      "attn  =  attnO/n\n",
+      " #output  of  the  final  stage\n",
+      "Vo  =  V1/(10**(attnO/20))\n",
+      " #voltage  output  of  the  third  stage\n",
+      "V3  =  V1/(10**(3*attn/20))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  attenuation  of  each  section  =  \",round(attn,2),\"  dB  \"\n",
+      "print  \"\\n  output  of  the  final  stage  is  \",round(Vo*1E6,2),\"uV  \"\n",
+      "print  \"\\n  voltage  output  of  the  third  stage  is  \",round(V3*1E3,3),\"mV  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  attenuation  of  each  section  =   14.0   dB  \n",
+        "\n",
+        "  output  of  the  final  stage  is   6.32 uV  \n",
+        "\n",
+        "  voltage  output  of  the  third  stage  is   0.159 mV  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 784</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r = 450; # in ohm\n",
+      "R0 = 450; # in ohms\n",
+      "x = 1/8\n",
+      "\n",
+      "#calculation:\n",
+      "N = 1/x\n",
+      "R1 = R0*(N-1)/(N+1)\n",
+      "R2 = R0*2*N/(N**2 - 1)\n",
+      "\n",
+      "Io = x*x\n",
+      "\n",
+      "attn  =  20*math.log10(N)\n",
+      "attnO = 4*attn\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  (a)For a T-section symmetrical attenuator pad, Resistance R1 is\",round(R1,2),\" ohm \"\n",
+      "print   \"and Resistance R2 is\",round(R2,0),\" ohm\"\n",
+      "print  \"\\n  (b)current flows in the load =  \",round(Io,2),\"of the original current.\"\n",
+      "print  \"\\n  (c)overall attenuation is  \",round(attnO,2),\"dB  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  (a)For a T-section symmetrical attenuator pad, Resistance R1 is 350.0  ohm \n",
+        "and Resistance R2 is 114.0  ohm\n",
+        "\n",
+        "  (b)current flows in the load =   0.02 of the original current.\n",
+        "\n",
+        "  (c)overall attenuation is   72.25 dB  \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint.ipynb
new file mode 100755
index 00000000..fcac8b6a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint.ipynb
@@ -0,0 +1,959 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 42: Filter networks</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 799</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L1  =  2*100E-3;#  in  Henry\n",
+      "C1  =  0.2E-6;#  in  Fareads\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "C2  =  2*200E-12;#  in  Fareads\n",
+      "\n",
+      "#calculation:\n",
+      " #cut-off  frequency\n",
+      "fc1  =  1/(math.pi*(L1*C1)**0.5)\n",
+      " #nominal  impedance\n",
+      "R01  =  (L1/C1)**0.5\n",
+      " #cut-off  frequency\n",
+      "fc2  =  1/(math.pi*(L2*C2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R02  =  (L2/C2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc1,2),\"  Hz  and  the  nominal  impedance  is  \",round(  R01,2),\"  ohm  \"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc2,2),\"  Hz  and  the  nominal  impedance  is  \",round(  R02,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  cut-off  frequency   1591.55   Hz  and  the  nominal  impedance  is   1000.0   ohm  \n",
+        "\n",
+        "  cut-off  frequency   25164.61   Hz  and  the  nominal  impedance  is   31622.78   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 801</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) a low-pass T section filter, and (b) a low-pass \u0003 section filter to meet these requirements.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  5E6;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  1/(math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L  =  R0/(math.pi*fc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"A  low-pass  T  section  filter  capcitance  is  \",round(C*1E12,2),\"pfarad  and  inductance  is\",round(  L/2*1E6,2),\"uHenry\"\n",
+      "print  \"A  low-pass  pi  section  filter  capcitance  is  \",round(C/2*1E12,2),\"pfarad  and  inductance  is\",round(  L*1E6,2),\"uHenry\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "A  low-pass  T  section  filter  capcitance  is   106.1 pfarad  and  inductance  is 19.1 uHenry\n",
+        "A  low-pass  pi  section  filter  capcitance  is   53.05 pfarad  and  inductance  is 38.2 uHenry\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 805</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of the characteristic impedance of the section at a frequency of 90 kHz, and \n",
+      "#(b) the value of the characteristic impedance of the equivalent low-pass T section filter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  500;#  in  ohm\n",
+      "fc  =  100000;#  in  Hz\n",
+      "f  =  90000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #characteristic  impedance  of  the  pi  section\n",
+      "Zpi  =  R0/(1  -  (f/fc)**2)**0.5\n",
+      " #characteristic  impedance  of  the  T  section\n",
+      "Zt  =    R0*(1  -  (f/fc)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncharacteristic  impedance  of  the  pi  section  is  \",round(Zpi,2),\"  ohm\"\n",
+      "print  \"\\ncharacteristic  impedance  of  the  T  section  is  \",round(Zt,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "characteristic  impedance  of  the  pi  section  is   1147.08   ohm\n",
+        "\n",
+        "characteristic  impedance  of  the  T  section  is   217.94   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 806</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the frequency at which the characteristic impedance of the section is (a) 600ohm (b) 1 kohm(c) 10 kohm\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  2E6;#  in  Hz\n",
+      "Z1  =  600;#  in  ohm\n",
+      "Z2  =  1000;#  in  ohm\n",
+      "Z3  =  10000;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #frequency\n",
+      "f1  =  fc*(1  -  (R0/Z1)**2)**0.5\n",
+      "f2  =  fc*(1  -  (R0/Z2)**2)**0.5\n",
+      "f3  =  fc*(1  -  (R0/Z3)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"frequency  at  which  the  characteristic  impedance  of  the  section  is  600  ohm  is  \",f1,\"  Hz \"\n",
+      "print  \"and  1000  Ohm  is  \",f2*1E-3,\"kHz  and  10000  ohm  is  \",round(f3*1E-3,2),\"kHz  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "frequency  at  which  the  characteristic  impedance  of  the  section  is  600  ohm  is   0.0   Hz \n",
+        "and  1000  Ohm  is   1600.0 kHz  and  10000  ohm  is   1996.4 kHz  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 809</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for each of the high-pass filter sections (i) the cut-off frequency, and (ii) the nominal impedance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L1  =  100*1E-3;#  in  Henry\n",
+      "C1  =  0.2*1E-6;#  in  Fareads\n",
+      "L2  =  200*1E-6;#  in  Henry\n",
+      "C2  =  4000*1E-12;#  in  Fareads\n",
+      "\n",
+      "#calculation:\n",
+      " #cut-off  frequency\n",
+      "fc1  =  1/(4*math.pi*(L1*C1/2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R01  =  (L1*2/C1)**0.5\n",
+      " #cut-off  frequency\n",
+      "fc2  =  1/(4*math.pi*(L2*C2/2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R02  =  (L2/(C2*2))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc1,0),\"  Hz  and  the  nominal  impedance  is  \",round(  R01,0),\"  ohm\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc2/1000,0),\"KHz  and  the  nominal  impedance  is  \",round(  R02,0),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  cut-off  frequency   796.0   Hz  and  the  nominal  impedance  is   1000.0   ohm\n",
+        "\n",
+        "  cut-off  frequency   126.0 KHz  and  the  nominal  impedance  is   158.0   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 811</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) a high-pass T section filter and (b) a high-pass pi-section filter to meet these requirements\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  25000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #capacitance\n",
+      "C1  =  2/(4*math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L1  =  R0/(4*math.pi*fc)\n",
+      " #capacitance\n",
+      "C2  =  1/(4*math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L2  =  2*R0/(4*math.pi*fc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  A  low-pass  T  section  filter  capcitance  is  \",round(C1*1E9,2),\"nfarad  and  inductance  is\",round(L1*1E3,2),\"mHenry\"\n",
+      "print  \"\\n  A  high-pass  pi  section  filter  capcitance  is  \",round(C2*1E9,3),\"nfarad  and  inductance  is\",round(L2*1E3,2),\"mHenry\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  A  low-pass  T  section  filter  capcitance  is   10.61 nfarad  and  inductance  is 1.91 mHenry\n",
+        "\n",
+        "  A  high-pass  pi  section  filter  capcitance  is   5.305 nfarad  and  inductance  is 3.82 mHenry"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 814</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the frequency at which the characteristic impedance of the section is (a) zero, (b) 300 ohm, (c) 590 ohm\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  500;#  in  Hz\n",
+      "Z1  =  0;#  in  ohm\n",
+      "Z2  =  300;#  in  ohm\n",
+      "Z3  =  590;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #frequency\n",
+      "f1  =  fc\n",
+      "f2  =  fc/(1  -  (Z2/R0)**2)**0.5\n",
+      "f3  =  fc/(1  -  (Z3/R0)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"requency  at  which  the  characteristic  impedance  of  the  section  is  0  ohm  is  \",f1,\"  Hz \"\n",
+      "print  \"and  300  Ohm  is  \",round(f2,2),\"  Hz  and  590  ohm  is  \",round(f3,2),\"  Hz  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "requency  at  which  the  characteristic  impedance  of  the  section  is  0  ohm  is   500   Hz \n",
+        "and  300  Ohm  is   577.35   Hz  and  590  ohm  is   2750.1   Hz  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 817</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for each (i) the attenuation coefficient, and (ii) the phase shift coefficient.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r1  =  1.25  + 0.52j;#  propagation  coefficients  \n",
+      "rr  =  1.794;#  propagation  coefficients  \n",
+      "thetar  =  -39.4;#  in  ddegrees\n",
+      "\n",
+      "#calculation:\n",
+      " #r\n",
+      "r2  =  rr*math.cos(thetar*math.pi/180)  +  1j*rr*math.sin(thetar*math.pi/180)\n",
+      " #attenuation  coefficient\n",
+      "a1  =  r1.real\n",
+      "a2  =  r2.real\n",
+      " #phase  shift  coefficient\n",
+      "b1  =  r1.imag\n",
+      "b2  =  r2.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  are  for  (a)  is  \",a1,\"  N  and  for  (b)  is  \",round(a2,2),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  are  for  (a)  is  \",b1,\"  rad  and  for  (b)  is  \",round(b2,2),\"  rad  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  are  for  (a)  is   1.25   N  and  for  (b)  is   1.39   N  \n",
+        "\n",
+        "phase  shift  coefficient  are  for  (a)  is   0.52   rad  and  for  (b)  is   -1.14   rad  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 818</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the section (a) the attenuation coefficient, \n",
+      "#(b) the phase shift coefficient, and (c) the propagation coefficient. \n",
+      "#(d) If five such sections are cascaded determine the output current of the fifth stage and \n",
+      "#the overall propagation constant of the network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri1  =  0.024;#  in  amperes\n",
+      "ri2  =  0.008;#  in  amperes\n",
+      "thetai1  =  10;#  in  ddegrees\n",
+      "thetai2  =  -45;#  in  ddegrees\n",
+      "\n",
+      "#calculation:\n",
+      " #currents\n",
+      "I1  =  ri1*math.cos(thetai1*math.pi/180)  +  1j*ri1*math.sin(thetai1*math.pi/180)\n",
+      "I2  =  ri2*math.cos(thetai2*math.pi/180)  +  1j*ri2*math.sin(thetai2*math.pi/180)\n",
+      " #ir\n",
+      "ir  =  I1/I2\n",
+      "irmag  =  ri1/ri2\n",
+      "thetai  =  thetai1-thetai2\n",
+      " #attenuation  coefficient\n",
+      "a  =  math.log(irmag)\n",
+      " #phase  shift  coefficient\n",
+      "b  =  thetai*math.pi/180\n",
+      " #propagation  coefficient  \n",
+      "r  =  a  +  1j*b\n",
+      " #output  current  of  the  fifth  stage\n",
+      "I6  =  I1/(ir**5)\n",
+      "x  =  ir**5\n",
+      "xmg  =  abs(x)\n",
+      " #overall  attenuation  coefficient\n",
+      "ad  =  math.log(xmg)\n",
+      " #overall  phase  shift  coefficient\n",
+      "bd  =  cmath.phase(complex(x.real,x.imag))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  is  \",round(a,3),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  is  \",round(b,3),\"  rad  \"\n",
+      "print  \"\\npropagation  coefficient  is  \",round(a,3),\"  +  (\",round(b,3),\")i  \"\n",
+      "print  \"\\nthe  output  current  of  the  fifth  stage  is  \",round(abs(I6*1E6),1),\"/_\",round(cmath.phase(complex(I6.real,I6.imag))*180/math.pi,2),\"deg   mA \"\n",
+      "print  \"and  the  overall  propagation  coefficient  is  \",round(ad,2),\"  +  (\",round(bd+(2*math.pi),2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  is   1.099   N  \n",
+        "\n",
+        "phase  shift  coefficient  is   0.96   rad  \n",
+        "\n",
+        "propagation  coefficient  is   1.099   +  ( 0.96 )i  \n",
+        "\n",
+        "the  output  current  of  the  fifth  stage  is   98.8 /_ 95.0 deg   mA \n",
+        "and  the  overall  propagation  coefficient  is   5.49   +  ( 4.8 )i\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 819</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the attenuation coefficient, (b) the phase shift coefficient and (c) the propagation coefficient gamma\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "XL  =  5j;#  in  ohms\n",
+      "Xc  =  -1j*10;#  in  ohms\n",
+      "RL  =  12;#  in  ohms\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      " #calculation:\n",
+      " #current  I2\n",
+      "I2  =  (Xc/(Xc  +  XL  +  RL))*I1\n",
+      " #current  ratio\n",
+      "Ir  =  I1/I2\n",
+      "Irmg  =  abs(Ir)\n",
+      " #attenuation  coefficient\n",
+      "a  =  math.log(Irmg)\n",
+      " #phase  shift  coefficient\n",
+      "b  =  cmath.phase(complex(Ir.real, Ir.imag))\n",
+      " #propagation  coefficient  \n",
+      "r  =  a  +  1j*b\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  is  \",round(a,2),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  is  \",round(b,2),\"  rad  \"\n",
+      "print  \"\\npropagation  coefficient  is  \",round(a,2),\"  +  (\",round(b,2),\")i  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  is   0.26   N  \n",
+        "\n",
+        "phase  shift  coefficient  is   1.18   rad  \n",
+        "\n",
+        "propagation  coefficient  is   0.26   +  ( 1.18 )i  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 823</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) the time delay for the signal to pass through the filter, assuming the phase shift is small, and \n",
+      "#(b) the time delay for a signal to pass through the section at the cut-off frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  2*0.5;#  in  Henry\n",
+      "C  =  2E-9;#  in  Farad\n",
+      "\n",
+      "#calculation:\n",
+      " #time  delay\n",
+      "t  =  (L*C)**0.5\n",
+      " #time  delay  at  the  cut-off  frequency\n",
+      "tfc  =  t*math.pi/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  time  delay  is  \",round(t*1E6,2),\"usec  \"\n",
+      "print  \"\\ntime  delay  at  the  cut-off  frequency  is  \",round(tfc*1E6,2),\"usec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  time  delay  is   44.72 usec  \n",
+        "\n",
+        "time  delay  at  the  cut-off  frequency  is   70.25 usec"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 824</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the values of the elements in each section, and (b) the value of n.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fc  =  500000;#  in  Hz\n",
+      "t1  =  9.55E-6;#  in  secs\n",
+      "R0  =  1000;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #for  a  low-pass  filter  section,  capacitance\n",
+      "C  =  1/(math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L  =  R0/(math.pi*fc)\n",
+      " #time  delay\n",
+      "t2  =  (L*C)**0.5\n",
+      " #number  of  cascaded  sections  required\n",
+      "n  =  t1/t2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for  low-pass  T  section  inductance  is  \",round(L/2*1E6,2),\"uH  and  capacitance  is  \",round(C*1E12,2),\"pF\"\n",
+      "print  \"\\n  for  low-pass  pi  section  inductance  is  \",round(L*1E6,2),\"uH  and  capacitance  is  \",round(C/2*1E12,2),\"pF\"\n",
+      "print  \"\\nnumber  of  cascaded  sections  required  is  \",round(n,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for  low-pass  T  section  inductance  is   318.31 uH  and  capacitance  is   636.62 pF\n",
+        "\n",
+        "  for  low-pass  pi  section  inductance  is   636.62 uH  and  capacitance  is   318.31 pF\n",
+        "\n",
+        "number  of  cascaded  sections  required  is   15.0"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 824</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the component values for each section if the filter is (a) a low-pass T network, and (b) a high-pass \u0003 network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "n  =  8;#  sections  in  cascade\n",
+      "R0  =  1000;#  in  ohm\n",
+      "t1  =  4E-6;#  in  secs\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      " #time  delay\n",
+      "t2  =  t1/n\n",
+      " #capacitance\n",
+      "C  =  t2/R0\n",
+      " #inductance\n",
+      "L  =  t2*R0\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for  low-pass  T  section  inductance  is  \",L/2*1E6,\"uH  and  capacitance  is  \",C*1E12,\"pF\"\n",
+      "print  \"\\n  for  high-pass  pi  section  inductance  is  \",2*L*1E6,\"uH  and  capacitance  is  \",C*1E12,\"pF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for  low-pass  T  section  inductance  is   250.0 uH  and  capacitance  is   500.0 pF\n",
+        "\n",
+        "  for  high-pass  pi  section  inductance  is   1000.0 uH  and  capacitance  is   500.0 pF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 829</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) an appropriate \u2018mderived\u2019 T section, and (b) an appropriate \u2018m-derived\u2019 pi section.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc = 5000; # in Hz\n",
+      "finf = 5500; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (fc/finf)**2)**0.5\n",
+      "C = 1/(math.pi*R0*fc)\n",
+      "L = R0/(math.pi*fc)\n",
+      "\n",
+      "LT = m*L/2\n",
+      "CT = m*C\n",
+      "Ls = (1- m**2)*L/(4*m)\n",
+      "\n",
+      "Cpi = m*C/2\n",
+      "Lpi = m*L\n",
+      "Cp = (1- (m**2))*C/(4*m)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for mderived  T  section  inductance  is  \",round(Ls*1000,2),\"mH  and  capacitance  is  \",round(CT*1E9,2),\"nF\"\n",
+      "print  \"\\n  for mderived  pi  section  inductance  is  \",round(Lpi*1000,2),\"mH  and  capacitance  is  \",round(Cp*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for mderived  T  section  inductance  is   18.94 mH  and  capacitance  is   44.2 nF\n",
+        "\n",
+        "  for mderived  pi  section  inductance  is   15.91 mH  and  capacitance  is   52.62 nF"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 832</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) a suitable \u2018m-derived\u2019 T section, and (b) a suitable \u2018m-derived\u2019 pi section having a cut-off frequency of 20 kHz,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  500;#  in  ohm\n",
+      "fc = 20000; # in Hz\n",
+      "finf = 16000; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (finf/fc)**2)**0.5\n",
+      "C = 1/(4*math.pi*R0*fc)\n",
+      "L = R0/(4*math.pi*fc)\n",
+      "\n",
+      "LT = L/m\n",
+      "CT = 4*m*C/(1- m**2)\n",
+      "Csa = 2*C/m\n",
+      "\n",
+      "Cpi = C/m\n",
+      "Lpi = 4*m*L/(1- m**2)\n",
+      "Lsa = 2*L/m\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  For an 'm-derived' high-pass T section: series arm contains a capacitance of  \",round(Csa*1E9,2),\"nF\"\n",
+      "print  \"the shunt arm contains an inductance of\",round(LT*1000,3),\" mH  in series with a capacitance of\",round(CT*1E9,2),\"nF\"\n",
+      "print  \"\\n  For an 'm-derived' high pass pi section: shunt arms each contain inductance of  \",round(Lsa*1000,2),\"mH\"\n",
+      "print  \"series arm contains a capacitance of  \",round(Cpi*1E9,2),\"nF in parallel with an inductance of\",round(Lpi*1E3,3),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  For an 'm-derived' high-pass T section: series arm contains a capacitance of   26.53 nF\n",
+        "the shunt arm contains an inductance of 3.316  mH  in series with a capacitance of 29.84 nF\n",
+        "\n",
+        "  For an 'm-derived' high pass pi section: shunt arms each contain inductance of   6.63 mH\n",
+        "series arm contains a capacitance of   13.26 nF in parallel with an inductance of 7.46 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 835</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the component values needed if the filter is to comprise a prototype T section, \n",
+      "#an \u2018m-derived\u2019 T section and two terminating \u2018m-derived\u2019 halfsections.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc = 10000; # in Hz\n",
+      "finf = 11800; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (fc/finf)**2)**0.5\n",
+      "C = 1/(math.pi*R0*fc)\n",
+      "L = R0/(math.pi*fc)\n",
+      "\n",
+      "LmT = (1- m**2)*L/(4*m)\n",
+      "\n",
+      "mH = 0.6\n",
+      "LmH = (1- mH**2)*L/(2*mH)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  For an Prototype T section: series arm contains a Inductance of  \",round(L*1000/2,1),\"mH\"\n",
+      "print   \"the shunt arm contains an Capacitance of\",round(C*1E6,4),\" uF\"\n",
+      "print  \"\\n  For an 'm-derived' T section: Series arms each contain inductance of  \",round(m*L*1000/2,2),\"mH \"\n",
+      "print   \"Shunt arm contains a capacitance of  \",round(m*C*1E6,4),\"uF in Series with an inductance of\",round(LmT*1E3,2),\"mH\"\n",
+      "print  \"\\n  For an 'm-derived' Half section: Series arms each contain inductance of  \",round(mH*L*1000/2,1),\"mH\"\n",
+      "print   \"Shunt arm contains a capacitance of  \",round(mH*C*1E6/2,4),\"uF in Series with an inductance of\",round(LmH*1E3,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  For an Prototype T section: series arm contains a Inductance of   9.5 mH\n",
+        "the shunt arm contains an Capacitance of 0.0531  uF\n",
+        "\n",
+        "  For an 'm-derived' T section: Series arms each contain inductance of   5.07 mH \n",
+        "Shunt arm contains a capacitance of   0.0282 uF in Series with an inductance of 6.46 mH\n",
+        "\n",
+        "  For an 'm-derived' Half section: Series arms each contain inductance of   5.7 mH\n",
+        "Shunt arm contains a capacitance of   0.0159 uF in Series with an inductance of 10.19 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_1.ipynb
new file mode 100755
index 00000000..fcac8b6a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_1.ipynb
@@ -0,0 +1,959 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 42: Filter networks</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 799</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L1  =  2*100E-3;#  in  Henry\n",
+      "C1  =  0.2E-6;#  in  Fareads\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "C2  =  2*200E-12;#  in  Fareads\n",
+      "\n",
+      "#calculation:\n",
+      " #cut-off  frequency\n",
+      "fc1  =  1/(math.pi*(L1*C1)**0.5)\n",
+      " #nominal  impedance\n",
+      "R01  =  (L1/C1)**0.5\n",
+      " #cut-off  frequency\n",
+      "fc2  =  1/(math.pi*(L2*C2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R02  =  (L2/C2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc1,2),\"  Hz  and  the  nominal  impedance  is  \",round(  R01,2),\"  ohm  \"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc2,2),\"  Hz  and  the  nominal  impedance  is  \",round(  R02,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  cut-off  frequency   1591.55   Hz  and  the  nominal  impedance  is   1000.0   ohm  \n",
+        "\n",
+        "  cut-off  frequency   25164.61   Hz  and  the  nominal  impedance  is   31622.78   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 801</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) a low-pass T section filter, and (b) a low-pass \u0003 section filter to meet these requirements.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  5E6;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  1/(math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L  =  R0/(math.pi*fc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"A  low-pass  T  section  filter  capcitance  is  \",round(C*1E12,2),\"pfarad  and  inductance  is\",round(  L/2*1E6,2),\"uHenry\"\n",
+      "print  \"A  low-pass  pi  section  filter  capcitance  is  \",round(C/2*1E12,2),\"pfarad  and  inductance  is\",round(  L*1E6,2),\"uHenry\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "A  low-pass  T  section  filter  capcitance  is   106.1 pfarad  and  inductance  is 19.1 uHenry\n",
+        "A  low-pass  pi  section  filter  capcitance  is   53.05 pfarad  and  inductance  is 38.2 uHenry\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 805</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of the characteristic impedance of the section at a frequency of 90 kHz, and \n",
+      "#(b) the value of the characteristic impedance of the equivalent low-pass T section filter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  500;#  in  ohm\n",
+      "fc  =  100000;#  in  Hz\n",
+      "f  =  90000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #characteristic  impedance  of  the  pi  section\n",
+      "Zpi  =  R0/(1  -  (f/fc)**2)**0.5\n",
+      " #characteristic  impedance  of  the  T  section\n",
+      "Zt  =    R0*(1  -  (f/fc)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncharacteristic  impedance  of  the  pi  section  is  \",round(Zpi,2),\"  ohm\"\n",
+      "print  \"\\ncharacteristic  impedance  of  the  T  section  is  \",round(Zt,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "characteristic  impedance  of  the  pi  section  is   1147.08   ohm\n",
+        "\n",
+        "characteristic  impedance  of  the  T  section  is   217.94   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 806</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the frequency at which the characteristic impedance of the section is (a) 600ohm (b) 1 kohm(c) 10 kohm\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  2E6;#  in  Hz\n",
+      "Z1  =  600;#  in  ohm\n",
+      "Z2  =  1000;#  in  ohm\n",
+      "Z3  =  10000;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #frequency\n",
+      "f1  =  fc*(1  -  (R0/Z1)**2)**0.5\n",
+      "f2  =  fc*(1  -  (R0/Z2)**2)**0.5\n",
+      "f3  =  fc*(1  -  (R0/Z3)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"frequency  at  which  the  characteristic  impedance  of  the  section  is  600  ohm  is  \",f1,\"  Hz \"\n",
+      "print  \"and  1000  Ohm  is  \",f2*1E-3,\"kHz  and  10000  ohm  is  \",round(f3*1E-3,2),\"kHz  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "frequency  at  which  the  characteristic  impedance  of  the  section  is  600  ohm  is   0.0   Hz \n",
+        "and  1000  Ohm  is   1600.0 kHz  and  10000  ohm  is   1996.4 kHz  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 809</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for each of the high-pass filter sections (i) the cut-off frequency, and (ii) the nominal impedance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L1  =  100*1E-3;#  in  Henry\n",
+      "C1  =  0.2*1E-6;#  in  Fareads\n",
+      "L2  =  200*1E-6;#  in  Henry\n",
+      "C2  =  4000*1E-12;#  in  Fareads\n",
+      "\n",
+      "#calculation:\n",
+      " #cut-off  frequency\n",
+      "fc1  =  1/(4*math.pi*(L1*C1/2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R01  =  (L1*2/C1)**0.5\n",
+      " #cut-off  frequency\n",
+      "fc2  =  1/(4*math.pi*(L2*C2/2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R02  =  (L2/(C2*2))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc1,0),\"  Hz  and  the  nominal  impedance  is  \",round(  R01,0),\"  ohm\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc2/1000,0),\"KHz  and  the  nominal  impedance  is  \",round(  R02,0),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  cut-off  frequency   796.0   Hz  and  the  nominal  impedance  is   1000.0   ohm\n",
+        "\n",
+        "  cut-off  frequency   126.0 KHz  and  the  nominal  impedance  is   158.0   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 811</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) a high-pass T section filter and (b) a high-pass pi-section filter to meet these requirements\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  25000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #capacitance\n",
+      "C1  =  2/(4*math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L1  =  R0/(4*math.pi*fc)\n",
+      " #capacitance\n",
+      "C2  =  1/(4*math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L2  =  2*R0/(4*math.pi*fc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  A  low-pass  T  section  filter  capcitance  is  \",round(C1*1E9,2),\"nfarad  and  inductance  is\",round(L1*1E3,2),\"mHenry\"\n",
+      "print  \"\\n  A  high-pass  pi  section  filter  capcitance  is  \",round(C2*1E9,3),\"nfarad  and  inductance  is\",round(L2*1E3,2),\"mHenry\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  A  low-pass  T  section  filter  capcitance  is   10.61 nfarad  and  inductance  is 1.91 mHenry\n",
+        "\n",
+        "  A  high-pass  pi  section  filter  capcitance  is   5.305 nfarad  and  inductance  is 3.82 mHenry"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 814</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the frequency at which the characteristic impedance of the section is (a) zero, (b) 300 ohm, (c) 590 ohm\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  500;#  in  Hz\n",
+      "Z1  =  0;#  in  ohm\n",
+      "Z2  =  300;#  in  ohm\n",
+      "Z3  =  590;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #frequency\n",
+      "f1  =  fc\n",
+      "f2  =  fc/(1  -  (Z2/R0)**2)**0.5\n",
+      "f3  =  fc/(1  -  (Z3/R0)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"requency  at  which  the  characteristic  impedance  of  the  section  is  0  ohm  is  \",f1,\"  Hz \"\n",
+      "print  \"and  300  Ohm  is  \",round(f2,2),\"  Hz  and  590  ohm  is  \",round(f3,2),\"  Hz  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "requency  at  which  the  characteristic  impedance  of  the  section  is  0  ohm  is   500   Hz \n",
+        "and  300  Ohm  is   577.35   Hz  and  590  ohm  is   2750.1   Hz  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 817</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for each (i) the attenuation coefficient, and (ii) the phase shift coefficient.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r1  =  1.25  + 0.52j;#  propagation  coefficients  \n",
+      "rr  =  1.794;#  propagation  coefficients  \n",
+      "thetar  =  -39.4;#  in  ddegrees\n",
+      "\n",
+      "#calculation:\n",
+      " #r\n",
+      "r2  =  rr*math.cos(thetar*math.pi/180)  +  1j*rr*math.sin(thetar*math.pi/180)\n",
+      " #attenuation  coefficient\n",
+      "a1  =  r1.real\n",
+      "a2  =  r2.real\n",
+      " #phase  shift  coefficient\n",
+      "b1  =  r1.imag\n",
+      "b2  =  r2.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  are  for  (a)  is  \",a1,\"  N  and  for  (b)  is  \",round(a2,2),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  are  for  (a)  is  \",b1,\"  rad  and  for  (b)  is  \",round(b2,2),\"  rad  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  are  for  (a)  is   1.25   N  and  for  (b)  is   1.39   N  \n",
+        "\n",
+        "phase  shift  coefficient  are  for  (a)  is   0.52   rad  and  for  (b)  is   -1.14   rad  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 818</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the section (a) the attenuation coefficient, \n",
+      "#(b) the phase shift coefficient, and (c) the propagation coefficient. \n",
+      "#(d) If five such sections are cascaded determine the output current of the fifth stage and \n",
+      "#the overall propagation constant of the network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri1  =  0.024;#  in  amperes\n",
+      "ri2  =  0.008;#  in  amperes\n",
+      "thetai1  =  10;#  in  ddegrees\n",
+      "thetai2  =  -45;#  in  ddegrees\n",
+      "\n",
+      "#calculation:\n",
+      " #currents\n",
+      "I1  =  ri1*math.cos(thetai1*math.pi/180)  +  1j*ri1*math.sin(thetai1*math.pi/180)\n",
+      "I2  =  ri2*math.cos(thetai2*math.pi/180)  +  1j*ri2*math.sin(thetai2*math.pi/180)\n",
+      " #ir\n",
+      "ir  =  I1/I2\n",
+      "irmag  =  ri1/ri2\n",
+      "thetai  =  thetai1-thetai2\n",
+      " #attenuation  coefficient\n",
+      "a  =  math.log(irmag)\n",
+      " #phase  shift  coefficient\n",
+      "b  =  thetai*math.pi/180\n",
+      " #propagation  coefficient  \n",
+      "r  =  a  +  1j*b\n",
+      " #output  current  of  the  fifth  stage\n",
+      "I6  =  I1/(ir**5)\n",
+      "x  =  ir**5\n",
+      "xmg  =  abs(x)\n",
+      " #overall  attenuation  coefficient\n",
+      "ad  =  math.log(xmg)\n",
+      " #overall  phase  shift  coefficient\n",
+      "bd  =  cmath.phase(complex(x.real,x.imag))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  is  \",round(a,3),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  is  \",round(b,3),\"  rad  \"\n",
+      "print  \"\\npropagation  coefficient  is  \",round(a,3),\"  +  (\",round(b,3),\")i  \"\n",
+      "print  \"\\nthe  output  current  of  the  fifth  stage  is  \",round(abs(I6*1E6),1),\"/_\",round(cmath.phase(complex(I6.real,I6.imag))*180/math.pi,2),\"deg   mA \"\n",
+      "print  \"and  the  overall  propagation  coefficient  is  \",round(ad,2),\"  +  (\",round(bd+(2*math.pi),2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  is   1.099   N  \n",
+        "\n",
+        "phase  shift  coefficient  is   0.96   rad  \n",
+        "\n",
+        "propagation  coefficient  is   1.099   +  ( 0.96 )i  \n",
+        "\n",
+        "the  output  current  of  the  fifth  stage  is   98.8 /_ 95.0 deg   mA \n",
+        "and  the  overall  propagation  coefficient  is   5.49   +  ( 4.8 )i\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 819</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the attenuation coefficient, (b) the phase shift coefficient and (c) the propagation coefficient gamma\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "XL  =  5j;#  in  ohms\n",
+      "Xc  =  -1j*10;#  in  ohms\n",
+      "RL  =  12;#  in  ohms\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      " #calculation:\n",
+      " #current  I2\n",
+      "I2  =  (Xc/(Xc  +  XL  +  RL))*I1\n",
+      " #current  ratio\n",
+      "Ir  =  I1/I2\n",
+      "Irmg  =  abs(Ir)\n",
+      " #attenuation  coefficient\n",
+      "a  =  math.log(Irmg)\n",
+      " #phase  shift  coefficient\n",
+      "b  =  cmath.phase(complex(Ir.real, Ir.imag))\n",
+      " #propagation  coefficient  \n",
+      "r  =  a  +  1j*b\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  is  \",round(a,2),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  is  \",round(b,2),\"  rad  \"\n",
+      "print  \"\\npropagation  coefficient  is  \",round(a,2),\"  +  (\",round(b,2),\")i  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  is   0.26   N  \n",
+        "\n",
+        "phase  shift  coefficient  is   1.18   rad  \n",
+        "\n",
+        "propagation  coefficient  is   0.26   +  ( 1.18 )i  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 823</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) the time delay for the signal to pass through the filter, assuming the phase shift is small, and \n",
+      "#(b) the time delay for a signal to pass through the section at the cut-off frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  2*0.5;#  in  Henry\n",
+      "C  =  2E-9;#  in  Farad\n",
+      "\n",
+      "#calculation:\n",
+      " #time  delay\n",
+      "t  =  (L*C)**0.5\n",
+      " #time  delay  at  the  cut-off  frequency\n",
+      "tfc  =  t*math.pi/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  time  delay  is  \",round(t*1E6,2),\"usec  \"\n",
+      "print  \"\\ntime  delay  at  the  cut-off  frequency  is  \",round(tfc*1E6,2),\"usec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  time  delay  is   44.72 usec  \n",
+        "\n",
+        "time  delay  at  the  cut-off  frequency  is   70.25 usec"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 824</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the values of the elements in each section, and (b) the value of n.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fc  =  500000;#  in  Hz\n",
+      "t1  =  9.55E-6;#  in  secs\n",
+      "R0  =  1000;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #for  a  low-pass  filter  section,  capacitance\n",
+      "C  =  1/(math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L  =  R0/(math.pi*fc)\n",
+      " #time  delay\n",
+      "t2  =  (L*C)**0.5\n",
+      " #number  of  cascaded  sections  required\n",
+      "n  =  t1/t2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for  low-pass  T  section  inductance  is  \",round(L/2*1E6,2),\"uH  and  capacitance  is  \",round(C*1E12,2),\"pF\"\n",
+      "print  \"\\n  for  low-pass  pi  section  inductance  is  \",round(L*1E6,2),\"uH  and  capacitance  is  \",round(C/2*1E12,2),\"pF\"\n",
+      "print  \"\\nnumber  of  cascaded  sections  required  is  \",round(n,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for  low-pass  T  section  inductance  is   318.31 uH  and  capacitance  is   636.62 pF\n",
+        "\n",
+        "  for  low-pass  pi  section  inductance  is   636.62 uH  and  capacitance  is   318.31 pF\n",
+        "\n",
+        "number  of  cascaded  sections  required  is   15.0"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 824</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the component values for each section if the filter is (a) a low-pass T network, and (b) a high-pass \u0003 network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "n  =  8;#  sections  in  cascade\n",
+      "R0  =  1000;#  in  ohm\n",
+      "t1  =  4E-6;#  in  secs\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      " #time  delay\n",
+      "t2  =  t1/n\n",
+      " #capacitance\n",
+      "C  =  t2/R0\n",
+      " #inductance\n",
+      "L  =  t2*R0\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for  low-pass  T  section  inductance  is  \",L/2*1E6,\"uH  and  capacitance  is  \",C*1E12,\"pF\"\n",
+      "print  \"\\n  for  high-pass  pi  section  inductance  is  \",2*L*1E6,\"uH  and  capacitance  is  \",C*1E12,\"pF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for  low-pass  T  section  inductance  is   250.0 uH  and  capacitance  is   500.0 pF\n",
+        "\n",
+        "  for  high-pass  pi  section  inductance  is   1000.0 uH  and  capacitance  is   500.0 pF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 829</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) an appropriate \u2018mderived\u2019 T section, and (b) an appropriate \u2018m-derived\u2019 pi section.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc = 5000; # in Hz\n",
+      "finf = 5500; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (fc/finf)**2)**0.5\n",
+      "C = 1/(math.pi*R0*fc)\n",
+      "L = R0/(math.pi*fc)\n",
+      "\n",
+      "LT = m*L/2\n",
+      "CT = m*C\n",
+      "Ls = (1- m**2)*L/(4*m)\n",
+      "\n",
+      "Cpi = m*C/2\n",
+      "Lpi = m*L\n",
+      "Cp = (1- (m**2))*C/(4*m)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for mderived  T  section  inductance  is  \",round(Ls*1000,2),\"mH  and  capacitance  is  \",round(CT*1E9,2),\"nF\"\n",
+      "print  \"\\n  for mderived  pi  section  inductance  is  \",round(Lpi*1000,2),\"mH  and  capacitance  is  \",round(Cp*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for mderived  T  section  inductance  is   18.94 mH  and  capacitance  is   44.2 nF\n",
+        "\n",
+        "  for mderived  pi  section  inductance  is   15.91 mH  and  capacitance  is   52.62 nF"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 832</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) a suitable \u2018m-derived\u2019 T section, and (b) a suitable \u2018m-derived\u2019 pi section having a cut-off frequency of 20 kHz,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  500;#  in  ohm\n",
+      "fc = 20000; # in Hz\n",
+      "finf = 16000; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (finf/fc)**2)**0.5\n",
+      "C = 1/(4*math.pi*R0*fc)\n",
+      "L = R0/(4*math.pi*fc)\n",
+      "\n",
+      "LT = L/m\n",
+      "CT = 4*m*C/(1- m**2)\n",
+      "Csa = 2*C/m\n",
+      "\n",
+      "Cpi = C/m\n",
+      "Lpi = 4*m*L/(1- m**2)\n",
+      "Lsa = 2*L/m\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  For an 'm-derived' high-pass T section: series arm contains a capacitance of  \",round(Csa*1E9,2),\"nF\"\n",
+      "print  \"the shunt arm contains an inductance of\",round(LT*1000,3),\" mH  in series with a capacitance of\",round(CT*1E9,2),\"nF\"\n",
+      "print  \"\\n  For an 'm-derived' high pass pi section: shunt arms each contain inductance of  \",round(Lsa*1000,2),\"mH\"\n",
+      "print  \"series arm contains a capacitance of  \",round(Cpi*1E9,2),\"nF in parallel with an inductance of\",round(Lpi*1E3,3),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  For an 'm-derived' high-pass T section: series arm contains a capacitance of   26.53 nF\n",
+        "the shunt arm contains an inductance of 3.316  mH  in series with a capacitance of 29.84 nF\n",
+        "\n",
+        "  For an 'm-derived' high pass pi section: shunt arms each contain inductance of   6.63 mH\n",
+        "series arm contains a capacitance of   13.26 nF in parallel with an inductance of 7.46 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 835</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the component values needed if the filter is to comprise a prototype T section, \n",
+      "#an \u2018m-derived\u2019 T section and two terminating \u2018m-derived\u2019 halfsections.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc = 10000; # in Hz\n",
+      "finf = 11800; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (fc/finf)**2)**0.5\n",
+      "C = 1/(math.pi*R0*fc)\n",
+      "L = R0/(math.pi*fc)\n",
+      "\n",
+      "LmT = (1- m**2)*L/(4*m)\n",
+      "\n",
+      "mH = 0.6\n",
+      "LmH = (1- mH**2)*L/(2*mH)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  For an Prototype T section: series arm contains a Inductance of  \",round(L*1000/2,1),\"mH\"\n",
+      "print   \"the shunt arm contains an Capacitance of\",round(C*1E6,4),\" uF\"\n",
+      "print  \"\\n  For an 'm-derived' T section: Series arms each contain inductance of  \",round(m*L*1000/2,2),\"mH \"\n",
+      "print   \"Shunt arm contains a capacitance of  \",round(m*C*1E6,4),\"uF in Series with an inductance of\",round(LmT*1E3,2),\"mH\"\n",
+      "print  \"\\n  For an 'm-derived' Half section: Series arms each contain inductance of  \",round(mH*L*1000/2,1),\"mH\"\n",
+      "print   \"Shunt arm contains a capacitance of  \",round(mH*C*1E6/2,4),\"uF in Series with an inductance of\",round(LmH*1E3,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  For an Prototype T section: series arm contains a Inductance of   9.5 mH\n",
+        "the shunt arm contains an Capacitance of 0.0531  uF\n",
+        "\n",
+        "  For an 'm-derived' T section: Series arms each contain inductance of   5.07 mH \n",
+        "Shunt arm contains a capacitance of   0.0282 uF in Series with an inductance of 6.46 mH\n",
+        "\n",
+        "  For an 'm-derived' Half section: Series arms each contain inductance of   5.7 mH\n",
+        "Shunt arm contains a capacitance of   0.0159 uF in Series with an inductance of 10.19 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_2.ipynb
new file mode 100755
index 00000000..fcac8b6a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_2.ipynb
@@ -0,0 +1,959 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 42: Filter networks</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 799</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L1  =  2*100E-3;#  in  Henry\n",
+      "C1  =  0.2E-6;#  in  Fareads\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "C2  =  2*200E-12;#  in  Fareads\n",
+      "\n",
+      "#calculation:\n",
+      " #cut-off  frequency\n",
+      "fc1  =  1/(math.pi*(L1*C1)**0.5)\n",
+      " #nominal  impedance\n",
+      "R01  =  (L1/C1)**0.5\n",
+      " #cut-off  frequency\n",
+      "fc2  =  1/(math.pi*(L2*C2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R02  =  (L2/C2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc1,2),\"  Hz  and  the  nominal  impedance  is  \",round(  R01,2),\"  ohm  \"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc2,2),\"  Hz  and  the  nominal  impedance  is  \",round(  R02,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  cut-off  frequency   1591.55   Hz  and  the  nominal  impedance  is   1000.0   ohm  \n",
+        "\n",
+        "  cut-off  frequency   25164.61   Hz  and  the  nominal  impedance  is   31622.78   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 801</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) a low-pass T section filter, and (b) a low-pass \u0003 section filter to meet these requirements.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  5E6;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  1/(math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L  =  R0/(math.pi*fc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"A  low-pass  T  section  filter  capcitance  is  \",round(C*1E12,2),\"pfarad  and  inductance  is\",round(  L/2*1E6,2),\"uHenry\"\n",
+      "print  \"A  low-pass  pi  section  filter  capcitance  is  \",round(C/2*1E12,2),\"pfarad  and  inductance  is\",round(  L*1E6,2),\"uHenry\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "A  low-pass  T  section  filter  capcitance  is   106.1 pfarad  and  inductance  is 19.1 uHenry\n",
+        "A  low-pass  pi  section  filter  capcitance  is   53.05 pfarad  and  inductance  is 38.2 uHenry\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 805</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the value of the characteristic impedance of the section at a frequency of 90 kHz, and \n",
+      "#(b) the value of the characteristic impedance of the equivalent low-pass T section filter.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  500;#  in  ohm\n",
+      "fc  =  100000;#  in  Hz\n",
+      "f  =  90000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #characteristic  impedance  of  the  pi  section\n",
+      "Zpi  =  R0/(1  -  (f/fc)**2)**0.5\n",
+      " #characteristic  impedance  of  the  T  section\n",
+      "Zt  =    R0*(1  -  (f/fc)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncharacteristic  impedance  of  the  pi  section  is  \",round(Zpi,2),\"  ohm\"\n",
+      "print  \"\\ncharacteristic  impedance  of  the  T  section  is  \",round(Zt,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "characteristic  impedance  of  the  pi  section  is   1147.08   ohm\n",
+        "\n",
+        "characteristic  impedance  of  the  T  section  is   217.94   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 806</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the frequency at which the characteristic impedance of the section is (a) 600ohm (b) 1 kohm(c) 10 kohm\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  2E6;#  in  Hz\n",
+      "Z1  =  600;#  in  ohm\n",
+      "Z2  =  1000;#  in  ohm\n",
+      "Z3  =  10000;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #frequency\n",
+      "f1  =  fc*(1  -  (R0/Z1)**2)**0.5\n",
+      "f2  =  fc*(1  -  (R0/Z2)**2)**0.5\n",
+      "f3  =  fc*(1  -  (R0/Z3)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"frequency  at  which  the  characteristic  impedance  of  the  section  is  600  ohm  is  \",f1,\"  Hz \"\n",
+      "print  \"and  1000  Ohm  is  \",f2*1E-3,\"kHz  and  10000  ohm  is  \",round(f3*1E-3,2),\"kHz  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "frequency  at  which  the  characteristic  impedance  of  the  section  is  600  ohm  is   0.0   Hz \n",
+        "and  1000  Ohm  is   1600.0 kHz  and  10000  ohm  is   1996.4 kHz  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 809</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for each of the high-pass filter sections (i) the cut-off frequency, and (ii) the nominal impedance\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L1  =  100*1E-3;#  in  Henry\n",
+      "C1  =  0.2*1E-6;#  in  Fareads\n",
+      "L2  =  200*1E-6;#  in  Henry\n",
+      "C2  =  4000*1E-12;#  in  Fareads\n",
+      "\n",
+      "#calculation:\n",
+      " #cut-off  frequency\n",
+      "fc1  =  1/(4*math.pi*(L1*C1/2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R01  =  (L1*2/C1)**0.5\n",
+      " #cut-off  frequency\n",
+      "fc2  =  1/(4*math.pi*(L2*C2/2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R02  =  (L2/(C2*2))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc1,0),\"  Hz  and  the  nominal  impedance  is  \",round(  R01,0),\"  ohm\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc2/1000,0),\"KHz  and  the  nominal  impedance  is  \",round(  R02,0),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  cut-off  frequency   796.0   Hz  and  the  nominal  impedance  is   1000.0   ohm\n",
+        "\n",
+        "  cut-off  frequency   126.0 KHz  and  the  nominal  impedance  is   158.0   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 811</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) a high-pass T section filter and (b) a high-pass pi-section filter to meet these requirements\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  25000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #capacitance\n",
+      "C1  =  2/(4*math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L1  =  R0/(4*math.pi*fc)\n",
+      " #capacitance\n",
+      "C2  =  1/(4*math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L2  =  2*R0/(4*math.pi*fc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  A  low-pass  T  section  filter  capcitance  is  \",round(C1*1E9,2),\"nfarad  and  inductance  is\",round(L1*1E3,2),\"mHenry\"\n",
+      "print  \"\\n  A  high-pass  pi  section  filter  capcitance  is  \",round(C2*1E9,3),\"nfarad  and  inductance  is\",round(L2*1E3,2),\"mHenry\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  A  low-pass  T  section  filter  capcitance  is   10.61 nfarad  and  inductance  is 1.91 mHenry\n",
+        "\n",
+        "  A  high-pass  pi  section  filter  capcitance  is   5.305 nfarad  and  inductance  is 3.82 mHenry"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 814</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the frequency at which the characteristic impedance of the section is (a) zero, (b) 300 ohm, (c) 590 ohm\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  500;#  in  Hz\n",
+      "Z1  =  0;#  in  ohm\n",
+      "Z2  =  300;#  in  ohm\n",
+      "Z3  =  590;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #frequency\n",
+      "f1  =  fc\n",
+      "f2  =  fc/(1  -  (Z2/R0)**2)**0.5\n",
+      "f3  =  fc/(1  -  (Z3/R0)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"requency  at  which  the  characteristic  impedance  of  the  section  is  0  ohm  is  \",f1,\"  Hz \"\n",
+      "print  \"and  300  Ohm  is  \",round(f2,2),\"  Hz  and  590  ohm  is  \",round(f3,2),\"  Hz  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "requency  at  which  the  characteristic  impedance  of  the  section  is  0  ohm  is   500   Hz \n",
+        "and  300  Ohm  is   577.35   Hz  and  590  ohm  is   2750.1   Hz  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 817</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for each (i) the attenuation coefficient, and (ii) the phase shift coefficient.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r1  =  1.25  + 0.52j;#  propagation  coefficients  \n",
+      "rr  =  1.794;#  propagation  coefficients  \n",
+      "thetar  =  -39.4;#  in  ddegrees\n",
+      "\n",
+      "#calculation:\n",
+      " #r\n",
+      "r2  =  rr*math.cos(thetar*math.pi/180)  +  1j*rr*math.sin(thetar*math.pi/180)\n",
+      " #attenuation  coefficient\n",
+      "a1  =  r1.real\n",
+      "a2  =  r2.real\n",
+      " #phase  shift  coefficient\n",
+      "b1  =  r1.imag\n",
+      "b2  =  r2.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  are  for  (a)  is  \",a1,\"  N  and  for  (b)  is  \",round(a2,2),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  are  for  (a)  is  \",b1,\"  rad  and  for  (b)  is  \",round(b2,2),\"  rad  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  are  for  (a)  is   1.25   N  and  for  (b)  is   1.39   N  \n",
+        "\n",
+        "phase  shift  coefficient  are  for  (a)  is   0.52   rad  and  for  (b)  is   -1.14   rad  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 818</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the section (a) the attenuation coefficient, \n",
+      "#(b) the phase shift coefficient, and (c) the propagation coefficient. \n",
+      "#(d) If five such sections are cascaded determine the output current of the fifth stage and \n",
+      "#the overall propagation constant of the network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri1  =  0.024;#  in  amperes\n",
+      "ri2  =  0.008;#  in  amperes\n",
+      "thetai1  =  10;#  in  ddegrees\n",
+      "thetai2  =  -45;#  in  ddegrees\n",
+      "\n",
+      "#calculation:\n",
+      " #currents\n",
+      "I1  =  ri1*math.cos(thetai1*math.pi/180)  +  1j*ri1*math.sin(thetai1*math.pi/180)\n",
+      "I2  =  ri2*math.cos(thetai2*math.pi/180)  +  1j*ri2*math.sin(thetai2*math.pi/180)\n",
+      " #ir\n",
+      "ir  =  I1/I2\n",
+      "irmag  =  ri1/ri2\n",
+      "thetai  =  thetai1-thetai2\n",
+      " #attenuation  coefficient\n",
+      "a  =  math.log(irmag)\n",
+      " #phase  shift  coefficient\n",
+      "b  =  thetai*math.pi/180\n",
+      " #propagation  coefficient  \n",
+      "r  =  a  +  1j*b\n",
+      " #output  current  of  the  fifth  stage\n",
+      "I6  =  I1/(ir**5)\n",
+      "x  =  ir**5\n",
+      "xmg  =  abs(x)\n",
+      " #overall  attenuation  coefficient\n",
+      "ad  =  math.log(xmg)\n",
+      " #overall  phase  shift  coefficient\n",
+      "bd  =  cmath.phase(complex(x.real,x.imag))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  is  \",round(a,3),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  is  \",round(b,3),\"  rad  \"\n",
+      "print  \"\\npropagation  coefficient  is  \",round(a,3),\"  +  (\",round(b,3),\")i  \"\n",
+      "print  \"\\nthe  output  current  of  the  fifth  stage  is  \",round(abs(I6*1E6),1),\"/_\",round(cmath.phase(complex(I6.real,I6.imag))*180/math.pi,2),\"deg   mA \"\n",
+      "print  \"and  the  overall  propagation  coefficient  is  \",round(ad,2),\"  +  (\",round(bd+(2*math.pi),2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  is   1.099   N  \n",
+        "\n",
+        "phase  shift  coefficient  is   0.96   rad  \n",
+        "\n",
+        "propagation  coefficient  is   1.099   +  ( 0.96 )i  \n",
+        "\n",
+        "the  output  current  of  the  fifth  stage  is   98.8 /_ 95.0 deg   mA \n",
+        "and  the  overall  propagation  coefficient  is   5.49   +  ( 4.8 )i\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 819</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the attenuation coefficient, (b) the phase shift coefficient and (c) the propagation coefficient gamma\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "XL  =  5j;#  in  ohms\n",
+      "Xc  =  -1j*10;#  in  ohms\n",
+      "RL  =  12;#  in  ohms\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      " #calculation:\n",
+      " #current  I2\n",
+      "I2  =  (Xc/(Xc  +  XL  +  RL))*I1\n",
+      " #current  ratio\n",
+      "Ir  =  I1/I2\n",
+      "Irmg  =  abs(Ir)\n",
+      " #attenuation  coefficient\n",
+      "a  =  math.log(Irmg)\n",
+      " #phase  shift  coefficient\n",
+      "b  =  cmath.phase(complex(Ir.real, Ir.imag))\n",
+      " #propagation  coefficient  \n",
+      "r  =  a  +  1j*b\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  is  \",round(a,2),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  is  \",round(b,2),\"  rad  \"\n",
+      "print  \"\\npropagation  coefficient  is  \",round(a,2),\"  +  (\",round(b,2),\")i  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  is   0.26   N  \n",
+        "\n",
+        "phase  shift  coefficient  is   1.18   rad  \n",
+        "\n",
+        "propagation  coefficient  is   0.26   +  ( 1.18 )i  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 823</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) the time delay for the signal to pass through the filter, assuming the phase shift is small, and \n",
+      "#(b) the time delay for a signal to pass through the section at the cut-off frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  2*0.5;#  in  Henry\n",
+      "C  =  2E-9;#  in  Farad\n",
+      "\n",
+      "#calculation:\n",
+      " #time  delay\n",
+      "t  =  (L*C)**0.5\n",
+      " #time  delay  at  the  cut-off  frequency\n",
+      "tfc  =  t*math.pi/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  time  delay  is  \",round(t*1E6,2),\"usec  \"\n",
+      "print  \"\\ntime  delay  at  the  cut-off  frequency  is  \",round(tfc*1E6,2),\"usec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  time  delay  is   44.72 usec  \n",
+        "\n",
+        "time  delay  at  the  cut-off  frequency  is   70.25 usec"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 824</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the values of the elements in each section, and (b) the value of n.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fc  =  500000;#  in  Hz\n",
+      "t1  =  9.55E-6;#  in  secs\n",
+      "R0  =  1000;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #for  a  low-pass  filter  section,  capacitance\n",
+      "C  =  1/(math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L  =  R0/(math.pi*fc)\n",
+      " #time  delay\n",
+      "t2  =  (L*C)**0.5\n",
+      " #number  of  cascaded  sections  required\n",
+      "n  =  t1/t2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for  low-pass  T  section  inductance  is  \",round(L/2*1E6,2),\"uH  and  capacitance  is  \",round(C*1E12,2),\"pF\"\n",
+      "print  \"\\n  for  low-pass  pi  section  inductance  is  \",round(L*1E6,2),\"uH  and  capacitance  is  \",round(C/2*1E12,2),\"pF\"\n",
+      "print  \"\\nnumber  of  cascaded  sections  required  is  \",round(n,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for  low-pass  T  section  inductance  is   318.31 uH  and  capacitance  is   636.62 pF\n",
+        "\n",
+        "  for  low-pass  pi  section  inductance  is   636.62 uH  and  capacitance  is   318.31 pF\n",
+        "\n",
+        "number  of  cascaded  sections  required  is   15.0"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 824</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the component values for each section if the filter is (a) a low-pass T network, and (b) a high-pass \u0003 network.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "n  =  8;#  sections  in  cascade\n",
+      "R0  =  1000;#  in  ohm\n",
+      "t1  =  4E-6;#  in  secs\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      " #time  delay\n",
+      "t2  =  t1/n\n",
+      " #capacitance\n",
+      "C  =  t2/R0\n",
+      " #inductance\n",
+      "L  =  t2*R0\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for  low-pass  T  section  inductance  is  \",L/2*1E6,\"uH  and  capacitance  is  \",C*1E12,\"pF\"\n",
+      "print  \"\\n  for  high-pass  pi  section  inductance  is  \",2*L*1E6,\"uH  and  capacitance  is  \",C*1E12,\"pF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for  low-pass  T  section  inductance  is   250.0 uH  and  capacitance  is   500.0 pF\n",
+        "\n",
+        "  for  high-pass  pi  section  inductance  is   1000.0 uH  and  capacitance  is   500.0 pF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 829</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) an appropriate \u2018mderived\u2019 T section, and (b) an appropriate \u2018m-derived\u2019 pi section.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc = 5000; # in Hz\n",
+      "finf = 5500; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (fc/finf)**2)**0.5\n",
+      "C = 1/(math.pi*R0*fc)\n",
+      "L = R0/(math.pi*fc)\n",
+      "\n",
+      "LT = m*L/2\n",
+      "CT = m*C\n",
+      "Ls = (1- m**2)*L/(4*m)\n",
+      "\n",
+      "Cpi = m*C/2\n",
+      "Lpi = m*L\n",
+      "Cp = (1- (m**2))*C/(4*m)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for mderived  T  section  inductance  is  \",round(Ls*1000,2),\"mH  and  capacitance  is  \",round(CT*1E9,2),\"nF\"\n",
+      "print  \"\\n  for mderived  pi  section  inductance  is  \",round(Lpi*1000,2),\"mH  and  capacitance  is  \",round(Cp*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for mderived  T  section  inductance  is   18.94 mH  and  capacitance  is   44.2 nF\n",
+        "\n",
+        "  for mderived  pi  section  inductance  is   15.91 mH  and  capacitance  is   52.62 nF"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 832</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Design (a) a suitable \u2018m-derived\u2019 T section, and (b) a suitable \u2018m-derived\u2019 pi section having a cut-off frequency of 20 kHz,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  500;#  in  ohm\n",
+      "fc = 20000; # in Hz\n",
+      "finf = 16000; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (finf/fc)**2)**0.5\n",
+      "C = 1/(4*math.pi*R0*fc)\n",
+      "L = R0/(4*math.pi*fc)\n",
+      "\n",
+      "LT = L/m\n",
+      "CT = 4*m*C/(1- m**2)\n",
+      "Csa = 2*C/m\n",
+      "\n",
+      "Cpi = C/m\n",
+      "Lpi = 4*m*L/(1- m**2)\n",
+      "Lsa = 2*L/m\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  For an 'm-derived' high-pass T section: series arm contains a capacitance of  \",round(Csa*1E9,2),\"nF\"\n",
+      "print  \"the shunt arm contains an inductance of\",round(LT*1000,3),\" mH  in series with a capacitance of\",round(CT*1E9,2),\"nF\"\n",
+      "print  \"\\n  For an 'm-derived' high pass pi section: shunt arms each contain inductance of  \",round(Lsa*1000,2),\"mH\"\n",
+      "print  \"series arm contains a capacitance of  \",round(Cpi*1E9,2),\"nF in parallel with an inductance of\",round(Lpi*1E3,3),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  For an 'm-derived' high-pass T section: series arm contains a capacitance of   26.53 nF\n",
+        "the shunt arm contains an inductance of 3.316  mH  in series with a capacitance of 29.84 nF\n",
+        "\n",
+        "  For an 'm-derived' high pass pi section: shunt arms each contain inductance of   6.63 mH\n",
+        "series arm contains a capacitance of   13.26 nF in parallel with an inductance of 7.46 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 835</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the component values needed if the filter is to comprise a prototype T section, \n",
+      "#an \u2018m-derived\u2019 T section and two terminating \u2018m-derived\u2019 halfsections.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc = 10000; # in Hz\n",
+      "finf = 11800; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (fc/finf)**2)**0.5\n",
+      "C = 1/(math.pi*R0*fc)\n",
+      "L = R0/(math.pi*fc)\n",
+      "\n",
+      "LmT = (1- m**2)*L/(4*m)\n",
+      "\n",
+      "mH = 0.6\n",
+      "LmH = (1- mH**2)*L/(2*mH)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  For an Prototype T section: series arm contains a Inductance of  \",round(L*1000/2,1),\"mH\"\n",
+      "print   \"the shunt arm contains an Capacitance of\",round(C*1E6,4),\" uF\"\n",
+      "print  \"\\n  For an 'm-derived' T section: Series arms each contain inductance of  \",round(m*L*1000/2,2),\"mH \"\n",
+      "print   \"Shunt arm contains a capacitance of  \",round(m*C*1E6,4),\"uF in Series with an inductance of\",round(LmT*1E3,2),\"mH\"\n",
+      "print  \"\\n  For an 'm-derived' Half section: Series arms each contain inductance of  \",round(mH*L*1000/2,1),\"mH\"\n",
+      "print   \"Shunt arm contains a capacitance of  \",round(mH*C*1E6/2,4),\"uF in Series with an inductance of\",round(LmH*1E3,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  For an Prototype T section: series arm contains a Inductance of   9.5 mH\n",
+        "the shunt arm contains an Capacitance of 0.0531  uF\n",
+        "\n",
+        "  For an 'm-derived' T section: Series arms each contain inductance of   5.07 mH \n",
+        "Shunt arm contains a capacitance of   0.0282 uF in Series with an inductance of 6.46 mH\n",
+        "\n",
+        "  For an 'm-derived' Half section: Series arms each contain inductance of   5.7 mH\n",
+        "Shunt arm contains a capacitance of   0.0159 uF in Series with an inductance of 10.19 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_3.ipynb
new file mode 100755
index 00000000..d475997f
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_3.ipynb
@@ -0,0 +1,954 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:7d58b74694fc58900e8e749aa15ae2b47c466e340431e05727905322512e69ef"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 42: Filter networks</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 799</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L1  =  2*100E-3;#  in  Henry\n",
+      "C1  =  0.2E-6;#  in  Fareads\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "C2  =  2*200E-12;#  in  Fareads\n",
+      "\n",
+      "#calculation:\n",
+      " #cut-off  frequency\n",
+      "fc1  =  1/(math.pi*(L1*C1)**0.5)\n",
+      " #nominal  impedance\n",
+      "R01  =  (L1/C1)**0.5\n",
+      " #cut-off  frequency\n",
+      "fc2  =  1/(math.pi*(L2*C2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R02  =  (L2/C2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc1,2),\"  Hz  and  the  nominal  impedance  is  \",round(  R01,2),\"  ohm  \"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc2,2),\"  Hz  and  the  nominal  impedance  is  \",round(  R02,2),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  cut-off  frequency   1591.55   Hz  and  the  nominal  impedance  is   1000.0   ohm  \n",
+        "\n",
+        "  cut-off  frequency   25164.61   Hz  and  the  nominal  impedance  is   31622.78   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 801</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  5E6;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitance\n",
+      "C  =  1/(math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L  =  R0/(math.pi*fc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"A  low-pass  T  section  filter  capcitance  is  \",round(C*1E12,2),\"pfarad  and  inductance  is\",round(  L/2*1E6,2),\"uHenry\"\n",
+      "print  \"A  low-pass  pi  section  filter  capcitance  is  \",round(C/2*1E12,2),\"pfarad  and  inductance  is\",round(  L*1E6,2),\"uHenry\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "A  low-pass  T  section  filter  capcitance  is   106.1 pfarad  and  inductance  is 19.1 uHenry\n",
+        "A  low-pass  pi  section  filter  capcitance  is   53.05 pfarad  and  inductance  is 38.2 uHenry\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 805</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  500;#  in  ohm\n",
+      "fc  =  100000;#  in  Hz\n",
+      "f  =  90000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #characteristic  impedance  of  the  pi  section\n",
+      "Zpi  =  R0/(1  -  (f/fc)**2)**0.5\n",
+      " #characteristic  impedance  of  the  T  section\n",
+      "Zt  =    R0*(1  -  (f/fc)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\ncharacteristic  impedance  of  the  pi  section  is  \",round(Zpi,2),\"  ohm\"\n",
+      "print  \"\\ncharacteristic  impedance  of  the  T  section  is  \",round(Zt,2),\"  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "characteristic  impedance  of  the  pi  section  is   1147.08   ohm\n",
+        "\n",
+        "characteristic  impedance  of  the  T  section  is   217.94   ohm"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 806</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  2E6;#  in  Hz\n",
+      "Z1  =  600;#  in  ohm\n",
+      "Z2  =  1000;#  in  ohm\n",
+      "Z3  =  10000;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #frequency\n",
+      "f1  =  fc*(1  -  (R0/Z1)**2)**0.5\n",
+      "f2  =  fc*(1  -  (R0/Z2)**2)**0.5\n",
+      "f3  =  fc*(1  -  (R0/Z3)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"frequency  at  which  the  characteristic  impedance  of  the  section  is  600  ohm  is  \",f1,\"  Hz \"\n",
+      "print  \"and  1000  Ohm  is  \",f2*1E-3,\"kHz  and  10000  ohm  is  \",round(f3*1E-3,2),\"kHz  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "frequency  at  which  the  characteristic  impedance  of  the  section  is  600  ohm  is   0.0   Hz \n",
+        "and  1000  Ohm  is   1600.0 kHz  and  10000  ohm  is   1996.4 kHz  "
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 809</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L1  =  100*1E-3;#  in  Henry\n",
+      "C1  =  0.2*1E-6;#  in  Fareads\n",
+      "L2  =  200*1E-6;#  in  Henry\n",
+      "C2  =  4000*1E-12;#  in  Fareads\n",
+      "\n",
+      "#calculation:\n",
+      " #cut-off  frequency\n",
+      "fc1  =  1/(4*math.pi*(L1*C1/2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R01  =  (L1*2/C1)**0.5\n",
+      " #cut-off  frequency\n",
+      "fc2  =  1/(4*math.pi*(L2*C2/2)**0.5)\n",
+      " #nominal  impedance\n",
+      "R02  =  (L2/(C2*2))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc1,0),\"  Hz  and  the  nominal  impedance  is  \",round(  R01,0),\"  ohm\"\n",
+      "print  \"\\n  cut-off  frequency  \",round(fc2/1000,0),\"KHz  and  the  nominal  impedance  is  \",round(  R02,0),\"  ohm  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  cut-off  frequency   796.0   Hz  and  the  nominal  impedance  is   1000.0   ohm\n",
+        "\n",
+        "  cut-off  frequency   126.0 KHz  and  the  nominal  impedance  is   158.0   ohm  "
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 811</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  25000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      " #capacitance\n",
+      "C1  =  2/(4*math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L1  =  R0/(4*math.pi*fc)\n",
+      " #capacitance\n",
+      "C2  =  1/(4*math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L2  =  2*R0/(4*math.pi*fc)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  A  low-pass  T  section  filter  capcitance  is  \",round(C1*1E9,2),\"nfarad  and  inductance  is\",round(L1*1E3,2),\"mHenry\"\n",
+      "print  \"\\n  A  high-pass  pi  section  filter  capcitance  is  \",round(C2*1E9,3),\"nfarad  and  inductance  is\",round(L2*1E3,2),\"mHenry\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  A  low-pass  T  section  filter  capcitance  is   10.61 nfarad  and  inductance  is 1.91 mHenry\n",
+        "\n",
+        "  A  high-pass  pi  section  filter  capcitance  is   5.305 nfarad  and  inductance  is 3.82 mHenry"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 814</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc  =  500;#  in  Hz\n",
+      "Z1  =  0;#  in  ohm\n",
+      "Z2  =  300;#  in  ohm\n",
+      "Z3  =  590;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #frequency\n",
+      "f1  =  fc\n",
+      "f2  =  fc/(1  -  (Z2/R0)**2)**0.5\n",
+      "f3  =  fc/(1  -  (Z3/R0)**2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"requency  at  which  the  characteristic  impedance  of  the  section  is  0  ohm  is  \",f1,\"  Hz \"\n",
+      "print  \"and  300  Ohm  is  \",round(f2,2),\"  Hz  and  590  ohm  is  \",round(f3,2),\"  Hz  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "requency  at  which  the  characteristic  impedance  of  the  section  is  0  ohm  is   500   Hz \n",
+        "and  300  Ohm  is   577.35   Hz  and  590  ohm  is   2750.1   Hz  "
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 817</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "r1  =  1.25  + 0.52j;#  propagation  coefficients  \n",
+      "rr  =  1.794;#  propagation  coefficients  \n",
+      "thetar  =  -39.4;#  in  ddegrees\n",
+      "\n",
+      "#calculation:\n",
+      " #r\n",
+      "r2  =  rr*math.cos(thetar*math.pi/180)  +  1j*rr*math.sin(thetar*math.pi/180)\n",
+      " #attenuation  coefficient\n",
+      "a1  =  r1.real\n",
+      "a2  =  r2.real\n",
+      " #phase  shift  coefficient\n",
+      "b1  =  r1.imag\n",
+      "b2  =  r2.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  are  for  (a)  is  \",a1,\"  N  and  for  (b)  is  \",round(a2,2),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  are  for  (a)  is  \",b1,\"  rad  and  for  (b)  is  \",round(b2,2),\"  rad  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  are  for  (a)  is   1.25   N  and  for  (b)  is   1.39   N  \n",
+        "\n",
+        "phase  shift  coefficient  are  for  (a)  is   0.52   rad  and  for  (b)  is   -1.14   rad  "
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 818</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "ri1  =  0.024;#  in  amperes\n",
+      "ri2  =  0.008;#  in  amperes\n",
+      "thetai1  =  10;#  in  ddegrees\n",
+      "thetai2  =  -45;#  in  ddegrees\n",
+      "\n",
+      "#calculation:\n",
+      " #currents\n",
+      "I1  =  ri1*math.cos(thetai1*math.pi/180)  +  1j*ri1*math.sin(thetai1*math.pi/180)\n",
+      "I2  =  ri2*math.cos(thetai2*math.pi/180)  +  1j*ri2*math.sin(thetai2*math.pi/180)\n",
+      " #ir\n",
+      "ir  =  I1/I2\n",
+      "irmag  =  ri1/ri2\n",
+      "thetai  =  thetai1-thetai2\n",
+      " #attenuation  coefficient\n",
+      "a  =  math.log(irmag)\n",
+      " #phase  shift  coefficient\n",
+      "b  =  thetai*math.pi/180\n",
+      " #propagation  coefficient  \n",
+      "r  =  a  +  1j*b\n",
+      " #output  current  of  the  fifth  stage\n",
+      "I6  =  I1/(ir**5)\n",
+      "x  =  ir**5\n",
+      "xmg  =  abs(x)\n",
+      " #overall  attenuation  coefficient\n",
+      "ad  =  math.log(xmg)\n",
+      " #overall  phase  shift  coefficient\n",
+      "bd  =  cmath.phase(complex(x.real,x.imag))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  is  \",round(a,3),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  is  \",round(b,3),\"  rad  \"\n",
+      "print  \"\\npropagation  coefficient  is  \",round(a,3),\"  +  (\",round(b,3),\")i  \"\n",
+      "print  \"\\nthe  output  current  of  the  fifth  stage  is  \",round(abs(I6*1E6),1),\"/_\",round(cmath.phase(complex(I6.real,I6.imag))*180/math.pi,2),\"deg   mA \"\n",
+      "print  \"and  the  overall  propagation  coefficient  is  \",round(ad,2),\"  +  (\",round(bd+(2*math.pi),2),\")i\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  is   1.099   N  \n",
+        "\n",
+        "phase  shift  coefficient  is   0.96   rad  \n",
+        "\n",
+        "propagation  coefficient  is   1.099   +  ( 0.96 )i  \n",
+        "\n",
+        "the  output  current  of  the  fifth  stage  is   98.8 /_ 95.0 deg   mA \n",
+        "and  the  overall  propagation  coefficient  is   5.49   +  ( 4.8 )i\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 819</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "XL  =  5j;#  in  ohms\n",
+      "Xc  =  -1j*10;#  in  ohms\n",
+      "RL  =  12;#  in  ohms\n",
+      "I1  =  1;#  in  amperes  (lets  say)\n",
+      "\n",
+      " #calculation:\n",
+      " #current  I2\n",
+      "I2  =  (Xc/(Xc  +  XL  +  RL))*I1\n",
+      " #current  ratio\n",
+      "Ir  =  I1/I2\n",
+      "Irmg  =  abs(Ir)\n",
+      " #attenuation  coefficient\n",
+      "a  =  math.log(Irmg)\n",
+      " #phase  shift  coefficient\n",
+      "b  =  cmath.phase(complex(Ir.real, Ir.imag))\n",
+      " #propagation  coefficient  \n",
+      "r  =  a  +  1j*b\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nattenuation  coefficient  is  \",round(a,2),\"  N  \"\n",
+      "print  \"\\nphase  shift  coefficient  is  \",round(b,2),\"  rad  \"\n",
+      "print  \"\\npropagation  coefficient  is  \",round(a,2),\"  +  (\",round(b,2),\")i  \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "attenuation  coefficient  is   0.26   N  \n",
+        "\n",
+        "phase  shift  coefficient  is   1.18   rad  \n",
+        "\n",
+        "propagation  coefficient  is   0.26   +  ( 1.18 )i  "
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 823</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "L  =  2*0.5;#  in  Henry\n",
+      "C  =  2E-9;#  in  Farad\n",
+      "\n",
+      "#calculation:\n",
+      " #time  delay\n",
+      "t  =  (L*C)**0.5\n",
+      " #time  delay  at  the  cut-off  frequency\n",
+      "tfc  =  t*math.pi/2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  time  delay  is  \",round(t*1E6,2),\"usec  \"\n",
+      "print  \"\\ntime  delay  at  the  cut-off  frequency  is  \",round(tfc*1E6,2),\"usec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  time  delay  is   44.72 usec  \n",
+        "\n",
+        "time  delay  at  the  cut-off  frequency  is   70.25 usec"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 824</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "#initializing  the  variables:\n",
+      "fc  =  500000;#  in  Hz\n",
+      "t1  =  9.55E-6;#  in  secs\n",
+      "R0  =  1000;#  in  ohm\n",
+      "\n",
+      "#calculation:\n",
+      " #for  a  low-pass  filter  section,  capacitance\n",
+      "C  =  1/(math.pi*R0*fc)\n",
+      " #inductance\n",
+      "L  =  R0/(math.pi*fc)\n",
+      " #time  delay\n",
+      "t2  =  (L*C)**0.5\n",
+      " #number  of  cascaded  sections  required\n",
+      "n  =  t1/t2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for  low-pass  T  section  inductance  is  \",round(L/2*1E6,2),\"uH  and  capacitance  is  \",round(C*1E12,2),\"pF\"\n",
+      "print  \"\\n  for  low-pass  pi  section  inductance  is  \",round(L*1E6,2),\"uH  and  capacitance  is  \",round(C/2*1E12,2),\"pF\"\n",
+      "print  \"\\nnumber  of  cascaded  sections  required  is  \",round(n,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for  low-pass  T  section  inductance  is   318.31 uH  and  capacitance  is   636.62 pF\n",
+        "\n",
+        "  for  low-pass  pi  section  inductance  is   636.62 uH  and  capacitance  is   318.31 pF\n",
+        "\n",
+        "number  of  cascaded  sections  required  is   15.0"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 824</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "n  =  8;#  sections  in  cascade\n",
+      "R0  =  1000;#  in  ohm\n",
+      "t1  =  4E-6;#  in  secs\n",
+      "\n",
+      "\n",
+      "#calculation:\n",
+      " #time  delay\n",
+      "t2  =  t1/n\n",
+      " #capacitance\n",
+      "C  =  t2/R0\n",
+      " #inductance\n",
+      "L  =  t2*R0\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for  low-pass  T  section  inductance  is  \",L/2*1E6,\"uH  and  capacitance  is  \",C*1E12,\"pF\"\n",
+      "print  \"\\n  for  high-pass  pi  section  inductance  is  \",2*L*1E6,\"uH  and  capacitance  is  \",C*1E12,\"pF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for  low-pass  T  section  inductance  is   250.0 uH  and  capacitance  is   500.0 pF\n",
+        "\n",
+        "  for  high-pass  pi  section  inductance  is   1000.0 uH  and  capacitance  is   500.0 pF"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 829</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc = 5000; # in Hz\n",
+      "finf = 5500; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (fc/finf)**2)**0.5\n",
+      "C = 1/(math.pi*R0*fc)\n",
+      "L = R0/(math.pi*fc)\n",
+      "\n",
+      "LT = m*L/2\n",
+      "CT = m*C\n",
+      "Ls = (1- m**2)*L/(4*m)\n",
+      "\n",
+      "Cpi = m*C/2\n",
+      "Lpi = m*L\n",
+      "Cp = (1- (m**2))*C/(4*m)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  for mderived  T  section  inductance  is  \",round(Ls*1000,2),\"mH  and  capacitance  is  \",round(CT*1E9,2),\"nF\"\n",
+      "print  \"\\n  for mderived  pi  section  inductance  is  \",round(Lpi*1000,2),\"mH  and  capacitance  is  \",round(Cp*1E9,2),\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  for mderived  T  section  inductance  is   18.94 mH  and  capacitance  is   44.2 nF\n",
+        "\n",
+        "  for mderived  pi  section  inductance  is   15.91 mH  and  capacitance  is   52.62 nF"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 832</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  500;#  in  ohm\n",
+      "fc = 20000; # in Hz\n",
+      "finf = 16000; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (finf/fc)**2)**0.5\n",
+      "C = 1/(4*math.pi*R0*fc)\n",
+      "L = R0/(4*math.pi*fc)\n",
+      "\n",
+      "LT = L/m\n",
+      "CT = 4*m*C/(1- m**2)\n",
+      "Csa = 2*C/m\n",
+      "\n",
+      "Cpi = C/m\n",
+      "Lpi = 4*m*L/(1- m**2)\n",
+      "Lsa = 2*L/m\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  For an 'm-derived' high-pass T section: series arm contains a capacitance of  \",round(Csa*1E9,2),\"nF\"\n",
+      "print  \"the shunt arm contains an inductance of\",round(LT*1000,3),\" mH  in series with a capacitance of\",round(CT*1E9,2),\"nF\"\n",
+      "print  \"\\n  For an 'm-derived' high pass pi section: shunt arms each contain inductance of  \",round(Lsa*1000,2),\"mH\"\n",
+      "print  \"series arm contains a capacitance of  \",round(Cpi*1E9,2),\"nF in parallel with an inductance of\",round(Lpi*1E3,3),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  For an 'm-derived' high-pass T section: series arm contains a capacitance of   26.53 nF\n",
+        "the shunt arm contains an inductance of 3.316  mH  in series with a capacitance of 29.84 nF\n",
+        "\n",
+        "  For an 'm-derived' high pass pi section: shunt arms each contain inductance of   6.63 mH\n",
+        "series arm contains a capacitance of   13.26 nF in parallel with an inductance of 7.46 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 835</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R0  =  600;#  in  ohm\n",
+      "fc = 10000; # in Hz\n",
+      "finf = 11800; #in Hz\n",
+      "\n",
+      "#calculation:\n",
+      "m = (1 - (fc/finf)**2)**0.5\n",
+      "C = 1/(math.pi*R0*fc)\n",
+      "L = R0/(math.pi*fc)\n",
+      "\n",
+      "LmT = (1- m**2)*L/(4*m)\n",
+      "\n",
+      "mH = 0.6\n",
+      "LmH = (1- mH**2)*L/(2*mH)\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  For an Prototype T section: series arm contains a Inductance of  \",round(L*1000/2,1),\"mH\"\n",
+      "print   \"the shunt arm contains an Capacitance of\",round(C*1E6,4),\" uF\"\n",
+      "print  \"\\n  For an 'm-derived' T section: Series arms each contain inductance of  \",round(m*L*1000/2,2),\"mH \"\n",
+      "print   \"Shunt arm contains a capacitance of  \",round(m*C*1E6,4),\"uF in Series with an inductance of\",round(LmT*1E3,2),\"mH\"\n",
+      "print  \"\\n  For an 'm-derived' Half section: Series arms each contain inductance of  \",round(mH*L*1000/2,1),\"mH\"\n",
+      "print   \"Shunt arm contains a capacitance of  \",round(mH*C*1E6/2,4),\"uF in Series with an inductance of\",round(LmH*1E3,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  For an Prototype T section: series arm contains a Inductance of   9.5 mH\n",
+        "the shunt arm contains an Capacitance of 0.0531  uF\n",
+        "\n",
+        "  For an 'm-derived' T section: Series arms each contain inductance of   5.07 mH \n",
+        "Shunt arm contains a capacitance of   0.0282 uF in Series with an inductance of 6.46 mH\n",
+        "\n",
+        "  For an 'm-derived' Half section: Series arms each contain inductance of   5.7 mH\n",
+        "Shunt arm contains a capacitance of   0.0159 uF in Series with an inductance of 10.19 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb
new file mode 100755
index 00000000..618fd99a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb
@@ -0,0 +1,1037 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 43: Magnetically coupled circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 842</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the self inductance of coil A, and (b) the mutual inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Na  =  1200;  \n",
+      "Nb  =  1000;\n",
+      "Ia  =  0.8;#  in  amperes\n",
+      "Phia  =  100E-6;#  in  Wb\n",
+      "xb  =  0.75;\n",
+      "\n",
+      " #calculation:\n",
+      " #self  inductance  of  coil  A\n",
+      "La  =  Na*Phia/Ia\n",
+      " #mutual  inductance,  M\n",
+      "Phib  =  xb*Phia\n",
+      "M  =  Nb*Phib/Ia\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  self  inductance  of  coil  A  is  \",La,\"  H\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M,\"H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  self  inductance  of  coil  A  is   0.15   H\n",
+        "\n",
+        "  mutual  inductance,  M  is   0.09375 H"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 843</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. induced in the secondary,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "M  =  600E-3;#  in  Henry\n",
+      "Ia  =  5;#  in  amperes\n",
+      "dt  =  0.2;#  in  secs\n",
+      "\n",
+      " #calculation:\n",
+      " #change  of  current\n",
+      "dIa  =  2*Ia  \n",
+      "dIadt  =  dIa/dt\n",
+      " #secondary  induced  e.m.f.,  E2\n",
+      "E2  =  -1*M*dIadt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  induced  e.m.f.,  E2  is  \",E2,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  induced  e.m.f.,  E2  is   -30.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 844</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnetic coupling coefficient of the pair of coils\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "La  =  250E-3;#  in  Henry\n",
+      "Lb  =  400E-3;#  in  Henry\n",
+      "M  =  80E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  coupling  coefficient,  is   0.253"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 845</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the mutual inductance between the coils, and (b) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Lx  =  80E-3;#  in  Henry\n",
+      "Ly  =  60E-3;#  in  Henry\n",
+      "Nx  =  200;#  turns\n",
+      "Ny  =  100;#  turns\n",
+      "Ix  =  4;#  in  Amperes\n",
+      "Phiy  =  0.005;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #mutual  inductance,  M\n",
+      "M  =  Ny*Phiy/(2*Ix)\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(Lx*Ly)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M*1E3,\"mH\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   62.5 mH\n",
+        "\n",
+        "  coupling  coefficient,  is   0.902"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 846</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the mutual inductance between the two coils, and (b) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "La  =  40E-3;#  in  Henry\n",
+      "Lb  =  10E-3;#  in  Henry\n",
+      "L  =  60E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L  -  La  -  Lb)/2\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M*1E3,\"mH\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",k"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   5.0 mH\n",
+        "\n",
+        "  coupling  coefficient,  is   0.25"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 847</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the mutual inductance between the coils and (b) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "Ra  =  5;#  in  Ohm\n",
+      "La  =  1;#  in  Henry\n",
+      "Rb  =  10;#  in  Ohm\n",
+      "Lb  =  5;#  in  Henry\n",
+      "I  =  8;#  in  amperes\n",
+      "dIdt  =  15;#  in  A/sec\n",
+      "\n",
+      " #calculation:\n",
+      " #Kirchhoff\u2019s  voltage  law\n",
+      "L  =  (V  -  I*(Ra  +  Rb))/dIdt\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L  -  La  -  Lb)/2\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M,\"H\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   1.0 H\n",
+        "\n",
+        "  coupling  coefficient,  is   0.447"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 848</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the self inductance of each coil, and (b) the mutual inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "k  =  0.7;#  coefficient  of  coupling\n",
+      "L1  =  15E-3;#  in  Henry\n",
+      "L2  =  10E-3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #L1  =  La  +  Lb  +  2*k*(La*Lb)**0.5\n",
+      " #L2  =  La  +  Lb  -  2*k*(La*Lb)**0.5\n",
+      " #self  inductance  of  coils\n",
+      "a  =  ((L1  -  (L1    +  L2)/2)/(2*k))**2\n",
+      "La1  =((L1    +  L2)/2  +  (((L1    +  L2)/2)**2  -  4*a)**0.5)/2\n",
+      "La2  =((L1    +  L2)/2  -  (((L1    +  L2)/2)**2  -  4*a)**0.5)/2\n",
+      "Lb1  =  (L1  +  L2)/2  -  La1\n",
+      "Lb2  =  (L1  +  L2)/2  -  La2\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L1  -  L2)/4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nself  inductance  of  coils  are  \",round(La1*1E3,2),\"mH  and  \",round(  Lb1*1E3,2),\"mH\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",round(M*1E3,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "self  inductance  of  coils  are   12.24 mH  and   0.26 mH\n",
+        "\n",
+        "  mutual  inductance,  M  is   1.25 mH"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 850</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the p.d. E2 which appears across the open-circuited secondary winding,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  8;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "w  =  2500;#  in  rad/sec\n",
+      "R  =  15;#  in  ohm\n",
+      "L  =  5E-3;#  in  Henry\n",
+      "M  =  0.1E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #Impedance  of  primary\n",
+      "Z1  =  R  + 1j*w*L\n",
+      " #Primary  current  I1\n",
+      "I1  =  E1/Z1\n",
+      " #E2\n",
+      "E2  =  1j*w*M*I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nE2  is  \",round(abs(E2),3),\"/_\",round(cmath.phase(complex(E2.real,E2.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "E2  is   0.102 /_ 50.19 deg  V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 850</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the open circuit e.m.f. induced in y.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Lx  =  20E-3;#  in  Henry\n",
+      "Ly  =  80E-3;#  in  Henry\n",
+      "k  =  0.75;#  coupling  coeff.\n",
+      "Ex  =  5;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #mutual  inductance\n",
+      "M  =  k*(Lx*Ly)**0.5\n",
+      " #magnitude  of  the  open  circuit  e.m.f.  induced\n",
+      "Ey  =  M*Ex/Lx\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance  is  \",M,\"  H\"\n",
+      "print  \"\\n  magnitude  of  the  open  circuit  e.m.f.  induced  is  \",Ey,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance  is   0.03   H\n",
+        "\n",
+        "  magnitude  of  the  open  circuit  e.m.f.  induced  is   7.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 852</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of the secondary current I2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  2;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "f  =  1000/math.pi;#  in  Hz\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  16;#  in  ohm\n",
+      "R3  =  16;#  in  ohm\n",
+      "R4  =  50;#  in  ohm\n",
+      "L  =  10E-3;#  in  Henry\n",
+      "M  =  2E-3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #R1e  is  the  real  part  of  Z1e\n",
+      "R1e  =  R1  +  R2  +  ((R3  +  R4)*(M**2)*(w**2))/((R3  +  R4)**2  +  (w*L)**2)\n",
+      " #X1e  is  the  imaginary  part  of  Z1e\n",
+      "X1e  =  w*L  -  (L*(M**2)*(w**3))/((R3  +  R4)**2  +  (w*L)**2)\n",
+      "Z1e  =  R1e  +  1j*X1e\n",
+      "Z2e  =  R3  +  R4  +  1j*w*L\n",
+      " #primary  current,  I1\n",
+      "I1  =  E1/Z1e\n",
+      " #E2\n",
+      "E2  =  1j*w*M*I1\n",
+      " #secondary  current  I2\n",
+      "I2  =  E2/Z2e\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"secondary  current  I2  is  \",round(abs(I2)*1E3,3),\"/_\", round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "secondary  current  I2  is   4.085 /_ 28.55 deg  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 853</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the self impedance of the primary circuit, \n",
+      "#(b) the self impedance of the secondary circuit, \n",
+      "#(c) the impedance reflected the primary circuit, \n",
+      "#(d) the effective primary impedance, \n",
+      "#(e) the primary current, and \n",
+      "#(f) the secondary current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "w  =  500;#  in  rad/sec\n",
+      "R1  =  300;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "L2  =  0.5;#  in  Henry\n",
+      "L3  =  0.3;#  in  Henry\n",
+      "R2  =  500;#  in  ohm\n",
+      "C  =  5E-6;#  in  farad\n",
+      "M  =  0.2;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #  Self  impedance  of  primary  circuit\n",
+      "Z1  =  R1  +  1j*w*(L1  +  L2)\n",
+      " #Self  impedance  of  secondary  circuit,\n",
+      "Z2  =  R2  +  1j*(w*L3  -  1/(w*C))\n",
+      " #reflected  impedance,  Zr\n",
+      "Zr  =  (w*M)**2/Z2\n",
+      " #Effective  primary  impedance,\n",
+      "Z1e  =  Z1  +  Zr\n",
+      " #Primary  current  I1  \n",
+      "I1  =  E1/Z1e\n",
+      " #Secondary  current  I2\n",
+      "E2  =  1j*w*M*I1\n",
+      "I2  =  E2/Z2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Self  impedance  of  primary  circuit,  Z1  is  \",Z1.real,\"  +  (\",  Z1.imag,\")i  ohm\"\n",
+      "print  \"\\n  Self  impedance  of  secondary  circuit,  Z2  is  \",Z2.real,\"  +  (\",  Z2.imag,\")i  ohm\"\n",
+      "print  \"\\n  reflected  impedance,  Zr  is  \",Zr.real,\"  +(\",  Zr.imag,\")i  ohm\"\n",
+      "print  \"\\n  Effective  primary  impedance  Z1(eff)  is  \",Z1e.real,\"  +(\",Z1e.imag,\")i  ohm\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Self  impedance  of  primary  circuit,  Z1  is   300.0   +  ( 350.0 )i  ohm\n",
+        "\n",
+        "  Self  impedance  of  secondary  circuit,  Z2  is   500.0   +  ( -250.0 )i  ohm\n",
+        "\n",
+        "  reflected  impedance,  Zr  is   16.0   +( 8.0 )i  ohm\n",
+        "\n",
+        "  Effective  primary  impedance  Z1(eff)  is   316.0   +( 358.0 )i  ohm\n",
+        "\n",
+        "  primary  current  I1  is   0.1 /_ -48.57 deg  A\n",
+        "\n",
+        "  secondary  current  I2  is   0.02 /_ 68.0 deg  A"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 855</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, \n",
+      "#(b) the value of capacitor C2, \n",
+      "#(c) the effective primary impedance, \n",
+      "#(d) the primary current, (e) the voltage across capacitor C2 and \n",
+      "#(f) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  20;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  15;#  in  ohm\n",
+      "C1  =  400E-12;#  in  farad\n",
+      "R2  =  30;#  in  ohm\n",
+      "L1  =  0.001;#  in  Henry\n",
+      "L2  =  0.0002;#  in  Henry\n",
+      "R3  =  50;#  in  ohm\n",
+      "M  =  10E-6;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #the  resonant  frequency,  fr  \n",
+      "fr  =  1/(2*math.pi*(L1*C1)**0.5)\n",
+      " #The  secondary  is  also  tuned  to  a  resonant  frequency\n",
+      " #capacitance,C2\n",
+      "C2  =  1/(L2*(2*math.pi*fr)**2)\n",
+      " #the  effective  primary  impedance  Z1eff\n",
+      "w = 2*math.pi*fr\n",
+      "Z1e  =  R1 + R2 +  ((w*M)**2)/R3\n",
+      " #Primary  current  I1  \n",
+      "I1  =  E1/Z1e\n",
+      " #Secondary  current  I2\n",
+      "E2  =  1j*w*M*I1\n",
+      "I2  =  E2/Z1e\n",
+      " #voltage  across  capacitor  C2\n",
+      "Vc2  =  I2*(-1*1j/(w*C2))\n",
+      " #coefficient  of  coupling,  k  \n",
+      "k  =  M/(L1*L2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  resonant  frequency,fr  is  \",round(fr/1000,2),\"KHz\"\n",
+      "print  \"\\n  capacitance,C2  is  \",round(C2*1E9,2),\"nF\"\n",
+      "print  \"\\n  Effective  primary  impedance  Z1(eff)  is  \",round(abs(Z1e),2),\" ohm\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag)),0),\"deg  A\"\n",
+      "print  \"\\n  voltage  across  capacitor  C2  is \",round(abs(Vc2),2),\"/_\",round(abs(cmath.phase(complex(Vc2.real,Vc2.imag))),0),\"deg  V\"\n",
+      "print  \"\\n  coefficient  of  coupling,  k    is  \",round(k,4),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  resonant  frequency,fr  is   251.65 KHz\n",
+        "\n",
+        "  capacitance,C2  is   2.0 nF\n",
+        "\n",
+        "  Effective  primary  impedance  Z1(eff)  is   50.0  ohm\n",
+        "\n",
+        "  primary  current  I1  is   0.4 /_ 0.0 deg  A\n",
+        "\n",
+        "  voltage  across  capacitor  C2  is  40.0 /_ 0.0 deg  V\n",
+        "\n",
+        "  coefficient  of  coupling,  k    is   0.0224 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 858</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of currents I1 and I2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  250;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  50j;#  in  ohm\n",
+      "R2  =  10;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "R4  =  50j;#  in  ohm\n",
+      "R5  =  50;#  in  ohm\n",
+      "M  =  10j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(R1  +  R2)*I1  -  M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*M*I1  +  (  R3  +  R4  +  R5)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((R1  +  R2)*(R3  +  R4  +  R5)/(-1*M)  +  (-1*M))\n",
+      "I1  =  I2*(R3  +  R4  +  R5)/(-1*M)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(I1.real,2),\"  +(\",round(  I1.imag,2),\")i  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(I2.real,2),\"  +(\",round(  I2.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  primary  current  I1  is   0.85   +( -4.77 )i  A\n",
+        "\n",
+        "  secondary  current  I2  is   -0.54   +( 0.31 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 859</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the circuit (a) the mutual inductance M, (b) the primary current I1 and (c) the secondary terminal p.d.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  40;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "L1  =  0.001;#  in  Henry\n",
+      "L2  =  0.006;#  in  Henry\n",
+      "R2  =  40;#  in  ohm\n",
+      "rzl  =  200;#  in  ohm\n",
+      "thetazl  =  -60;#  in  degrees\n",
+      "k  =  0.70\n",
+      "f  =  20000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #voltage\n",
+      "#E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #impedance\n",
+      "ZL  =  rzl*math.cos(thetazl*math.pi/180)  +  1j*rzl*math.sin(thetazl*math.pi/180)\n",
+      " #mutual  inductance,  M\n",
+      "M  =  k*(L1*L2)**0.5\n",
+      " #Applying  Kirchhoff\u00e2\u20ac\u2122s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(R1  +  1j*w*L1)*I1  -  1j*w*M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u00e2\u20ac\u2122s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1j*w*M*I1  +  (  R2  +  ZL  +  1j*w*L2)*I2  =  0\n",
+      " #solving  these  two\n",
+      "\n",
+      "a = R1  +  1j*w*L1\n",
+      "b = 1j*w*M\n",
+      "c = R2  +  ZL  +  1j*w*L2\n",
+      "I1  =  E1/(1*a - (b**2)/c)\n",
+      "d = -1*cmath.phase(complex(I1.real,I1.imag))\n",
+      "e = abs(I1)\n",
+      "I2  =  (b/c)*(e*math.cos(d)  +  1j*e*math.sin(d))\n",
+      "pd2 = I2*ZL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  induction  M  is  \",round(M*1E3,3),\"mH\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),3),\"/_\",round(-1*cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(abs(pd2),1),\"/_\",round(cmath.phase(complex(pd2.real,pd2.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  induction  M  is   1.715 mH\n",
+        "\n",
+        "  primary  current  I1  is   0.724 /_ 65.15 deg  A\n",
+        "\n",
+        "  secondary  current  I2  is   52.2 /_ 18.7 deg  V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 860</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the generator current I1 and (b) the load current I2.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "r  =  5;#  in  ohm\n",
+      "R1  =  20;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "R2  =  25;#  in  ohm\n",
+      "RL  =  20;#  in  ohm\n",
+      "M  =  0.1;#  in  Henry\n",
+      "f  =  75/math.pi;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1)*I1  -  1j*w*M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*1j*w*M*I1  +  (  R2  +  RL  +  1j*w*L2)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((r  +  R1  +  1j*w*L1)*(R2  +  RL  +  1j*w*L2)/(1j*w*M)  +  (-1*1j*w*M))\n",
+      "I1  =  I2*(R2  +  RL  +  1j*w*L2)/(1j*w*M)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  load  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  primary  current  I1  is   1.3 /_ -45.84 deg  A\n",
+        "\n",
+        "  load  current  I2  is   0.26 /_ -8.97 deg  A"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 862</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the source and load currents for (a) the windings as shown , and (b) with one winding reversed\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "r  =  5;#  in  ohm\n",
+      "R1  =  20;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "R  =  8;#  in  ohm\n",
+      "L  =  0.1;#  in  Henry\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "R2  =  25;#  in  ohm\n",
+      "RL  =  20;#  in  ohm\n",
+      "M  =  0.1;#  in  Henry\n",
+      "f  =  75/math.pi;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*I1  -  (1j*w*M  +  R  +  1j*w*L)*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*(1j*w*M  +  R  +  1j*w*L)*I1  +  (R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(1j*w*M  +  R  +  1j*w*L)  +  (-1*(1j*w*M  +  R  +  1j*w*L)))\n",
+      "I1  =  I2*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(1j*w*M  +  R  +  1j*w*L)\n",
+      " #reversing\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*I1r  -  (-1*1j*w*M  +  R  +  1j*w*L)*I2r  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*(-1*1j*w*M  +  R  +  1j*w*L)*I1r  +  (R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)*I2r  =  0\n",
+      " #solving  these  two\n",
+      "I2r  =  E1/((r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(-1*1j*w*M  +  R  +  1j*w*L)  +  (-1*(-1*1j*w*M  +  R  +  1j*w*L)))\n",
+      "I1r  =  I2r*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(-1*1j*w*M  +  R  +  1j*w*L)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"load  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg   A\"\n",
+      "print  \"reversed  primary  current  I1r  is  \",round(abs(I1r),2),\"/_\",round(cmath.phase(complex(I1r.real,I1r.imag))*180/math.pi,2),\"deg   A\"\n",
+      "print  \"reversed  load  current  I2r  is  \",round(abs(I2r),2),\"/_\",round(cmath.phase(complex(I2r.real,I2r.imag))*180/math.pi,2),\"deg   A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "primary  current  I1  is   1.03 /_ -45.47 deg  A\n",
+        "load  current  I2  is   0.35 /_ -25.16 deg   A\n",
+        "reversed  primary  current  I1r  is   0.89 /_ -54.42 deg   A\n",
+        "reversed  load  current  I2r  is   0.08 /_ -109.17 deg   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_1.ipynb
new file mode 100755
index 00000000..618fd99a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_1.ipynb
@@ -0,0 +1,1037 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 43: Magnetically coupled circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 842</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the self inductance of coil A, and (b) the mutual inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Na  =  1200;  \n",
+      "Nb  =  1000;\n",
+      "Ia  =  0.8;#  in  amperes\n",
+      "Phia  =  100E-6;#  in  Wb\n",
+      "xb  =  0.75;\n",
+      "\n",
+      " #calculation:\n",
+      " #self  inductance  of  coil  A\n",
+      "La  =  Na*Phia/Ia\n",
+      " #mutual  inductance,  M\n",
+      "Phib  =  xb*Phia\n",
+      "M  =  Nb*Phib/Ia\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  self  inductance  of  coil  A  is  \",La,\"  H\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M,\"H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  self  inductance  of  coil  A  is   0.15   H\n",
+        "\n",
+        "  mutual  inductance,  M  is   0.09375 H"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 843</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. induced in the secondary,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "M  =  600E-3;#  in  Henry\n",
+      "Ia  =  5;#  in  amperes\n",
+      "dt  =  0.2;#  in  secs\n",
+      "\n",
+      " #calculation:\n",
+      " #change  of  current\n",
+      "dIa  =  2*Ia  \n",
+      "dIadt  =  dIa/dt\n",
+      " #secondary  induced  e.m.f.,  E2\n",
+      "E2  =  -1*M*dIadt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  induced  e.m.f.,  E2  is  \",E2,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  induced  e.m.f.,  E2  is   -30.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 844</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnetic coupling coefficient of the pair of coils\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "La  =  250E-3;#  in  Henry\n",
+      "Lb  =  400E-3;#  in  Henry\n",
+      "M  =  80E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  coupling  coefficient,  is   0.253"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 845</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the mutual inductance between the coils, and (b) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Lx  =  80E-3;#  in  Henry\n",
+      "Ly  =  60E-3;#  in  Henry\n",
+      "Nx  =  200;#  turns\n",
+      "Ny  =  100;#  turns\n",
+      "Ix  =  4;#  in  Amperes\n",
+      "Phiy  =  0.005;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #mutual  inductance,  M\n",
+      "M  =  Ny*Phiy/(2*Ix)\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(Lx*Ly)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M*1E3,\"mH\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   62.5 mH\n",
+        "\n",
+        "  coupling  coefficient,  is   0.902"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 846</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the mutual inductance between the two coils, and (b) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "La  =  40E-3;#  in  Henry\n",
+      "Lb  =  10E-3;#  in  Henry\n",
+      "L  =  60E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L  -  La  -  Lb)/2\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M*1E3,\"mH\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",k"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   5.0 mH\n",
+        "\n",
+        "  coupling  coefficient,  is   0.25"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 847</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the mutual inductance between the coils and (b) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "Ra  =  5;#  in  Ohm\n",
+      "La  =  1;#  in  Henry\n",
+      "Rb  =  10;#  in  Ohm\n",
+      "Lb  =  5;#  in  Henry\n",
+      "I  =  8;#  in  amperes\n",
+      "dIdt  =  15;#  in  A/sec\n",
+      "\n",
+      " #calculation:\n",
+      " #Kirchhoff\u2019s  voltage  law\n",
+      "L  =  (V  -  I*(Ra  +  Rb))/dIdt\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L  -  La  -  Lb)/2\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M,\"H\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   1.0 H\n",
+        "\n",
+        "  coupling  coefficient,  is   0.447"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 848</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the self inductance of each coil, and (b) the mutual inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "k  =  0.7;#  coefficient  of  coupling\n",
+      "L1  =  15E-3;#  in  Henry\n",
+      "L2  =  10E-3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #L1  =  La  +  Lb  +  2*k*(La*Lb)**0.5\n",
+      " #L2  =  La  +  Lb  -  2*k*(La*Lb)**0.5\n",
+      " #self  inductance  of  coils\n",
+      "a  =  ((L1  -  (L1    +  L2)/2)/(2*k))**2\n",
+      "La1  =((L1    +  L2)/2  +  (((L1    +  L2)/2)**2  -  4*a)**0.5)/2\n",
+      "La2  =((L1    +  L2)/2  -  (((L1    +  L2)/2)**2  -  4*a)**0.5)/2\n",
+      "Lb1  =  (L1  +  L2)/2  -  La1\n",
+      "Lb2  =  (L1  +  L2)/2  -  La2\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L1  -  L2)/4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nself  inductance  of  coils  are  \",round(La1*1E3,2),\"mH  and  \",round(  Lb1*1E3,2),\"mH\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",round(M*1E3,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "self  inductance  of  coils  are   12.24 mH  and   0.26 mH\n",
+        "\n",
+        "  mutual  inductance,  M  is   1.25 mH"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 850</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the p.d. E2 which appears across the open-circuited secondary winding,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  8;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "w  =  2500;#  in  rad/sec\n",
+      "R  =  15;#  in  ohm\n",
+      "L  =  5E-3;#  in  Henry\n",
+      "M  =  0.1E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #Impedance  of  primary\n",
+      "Z1  =  R  + 1j*w*L\n",
+      " #Primary  current  I1\n",
+      "I1  =  E1/Z1\n",
+      " #E2\n",
+      "E2  =  1j*w*M*I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nE2  is  \",round(abs(E2),3),\"/_\",round(cmath.phase(complex(E2.real,E2.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "E2  is   0.102 /_ 50.19 deg  V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 850</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the open circuit e.m.f. induced in y.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Lx  =  20E-3;#  in  Henry\n",
+      "Ly  =  80E-3;#  in  Henry\n",
+      "k  =  0.75;#  coupling  coeff.\n",
+      "Ex  =  5;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #mutual  inductance\n",
+      "M  =  k*(Lx*Ly)**0.5\n",
+      " #magnitude  of  the  open  circuit  e.m.f.  induced\n",
+      "Ey  =  M*Ex/Lx\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance  is  \",M,\"  H\"\n",
+      "print  \"\\n  magnitude  of  the  open  circuit  e.m.f.  induced  is  \",Ey,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance  is   0.03   H\n",
+        "\n",
+        "  magnitude  of  the  open  circuit  e.m.f.  induced  is   7.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 852</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of the secondary current I2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  2;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "f  =  1000/math.pi;#  in  Hz\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  16;#  in  ohm\n",
+      "R3  =  16;#  in  ohm\n",
+      "R4  =  50;#  in  ohm\n",
+      "L  =  10E-3;#  in  Henry\n",
+      "M  =  2E-3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #R1e  is  the  real  part  of  Z1e\n",
+      "R1e  =  R1  +  R2  +  ((R3  +  R4)*(M**2)*(w**2))/((R3  +  R4)**2  +  (w*L)**2)\n",
+      " #X1e  is  the  imaginary  part  of  Z1e\n",
+      "X1e  =  w*L  -  (L*(M**2)*(w**3))/((R3  +  R4)**2  +  (w*L)**2)\n",
+      "Z1e  =  R1e  +  1j*X1e\n",
+      "Z2e  =  R3  +  R4  +  1j*w*L\n",
+      " #primary  current,  I1\n",
+      "I1  =  E1/Z1e\n",
+      " #E2\n",
+      "E2  =  1j*w*M*I1\n",
+      " #secondary  current  I2\n",
+      "I2  =  E2/Z2e\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"secondary  current  I2  is  \",round(abs(I2)*1E3,3),\"/_\", round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "secondary  current  I2  is   4.085 /_ 28.55 deg  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 853</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the self impedance of the primary circuit, \n",
+      "#(b) the self impedance of the secondary circuit, \n",
+      "#(c) the impedance reflected the primary circuit, \n",
+      "#(d) the effective primary impedance, \n",
+      "#(e) the primary current, and \n",
+      "#(f) the secondary current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "w  =  500;#  in  rad/sec\n",
+      "R1  =  300;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "L2  =  0.5;#  in  Henry\n",
+      "L3  =  0.3;#  in  Henry\n",
+      "R2  =  500;#  in  ohm\n",
+      "C  =  5E-6;#  in  farad\n",
+      "M  =  0.2;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #  Self  impedance  of  primary  circuit\n",
+      "Z1  =  R1  +  1j*w*(L1  +  L2)\n",
+      " #Self  impedance  of  secondary  circuit,\n",
+      "Z2  =  R2  +  1j*(w*L3  -  1/(w*C))\n",
+      " #reflected  impedance,  Zr\n",
+      "Zr  =  (w*M)**2/Z2\n",
+      " #Effective  primary  impedance,\n",
+      "Z1e  =  Z1  +  Zr\n",
+      " #Primary  current  I1  \n",
+      "I1  =  E1/Z1e\n",
+      " #Secondary  current  I2\n",
+      "E2  =  1j*w*M*I1\n",
+      "I2  =  E2/Z2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Self  impedance  of  primary  circuit,  Z1  is  \",Z1.real,\"  +  (\",  Z1.imag,\")i  ohm\"\n",
+      "print  \"\\n  Self  impedance  of  secondary  circuit,  Z2  is  \",Z2.real,\"  +  (\",  Z2.imag,\")i  ohm\"\n",
+      "print  \"\\n  reflected  impedance,  Zr  is  \",Zr.real,\"  +(\",  Zr.imag,\")i  ohm\"\n",
+      "print  \"\\n  Effective  primary  impedance  Z1(eff)  is  \",Z1e.real,\"  +(\",Z1e.imag,\")i  ohm\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Self  impedance  of  primary  circuit,  Z1  is   300.0   +  ( 350.0 )i  ohm\n",
+        "\n",
+        "  Self  impedance  of  secondary  circuit,  Z2  is   500.0   +  ( -250.0 )i  ohm\n",
+        "\n",
+        "  reflected  impedance,  Zr  is   16.0   +( 8.0 )i  ohm\n",
+        "\n",
+        "  Effective  primary  impedance  Z1(eff)  is   316.0   +( 358.0 )i  ohm\n",
+        "\n",
+        "  primary  current  I1  is   0.1 /_ -48.57 deg  A\n",
+        "\n",
+        "  secondary  current  I2  is   0.02 /_ 68.0 deg  A"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 855</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, \n",
+      "#(b) the value of capacitor C2, \n",
+      "#(c) the effective primary impedance, \n",
+      "#(d) the primary current, (e) the voltage across capacitor C2 and \n",
+      "#(f) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  20;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  15;#  in  ohm\n",
+      "C1  =  400E-12;#  in  farad\n",
+      "R2  =  30;#  in  ohm\n",
+      "L1  =  0.001;#  in  Henry\n",
+      "L2  =  0.0002;#  in  Henry\n",
+      "R3  =  50;#  in  ohm\n",
+      "M  =  10E-6;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #the  resonant  frequency,  fr  \n",
+      "fr  =  1/(2*math.pi*(L1*C1)**0.5)\n",
+      " #The  secondary  is  also  tuned  to  a  resonant  frequency\n",
+      " #capacitance,C2\n",
+      "C2  =  1/(L2*(2*math.pi*fr)**2)\n",
+      " #the  effective  primary  impedance  Z1eff\n",
+      "w = 2*math.pi*fr\n",
+      "Z1e  =  R1 + R2 +  ((w*M)**2)/R3\n",
+      " #Primary  current  I1  \n",
+      "I1  =  E1/Z1e\n",
+      " #Secondary  current  I2\n",
+      "E2  =  1j*w*M*I1\n",
+      "I2  =  E2/Z1e\n",
+      " #voltage  across  capacitor  C2\n",
+      "Vc2  =  I2*(-1*1j/(w*C2))\n",
+      " #coefficient  of  coupling,  k  \n",
+      "k  =  M/(L1*L2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  resonant  frequency,fr  is  \",round(fr/1000,2),\"KHz\"\n",
+      "print  \"\\n  capacitance,C2  is  \",round(C2*1E9,2),\"nF\"\n",
+      "print  \"\\n  Effective  primary  impedance  Z1(eff)  is  \",round(abs(Z1e),2),\" ohm\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag)),0),\"deg  A\"\n",
+      "print  \"\\n  voltage  across  capacitor  C2  is \",round(abs(Vc2),2),\"/_\",round(abs(cmath.phase(complex(Vc2.real,Vc2.imag))),0),\"deg  V\"\n",
+      "print  \"\\n  coefficient  of  coupling,  k    is  \",round(k,4),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  resonant  frequency,fr  is   251.65 KHz\n",
+        "\n",
+        "  capacitance,C2  is   2.0 nF\n",
+        "\n",
+        "  Effective  primary  impedance  Z1(eff)  is   50.0  ohm\n",
+        "\n",
+        "  primary  current  I1  is   0.4 /_ 0.0 deg  A\n",
+        "\n",
+        "  voltage  across  capacitor  C2  is  40.0 /_ 0.0 deg  V\n",
+        "\n",
+        "  coefficient  of  coupling,  k    is   0.0224 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 858</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of currents I1 and I2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  250;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  50j;#  in  ohm\n",
+      "R2  =  10;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "R4  =  50j;#  in  ohm\n",
+      "R5  =  50;#  in  ohm\n",
+      "M  =  10j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(R1  +  R2)*I1  -  M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*M*I1  +  (  R3  +  R4  +  R5)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((R1  +  R2)*(R3  +  R4  +  R5)/(-1*M)  +  (-1*M))\n",
+      "I1  =  I2*(R3  +  R4  +  R5)/(-1*M)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(I1.real,2),\"  +(\",round(  I1.imag,2),\")i  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(I2.real,2),\"  +(\",round(  I2.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  primary  current  I1  is   0.85   +( -4.77 )i  A\n",
+        "\n",
+        "  secondary  current  I2  is   -0.54   +( 0.31 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 859</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the circuit (a) the mutual inductance M, (b) the primary current I1 and (c) the secondary terminal p.d.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  40;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "L1  =  0.001;#  in  Henry\n",
+      "L2  =  0.006;#  in  Henry\n",
+      "R2  =  40;#  in  ohm\n",
+      "rzl  =  200;#  in  ohm\n",
+      "thetazl  =  -60;#  in  degrees\n",
+      "k  =  0.70\n",
+      "f  =  20000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #voltage\n",
+      "#E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #impedance\n",
+      "ZL  =  rzl*math.cos(thetazl*math.pi/180)  +  1j*rzl*math.sin(thetazl*math.pi/180)\n",
+      " #mutual  inductance,  M\n",
+      "M  =  k*(L1*L2)**0.5\n",
+      " #Applying  Kirchhoff\u00e2\u20ac\u2122s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(R1  +  1j*w*L1)*I1  -  1j*w*M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u00e2\u20ac\u2122s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1j*w*M*I1  +  (  R2  +  ZL  +  1j*w*L2)*I2  =  0\n",
+      " #solving  these  two\n",
+      "\n",
+      "a = R1  +  1j*w*L1\n",
+      "b = 1j*w*M\n",
+      "c = R2  +  ZL  +  1j*w*L2\n",
+      "I1  =  E1/(1*a - (b**2)/c)\n",
+      "d = -1*cmath.phase(complex(I1.real,I1.imag))\n",
+      "e = abs(I1)\n",
+      "I2  =  (b/c)*(e*math.cos(d)  +  1j*e*math.sin(d))\n",
+      "pd2 = I2*ZL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  induction  M  is  \",round(M*1E3,3),\"mH\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),3),\"/_\",round(-1*cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(abs(pd2),1),\"/_\",round(cmath.phase(complex(pd2.real,pd2.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  induction  M  is   1.715 mH\n",
+        "\n",
+        "  primary  current  I1  is   0.724 /_ 65.15 deg  A\n",
+        "\n",
+        "  secondary  current  I2  is   52.2 /_ 18.7 deg  V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 860</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the generator current I1 and (b) the load current I2.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "r  =  5;#  in  ohm\n",
+      "R1  =  20;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "R2  =  25;#  in  ohm\n",
+      "RL  =  20;#  in  ohm\n",
+      "M  =  0.1;#  in  Henry\n",
+      "f  =  75/math.pi;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1)*I1  -  1j*w*M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*1j*w*M*I1  +  (  R2  +  RL  +  1j*w*L2)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((r  +  R1  +  1j*w*L1)*(R2  +  RL  +  1j*w*L2)/(1j*w*M)  +  (-1*1j*w*M))\n",
+      "I1  =  I2*(R2  +  RL  +  1j*w*L2)/(1j*w*M)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  load  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  primary  current  I1  is   1.3 /_ -45.84 deg  A\n",
+        "\n",
+        "  load  current  I2  is   0.26 /_ -8.97 deg  A"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 862</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the source and load currents for (a) the windings as shown , and (b) with one winding reversed\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "r  =  5;#  in  ohm\n",
+      "R1  =  20;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "R  =  8;#  in  ohm\n",
+      "L  =  0.1;#  in  Henry\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "R2  =  25;#  in  ohm\n",
+      "RL  =  20;#  in  ohm\n",
+      "M  =  0.1;#  in  Henry\n",
+      "f  =  75/math.pi;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*I1  -  (1j*w*M  +  R  +  1j*w*L)*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*(1j*w*M  +  R  +  1j*w*L)*I1  +  (R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(1j*w*M  +  R  +  1j*w*L)  +  (-1*(1j*w*M  +  R  +  1j*w*L)))\n",
+      "I1  =  I2*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(1j*w*M  +  R  +  1j*w*L)\n",
+      " #reversing\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*I1r  -  (-1*1j*w*M  +  R  +  1j*w*L)*I2r  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*(-1*1j*w*M  +  R  +  1j*w*L)*I1r  +  (R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)*I2r  =  0\n",
+      " #solving  these  two\n",
+      "I2r  =  E1/((r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(-1*1j*w*M  +  R  +  1j*w*L)  +  (-1*(-1*1j*w*M  +  R  +  1j*w*L)))\n",
+      "I1r  =  I2r*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(-1*1j*w*M  +  R  +  1j*w*L)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"load  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg   A\"\n",
+      "print  \"reversed  primary  current  I1r  is  \",round(abs(I1r),2),\"/_\",round(cmath.phase(complex(I1r.real,I1r.imag))*180/math.pi,2),\"deg   A\"\n",
+      "print  \"reversed  load  current  I2r  is  \",round(abs(I2r),2),\"/_\",round(cmath.phase(complex(I2r.real,I2r.imag))*180/math.pi,2),\"deg   A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "primary  current  I1  is   1.03 /_ -45.47 deg  A\n",
+        "load  current  I2  is   0.35 /_ -25.16 deg   A\n",
+        "reversed  primary  current  I1r  is   0.89 /_ -54.42 deg   A\n",
+        "reversed  load  current  I2r  is   0.08 /_ -109.17 deg   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_2.ipynb
new file mode 100755
index 00000000..618fd99a
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_2.ipynb
@@ -0,0 +1,1037 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 43: Magnetically coupled circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 842</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the self inductance of coil A, and (b) the mutual inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Na  =  1200;  \n",
+      "Nb  =  1000;\n",
+      "Ia  =  0.8;#  in  amperes\n",
+      "Phia  =  100E-6;#  in  Wb\n",
+      "xb  =  0.75;\n",
+      "\n",
+      " #calculation:\n",
+      " #self  inductance  of  coil  A\n",
+      "La  =  Na*Phia/Ia\n",
+      " #mutual  inductance,  M\n",
+      "Phib  =  xb*Phia\n",
+      "M  =  Nb*Phib/Ia\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  self  inductance  of  coil  A  is  \",La,\"  H\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M,\"H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  self  inductance  of  coil  A  is   0.15   H\n",
+        "\n",
+        "  mutual  inductance,  M  is   0.09375 H"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 843</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the e.m.f. induced in the secondary,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "M  =  600E-3;#  in  Henry\n",
+      "Ia  =  5;#  in  amperes\n",
+      "dt  =  0.2;#  in  secs\n",
+      "\n",
+      " #calculation:\n",
+      " #change  of  current\n",
+      "dIa  =  2*Ia  \n",
+      "dIadt  =  dIa/dt\n",
+      " #secondary  induced  e.m.f.,  E2\n",
+      "E2  =  -1*M*dIadt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  induced  e.m.f.,  E2  is  \",E2,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  induced  e.m.f.,  E2  is   -30.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 844</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnetic coupling coefficient of the pair of coils\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "La  =  250E-3;#  in  Henry\n",
+      "Lb  =  400E-3;#  in  Henry\n",
+      "M  =  80E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  coupling  coefficient,  is   0.253"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 845</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the mutual inductance between the coils, and (b) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Lx  =  80E-3;#  in  Henry\n",
+      "Ly  =  60E-3;#  in  Henry\n",
+      "Nx  =  200;#  turns\n",
+      "Ny  =  100;#  turns\n",
+      "Ix  =  4;#  in  Amperes\n",
+      "Phiy  =  0.005;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #mutual  inductance,  M\n",
+      "M  =  Ny*Phiy/(2*Ix)\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(Lx*Ly)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M*1E3,\"mH\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   62.5 mH\n",
+        "\n",
+        "  coupling  coefficient,  is   0.902"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 846</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the mutual inductance between the two coils, and (b) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "La  =  40E-3;#  in  Henry\n",
+      "Lb  =  10E-3;#  in  Henry\n",
+      "L  =  60E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L  -  La  -  Lb)/2\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M*1E3,\"mH\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",k"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   5.0 mH\n",
+        "\n",
+        "  coupling  coefficient,  is   0.25"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 847</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the mutual inductance between the coils and (b) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "Ra  =  5;#  in  Ohm\n",
+      "La  =  1;#  in  Henry\n",
+      "Rb  =  10;#  in  Ohm\n",
+      "Lb  =  5;#  in  Henry\n",
+      "I  =  8;#  in  amperes\n",
+      "dIdt  =  15;#  in  A/sec\n",
+      "\n",
+      " #calculation:\n",
+      " #Kirchhoff\u2019s  voltage  law\n",
+      "L  =  (V  -  I*(Ra  +  Rb))/dIdt\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L  -  La  -  Lb)/2\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M,\"H\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   1.0 H\n",
+        "\n",
+        "  coupling  coefficient,  is   0.447"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 848</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the self inductance of each coil, and (b) the mutual inductance.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "k  =  0.7;#  coefficient  of  coupling\n",
+      "L1  =  15E-3;#  in  Henry\n",
+      "L2  =  10E-3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #L1  =  La  +  Lb  +  2*k*(La*Lb)**0.5\n",
+      " #L2  =  La  +  Lb  -  2*k*(La*Lb)**0.5\n",
+      " #self  inductance  of  coils\n",
+      "a  =  ((L1  -  (L1    +  L2)/2)/(2*k))**2\n",
+      "La1  =((L1    +  L2)/2  +  (((L1    +  L2)/2)**2  -  4*a)**0.5)/2\n",
+      "La2  =((L1    +  L2)/2  -  (((L1    +  L2)/2)**2  -  4*a)**0.5)/2\n",
+      "Lb1  =  (L1  +  L2)/2  -  La1\n",
+      "Lb2  =  (L1  +  L2)/2  -  La2\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L1  -  L2)/4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nself  inductance  of  coils  are  \",round(La1*1E3,2),\"mH  and  \",round(  Lb1*1E3,2),\"mH\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",round(M*1E3,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "self  inductance  of  coils  are   12.24 mH  and   0.26 mH\n",
+        "\n",
+        "  mutual  inductance,  M  is   1.25 mH"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 850</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the p.d. E2 which appears across the open-circuited secondary winding,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  8;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "w  =  2500;#  in  rad/sec\n",
+      "R  =  15;#  in  ohm\n",
+      "L  =  5E-3;#  in  Henry\n",
+      "M  =  0.1E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #Impedance  of  primary\n",
+      "Z1  =  R  + 1j*w*L\n",
+      " #Primary  current  I1\n",
+      "I1  =  E1/Z1\n",
+      " #E2\n",
+      "E2  =  1j*w*M*I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nE2  is  \",round(abs(E2),3),\"/_\",round(cmath.phase(complex(E2.real,E2.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "E2  is   0.102 /_ 50.19 deg  V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 850</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the magnitude of the open circuit e.m.f. induced in y.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Lx  =  20E-3;#  in  Henry\n",
+      "Ly  =  80E-3;#  in  Henry\n",
+      "k  =  0.75;#  coupling  coeff.\n",
+      "Ex  =  5;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #mutual  inductance\n",
+      "M  =  k*(Lx*Ly)**0.5\n",
+      " #magnitude  of  the  open  circuit  e.m.f.  induced\n",
+      "Ey  =  M*Ex/Lx\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance  is  \",M,\"  H\"\n",
+      "print  \"\\n  magnitude  of  the  open  circuit  e.m.f.  induced  is  \",Ey,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance  is   0.03   H\n",
+        "\n",
+        "  magnitude  of  the  open  circuit  e.m.f.  induced  is   7.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 852</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of the secondary current I2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  2;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "f  =  1000/math.pi;#  in  Hz\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  16;#  in  ohm\n",
+      "R3  =  16;#  in  ohm\n",
+      "R4  =  50;#  in  ohm\n",
+      "L  =  10E-3;#  in  Henry\n",
+      "M  =  2E-3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #R1e  is  the  real  part  of  Z1e\n",
+      "R1e  =  R1  +  R2  +  ((R3  +  R4)*(M**2)*(w**2))/((R3  +  R4)**2  +  (w*L)**2)\n",
+      " #X1e  is  the  imaginary  part  of  Z1e\n",
+      "X1e  =  w*L  -  (L*(M**2)*(w**3))/((R3  +  R4)**2  +  (w*L)**2)\n",
+      "Z1e  =  R1e  +  1j*X1e\n",
+      "Z2e  =  R3  +  R4  +  1j*w*L\n",
+      " #primary  current,  I1\n",
+      "I1  =  E1/Z1e\n",
+      " #E2\n",
+      "E2  =  1j*w*M*I1\n",
+      " #secondary  current  I2\n",
+      "I2  =  E2/Z2e\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"secondary  current  I2  is  \",round(abs(I2)*1E3,3),\"/_\", round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "secondary  current  I2  is   4.085 /_ 28.55 deg  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 853</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate (a) the self impedance of the primary circuit, \n",
+      "#(b) the self impedance of the secondary circuit, \n",
+      "#(c) the impedance reflected the primary circuit, \n",
+      "#(d) the effective primary impedance, \n",
+      "#(e) the primary current, and \n",
+      "#(f) the secondary current\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "w  =  500;#  in  rad/sec\n",
+      "R1  =  300;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "L2  =  0.5;#  in  Henry\n",
+      "L3  =  0.3;#  in  Henry\n",
+      "R2  =  500;#  in  ohm\n",
+      "C  =  5E-6;#  in  farad\n",
+      "M  =  0.2;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #  Self  impedance  of  primary  circuit\n",
+      "Z1  =  R1  +  1j*w*(L1  +  L2)\n",
+      " #Self  impedance  of  secondary  circuit,\n",
+      "Z2  =  R2  +  1j*(w*L3  -  1/(w*C))\n",
+      " #reflected  impedance,  Zr\n",
+      "Zr  =  (w*M)**2/Z2\n",
+      " #Effective  primary  impedance,\n",
+      "Z1e  =  Z1  +  Zr\n",
+      " #Primary  current  I1  \n",
+      "I1  =  E1/Z1e\n",
+      " #Secondary  current  I2\n",
+      "E2  =  1j*w*M*I1\n",
+      "I2  =  E2/Z2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Self  impedance  of  primary  circuit,  Z1  is  \",Z1.real,\"  +  (\",  Z1.imag,\")i  ohm\"\n",
+      "print  \"\\n  Self  impedance  of  secondary  circuit,  Z2  is  \",Z2.real,\"  +  (\",  Z2.imag,\")i  ohm\"\n",
+      "print  \"\\n  reflected  impedance,  Zr  is  \",Zr.real,\"  +(\",  Zr.imag,\")i  ohm\"\n",
+      "print  \"\\n  Effective  primary  impedance  Z1(eff)  is  \",Z1e.real,\"  +(\",Z1e.imag,\")i  ohm\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Self  impedance  of  primary  circuit,  Z1  is   300.0   +  ( 350.0 )i  ohm\n",
+        "\n",
+        "  Self  impedance  of  secondary  circuit,  Z2  is   500.0   +  ( -250.0 )i  ohm\n",
+        "\n",
+        "  reflected  impedance,  Zr  is   16.0   +( 8.0 )i  ohm\n",
+        "\n",
+        "  Effective  primary  impedance  Z1(eff)  is   316.0   +( 358.0 )i  ohm\n",
+        "\n",
+        "  primary  current  I1  is   0.1 /_ -48.57 deg  A\n",
+        "\n",
+        "  secondary  current  I2  is   0.02 /_ 68.0 deg  A"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 855</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the resonant frequency, \n",
+      "#(b) the value of capacitor C2, \n",
+      "#(c) the effective primary impedance, \n",
+      "#(d) the primary current, (e) the voltage across capacitor C2 and \n",
+      "#(f) the coefficient of coupling.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  20;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  15;#  in  ohm\n",
+      "C1  =  400E-12;#  in  farad\n",
+      "R2  =  30;#  in  ohm\n",
+      "L1  =  0.001;#  in  Henry\n",
+      "L2  =  0.0002;#  in  Henry\n",
+      "R3  =  50;#  in  ohm\n",
+      "M  =  10E-6;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #the  resonant  frequency,  fr  \n",
+      "fr  =  1/(2*math.pi*(L1*C1)**0.5)\n",
+      " #The  secondary  is  also  tuned  to  a  resonant  frequency\n",
+      " #capacitance,C2\n",
+      "C2  =  1/(L2*(2*math.pi*fr)**2)\n",
+      " #the  effective  primary  impedance  Z1eff\n",
+      "w = 2*math.pi*fr\n",
+      "Z1e  =  R1 + R2 +  ((w*M)**2)/R3\n",
+      " #Primary  current  I1  \n",
+      "I1  =  E1/Z1e\n",
+      " #Secondary  current  I2\n",
+      "E2  =  1j*w*M*I1\n",
+      "I2  =  E2/Z1e\n",
+      " #voltage  across  capacitor  C2\n",
+      "Vc2  =  I2*(-1*1j/(w*C2))\n",
+      " #coefficient  of  coupling,  k  \n",
+      "k  =  M/(L1*L2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  resonant  frequency,fr  is  \",round(fr/1000,2),\"KHz\"\n",
+      "print  \"\\n  capacitance,C2  is  \",round(C2*1E9,2),\"nF\"\n",
+      "print  \"\\n  Effective  primary  impedance  Z1(eff)  is  \",round(abs(Z1e),2),\" ohm\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag)),0),\"deg  A\"\n",
+      "print  \"\\n  voltage  across  capacitor  C2  is \",round(abs(Vc2),2),\"/_\",round(abs(cmath.phase(complex(Vc2.real,Vc2.imag))),0),\"deg  V\"\n",
+      "print  \"\\n  coefficient  of  coupling,  k    is  \",round(k,4),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  resonant  frequency,fr  is   251.65 KHz\n",
+        "\n",
+        "  capacitance,C2  is   2.0 nF\n",
+        "\n",
+        "  Effective  primary  impedance  Z1(eff)  is   50.0  ohm\n",
+        "\n",
+        "  primary  current  I1  is   0.4 /_ 0.0 deg  A\n",
+        "\n",
+        "  voltage  across  capacitor  C2  is  40.0 /_ 0.0 deg  V\n",
+        "\n",
+        "  coefficient  of  coupling,  k    is   0.0224 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 858</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the values of currents I1 and I2\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  250;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  50j;#  in  ohm\n",
+      "R2  =  10;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "R4  =  50j;#  in  ohm\n",
+      "R5  =  50;#  in  ohm\n",
+      "M  =  10j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(R1  +  R2)*I1  -  M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*M*I1  +  (  R3  +  R4  +  R5)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((R1  +  R2)*(R3  +  R4  +  R5)/(-1*M)  +  (-1*M))\n",
+      "I1  =  I2*(R3  +  R4  +  R5)/(-1*M)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(I1.real,2),\"  +(\",round(  I1.imag,2),\")i  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(I2.real,2),\"  +(\",round(  I2.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  primary  current  I1  is   0.85   +( -4.77 )i  A\n",
+        "\n",
+        "  secondary  current  I2  is   -0.54   +( 0.31 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 859</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the circuit (a) the mutual inductance M, (b) the primary current I1 and (c) the secondary terminal p.d.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  40;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "L1  =  0.001;#  in  Henry\n",
+      "L2  =  0.006;#  in  Henry\n",
+      "R2  =  40;#  in  ohm\n",
+      "rzl  =  200;#  in  ohm\n",
+      "thetazl  =  -60;#  in  degrees\n",
+      "k  =  0.70\n",
+      "f  =  20000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #voltage\n",
+      "#E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #impedance\n",
+      "ZL  =  rzl*math.cos(thetazl*math.pi/180)  +  1j*rzl*math.sin(thetazl*math.pi/180)\n",
+      " #mutual  inductance,  M\n",
+      "M  =  k*(L1*L2)**0.5\n",
+      " #Applying  Kirchhoff\u00e2\u20ac\u2122s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(R1  +  1j*w*L1)*I1  -  1j*w*M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u00e2\u20ac\u2122s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1j*w*M*I1  +  (  R2  +  ZL  +  1j*w*L2)*I2  =  0\n",
+      " #solving  these  two\n",
+      "\n",
+      "a = R1  +  1j*w*L1\n",
+      "b = 1j*w*M\n",
+      "c = R2  +  ZL  +  1j*w*L2\n",
+      "I1  =  E1/(1*a - (b**2)/c)\n",
+      "d = -1*cmath.phase(complex(I1.real,I1.imag))\n",
+      "e = abs(I1)\n",
+      "I2  =  (b/c)*(e*math.cos(d)  +  1j*e*math.sin(d))\n",
+      "pd2 = I2*ZL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  induction  M  is  \",round(M*1E3,3),\"mH\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),3),\"/_\",round(-1*cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(abs(pd2),1),\"/_\",round(cmath.phase(complex(pd2.real,pd2.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  induction  M  is   1.715 mH\n",
+        "\n",
+        "  primary  current  I1  is   0.724 /_ 65.15 deg  A\n",
+        "\n",
+        "  secondary  current  I2  is   52.2 /_ 18.7 deg  V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 860</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the generator current I1 and (b) the load current I2.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "r  =  5;#  in  ohm\n",
+      "R1  =  20;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "R2  =  25;#  in  ohm\n",
+      "RL  =  20;#  in  ohm\n",
+      "M  =  0.1;#  in  Henry\n",
+      "f  =  75/math.pi;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1)*I1  -  1j*w*M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*1j*w*M*I1  +  (  R2  +  RL  +  1j*w*L2)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((r  +  R1  +  1j*w*L1)*(R2  +  RL  +  1j*w*L2)/(1j*w*M)  +  (-1*1j*w*M))\n",
+      "I1  =  I2*(R2  +  RL  +  1j*w*L2)/(1j*w*M)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  load  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  primary  current  I1  is   1.3 /_ -45.84 deg  A\n",
+        "\n",
+        "  load  current  I2  is   0.26 /_ -8.97 deg  A"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 862</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the source and load currents for (a) the windings as shown , and (b) with one winding reversed\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "r  =  5;#  in  ohm\n",
+      "R1  =  20;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "R  =  8;#  in  ohm\n",
+      "L  =  0.1;#  in  Henry\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "R2  =  25;#  in  ohm\n",
+      "RL  =  20;#  in  ohm\n",
+      "M  =  0.1;#  in  Henry\n",
+      "f  =  75/math.pi;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*I1  -  (1j*w*M  +  R  +  1j*w*L)*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*(1j*w*M  +  R  +  1j*w*L)*I1  +  (R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(1j*w*M  +  R  +  1j*w*L)  +  (-1*(1j*w*M  +  R  +  1j*w*L)))\n",
+      "I1  =  I2*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(1j*w*M  +  R  +  1j*w*L)\n",
+      " #reversing\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*I1r  -  (-1*1j*w*M  +  R  +  1j*w*L)*I2r  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*(-1*1j*w*M  +  R  +  1j*w*L)*I1r  +  (R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)*I2r  =  0\n",
+      " #solving  these  two\n",
+      "I2r  =  E1/((r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(-1*1j*w*M  +  R  +  1j*w*L)  +  (-1*(-1*1j*w*M  +  R  +  1j*w*L)))\n",
+      "I1r  =  I2r*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(-1*1j*w*M  +  R  +  1j*w*L)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"load  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg   A\"\n",
+      "print  \"reversed  primary  current  I1r  is  \",round(abs(I1r),2),\"/_\",round(cmath.phase(complex(I1r.real,I1r.imag))*180/math.pi,2),\"deg   A\"\n",
+      "print  \"reversed  load  current  I2r  is  \",round(abs(I2r),2),\"/_\",round(cmath.phase(complex(I2r.real,I2r.imag))*180/math.pi,2),\"deg   A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "primary  current  I1  is   1.03 /_ -45.47 deg  A\n",
+        "load  current  I2  is   0.35 /_ -25.16 deg   A\n",
+        "reversed  primary  current  I1r  is   0.89 /_ -54.42 deg   A\n",
+        "reversed  load  current  I2r  is   0.08 /_ -109.17 deg   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_3.ipynb
new file mode 100755
index 00000000..628db606
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint_3.ipynb
@@ -0,0 +1,1029 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:46dd88e506b63134e24582a78d5a791547c8824340e8bb4b0b9ee7239c438783"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 43: Magnetically coupled circuits</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 842</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Na  =  1200;  \n",
+      "Nb  =  1000;\n",
+      "Ia  =  0.8;#  in  amperes\n",
+      "Phia  =  100E-6;#  in  Wb\n",
+      "xb  =  0.75;\n",
+      "\n",
+      " #calculation:\n",
+      " #self  inductance  of  coil  A\n",
+      "La  =  Na*Phia/Ia\n",
+      " #mutual  inductance,  M\n",
+      "Phib  =  xb*Phia\n",
+      "M  =  Nb*Phib/Ia\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  self  inductance  of  coil  A  is  \",La,\"  H\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M,\"H\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  self  inductance  of  coil  A  is   0.15   H\n",
+        "\n",
+        "  mutual  inductance,  M  is   0.09375 H"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 843</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "M  =  600E-3;#  in  Henry\n",
+      "Ia  =  5;#  in  amperes\n",
+      "dt  =  0.2;#  in  secs\n",
+      "\n",
+      " #calculation:\n",
+      " #change  of  current\n",
+      "dIa  =  2*Ia  \n",
+      "dIadt  =  dIa/dt\n",
+      " #secondary  induced  e.m.f.,  E2\n",
+      "E2  =  -1*M*dIadt\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  secondary  induced  e.m.f.,  E2  is  \",E2,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  secondary  induced  e.m.f.,  E2  is   -30.0   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 844</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "La  =  250E-3;#  in  Henry\n",
+      "Lb  =  400E-3;#  in  Henry\n",
+      "M  =  80E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  coupling  coefficient,  is   0.253"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 845</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Lx  =  80E-3;#  in  Henry\n",
+      "Ly  =  60E-3;#  in  Henry\n",
+      "Nx  =  200;#  turns\n",
+      "Ny  =  100;#  turns\n",
+      "Ix  =  4;#  in  Amperes\n",
+      "Phiy  =  0.005;#  in  Wb\n",
+      "\n",
+      "#calculation:\n",
+      " #mutual  inductance,  M\n",
+      "M  =  Ny*Phiy/(2*Ix)\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(Lx*Ly)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M*1E3,\"mH\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   62.5 mH\n",
+        "\n",
+        "  coupling  coefficient,  is   0.902"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 846</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "La  =  40E-3;#  in  Henry\n",
+      "Lb  =  10E-3;#  in  Henry\n",
+      "L  =  60E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L  -  La  -  Lb)/2\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M*1E3,\"mH\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",k"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   5.0 mH\n",
+        "\n",
+        "  coupling  coefficient,  is   0.25"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 847</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "V  =  240;#  in  Volts\n",
+      "Ra  =  5;#  in  Ohm\n",
+      "La  =  1;#  in  Henry\n",
+      "Rb  =  10;#  in  Ohm\n",
+      "Lb  =  5;#  in  Henry\n",
+      "I  =  8;#  in  amperes\n",
+      "dIdt  =  15;#  in  A/sec\n",
+      "\n",
+      " #calculation:\n",
+      " #Kirchhoff\u2019s  voltage  law\n",
+      "L  =  (V  -  I*(Ra  +  Rb))/dIdt\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L  -  La  -  Lb)/2\n",
+      " #coupling  coefficient,\n",
+      "k  =  M/(La*Lb)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",M,\"H\"\n",
+      "print  \"\\n  coupling  coefficient,  is  \",round(k,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance,  M  is   1.0 H\n",
+        "\n",
+        "  coupling  coefficient,  is   0.447"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 848</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "k  =  0.7;#  coefficient  of  coupling\n",
+      "L1  =  15E-3;#  in  Henry\n",
+      "L2  =  10E-3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #L1  =  La  +  Lb  +  2*k*(La*Lb)**0.5\n",
+      " #L2  =  La  +  Lb  -  2*k*(La*Lb)**0.5\n",
+      " #self  inductance  of  coils\n",
+      "a  =  ((L1  -  (L1    +  L2)/2)/(2*k))**2\n",
+      "La1  =((L1    +  L2)/2  +  (((L1    +  L2)/2)**2  -  4*a)**0.5)/2\n",
+      "La2  =((L1    +  L2)/2  -  (((L1    +  L2)/2)**2  -  4*a)**0.5)/2\n",
+      "Lb1  =  (L1  +  L2)/2  -  La1\n",
+      "Lb2  =  (L1  +  L2)/2  -  La2\n",
+      " #mutual  inductance,  M\n",
+      "M  =  (L1  -  L2)/4\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nself  inductance  of  coils  are  \",round(La1*1E3,2),\"mH  and  \",round(  Lb1*1E3,2),\"mH\"\n",
+      "print  \"\\n  mutual  inductance,  M  is  \",round(M*1E3,2),\"mH\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "self  inductance  of  coils  are   12.24 mH  and   0.26 mH\n",
+        "\n",
+        "  mutual  inductance,  M  is   1.25 mH\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 850</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  8;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "w  =  2500;#  in  rad/sec\n",
+      "R  =  15;#  in  ohm\n",
+      "L  =  5E-3;#  in  Henry\n",
+      "M  =  0.1E-3;#  in  Henry\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #Impedance  of  primary\n",
+      "Z1  =  R  + 1j*w*L\n",
+      " #Primary  current  I1\n",
+      "I1  =  E1/Z1\n",
+      " #E2\n",
+      "E2  =  1j*w*M*I1\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\nE2  is  \",round(abs(E2),3),\"/_\",round(cmath.phase(complex(E2.real,E2.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "E2  is   0.102 /_ 50.19 deg  V"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 850</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Lx  =  20E-3;#  in  Henry\n",
+      "Ly  =  80E-3;#  in  Henry\n",
+      "k  =  0.75;#  coupling  coeff.\n",
+      "Ex  =  5;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #mutual  inductance\n",
+      "M  =  k*(Lx*Ly)**0.5\n",
+      " #magnitude  of  the  open  circuit  e.m.f.  induced\n",
+      "Ey  =  M*Ex/Lx\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  inductance  is  \",M,\"  H\"\n",
+      "print  \"\\n  magnitude  of  the  open  circuit  e.m.f.  induced  is  \",Ey,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  inductance  is   0.03   H\n",
+        "\n",
+        "  magnitude  of  the  open  circuit  e.m.f.  induced  is   7.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 852</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  2;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "f  =  1000/math.pi;#  in  Hz\n",
+      "R1  =  4;#  in  ohm\n",
+      "R2  =  16;#  in  ohm\n",
+      "R3  =  16;#  in  ohm\n",
+      "R4  =  50;#  in  ohm\n",
+      "L  =  10E-3;#  in  Henry\n",
+      "M  =  2E-3;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #R1e  is  the  real  part  of  Z1e\n",
+      "R1e  =  R1  +  R2  +  ((R3  +  R4)*(M**2)*(w**2))/((R3  +  R4)**2  +  (w*L)**2)\n",
+      " #X1e  is  the  imaginary  part  of  Z1e\n",
+      "X1e  =  w*L  -  (L*(M**2)*(w**3))/((R3  +  R4)**2  +  (w*L)**2)\n",
+      "Z1e  =  R1e  +  1j*X1e\n",
+      "Z2e  =  R3  +  R4  +  1j*w*L\n",
+      " #primary  current,  I1\n",
+      "I1  =  E1/Z1e\n",
+      " #E2\n",
+      "E2  =  1j*w*M*I1\n",
+      " #secondary  current  I2\n",
+      "I2  =  E2/Z2e\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"secondary  current  I2  is  \",round(abs(I2)*1E3,3),\"/_\", round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "secondary  current  I2  is   4.085 /_ 28.55 deg  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 853</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "w  =  500;#  in  rad/sec\n",
+      "R1  =  300;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "L2  =  0.5;#  in  Henry\n",
+      "L3  =  0.3;#  in  Henry\n",
+      "R2  =  500;#  in  ohm\n",
+      "C  =  5E-6;#  in  farad\n",
+      "M  =  0.2;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #  Self  impedance  of  primary  circuit\n",
+      "Z1  =  R1  +  1j*w*(L1  +  L2)\n",
+      " #Self  impedance  of  secondary  circuit,\n",
+      "Z2  =  R2  +  1j*(w*L3  -  1/(w*C))\n",
+      " #reflected  impedance,  Zr\n",
+      "Zr  =  (w*M)**2/Z2\n",
+      " #Effective  primary  impedance,\n",
+      "Z1e  =  Z1  +  Zr\n",
+      " #Primary  current  I1  \n",
+      "I1  =  E1/Z1e\n",
+      " #Secondary  current  I2\n",
+      "E2  =  1j*w*M*I1\n",
+      "I2  =  E2/Z2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Self  impedance  of  primary  circuit,  Z1  is  \",Z1.real,\"  +  (\",  Z1.imag,\")i  ohm\"\n",
+      "print  \"\\n  Self  impedance  of  secondary  circuit,  Z2  is  \",Z2.real,\"  +  (\",  Z2.imag,\")i  ohm\"\n",
+      "print  \"\\n  reflected  impedance,  Zr  is  \",Zr.real,\"  +(\",  Zr.imag,\")i  ohm\"\n",
+      "print  \"\\n  Effective  primary  impedance  Z1(eff)  is  \",Z1e.real,\"  +(\",Z1e.imag,\")i  ohm\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Self  impedance  of  primary  circuit,  Z1  is   300.0   +  ( 350.0 )i  ohm\n",
+        "\n",
+        "  Self  impedance  of  secondary  circuit,  Z2  is   500.0   +  ( -250.0 )i  ohm\n",
+        "\n",
+        "  reflected  impedance,  Zr  is   16.0   +( 8.0 )i  ohm\n",
+        "\n",
+        "  Effective  primary  impedance  Z1(eff)  is   316.0   +( 358.0 )i  ohm\n",
+        "\n",
+        "  primary  current  I1  is   0.1 /_ -48.57 deg  A\n",
+        "\n",
+        "  secondary  current  I2  is   0.02 /_ 68.0 deg  A"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 855</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  20;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  15;#  in  ohm\n",
+      "C1  =  400E-12;#  in  farad\n",
+      "R2  =  30;#  in  ohm\n",
+      "L1  =  0.001;#  in  Henry\n",
+      "L2  =  0.0002;#  in  Henry\n",
+      "R3  =  50;#  in  ohm\n",
+      "M  =  10E-6;#  in  Henry\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #the  resonant  frequency,  fr  \n",
+      "fr  =  1/(2*math.pi*(L1*C1)**0.5)\n",
+      " #The  secondary  is  also  tuned  to  a  resonant  frequency\n",
+      " #capacitance,C2\n",
+      "C2  =  1/(L2*(2*math.pi*fr)**2)\n",
+      " #the  effective  primary  impedance  Z1eff\n",
+      "w = 2*math.pi*fr\n",
+      "Z1e  =  R1 + R2 +  ((w*M)**2)/R3\n",
+      " #Primary  current  I1  \n",
+      "I1  =  E1/Z1e\n",
+      " #Secondary  current  I2\n",
+      "E2  =  1j*w*M*I1\n",
+      "I2  =  E2/Z1e\n",
+      " #voltage  across  capacitor  C2\n",
+      "Vc2  =  I2*(-1*1j/(w*C2))\n",
+      " #coefficient  of  coupling,  k  \n",
+      "k  =  M/(L1*L2)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  resonant  frequency,fr  is  \",round(fr/1000,2),\"KHz\"\n",
+      "print  \"\\n  capacitance,C2  is  \",round(C2*1E9,2),\"nF\"\n",
+      "print  \"\\n  Effective  primary  impedance  Z1(eff)  is  \",round(abs(Z1e),2),\" ohm\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag)),0),\"deg  A\"\n",
+      "print  \"\\n  voltage  across  capacitor  C2  is \",round(abs(Vc2),2),\"/_\",round(abs(cmath.phase(complex(Vc2.real,Vc2.imag))),0),\"deg  V\"\n",
+      "print  \"\\n  coefficient  of  coupling,  k    is  \",round(k,4),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  resonant  frequency,fr  is   251.65 KHz\n",
+        "\n",
+        "  capacitance,C2  is   2.0 nF\n",
+        "\n",
+        "  Effective  primary  impedance  Z1(eff)  is   50.0  ohm\n",
+        "\n",
+        "  primary  current  I1  is   0.4 /_ 0.0 deg  A\n",
+        "\n",
+        "  voltage  across  capacitor  C2  is  40.0 /_ 0.0 deg  V\n",
+        "\n",
+        "  coefficient  of  coupling,  k    is   0.0224 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 858</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  250;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  50j;#  in  ohm\n",
+      "R2  =  10;#  in  ohm\n",
+      "R3  =  10;#  in  ohm\n",
+      "R4  =  50j;#  in  ohm\n",
+      "R5  =  50;#  in  ohm\n",
+      "M  =  10j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(R1  +  R2)*I1  -  M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*M*I1  +  (  R3  +  R4  +  R5)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((R1  +  R2)*(R3  +  R4  +  R5)/(-1*M)  +  (-1*M))\n",
+      "I1  =  I2*(R3  +  R4  +  R5)/(-1*M)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(I1.real,2),\"  +(\",round(  I1.imag,2),\")i  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(I2.real,2),\"  +(\",round(  I2.imag,2),\")i  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  primary  current  I1  is   0.85   +( -4.77 )i  A\n",
+        "\n",
+        "  secondary  current  I2  is   -0.54   +( 0.31 )i  A"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 859</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  40;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "R1  =  5;#  in  ohm\n",
+      "L1  =  0.001;#  in  Henry\n",
+      "L2  =  0.006;#  in  Henry\n",
+      "R2  =  40;#  in  ohm\n",
+      "rzl  =  200;#  in  ohm\n",
+      "thetazl  =  -60;#  in  degrees\n",
+      "k  =  0.70\n",
+      "f  =  20000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #voltage\n",
+      "#E1  =  E1*math.cos(thetae1*math.pi/180)  +  1j*E1*math.sin(thetae1*math.pi/180)\n",
+      " #impedance\n",
+      "ZL  =  rzl*math.cos(thetazl*math.pi/180)  +  1j*rzl*math.sin(thetazl*math.pi/180)\n",
+      " #mutual  inductance,  M\n",
+      "M  =  k*(L1*L2)**0.5\n",
+      " #Applying  Kirchhoff\u00e2\u20ac\u2122s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(R1  +  1j*w*L1)*I1  -  1j*w*M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u00e2\u20ac\u2122s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1j*w*M*I1  +  (  R2  +  ZL  +  1j*w*L2)*I2  =  0\n",
+      " #solving  these  two\n",
+      "\n",
+      "a = R1  +  1j*w*L1\n",
+      "b = 1j*w*M\n",
+      "c = R2  +  ZL  +  1j*w*L2\n",
+      "I1  =  E1/(1*a - (b**2)/c)\n",
+      "d = -1*cmath.phase(complex(I1.real,I1.imag))\n",
+      "e = abs(I1)\n",
+      "I2  =  (b/c)*(e*math.cos(d)  +  1j*e*math.sin(d))\n",
+      "pd2 = I2*ZL\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  mutual  induction  M  is  \",round(M*1E3,3),\"mH\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),3),\"/_\",round(-1*cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  secondary  current  I2  is  \",round(abs(pd2),1),\"/_\",round(cmath.phase(complex(pd2.real,pd2.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  mutual  induction  M  is   1.715 mH\n",
+        "\n",
+        "  primary  current  I1  is   0.724 /_ 65.15 deg  A\n",
+        "\n",
+        "  secondary  current  I2  is   52.2 /_ 18.7 deg  V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 860</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "r  =  5;#  in  ohm\n",
+      "R1  =  20;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "R2  =  25;#  in  ohm\n",
+      "RL  =  20;#  in  ohm\n",
+      "M  =  0.1;#  in  Henry\n",
+      "f  =  75/math.pi;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1)*I1  -  1j*w*M*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*1j*w*M*I1  +  (  R2  +  RL  +  1j*w*L2)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((r  +  R1  +  1j*w*L1)*(R2  +  RL  +  1j*w*L2)/(1j*w*M)  +  (-1*1j*w*M))\n",
+      "I1  =  I2*(R2  +  RL  +  1j*w*L2)/(1j*w*M)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"\\n  load  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  primary  current  I1  is   1.3 /_ -45.84 deg  A\n",
+        "\n",
+        "  load  current  I2  is   0.26 /_ -8.97 deg  A"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 862</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "E1  =  50;#  in  Volts\n",
+      "thetae1  =  0;#  in  degrees\n",
+      "r  =  5;#  in  ohm\n",
+      "R1  =  20;#  in  ohm\n",
+      "L1  =  0.2;#  in  Henry\n",
+      "R  =  8;#  in  ohm\n",
+      "L  =  0.1;#  in  Henry\n",
+      "L2  =  0.4;#  in  Henry\n",
+      "R2  =  25;#  in  ohm\n",
+      "RL  =  20;#  in  ohm\n",
+      "M  =  0.1;#  in  Henry\n",
+      "f  =  75/math.pi;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*I1  -  (1j*w*M  +  R  +  1j*w*L)*I2  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*(1j*w*M  +  R  +  1j*w*L)*I1  +  (R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)*I2  =  0\n",
+      " #solving  these  two\n",
+      "I2  =  E1/((r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(1j*w*M  +  R  +  1j*w*L)  +  (-1*(1j*w*M  +  R  +  1j*w*L)))\n",
+      "I1  =  I2*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(1j*w*M  +  R  +  1j*w*L)\n",
+      " #reversing\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  primary  circuit  gives\n",
+      " #(r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*I1r  -  (-1*1j*w*M  +  R  +  1j*w*L)*I2r  =  E1\n",
+      " #Applying  Kirchhoff\u2019s  voltage  law  to  the  secondary  circuit  gives\n",
+      " #-1*(-1*1j*w*M  +  R  +  1j*w*L)*I1r  +  (R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)*I2r  =  0\n",
+      " #solving  these  two\n",
+      "I2r  =  E1/((r  +  R1  +  1j*w*L1  +  R  +  1j*w*L)*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(-1*1j*w*M  +  R  +  1j*w*L)  +  (-1*(-1*1j*w*M  +  R  +  1j*w*L)))\n",
+      "I1r  =  I2r*(R2  +  RL  +  1j*w*L2  +  R  +  1j*w*L)/(-1*1j*w*M  +  R  +  1j*w*L)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"primary  current  I1  is  \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg  A\"\n",
+      "print  \"load  current  I2  is  \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg   A\"\n",
+      "print  \"reversed  primary  current  I1r  is  \",round(abs(I1r),2),\"/_\",round(cmath.phase(complex(I1r.real,I1r.imag))*180/math.pi,2),\"deg   A\"\n",
+      "print  \"reversed  load  current  I2r  is  \",round(abs(I2r),2),\"/_\",round(cmath.phase(complex(I2r.real,I2r.imag))*180/math.pi,2),\"deg   A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "primary  current  I1  is   1.03 /_ -45.47 deg  A\n",
+        "load  current  I2  is   0.35 /_ -25.16 deg   A\n",
+        "reversed  primary  current  I1r  is   0.89 /_ -54.42 deg   A\n",
+        "reversed  load  current  I2r  is   0.08 /_ -109.17 deg   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint.ipynb
new file mode 100755
index 00000000..03d830a4
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint.ipynb
@@ -0,0 +1,1093 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 44: Transmission lines</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 873</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the wavelength on the line, and (b) the speed of transmission of a signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "f  =  1910;#  in  Hz\n",
+      "b  =  0.05;#  in  rad/km\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #wavelength  \n",
+      "Y  =  2*math.pi/b\n",
+      " #speed  of  transmission\n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y,1),\"  km\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,1),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  wavelength  Y  is   125.7   km\n",
+        "\n",
+        "  speed  of  transmission   240017.7 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 873</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for a frequency of operation of 1 kHz, \n",
+      "#(a) the phase delay, \n",
+      "#(b) the wavelength on the line, and \n",
+      "#(c) the velocity of propagation (in metres per second) of the signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.004;#  in  Henry/loop\n",
+      "C  =  0.004E-6;#  in  F/loop\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #phase  delay\n",
+      "b  =  w*(L*C)**0.5\n",
+      " #wavelength  \n",
+      "Y  =  2*math.pi/b\n",
+      " #speed  of  transmission\n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  phase  delay  is  \",round(b,3),\"  rad/km\"\n",
+      "print  \"\\n  wavelength  Y  is  \",Y,\"  km\"\n",
+      "print  \"\\n  speed  of  transmission  \",u,\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  phase  delay  is   0.025   rad/km\n",
+        "\n",
+        "  wavelength  Y  is   250.0   km\n",
+        "\n",
+        "  speed  of  transmission   250000.0 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 874</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the voltage at a point 10 km down the line,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "a  =  0.25;#  in  Np/km\n",
+      "b  =  0.20;#  in  rad/km\n",
+      "Vs  =  5;#  in  Volts\n",
+      "n  =  10;#  in  km\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #the  voltage  10  km  down  the  line\n",
+      "r  =  a  +  1j*b\n",
+      "VR  =  Vs*cmath.e**(-1*n*r)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n Result  \\n\\n\"\n",
+      "print  \"voltage  10  km  down  the  line is \",round(abs(VR),2),\"/_\",round(cmath.phase(complex(VR.real,VR.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Result  \n",
+        "\n",
+        "\n",
+        "voltage  10  km  down  the  line is  0.41 /_ -114.59 deg  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 875</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnitude and phase of the current at the receiving end,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "a  =  0.5;#  in  Np/km\n",
+      "b  =  0.25;#  in  rad/km\n",
+      "rvs  =  2;#  in  Volts\n",
+      "thetavs  =  0;#  in  degrees\n",
+      "rzo  =  800;#  in  ohm\n",
+      "thetazo  =  -25;#  in  degrees\n",
+      "n  =  5;#  in  km\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "Vs  =  rvs*math.cos(thetavs*math.pi/180)  +  1j*rvs*math.sin(thetavs*math.pi/180)\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #  receiving  end  voltage\n",
+      "r  =  a  +  1j*b\n",
+      "VR  =  Vs*cmath.e**(-1*n*r)\n",
+      " #Receiving  end  current,\n",
+      "IR  =  VR/Zo\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"Receiving end current, IR is  \",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg  mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "Receiving end current, IR is   0.205 /_ -46.62 deg  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 875</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the output voltage if the length of the line is doubled.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vs  =  8;#  in  Volts\n",
+      "VR  =  2;#  in  Volts\n",
+      "x  =  2;  \n",
+      "\n",
+      "#calculation:\n",
+      " #  receiving  end  voltage  VR  =  Vs*e**(-nr)\n",
+      " #e**-nr  =  p\n",
+      "p  =  VR/Vs\n",
+      " #If  the  line  is  doubled  in  length,  then\n",
+      "VR  =  Vs*(p)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Receiving  end  voltage  If  the  line  is  doubled  in  length,  VR  is  \",abs(VR),\" V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Receiving  end  voltage  If  the  line  is  doubled  in  length,  VR  is   0.5  V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 876</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the characteristic impedance of the line at this frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzoc  =  800;#  in  ohm\n",
+      "thetazoc  =  -50;#  in  degrees\n",
+      "rzsc  =  413;#  in  ohm\n",
+      "thetazsc  =  -20;#  in  degrees\n",
+      "f  =  1500;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #open  circuit  impedance\n",
+      "Zoc  =  rzoc*math.cos(thetazoc*math.pi/180)  +  1j*rzoc*math.sin(thetazoc*math.pi/180)\n",
+      " #short  circuit  impedance\n",
+      "Zsc  =  rzsc*math.cos(thetazsc*math.pi/180)  +  1j*rzsc*math.sin(thetazsc*math.pi/180)\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  (Zoc*Zsc)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo is\",round(abs(Zo)),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo is 575.0 /_ -35.0 deg  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 877</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the characteristic impedance of the line when the frequency is 2 kHz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohm/loop  km\n",
+      "L  =  0.0034;#  in  H/loop  km\n",
+      "C  =  10E-9;#  in  F/km\n",
+      "G  =  3E-6;#  in  S/km\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo  is  \",round(abs(Zo),0),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo  is   600.0 /_ -8.99 deg  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 879</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, at an operating frequency of 400 kHz, (a) the characteristic impedance,\n",
+      "#(b) the propagation coefficient, (c) the wavelength on the line, and \n",
+      "#(d) the velocity of propagation, in metres per second, of a signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.0005;#  in  H/loop  km\n",
+      "C  =  0.12E-6;#  in  F/km\n",
+      "f  =  400000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  (L/C)**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  1j*w*(L*C)**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      " #wavelength\n",
+      "Y  =  2*math.pi/b\n",
+      " #velocity  of  propagation  \n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  characteristic  impedance  Zo  is  \",abs(Zo),\"ohm\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",a,\"  +(\",round(b,2),\")i\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y*1E3,0),\"m\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,2),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  characteristic  impedance  Zo  is   64.5497224368 ohm\n",
+        "\n",
+        "  propagation  coefficient  is   0.0   +( 19.47 )i\n",
+        "\n",
+        "  wavelength  Y  is   323.0 m\n",
+        "\n",
+        "  speed  of  transmission   129099.44 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 880</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the line (a) the characteristic impedance,\n",
+      "#(b) the propagation coefficient, (c) the attenuation coefficient and\n",
+      "#(d) the phase-shift coefficient\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohm/loop  km\n",
+      "L  =  0.005;#  in  H/loop  km\n",
+      "C  =  0.04E-6;#  in  F/km\n",
+      "G  =  80E-6;#  in  S/km\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  ((R  +  1j*w*L)*(G  +  1j*w*C))**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo  is\",round(abs(Zo),2),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",round(abs(r),4),\"/_\",round(cmath.phase(complex(a,b))*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  attenuation  coefficient  is  \",round(a,4),\"  Np/km\"\n",
+      "print  \"\\n  the  phase-shift  coefficient  \",round(b,4),\"  rad/km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo  is 390.16 /_ -10.43 deg  ohm\n",
+        "\n",
+        "  propagation  coefficient  is   0.1029 /_ 61.92 deg\n",
+        "\n",
+        "  attenuation  coefficient  is   0.0484   Np/km\n",
+        "\n",
+        "  the  phase-shift  coefficient   0.0908   rad/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 881</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the characteristic impedance, \n",
+      "#(b) the propagation coefficient, \n",
+      "#(c) the attenuation and phase-shift coefficients, \n",
+      "#(d) the sending-end current, \n",
+      "#(e) the receiving-end current, \n",
+      "#(f) the wavelength on the line, and \n",
+      "#(g) the speed of transmission of signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8;#  in  ohm/loop  km\n",
+      "L  =  0.003;#  in  H/loop  km\n",
+      "C  =  7500E-12;#  in  F/km\n",
+      "G  =  0.25E-6;#  in  S/km\n",
+      "f  =  1000;#  in  Hz\n",
+      "n  =  300;#  in  km\n",
+      "Zg  =  400  +  1j*0;#  in  ohm\n",
+      "Vg  =  10;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  ((R  +  1j*w*L)*(G  +  1j*w*C))**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      " #the  sending-end  current,\n",
+      "Is  =  Vg/(Zg  +  Zo)\n",
+      " #the  receiving-end  current,\n",
+      "IR  =  Is*cmath.e**(-1*n*r)\n",
+      " #wavelength\n",
+      "Y  =  2*math.pi/b\n",
+      " #velocity  of  propagation  \n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo is\",round(abs(Zo),1),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\"\n",
+      "print  \"propagation  coefficient  is  \",round(abs(r),5),\"/_\",round(cmath.phase(complex(r.real,r.imag))*180/math.pi,2),\"deg\"\n",
+      "print  \"attenuation  coefficient  is  \",round(a,5),\"  Np/km  and  the  phaseshift  coefficient  \",round(b,5),\"  rad/km\"\n",
+      "print  \"sending-end  current  Is  is  \",round(abs(Is)*1E3,3),\"/_\",round(cmath.phase(complex(Is.real,Is.imag))*180/math.pi,2),\"deg  mA\"\n",
+      "print  \"receiving-end  current  IR is\",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg  mA\"\n",
+      "print  \"wavelength  Y  is  \",round(Y,1),\"  km\"\n",
+      "print  \"speed  of  transmission  \",round(u,1),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo is 659.2 /_ -11.35 deg  ohm\n",
+        "propagation  coefficient  is   0.03106 /_ 78.35 deg\n",
+        "attenuation  coefficient  is   0.00627   Np/km  and  the  phaseshift  coefficient   0.03042   rad/km\n",
+        "sending-end  current  Is  is   9.485 /_ 7.07 deg  mA\n",
+        "receiving-end  current  IR is 1.445 /_ -155.88 deg  mA\n",
+        "wavelength  Y  is   206.5   km\n",
+        "speed  of  transmission   206521.1 km/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 884</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine by how much the inductance should be increased to satisfy the condition for minimum distortion.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohm/loop  km\n",
+      "L  =  0.0015;#  in  H/loop  km\n",
+      "C  =  0.06E-6;#  in  F/km\n",
+      "G  =  1.2E-6;#  in  S/km\n",
+      "\n",
+      " #calculation:\n",
+      " #the  condition  for  minimum  distortion  is  given  by  LG  =  CR,  from  which,\n",
+      "Lm  =  C*R/G\n",
+      "dL  =  Lm  -  L\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  should  be  increased  by  \",round(dL*1E3,1),\"mH/loop  km  for  minimum  distortion\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  should  be  increased  by   498.5 mH/loop  km  for  minimum  distortion"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 884</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for minimum distortion at a frequency of 1.5 kHz \n",
+      "#(a) the value of inductance per loop kilometre required, \n",
+      "#(b) the propagation coefficient, \n",
+      "#(c) the velocity of propagation of signal, and \n",
+      "#(d) the wavelength on the line\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  ohm/loop  km\n",
+      "C  =  5E-9;#  in  F/km\n",
+      "G  =  2E-6;#  in  S/km\n",
+      "f  =  1500;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #the  condition  for  minimum  distortion  is  given  by  LG  =  CR,  from  which,  inductance\n",
+      "L  =  C*R/G\n",
+      " #attenuation  coefficient,\n",
+      "a  =  (R*G)**0.5\n",
+      " #phase  shift  coefficient,\n",
+      "b  =  w*(L*C)**0.5\n",
+      " #propagation  coefficient,\n",
+      "r  =  a  +  1j*b\n",
+      " #velocity  of  propagation,\n",
+      "u  =  1/(L*C)**0.5\n",
+      " #wavelength\n",
+      "Y  =  u/f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  is  \",round(L,2),\"  H\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",round(a,2),\"  +(\",round(b,2),\")i\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,2),\"km/sec\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y,2),\"  km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  is   0.2   H\n",
+        "\n",
+        "  propagation  coefficient  is   0.01   +( 0.3 )i\n",
+        "\n",
+        "  speed  of  transmission   31622.78 km/sec\n",
+        "\n",
+        "  wavelength  Y  is   21.08   km\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 888</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the value of (a) the reflection coefficient for the line, (b) the incident current, \n",
+      "#(c) the incident voltage, (d) the reflected current, and (e) the reflected voltage \n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  75;#  in  ohm\n",
+      "ZR  =  250;#  in  ohm\n",
+      "VR  =  10;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #reflection  coefficient\n",
+      "p  =  (Zo  -  ZR)/(Zo  +  ZR)\n",
+      " #Current  flowing  in  the  terminating  load\n",
+      "IR  =  VR/ZR\n",
+      " #incident  current,  Ii\n",
+      "Ii  =  IR/(1  +  p)\n",
+      " #incident  voltage,  Vi    \n",
+      "Vi  =  Ii*Zo\n",
+      " #reflected  current,  Ir\n",
+      "Ir  =  IR  -  Ii\n",
+      " #reflected  voltage,  Vr\n",
+      "Vr  =  -1*Ir*Zo\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient  is  \",round(p,3),\"\"\n",
+      "print  \"\\n  incident  current,  Ii  is  \",round(Ii,4),\"  A\"\n",
+      "print  \"\\n  incident  voltage,  Vi  is  \",round(Vi,2),\"  V\"\n",
+      "print  \"\\n  reflected  current,  Ir  is  \",round(Ir,4),\"  A\"\n",
+      "print  \"\\n  reflected  voltage,  Vr  is  \",round(Vr,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient  is   -0.538 \n",
+        "\n",
+        "  incident  current,  Ii  is   0.0867   A\n",
+        "\n",
+        "  incident  voltage,  Vi  is   6.5   V\n",
+        "\n",
+        "  reflected  current,  Ir  is   -0.0467   A\n",
+        "\n",
+        "  reflected  voltage,  Vr  is   3.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 889</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnitude of the reflection coefficient in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  500  -  1j*40;#  in  ohm\n",
+      "ZR1  =  500  +  1j*40;#  in  ohm\n",
+      "ZR2  =  600  +  1j*0;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #reflection  coefficient\n",
+      "p1  =  (Zo  -  ZR1)/(Zo  +  ZR1)\n",
+      "p2  =  (Zo  -  ZR2)/(Zo  +  ZR2)\n",
+      "p1mag  =  abs(p1)\n",
+      "p2mag  =  abs(p2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient  (a)\",p1mag,\"  and  (b)\", round(p2mag,2),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient  (a) 0.08   and  (b) 0.1 "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 890</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the magnitude of the ratio of the reflected to the incident voltage wave, and \n",
+      "#(b) the incident voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzo  =  500;#  in  ohm\n",
+      "thetazo  =  0;#  in  degrees\n",
+      "ZR  =  320  +  1j*240;#  in  ohm\n",
+      "rvr  =  20;#  in  volts\n",
+      "thetavr  =  35;#  in  degrees\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "VR  =  rvr*math.cos(thetavr*math.pi/180)  +  1j*rvr*math.sin(thetavr*math.pi/180)\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #the  ratio  of  the  reflected  to  the  incident  voltage  \n",
+      " #vr  =  VR/Vi\n",
+      "vr  =  (ZR  -  Zo)/(Zo  +  ZR)\n",
+      "vrmag  =  abs(vr)\n",
+      " #incident  voltage,  Vi\n",
+      "Vi  =  VR/vr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  ratio  Vr  :  Vi  is  \",round(vrmag,3),\"\"\n",
+      "print  \"\\n  incident  voltage,  Vi  is  \",round(abs(Vi),1),\"/_\",round(cmath.phase(complex(Vi.real,Vi.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  ratio  Vr  :  Vi  is   0.351 \n",
+        "\n",
+        "  incident  voltage,  Vi  is   57.0 /_ -75.56 deg  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 895</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the reflection coefficient and (b) the standing-wave ratio.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzo  =  600;#  in  ohm\n",
+      "thetazo  =  0;#  in  degrees\n",
+      "ZR  =  400  +  250j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #reflection  coefficient\n",
+      "p  =  (Zo  -  ZR)/(Zo  +  ZR)\n",
+      "pmag  =  abs(p)\n",
+      " #standing-wave  ratio,\n",
+      "s  =  (1  +  pmag)/(1  -  pmag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient,  is  \",round(abs(p),4),\"/_\",round(cmath.phase(complex(p.real,p.imag))*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  standing-wave  ratio,  s  is  \",round(s,3),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient,  is   0.3106 /_ -65.38 deg\n",
+        "\n",
+        "  standing-wave  ratio,  s  is   1.901 "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 896</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the standing-wave ratio, \n",
+      "#(b) the load impedance, and\n",
+      "#(c) the incident current flowing if the reflected current is 10 mA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rp  =  0.2;  \n",
+      "thetap  =  -120;#  in  degrees\n",
+      "Zo  =  80;#  in  ohm\n",
+      "Ir  =  0.01;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #reflection  coefficient\n",
+      "p  =  rp*math.cos(thetap*math.pi/180)  +  1j*rp*math.sin(thetap*math.pi/180)\n",
+      " #standing-wave  ratio,\n",
+      "s  =  (1  +  rp)/(1  -  rp)\n",
+      " #load  impedance  ZR  \n",
+      "ZR  =  Zo*(1  -  p)/(1  +  p)\n",
+      " #incident  current\n",
+      "Ii  =  Ir*(s  +  1)/(s  -  1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  standing-wave  ratio,  s  is  \",s,\"\"\n",
+      "print  \"\\n  load  impedance  ZR  is  \",round(ZR.real,2),\"  +(\",round(ZR.imag,1),\")i  ohm\"\n",
+      "print  \"\\n  incident  current  is  \",Ii,\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  standing-wave  ratio,  s  is   1.5 \n",
+        "\n",
+        "  load  impedance  ZR  is   91.43   +( 33.0 )i  ohm\n",
+        "\n",
+        "  incident  current  is   0.05   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 897</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of the reflected power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "s  =  1.6;\n",
+      "Pi  =  0.2;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #reflected  power,  Pr\n",
+      "Pr  =  Pi*((s  -  1)/(s  +  1))**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflected  power,  Pr  is  \",round(Pr*1E3,2),\" mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflected  power,  Pr  is   10.65  mW"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_1.ipynb
new file mode 100755
index 00000000..03d830a4
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_1.ipynb
@@ -0,0 +1,1093 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 44: Transmission lines</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 873</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the wavelength on the line, and (b) the speed of transmission of a signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "f  =  1910;#  in  Hz\n",
+      "b  =  0.05;#  in  rad/km\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #wavelength  \n",
+      "Y  =  2*math.pi/b\n",
+      " #speed  of  transmission\n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y,1),\"  km\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,1),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  wavelength  Y  is   125.7   km\n",
+        "\n",
+        "  speed  of  transmission   240017.7 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 873</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for a frequency of operation of 1 kHz, \n",
+      "#(a) the phase delay, \n",
+      "#(b) the wavelength on the line, and \n",
+      "#(c) the velocity of propagation (in metres per second) of the signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.004;#  in  Henry/loop\n",
+      "C  =  0.004E-6;#  in  F/loop\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #phase  delay\n",
+      "b  =  w*(L*C)**0.5\n",
+      " #wavelength  \n",
+      "Y  =  2*math.pi/b\n",
+      " #speed  of  transmission\n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  phase  delay  is  \",round(b,3),\"  rad/km\"\n",
+      "print  \"\\n  wavelength  Y  is  \",Y,\"  km\"\n",
+      "print  \"\\n  speed  of  transmission  \",u,\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  phase  delay  is   0.025   rad/km\n",
+        "\n",
+        "  wavelength  Y  is   250.0   km\n",
+        "\n",
+        "  speed  of  transmission   250000.0 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 874</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the voltage at a point 10 km down the line,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "a  =  0.25;#  in  Np/km\n",
+      "b  =  0.20;#  in  rad/km\n",
+      "Vs  =  5;#  in  Volts\n",
+      "n  =  10;#  in  km\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #the  voltage  10  km  down  the  line\n",
+      "r  =  a  +  1j*b\n",
+      "VR  =  Vs*cmath.e**(-1*n*r)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n Result  \\n\\n\"\n",
+      "print  \"voltage  10  km  down  the  line is \",round(abs(VR),2),\"/_\",round(cmath.phase(complex(VR.real,VR.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Result  \n",
+        "\n",
+        "\n",
+        "voltage  10  km  down  the  line is  0.41 /_ -114.59 deg  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 875</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnitude and phase of the current at the receiving end,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "a  =  0.5;#  in  Np/km\n",
+      "b  =  0.25;#  in  rad/km\n",
+      "rvs  =  2;#  in  Volts\n",
+      "thetavs  =  0;#  in  degrees\n",
+      "rzo  =  800;#  in  ohm\n",
+      "thetazo  =  -25;#  in  degrees\n",
+      "n  =  5;#  in  km\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "Vs  =  rvs*math.cos(thetavs*math.pi/180)  +  1j*rvs*math.sin(thetavs*math.pi/180)\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #  receiving  end  voltage\n",
+      "r  =  a  +  1j*b\n",
+      "VR  =  Vs*cmath.e**(-1*n*r)\n",
+      " #Receiving  end  current,\n",
+      "IR  =  VR/Zo\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"Receiving end current, IR is  \",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg  mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "Receiving end current, IR is   0.205 /_ -46.62 deg  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 875</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the output voltage if the length of the line is doubled.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vs  =  8;#  in  Volts\n",
+      "VR  =  2;#  in  Volts\n",
+      "x  =  2;  \n",
+      "\n",
+      "#calculation:\n",
+      " #  receiving  end  voltage  VR  =  Vs*e**(-nr)\n",
+      " #e**-nr  =  p\n",
+      "p  =  VR/Vs\n",
+      " #If  the  line  is  doubled  in  length,  then\n",
+      "VR  =  Vs*(p)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Receiving  end  voltage  If  the  line  is  doubled  in  length,  VR  is  \",abs(VR),\" V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Receiving  end  voltage  If  the  line  is  doubled  in  length,  VR  is   0.5  V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 876</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the characteristic impedance of the line at this frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzoc  =  800;#  in  ohm\n",
+      "thetazoc  =  -50;#  in  degrees\n",
+      "rzsc  =  413;#  in  ohm\n",
+      "thetazsc  =  -20;#  in  degrees\n",
+      "f  =  1500;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #open  circuit  impedance\n",
+      "Zoc  =  rzoc*math.cos(thetazoc*math.pi/180)  +  1j*rzoc*math.sin(thetazoc*math.pi/180)\n",
+      " #short  circuit  impedance\n",
+      "Zsc  =  rzsc*math.cos(thetazsc*math.pi/180)  +  1j*rzsc*math.sin(thetazsc*math.pi/180)\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  (Zoc*Zsc)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo is\",round(abs(Zo)),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo is 575.0 /_ -35.0 deg  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 877</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the characteristic impedance of the line when the frequency is 2 kHz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohm/loop  km\n",
+      "L  =  0.0034;#  in  H/loop  km\n",
+      "C  =  10E-9;#  in  F/km\n",
+      "G  =  3E-6;#  in  S/km\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo  is  \",round(abs(Zo),0),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo  is   600.0 /_ -8.99 deg  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 879</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, at an operating frequency of 400 kHz, (a) the characteristic impedance,\n",
+      "#(b) the propagation coefficient, (c) the wavelength on the line, and \n",
+      "#(d) the velocity of propagation, in metres per second, of a signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.0005;#  in  H/loop  km\n",
+      "C  =  0.12E-6;#  in  F/km\n",
+      "f  =  400000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  (L/C)**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  1j*w*(L*C)**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      " #wavelength\n",
+      "Y  =  2*math.pi/b\n",
+      " #velocity  of  propagation  \n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  characteristic  impedance  Zo  is  \",abs(Zo),\"ohm\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",a,\"  +(\",round(b,2),\")i\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y*1E3,0),\"m\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,2),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  characteristic  impedance  Zo  is   64.5497224368 ohm\n",
+        "\n",
+        "  propagation  coefficient  is   0.0   +( 19.47 )i\n",
+        "\n",
+        "  wavelength  Y  is   323.0 m\n",
+        "\n",
+        "  speed  of  transmission   129099.44 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 880</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the line (a) the characteristic impedance,\n",
+      "#(b) the propagation coefficient, (c) the attenuation coefficient and\n",
+      "#(d) the phase-shift coefficient\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohm/loop  km\n",
+      "L  =  0.005;#  in  H/loop  km\n",
+      "C  =  0.04E-6;#  in  F/km\n",
+      "G  =  80E-6;#  in  S/km\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  ((R  +  1j*w*L)*(G  +  1j*w*C))**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo  is\",round(abs(Zo),2),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",round(abs(r),4),\"/_\",round(cmath.phase(complex(a,b))*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  attenuation  coefficient  is  \",round(a,4),\"  Np/km\"\n",
+      "print  \"\\n  the  phase-shift  coefficient  \",round(b,4),\"  rad/km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo  is 390.16 /_ -10.43 deg  ohm\n",
+        "\n",
+        "  propagation  coefficient  is   0.1029 /_ 61.92 deg\n",
+        "\n",
+        "  attenuation  coefficient  is   0.0484   Np/km\n",
+        "\n",
+        "  the  phase-shift  coefficient   0.0908   rad/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 881</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the characteristic impedance, \n",
+      "#(b) the propagation coefficient, \n",
+      "#(c) the attenuation and phase-shift coefficients, \n",
+      "#(d) the sending-end current, \n",
+      "#(e) the receiving-end current, \n",
+      "#(f) the wavelength on the line, and \n",
+      "#(g) the speed of transmission of signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8;#  in  ohm/loop  km\n",
+      "L  =  0.003;#  in  H/loop  km\n",
+      "C  =  7500E-12;#  in  F/km\n",
+      "G  =  0.25E-6;#  in  S/km\n",
+      "f  =  1000;#  in  Hz\n",
+      "n  =  300;#  in  km\n",
+      "Zg  =  400  +  1j*0;#  in  ohm\n",
+      "Vg  =  10;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  ((R  +  1j*w*L)*(G  +  1j*w*C))**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      " #the  sending-end  current,\n",
+      "Is  =  Vg/(Zg  +  Zo)\n",
+      " #the  receiving-end  current,\n",
+      "IR  =  Is*cmath.e**(-1*n*r)\n",
+      " #wavelength\n",
+      "Y  =  2*math.pi/b\n",
+      " #velocity  of  propagation  \n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo is\",round(abs(Zo),1),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\"\n",
+      "print  \"propagation  coefficient  is  \",round(abs(r),5),\"/_\",round(cmath.phase(complex(r.real,r.imag))*180/math.pi,2),\"deg\"\n",
+      "print  \"attenuation  coefficient  is  \",round(a,5),\"  Np/km  and  the  phaseshift  coefficient  \",round(b,5),\"  rad/km\"\n",
+      "print  \"sending-end  current  Is  is  \",round(abs(Is)*1E3,3),\"/_\",round(cmath.phase(complex(Is.real,Is.imag))*180/math.pi,2),\"deg  mA\"\n",
+      "print  \"receiving-end  current  IR is\",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg  mA\"\n",
+      "print  \"wavelength  Y  is  \",round(Y,1),\"  km\"\n",
+      "print  \"speed  of  transmission  \",round(u,1),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo is 659.2 /_ -11.35 deg  ohm\n",
+        "propagation  coefficient  is   0.03106 /_ 78.35 deg\n",
+        "attenuation  coefficient  is   0.00627   Np/km  and  the  phaseshift  coefficient   0.03042   rad/km\n",
+        "sending-end  current  Is  is   9.485 /_ 7.07 deg  mA\n",
+        "receiving-end  current  IR is 1.445 /_ -155.88 deg  mA\n",
+        "wavelength  Y  is   206.5   km\n",
+        "speed  of  transmission   206521.1 km/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 884</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine by how much the inductance should be increased to satisfy the condition for minimum distortion.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohm/loop  km\n",
+      "L  =  0.0015;#  in  H/loop  km\n",
+      "C  =  0.06E-6;#  in  F/km\n",
+      "G  =  1.2E-6;#  in  S/km\n",
+      "\n",
+      " #calculation:\n",
+      " #the  condition  for  minimum  distortion  is  given  by  LG  =  CR,  from  which,\n",
+      "Lm  =  C*R/G\n",
+      "dL  =  Lm  -  L\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  should  be  increased  by  \",round(dL*1E3,1),\"mH/loop  km  for  minimum  distortion\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  should  be  increased  by   498.5 mH/loop  km  for  minimum  distortion"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 884</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for minimum distortion at a frequency of 1.5 kHz \n",
+      "#(a) the value of inductance per loop kilometre required, \n",
+      "#(b) the propagation coefficient, \n",
+      "#(c) the velocity of propagation of signal, and \n",
+      "#(d) the wavelength on the line\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  ohm/loop  km\n",
+      "C  =  5E-9;#  in  F/km\n",
+      "G  =  2E-6;#  in  S/km\n",
+      "f  =  1500;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #the  condition  for  minimum  distortion  is  given  by  LG  =  CR,  from  which,  inductance\n",
+      "L  =  C*R/G\n",
+      " #attenuation  coefficient,\n",
+      "a  =  (R*G)**0.5\n",
+      " #phase  shift  coefficient,\n",
+      "b  =  w*(L*C)**0.5\n",
+      " #propagation  coefficient,\n",
+      "r  =  a  +  1j*b\n",
+      " #velocity  of  propagation,\n",
+      "u  =  1/(L*C)**0.5\n",
+      " #wavelength\n",
+      "Y  =  u/f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  is  \",round(L,2),\"  H\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",round(a,2),\"  +(\",round(b,2),\")i\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,2),\"km/sec\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y,2),\"  km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  is   0.2   H\n",
+        "\n",
+        "  propagation  coefficient  is   0.01   +( 0.3 )i\n",
+        "\n",
+        "  speed  of  transmission   31622.78 km/sec\n",
+        "\n",
+        "  wavelength  Y  is   21.08   km\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 888</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the value of (a) the reflection coefficient for the line, (b) the incident current, \n",
+      "#(c) the incident voltage, (d) the reflected current, and (e) the reflected voltage \n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  75;#  in  ohm\n",
+      "ZR  =  250;#  in  ohm\n",
+      "VR  =  10;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #reflection  coefficient\n",
+      "p  =  (Zo  -  ZR)/(Zo  +  ZR)\n",
+      " #Current  flowing  in  the  terminating  load\n",
+      "IR  =  VR/ZR\n",
+      " #incident  current,  Ii\n",
+      "Ii  =  IR/(1  +  p)\n",
+      " #incident  voltage,  Vi    \n",
+      "Vi  =  Ii*Zo\n",
+      " #reflected  current,  Ir\n",
+      "Ir  =  IR  -  Ii\n",
+      " #reflected  voltage,  Vr\n",
+      "Vr  =  -1*Ir*Zo\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient  is  \",round(p,3),\"\"\n",
+      "print  \"\\n  incident  current,  Ii  is  \",round(Ii,4),\"  A\"\n",
+      "print  \"\\n  incident  voltage,  Vi  is  \",round(Vi,2),\"  V\"\n",
+      "print  \"\\n  reflected  current,  Ir  is  \",round(Ir,4),\"  A\"\n",
+      "print  \"\\n  reflected  voltage,  Vr  is  \",round(Vr,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient  is   -0.538 \n",
+        "\n",
+        "  incident  current,  Ii  is   0.0867   A\n",
+        "\n",
+        "  incident  voltage,  Vi  is   6.5   V\n",
+        "\n",
+        "  reflected  current,  Ir  is   -0.0467   A\n",
+        "\n",
+        "  reflected  voltage,  Vr  is   3.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 889</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnitude of the reflection coefficient in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  500  -  1j*40;#  in  ohm\n",
+      "ZR1  =  500  +  1j*40;#  in  ohm\n",
+      "ZR2  =  600  +  1j*0;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #reflection  coefficient\n",
+      "p1  =  (Zo  -  ZR1)/(Zo  +  ZR1)\n",
+      "p2  =  (Zo  -  ZR2)/(Zo  +  ZR2)\n",
+      "p1mag  =  abs(p1)\n",
+      "p2mag  =  abs(p2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient  (a)\",p1mag,\"  and  (b)\", round(p2mag,2),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient  (a) 0.08   and  (b) 0.1 "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 890</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the magnitude of the ratio of the reflected to the incident voltage wave, and \n",
+      "#(b) the incident voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzo  =  500;#  in  ohm\n",
+      "thetazo  =  0;#  in  degrees\n",
+      "ZR  =  320  +  1j*240;#  in  ohm\n",
+      "rvr  =  20;#  in  volts\n",
+      "thetavr  =  35;#  in  degrees\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "VR  =  rvr*math.cos(thetavr*math.pi/180)  +  1j*rvr*math.sin(thetavr*math.pi/180)\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #the  ratio  of  the  reflected  to  the  incident  voltage  \n",
+      " #vr  =  VR/Vi\n",
+      "vr  =  (ZR  -  Zo)/(Zo  +  ZR)\n",
+      "vrmag  =  abs(vr)\n",
+      " #incident  voltage,  Vi\n",
+      "Vi  =  VR/vr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  ratio  Vr  :  Vi  is  \",round(vrmag,3),\"\"\n",
+      "print  \"\\n  incident  voltage,  Vi  is  \",round(abs(Vi),1),\"/_\",round(cmath.phase(complex(Vi.real,Vi.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  ratio  Vr  :  Vi  is   0.351 \n",
+        "\n",
+        "  incident  voltage,  Vi  is   57.0 /_ -75.56 deg  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 895</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the reflection coefficient and (b) the standing-wave ratio.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzo  =  600;#  in  ohm\n",
+      "thetazo  =  0;#  in  degrees\n",
+      "ZR  =  400  +  250j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #reflection  coefficient\n",
+      "p  =  (Zo  -  ZR)/(Zo  +  ZR)\n",
+      "pmag  =  abs(p)\n",
+      " #standing-wave  ratio,\n",
+      "s  =  (1  +  pmag)/(1  -  pmag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient,  is  \",round(abs(p),4),\"/_\",round(cmath.phase(complex(p.real,p.imag))*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  standing-wave  ratio,  s  is  \",round(s,3),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient,  is   0.3106 /_ -65.38 deg\n",
+        "\n",
+        "  standing-wave  ratio,  s  is   1.901 "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 896</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the standing-wave ratio, \n",
+      "#(b) the load impedance, and\n",
+      "#(c) the incident current flowing if the reflected current is 10 mA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rp  =  0.2;  \n",
+      "thetap  =  -120;#  in  degrees\n",
+      "Zo  =  80;#  in  ohm\n",
+      "Ir  =  0.01;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #reflection  coefficient\n",
+      "p  =  rp*math.cos(thetap*math.pi/180)  +  1j*rp*math.sin(thetap*math.pi/180)\n",
+      " #standing-wave  ratio,\n",
+      "s  =  (1  +  rp)/(1  -  rp)\n",
+      " #load  impedance  ZR  \n",
+      "ZR  =  Zo*(1  -  p)/(1  +  p)\n",
+      " #incident  current\n",
+      "Ii  =  Ir*(s  +  1)/(s  -  1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  standing-wave  ratio,  s  is  \",s,\"\"\n",
+      "print  \"\\n  load  impedance  ZR  is  \",round(ZR.real,2),\"  +(\",round(ZR.imag,1),\")i  ohm\"\n",
+      "print  \"\\n  incident  current  is  \",Ii,\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  standing-wave  ratio,  s  is   1.5 \n",
+        "\n",
+        "  load  impedance  ZR  is   91.43   +( 33.0 )i  ohm\n",
+        "\n",
+        "  incident  current  is   0.05   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 897</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of the reflected power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "s  =  1.6;\n",
+      "Pi  =  0.2;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #reflected  power,  Pr\n",
+      "Pr  =  Pi*((s  -  1)/(s  +  1))**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflected  power,  Pr  is  \",round(Pr*1E3,2),\" mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflected  power,  Pr  is   10.65  mW"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_2.ipynb
new file mode 100755
index 00000000..03d830a4
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_2.ipynb
@@ -0,0 +1,1093 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 44: Transmission lines</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 873</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the wavelength on the line, and (b) the speed of transmission of a signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "f  =  1910;#  in  Hz\n",
+      "b  =  0.05;#  in  rad/km\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #wavelength  \n",
+      "Y  =  2*math.pi/b\n",
+      " #speed  of  transmission\n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y,1),\"  km\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,1),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  wavelength  Y  is   125.7   km\n",
+        "\n",
+        "  speed  of  transmission   240017.7 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 873</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for a frequency of operation of 1 kHz, \n",
+      "#(a) the phase delay, \n",
+      "#(b) the wavelength on the line, and \n",
+      "#(c) the velocity of propagation (in metres per second) of the signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.004;#  in  Henry/loop\n",
+      "C  =  0.004E-6;#  in  F/loop\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #phase  delay\n",
+      "b  =  w*(L*C)**0.5\n",
+      " #wavelength  \n",
+      "Y  =  2*math.pi/b\n",
+      " #speed  of  transmission\n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  phase  delay  is  \",round(b,3),\"  rad/km\"\n",
+      "print  \"\\n  wavelength  Y  is  \",Y,\"  km\"\n",
+      "print  \"\\n  speed  of  transmission  \",u,\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  phase  delay  is   0.025   rad/km\n",
+        "\n",
+        "  wavelength  Y  is   250.0   km\n",
+        "\n",
+        "  speed  of  transmission   250000.0 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 874</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the voltage at a point 10 km down the line,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "a  =  0.25;#  in  Np/km\n",
+      "b  =  0.20;#  in  rad/km\n",
+      "Vs  =  5;#  in  Volts\n",
+      "n  =  10;#  in  km\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #the  voltage  10  km  down  the  line\n",
+      "r  =  a  +  1j*b\n",
+      "VR  =  Vs*cmath.e**(-1*n*r)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n Result  \\n\\n\"\n",
+      "print  \"voltage  10  km  down  the  line is \",round(abs(VR),2),\"/_\",round(cmath.phase(complex(VR.real,VR.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Result  \n",
+        "\n",
+        "\n",
+        "voltage  10  km  down  the  line is  0.41 /_ -114.59 deg  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 875</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnitude and phase of the current at the receiving end,\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "a  =  0.5;#  in  Np/km\n",
+      "b  =  0.25;#  in  rad/km\n",
+      "rvs  =  2;#  in  Volts\n",
+      "thetavs  =  0;#  in  degrees\n",
+      "rzo  =  800;#  in  ohm\n",
+      "thetazo  =  -25;#  in  degrees\n",
+      "n  =  5;#  in  km\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "Vs  =  rvs*math.cos(thetavs*math.pi/180)  +  1j*rvs*math.sin(thetavs*math.pi/180)\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #  receiving  end  voltage\n",
+      "r  =  a  +  1j*b\n",
+      "VR  =  Vs*cmath.e**(-1*n*r)\n",
+      " #Receiving  end  current,\n",
+      "IR  =  VR/Zo\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"Receiving end current, IR is  \",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg  mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "Receiving end current, IR is   0.205 /_ -46.62 deg  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 875</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the output voltage if the length of the line is doubled.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vs  =  8;#  in  Volts\n",
+      "VR  =  2;#  in  Volts\n",
+      "x  =  2;  \n",
+      "\n",
+      "#calculation:\n",
+      " #  receiving  end  voltage  VR  =  Vs*e**(-nr)\n",
+      " #e**-nr  =  p\n",
+      "p  =  VR/Vs\n",
+      " #If  the  line  is  doubled  in  length,  then\n",
+      "VR  =  Vs*(p)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Receiving  end  voltage  If  the  line  is  doubled  in  length,  VR  is  \",abs(VR),\" V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Receiving  end  voltage  If  the  line  is  doubled  in  length,  VR  is   0.5  V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 876</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the characteristic impedance of the line at this frequency.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzoc  =  800;#  in  ohm\n",
+      "thetazoc  =  -50;#  in  degrees\n",
+      "rzsc  =  413;#  in  ohm\n",
+      "thetazsc  =  -20;#  in  degrees\n",
+      "f  =  1500;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #open  circuit  impedance\n",
+      "Zoc  =  rzoc*math.cos(thetazoc*math.pi/180)  +  1j*rzoc*math.sin(thetazoc*math.pi/180)\n",
+      " #short  circuit  impedance\n",
+      "Zsc  =  rzsc*math.cos(thetazsc*math.pi/180)  +  1j*rzsc*math.sin(thetazsc*math.pi/180)\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  (Zoc*Zsc)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo is\",round(abs(Zo)),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo is 575.0 /_ -35.0 deg  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 877</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the characteristic impedance of the line when the frequency is 2 kHz.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohm/loop  km\n",
+      "L  =  0.0034;#  in  H/loop  km\n",
+      "C  =  10E-9;#  in  F/km\n",
+      "G  =  3E-6;#  in  S/km\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo  is  \",round(abs(Zo),0),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo  is   600.0 /_ -8.99 deg  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 879</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, at an operating frequency of 400 kHz, (a) the characteristic impedance,\n",
+      "#(b) the propagation coefficient, (c) the wavelength on the line, and \n",
+      "#(d) the velocity of propagation, in metres per second, of a signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.0005;#  in  H/loop  km\n",
+      "C  =  0.12E-6;#  in  F/km\n",
+      "f  =  400000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  (L/C)**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  1j*w*(L*C)**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      " #wavelength\n",
+      "Y  =  2*math.pi/b\n",
+      " #velocity  of  propagation  \n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  characteristic  impedance  Zo  is  \",abs(Zo),\"ohm\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",a,\"  +(\",round(b,2),\")i\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y*1E3,0),\"m\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,2),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  characteristic  impedance  Zo  is   64.5497224368 ohm\n",
+        "\n",
+        "  propagation  coefficient  is   0.0   +( 19.47 )i\n",
+        "\n",
+        "  wavelength  Y  is   323.0 m\n",
+        "\n",
+        "  speed  of  transmission   129099.44 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 880</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine for the line (a) the characteristic impedance,\n",
+      "#(b) the propagation coefficient, (c) the attenuation coefficient and\n",
+      "#(d) the phase-shift coefficient\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohm/loop  km\n",
+      "L  =  0.005;#  in  H/loop  km\n",
+      "C  =  0.04E-6;#  in  F/km\n",
+      "G  =  80E-6;#  in  S/km\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  ((R  +  1j*w*L)*(G  +  1j*w*C))**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo  is\",round(abs(Zo),2),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",round(abs(r),4),\"/_\",round(cmath.phase(complex(a,b))*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  attenuation  coefficient  is  \",round(a,4),\"  Np/km\"\n",
+      "print  \"\\n  the  phase-shift  coefficient  \",round(b,4),\"  rad/km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo  is 390.16 /_ -10.43 deg  ohm\n",
+        "\n",
+        "  propagation  coefficient  is   0.1029 /_ 61.92 deg\n",
+        "\n",
+        "  attenuation  coefficient  is   0.0484   Np/km\n",
+        "\n",
+        "  the  phase-shift  coefficient   0.0908   rad/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 881</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the characteristic impedance, \n",
+      "#(b) the propagation coefficient, \n",
+      "#(c) the attenuation and phase-shift coefficients, \n",
+      "#(d) the sending-end current, \n",
+      "#(e) the receiving-end current, \n",
+      "#(f) the wavelength on the line, and \n",
+      "#(g) the speed of transmission of signal.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8;#  in  ohm/loop  km\n",
+      "L  =  0.003;#  in  H/loop  km\n",
+      "C  =  7500E-12;#  in  F/km\n",
+      "G  =  0.25E-6;#  in  S/km\n",
+      "f  =  1000;#  in  Hz\n",
+      "n  =  300;#  in  km\n",
+      "Zg  =  400  +  1j*0;#  in  ohm\n",
+      "Vg  =  10;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  ((R  +  1j*w*L)*(G  +  1j*w*C))**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      " #the  sending-end  current,\n",
+      "Is  =  Vg/(Zg  +  Zo)\n",
+      " #the  receiving-end  current,\n",
+      "IR  =  Is*cmath.e**(-1*n*r)\n",
+      " #wavelength\n",
+      "Y  =  2*math.pi/b\n",
+      " #velocity  of  propagation  \n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo is\",round(abs(Zo),1),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\"\n",
+      "print  \"propagation  coefficient  is  \",round(abs(r),5),\"/_\",round(cmath.phase(complex(r.real,r.imag))*180/math.pi,2),\"deg\"\n",
+      "print  \"attenuation  coefficient  is  \",round(a,5),\"  Np/km  and  the  phaseshift  coefficient  \",round(b,5),\"  rad/km\"\n",
+      "print  \"sending-end  current  Is  is  \",round(abs(Is)*1E3,3),\"/_\",round(cmath.phase(complex(Is.real,Is.imag))*180/math.pi,2),\"deg  mA\"\n",
+      "print  \"receiving-end  current  IR is\",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg  mA\"\n",
+      "print  \"wavelength  Y  is  \",round(Y,1),\"  km\"\n",
+      "print  \"speed  of  transmission  \",round(u,1),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo is 659.2 /_ -11.35 deg  ohm\n",
+        "propagation  coefficient  is   0.03106 /_ 78.35 deg\n",
+        "attenuation  coefficient  is   0.00627   Np/km  and  the  phaseshift  coefficient   0.03042   rad/km\n",
+        "sending-end  current  Is  is   9.485 /_ 7.07 deg  mA\n",
+        "receiving-end  current  IR is 1.445 /_ -155.88 deg  mA\n",
+        "wavelength  Y  is   206.5   km\n",
+        "speed  of  transmission   206521.1 km/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 884</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine by how much the inductance should be increased to satisfy the condition for minimum distortion.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohm/loop  km\n",
+      "L  =  0.0015;#  in  H/loop  km\n",
+      "C  =  0.06E-6;#  in  F/km\n",
+      "G  =  1.2E-6;#  in  S/km\n",
+      "\n",
+      " #calculation:\n",
+      " #the  condition  for  minimum  distortion  is  given  by  LG  =  CR,  from  which,\n",
+      "Lm  =  C*R/G\n",
+      "dL  =  Lm  -  L\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  should  be  increased  by  \",round(dL*1E3,1),\"mH/loop  km  for  minimum  distortion\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  should  be  increased  by   498.5 mH/loop  km  for  minimum  distortion"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 884</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine, for minimum distortion at a frequency of 1.5 kHz \n",
+      "#(a) the value of inductance per loop kilometre required, \n",
+      "#(b) the propagation coefficient, \n",
+      "#(c) the velocity of propagation of signal, and \n",
+      "#(d) the wavelength on the line\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  ohm/loop  km\n",
+      "C  =  5E-9;#  in  F/km\n",
+      "G  =  2E-6;#  in  S/km\n",
+      "f  =  1500;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #the  condition  for  minimum  distortion  is  given  by  LG  =  CR,  from  which,  inductance\n",
+      "L  =  C*R/G\n",
+      " #attenuation  coefficient,\n",
+      "a  =  (R*G)**0.5\n",
+      " #phase  shift  coefficient,\n",
+      "b  =  w*(L*C)**0.5\n",
+      " #propagation  coefficient,\n",
+      "r  =  a  +  1j*b\n",
+      " #velocity  of  propagation,\n",
+      "u  =  1/(L*C)**0.5\n",
+      " #wavelength\n",
+      "Y  =  u/f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  is  \",round(L,2),\"  H\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",round(a,2),\"  +(\",round(b,2),\")i\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,2),\"km/sec\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y,2),\"  km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  is   0.2   H\n",
+        "\n",
+        "  propagation  coefficient  is   0.01   +( 0.3 )i\n",
+        "\n",
+        "  speed  of  transmission   31622.78 km/sec\n",
+        "\n",
+        "  wavelength  Y  is   21.08   km\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 888</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#calculate the value of (a) the reflection coefficient for the line, (b) the incident current, \n",
+      "#(c) the incident voltage, (d) the reflected current, and (e) the reflected voltage \n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  75;#  in  ohm\n",
+      "ZR  =  250;#  in  ohm\n",
+      "VR  =  10;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #reflection  coefficient\n",
+      "p  =  (Zo  -  ZR)/(Zo  +  ZR)\n",
+      " #Current  flowing  in  the  terminating  load\n",
+      "IR  =  VR/ZR\n",
+      " #incident  current,  Ii\n",
+      "Ii  =  IR/(1  +  p)\n",
+      " #incident  voltage,  Vi    \n",
+      "Vi  =  Ii*Zo\n",
+      " #reflected  current,  Ir\n",
+      "Ir  =  IR  -  Ii\n",
+      " #reflected  voltage,  Vr\n",
+      "Vr  =  -1*Ir*Zo\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient  is  \",round(p,3),\"\"\n",
+      "print  \"\\n  incident  current,  Ii  is  \",round(Ii,4),\"  A\"\n",
+      "print  \"\\n  incident  voltage,  Vi  is  \",round(Vi,2),\"  V\"\n",
+      "print  \"\\n  reflected  current,  Ir  is  \",round(Ir,4),\"  A\"\n",
+      "print  \"\\n  reflected  voltage,  Vr  is  \",round(Vr,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient  is   -0.538 \n",
+        "\n",
+        "  incident  current,  Ii  is   0.0867   A\n",
+        "\n",
+        "  incident  voltage,  Vi  is   6.5   V\n",
+        "\n",
+        "  reflected  current,  Ir  is   -0.0467   A\n",
+        "\n",
+        "  reflected  voltage,  Vr  is   3.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 889</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the magnitude of the reflection coefficient in each case.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  500  -  1j*40;#  in  ohm\n",
+      "ZR1  =  500  +  1j*40;#  in  ohm\n",
+      "ZR2  =  600  +  1j*0;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #reflection  coefficient\n",
+      "p1  =  (Zo  -  ZR1)/(Zo  +  ZR1)\n",
+      "p2  =  (Zo  -  ZR2)/(Zo  +  ZR2)\n",
+      "p1mag  =  abs(p1)\n",
+      "p2mag  =  abs(p2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient  (a)\",p1mag,\"  and  (b)\", round(p2mag,2),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient  (a) 0.08   and  (b) 0.1 "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 890</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the magnitude of the ratio of the reflected to the incident voltage wave, and \n",
+      "#(b) the incident voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzo  =  500;#  in  ohm\n",
+      "thetazo  =  0;#  in  degrees\n",
+      "ZR  =  320  +  1j*240;#  in  ohm\n",
+      "rvr  =  20;#  in  volts\n",
+      "thetavr  =  35;#  in  degrees\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "VR  =  rvr*math.cos(thetavr*math.pi/180)  +  1j*rvr*math.sin(thetavr*math.pi/180)\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #the  ratio  of  the  reflected  to  the  incident  voltage  \n",
+      " #vr  =  VR/Vi\n",
+      "vr  =  (ZR  -  Zo)/(Zo  +  ZR)\n",
+      "vrmag  =  abs(vr)\n",
+      " #incident  voltage,  Vi\n",
+      "Vi  =  VR/vr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  ratio  Vr  :  Vi  is  \",round(vrmag,3),\"\"\n",
+      "print  \"\\n  incident  voltage,  Vi  is  \",round(abs(Vi),1),\"/_\",round(cmath.phase(complex(Vi.real,Vi.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  ratio  Vr  :  Vi  is   0.351 \n",
+        "\n",
+        "  incident  voltage,  Vi  is   57.0 /_ -75.56 deg  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 895</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine (a) the reflection coefficient and (b) the standing-wave ratio.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzo  =  600;#  in  ohm\n",
+      "thetazo  =  0;#  in  degrees\n",
+      "ZR  =  400  +  250j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #reflection  coefficient\n",
+      "p  =  (Zo  -  ZR)/(Zo  +  ZR)\n",
+      "pmag  =  abs(p)\n",
+      " #standing-wave  ratio,\n",
+      "s  =  (1  +  pmag)/(1  -  pmag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient,  is  \",round(abs(p),4),\"/_\",round(cmath.phase(complex(p.real,p.imag))*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  standing-wave  ratio,  s  is  \",round(s,3),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient,  is   0.3106 /_ -65.38 deg\n",
+        "\n",
+        "  standing-wave  ratio,  s  is   1.901 "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 896</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the standing-wave ratio, \n",
+      "#(b) the load impedance, and\n",
+      "#(c) the incident current flowing if the reflected current is 10 mA.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rp  =  0.2;  \n",
+      "thetap  =  -120;#  in  degrees\n",
+      "Zo  =  80;#  in  ohm\n",
+      "Ir  =  0.01;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #reflection  coefficient\n",
+      "p  =  rp*math.cos(thetap*math.pi/180)  +  1j*rp*math.sin(thetap*math.pi/180)\n",
+      " #standing-wave  ratio,\n",
+      "s  =  (1  +  rp)/(1  -  rp)\n",
+      " #load  impedance  ZR  \n",
+      "ZR  =  Zo*(1  -  p)/(1  +  p)\n",
+      " #incident  current\n",
+      "Ii  =  Ir*(s  +  1)/(s  -  1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  standing-wave  ratio,  s  is  \",s,\"\"\n",
+      "print  \"\\n  load  impedance  ZR  is  \",round(ZR.real,2),\"  +(\",round(ZR.imag,1),\")i  ohm\"\n",
+      "print  \"\\n  incident  current  is  \",Ii,\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  standing-wave  ratio,  s  is   1.5 \n",
+        "\n",
+        "  load  impedance  ZR  is   91.43   +( 33.0 )i  ohm\n",
+        "\n",
+        "  incident  current  is   0.05   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 897</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#determine the value of the reflected power.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "s  =  1.6;\n",
+      "Pi  =  0.2;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #reflected  power,  Pr\n",
+      "Pr  =  Pi*((s  -  1)/(s  +  1))**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflected  power,  Pr  is  \",round(Pr*1E3,2),\" mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflected  power,  Pr  is   10.65  mW"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_3.ipynb
new file mode 100755
index 00000000..61794674
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_44-checkpoint_3.ipynb
@@ -0,0 +1,1074 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:09f43d439565c21fbc9d48d0b329d310879f650b07dbafd58f9c9e880189ce2d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 44: Transmission lines</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 873</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "f  =  1910;#  in  Hz\n",
+      "b  =  0.05;#  in  rad/km\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #wavelength  \n",
+      "Y  =  2*math.pi/b\n",
+      " #speed  of  transmission\n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y,1),\"  km\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,1),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  wavelength  Y  is   125.7   km\n",
+        "\n",
+        "  speed  of  transmission   240017.7 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 873</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.004;#  in  Henry/loop\n",
+      "C  =  0.004E-6;#  in  F/loop\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #phase  delay\n",
+      "b  =  w*(L*C)**0.5\n",
+      " #wavelength  \n",
+      "Y  =  2*math.pi/b\n",
+      " #speed  of  transmission\n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  phase  delay  is  \",round(b,3),\"  rad/km\"\n",
+      "print  \"\\n  wavelength  Y  is  \",Y,\"  km\"\n",
+      "print  \"\\n  speed  of  transmission  \",u,\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  phase  delay  is   0.025   rad/km\n",
+        "\n",
+        "  wavelength  Y  is   250.0   km\n",
+        "\n",
+        "  speed  of  transmission   250000.0 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 874</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "a  =  0.25;#  in  Np/km\n",
+      "b  =  0.20;#  in  rad/km\n",
+      "Vs  =  5;#  in  Volts\n",
+      "n  =  10;#  in  km\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #the  voltage  10  km  down  the  line\n",
+      "r  =  a  +  1j*b\n",
+      "VR  =  Vs*cmath.e**(-1*n*r)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n Result  \\n\\n\"\n",
+      "print  \"voltage  10  km  down  the  line is \",round(abs(VR),2),\"/_\",round(cmath.phase(complex(VR.real,VR.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Result  \n",
+        "\n",
+        "\n",
+        "voltage  10  km  down  the  line is  0.41 /_ -114.59 deg  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 875</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "a  =  0.5;#  in  Np/km\n",
+      "b  =  0.25;#  in  rad/km\n",
+      "rvs  =  2;#  in  Volts\n",
+      "thetavs  =  0;#  in  degrees\n",
+      "rzo  =  800;#  in  ohm\n",
+      "thetazo  =  -25;#  in  degrees\n",
+      "n  =  5;#  in  km\n",
+      "\n",
+      "#calculation:\n",
+      " #voltage\n",
+      "Vs  =  rvs*math.cos(thetavs*math.pi/180)  +  1j*rvs*math.sin(thetavs*math.pi/180)\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #  receiving  end  voltage\n",
+      "r  =  a  +  1j*b\n",
+      "VR  =  Vs*cmath.e**(-1*n*r)\n",
+      " #Receiving  end  current,\n",
+      "IR  =  VR/Zo\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"Receiving end current, IR is  \",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg  mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "Receiving end current, IR is   0.205 /_ -46.62 deg  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 875</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Vs  =  8;#  in  Volts\n",
+      "VR  =  2;#  in  Volts\n",
+      "x  =  2;  \n",
+      "\n",
+      "#calculation:\n",
+      " #  receiving  end  voltage  VR  =  Vs*e**(-nr)\n",
+      " #e**-nr  =  p\n",
+      "p  =  VR/Vs\n",
+      " #If  the  line  is  doubled  in  length,  then\n",
+      "VR  =  Vs*(p)**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  Receiving  end  voltage  If  the  line  is  doubled  in  length,  VR  is  \",abs(VR),\" V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  Receiving  end  voltage  If  the  line  is  doubled  in  length,  VR  is   0.5  V"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 876</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzoc  =  800;#  in  ohm\n",
+      "thetazoc  =  -50;#  in  degrees\n",
+      "rzsc  =  413;#  in  ohm\n",
+      "thetazsc  =  -20;#  in  degrees\n",
+      "f  =  1500;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      " #open  circuit  impedance\n",
+      "Zoc  =  rzoc*math.cos(thetazoc*math.pi/180)  +  1j*rzoc*math.sin(thetazoc*math.pi/180)\n",
+      " #short  circuit  impedance\n",
+      "Zsc  =  rzsc*math.cos(thetazsc*math.pi/180)  +  1j*rzsc*math.sin(thetazsc*math.pi/180)\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  (Zoc*Zsc)**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo is\",round(abs(Zo)),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo is 575.0 /_ -35.0 deg  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 877</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  15;#  in  ohm/loop  km\n",
+      "L  =  0.0034;#  in  H/loop  km\n",
+      "C  =  10E-9;#  in  F/km\n",
+      "G  =  3E-6;#  in  S/km\n",
+      "f  =  2000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo  is  \",round(abs(Zo),0),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo  is   600.0 /_ -8.99 deg  ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 8, page no. 879</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.0005;#  in  H/loop  km\n",
+      "C  =  0.12E-6;#  in  F/km\n",
+      "f  =  400000;#  in  Hz\n",
+      "\n",
+      "#calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  (L/C)**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  1j*w*(L*C)**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      " #wavelength\n",
+      "Y  =  2*math.pi/b\n",
+      " #velocity  of  propagation  \n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  characteristic  impedance  Zo  is  \",abs(Zo),\"ohm\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",a,\"  +(\",round(b,2),\")i\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y*1E3,0),\"m\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,2),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  characteristic  impedance  Zo  is   64.5497224368 ohm\n",
+        "\n",
+        "  propagation  coefficient  is   0.0   +( 19.47 )i\n",
+        "\n",
+        "  wavelength  Y  is   323.0 m\n",
+        "\n",
+        "  speed  of  transmission   129099.44 km/sec"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 9, page no. 880</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  25;#  in  ohm/loop  km\n",
+      "L  =  0.005;#  in  H/loop  km\n",
+      "C  =  0.04E-6;#  in  F/km\n",
+      "G  =  80E-6;#  in  S/km\n",
+      "f  =  1000;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  ((R  +  1j*w*L)*(G  +  1j*w*C))**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo  is\",round(abs(Zo),2),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",round(abs(r),4),\"/_\",round(cmath.phase(complex(a,b))*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  attenuation  coefficient  is  \",round(a,4),\"  Np/km\"\n",
+      "print  \"\\n  the  phase-shift  coefficient  \",round(b,4),\"  rad/km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo  is 390.16 /_ -10.43 deg  ohm\n",
+        "\n",
+        "  propagation  coefficient  is   0.1029 /_ 61.92 deg\n",
+        "\n",
+        "  attenuation  coefficient  is   0.0484   Np/km\n",
+        "\n",
+        "  the  phase-shift  coefficient   0.0908   rad/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 10, page no. 881</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  8;#  in  ohm/loop  km\n",
+      "L  =  0.003;#  in  H/loop  km\n",
+      "C  =  7500E-12;#  in  F/km\n",
+      "G  =  0.25E-6;#  in  S/km\n",
+      "f  =  1000;#  in  Hz\n",
+      "n  =  300;#  in  km\n",
+      "Zg  =  400  +  1j*0;#  in  ohm\n",
+      "Vg  =  10;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #characteristic  impedance  Zo\n",
+      "Zo  =  ((R  +  1j*w*L)/(G  +  1j*w*C))**0.5\n",
+      " #the  propagation  coefficient\n",
+      "r  =  ((R  +  1j*w*L)*(G  +  1j*w*C))**0.5\n",
+      " #the  attenuation  coefficient  \n",
+      "a  =  r.real\n",
+      " #the  phaseshift  coefficient\n",
+      "b  =  r.imag\n",
+      " #the  sending-end  current,\n",
+      "Is  =  Vg/(Zg  +  Zo)\n",
+      " #the  receiving-end  current,\n",
+      "IR  =  Is*cmath.e**(-1*n*r)\n",
+      " #wavelength\n",
+      "Y  =  2*math.pi/b\n",
+      " #velocity  of  propagation  \n",
+      "u  =  f*Y\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"characteristic impedance Zo is\",round(abs(Zo),1),\"/_\",round(cmath.phase(complex(Zo.real,Zo.imag))*180/math.pi,2),\"deg  ohm\"\n",
+      "print  \"propagation  coefficient  is  \",round(abs(r),5),\"/_\",round(cmath.phase(complex(r.real,r.imag))*180/math.pi,2),\"deg\"\n",
+      "print  \"attenuation  coefficient  is  \",round(a,5),\"  Np/km  and  the  phaseshift  coefficient  \",round(b,5),\"  rad/km\"\n",
+      "print  \"sending-end  current  Is  is  \",round(abs(Is)*1E3,3),\"/_\",round(cmath.phase(complex(Is.real,Is.imag))*180/math.pi,2),\"deg  mA\"\n",
+      "print  \"receiving-end  current  IR is\",round(abs(IR)*1E3,3),\"/_\",round(cmath.phase(complex(IR.real,IR.imag))*180/math.pi,2),\"deg  mA\"\n",
+      "print  \"wavelength  Y  is  \",round(Y,1),\"  km\"\n",
+      "print  \"speed  of  transmission  \",round(u,1),\"km/sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "characteristic impedance Zo is 659.2 /_ -11.35 deg  ohm\n",
+        "propagation  coefficient  is   0.03106 /_ 78.35 deg\n",
+        "attenuation  coefficient  is   0.00627   Np/km  and  the  phaseshift  coefficient   0.03042   rad/km\n",
+        "sending-end  current  Is  is   9.485 /_ 7.07 deg  mA\n",
+        "receiving-end  current  IR is 1.445 /_ -155.88 deg  mA\n",
+        "wavelength  Y  is   206.5   km\n",
+        "speed  of  transmission   206521.1 km/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 11, page no. 884</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  10;#  in  ohm/loop  km\n",
+      "L  =  0.0015;#  in  H/loop  km\n",
+      "C  =  0.06E-6;#  in  F/km\n",
+      "G  =  1.2E-6;#  in  S/km\n",
+      "\n",
+      " #calculation:\n",
+      " #the  condition  for  minimum  distortion  is  given  by  LG  =  CR,  from  which,\n",
+      "Lm  =  C*R/G\n",
+      "dL  =  Lm  -  L\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  should  be  increased  by  \",round(dL*1E3,1),\"mH/loop  km  for  minimum  distortion\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  should  be  increased  by   498.5 mH/loop  km  for  minimum  distortion"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 12, page no. 884</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "R  =  80;#  in  ohm/loop  km\n",
+      "C  =  5E-9;#  in  F/km\n",
+      "G  =  2E-6;#  in  S/km\n",
+      "f  =  1500;#  in  Hz\n",
+      "\n",
+      " #calculation:\n",
+      "w  =  2*math.pi*f\n",
+      " #the  condition  for  minimum  distortion  is  given  by  LG  =  CR,  from  which,  inductance\n",
+      "L  =  C*R/G\n",
+      " #attenuation  coefficient,\n",
+      "a  =  (R*G)**0.5\n",
+      " #phase  shift  coefficient,\n",
+      "b  =  w*(L*C)**0.5\n",
+      " #propagation  coefficient,\n",
+      "r  =  a  +  1j*b\n",
+      " #velocity  of  propagation,\n",
+      "u  =  1/(L*C)**0.5\n",
+      " #wavelength\n",
+      "Y  =  u/f\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  inductance  is  \",round(L,2),\"  H\"\n",
+      "print  \"\\n  propagation  coefficient  is  \",round(a,2),\"  +(\",round(b,2),\")i\"\n",
+      "print  \"\\n  speed  of  transmission  \",round(u,2),\"km/sec\"\n",
+      "print  \"\\n  wavelength  Y  is  \",round(Y,2),\"  km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  inductance  is   0.2   H\n",
+        "\n",
+        "  propagation  coefficient  is   0.01   +( 0.3 )i\n",
+        "\n",
+        "  speed  of  transmission   31622.78 km/sec\n",
+        "\n",
+        "  wavelength  Y  is   21.08   km\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 13, page no. 888</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      " \n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  75;#  in  ohm\n",
+      "ZR  =  250;#  in  ohm\n",
+      "VR  =  10;#  in  Volts\n",
+      "\n",
+      "#calculation:\n",
+      " #reflection  coefficient\n",
+      "p  =  (Zo  -  ZR)/(Zo  +  ZR)\n",
+      " #Current  flowing  in  the  terminating  load\n",
+      "IR  =  VR/ZR\n",
+      " #incident  current,  Ii\n",
+      "Ii  =  IR/(1  +  p)\n",
+      " #incident  voltage,  Vi    \n",
+      "Vi  =  Ii*Zo\n",
+      " #reflected  current,  Ir\n",
+      "Ir  =  IR  -  Ii\n",
+      " #reflected  voltage,  Vr\n",
+      "Vr  =  -1*Ir*Zo\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient  is  \",round(p,3),\"\"\n",
+      "print  \"\\n  incident  current,  Ii  is  \",round(Ii,4),\"  A\"\n",
+      "print  \"\\n  incident  voltage,  Vi  is  \",round(Vi,2),\"  V\"\n",
+      "print  \"\\n  reflected  current,  Ir  is  \",round(Ir,4),\"  A\"\n",
+      "print  \"\\n  reflected  voltage,  Vr  is  \",round(Vr,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient  is   -0.538 \n",
+        "\n",
+        "  incident  current,  Ii  is   0.0867   A\n",
+        "\n",
+        "  incident  voltage,  Vi  is   6.5   V\n",
+        "\n",
+        "  reflected  current,  Ir  is   -0.0467   A\n",
+        "\n",
+        "  reflected  voltage,  Vr  is   3.5   V"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 14, page no. 889</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "Zo  =  500  -  1j*40;#  in  ohm\n",
+      "ZR1  =  500  +  1j*40;#  in  ohm\n",
+      "ZR2  =  600  +  1j*0;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #reflection  coefficient\n",
+      "p1  =  (Zo  -  ZR1)/(Zo  +  ZR1)\n",
+      "p2  =  (Zo  -  ZR2)/(Zo  +  ZR2)\n",
+      "p1mag  =  abs(p1)\n",
+      "p2mag  =  abs(p2)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient  (a)\",p1mag,\"  and  (b)\", round(p2mag,2),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient  (a) 0.08   and  (b) 0.1 "
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 15, page no. 890</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#(b) the incident voltage\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzo  =  500;#  in  ohm\n",
+      "thetazo  =  0;#  in  degrees\n",
+      "ZR  =  320  +  1j*240;#  in  ohm\n",
+      "rvr  =  20;#  in  volts\n",
+      "thetavr  =  35;#  in  degrees\n",
+      "\n",
+      " #calculation:\n",
+      " #voltage\n",
+      "VR  =  rvr*math.cos(thetavr*math.pi/180)  +  1j*rvr*math.sin(thetavr*math.pi/180)\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #the  ratio  of  the  reflected  to  the  incident  voltage  \n",
+      " #vr  =  VR/Vi\n",
+      "vr  =  (ZR  -  Zo)/(Zo  +  ZR)\n",
+      "vrmag  =  abs(vr)\n",
+      " #incident  voltage,  Vi\n",
+      "Vi  =  VR/vr\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  the  magnitude  of  the  ratio  Vr  :  Vi  is  \",round(vrmag,3),\"\"\n",
+      "print  \"\\n  incident  voltage,  Vi  is  \",round(abs(Vi),1),\"/_\",round(cmath.phase(complex(Vi.real,Vi.imag))*180/math.pi,2),\"deg  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  the  magnitude  of  the  ratio  Vr  :  Vi  is   0.351 \n",
+        "\n",
+        "  incident  voltage,  Vi  is   57.0 /_ -75.56 deg  V\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 16, page no. 895</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rzo  =  600;#  in  ohm\n",
+      "thetazo  =  0;#  in  degrees\n",
+      "ZR  =  400  +  250j;#  in  ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #characteristic  impedance\n",
+      "Zo  =  rzo*math.cos(thetazo*math.pi/180)  +  1j*rzo*math.sin(thetazo*math.pi/180)\n",
+      " #reflection  coefficient\n",
+      "p  =  (Zo  -  ZR)/(Zo  +  ZR)\n",
+      "pmag  =  abs(p)\n",
+      " #standing-wave  ratio,\n",
+      "s  =  (1  +  pmag)/(1  -  pmag)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflection  coefficient,  is  \",round(abs(p),4),\"/_\",round(cmath.phase(complex(p.real,p.imag))*180/math.pi,2),\"deg\"\n",
+      "print  \"\\n  standing-wave  ratio,  s  is  \",round(s,3),\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflection  coefficient,  is   0.3106 /_ -65.38 deg\n",
+        "\n",
+        "  standing-wave  ratio,  s  is   1.901 "
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 17, page no. 896</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "rp  =  0.2;  \n",
+      "thetap  =  -120;#  in  degrees\n",
+      "Zo  =  80;#  in  ohm\n",
+      "Ir  =  0.01;#  in  Amperes\n",
+      "\n",
+      "#calculation:\n",
+      " #reflection  coefficient\n",
+      "p  =  rp*math.cos(thetap*math.pi/180)  +  1j*rp*math.sin(thetap*math.pi/180)\n",
+      " #standing-wave  ratio,\n",
+      "s  =  (1  +  rp)/(1  -  rp)\n",
+      " #load  impedance  ZR  \n",
+      "ZR  =  Zo*(1  -  p)/(1  +  p)\n",
+      " #incident  current\n",
+      "Ii  =  Ir*(s  +  1)/(s  -  1)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  standing-wave  ratio,  s  is  \",s,\"\"\n",
+      "print  \"\\n  load  impedance  ZR  is  \",round(ZR.real,2),\"  +(\",round(ZR.imag,1),\")i  ohm\"\n",
+      "print  \"\\n  incident  current  is  \",Ii,\"  A\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  standing-wave  ratio,  s  is   1.5 \n",
+        "\n",
+        "  load  impedance  ZR  is   91.43   +( 33.0 )i  ohm\n",
+        "\n",
+        "  incident  current  is   0.05   A\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 18, page no. 897</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "s  =  1.6;\n",
+      "Pi  =  0.2;#  in  Watts\n",
+      "\n",
+      "#calculation:\n",
+      " #reflected  power,  Pr\n",
+      "Pr  =  Pi*((s  -  1)/(s  +  1))**2\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  reflected  power,  Pr  is  \",round(Pr*1E3,2),\" mW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  reflected  power,  Pr  is   10.65  mW"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb
new file mode 100755
index 00000000..f30323f6
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb
@@ -0,0 +1,435 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 45: Transients and Laplace\n",
+      "transforms</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 903</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the initial value of current flowing, \n",
+      "#(b) the value of current 150 ms after connection, \n",
+      "#(c) the value of capacitor voltage 80 ms after connection, and \n",
+      "#(d) the time after connection when the resistor voltage is 35 V.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  500E-9;#  in  Farad\n",
+      "R  =  100000;#  in  Ohm\n",
+      "V  =  50;#  in  VOlts\n",
+      "ti  =  0.15;#  in  sec\n",
+      "tc  =  0.08;#  in  sec\n",
+      "Vrt  =  35;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #Initial  current,  \n",
+      "i0  =  (V/R)\n",
+      " #when  time  t  =  150ms  current  is\n",
+      "i150  =  (V/R)*math.e**(-1*ti/(R*C))\n",
+      " #capacitor  voltage,  Vc\n",
+      "Vc  =  V*(1  -  math.e**(-1*tc/(R*C)))\n",
+      " #time,  t\n",
+      "tvr  =  -1*R*C*math.log(Vrt/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  initial  value  of  current  flowing  is  \",round(i0*1E3,2),\"mA\"\n",
+      "print  \"\\n  current  flowing  at  t  =  150ms  is  \",round(i150*1E6,2),\"uA\"\n",
+      "print  \"\\n    value  of  capacitor  voltage  at  t  =  80ms  is  \",round(Vc,2),\"  V\"\n",
+      "print  \"\\n    the  time  after  connection  when  the  resistor  voltage  is  35  V  is  \",round(tvr*1E3,2),\"msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  initial  value  of  current  flowing  is   0.5 mA\n",
+        "\n",
+        "  current  flowing  at  t  =  150ms  is   24.89 uA\n",
+        "\n",
+        "    value  of  capacitor  voltage  at  t  =  80ms  is   39.91   V\n",
+        "\n",
+        "    the  time  after  connection  when  the  resistor  voltage  is  35  V  is   17.83 msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 905</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the p.d. across the capacitor after 20 s\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  5E-6;#  in  Farad\n",
+      "R  =  2000000;#  in  Ohm\n",
+      "V  =  200;#  in  VOlts\n",
+      "tc  =  20;#  in  sec\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitor  voltage,  Vc\n",
+      "Vc  =  V*(math.e**(-1*tc/(R*C)))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    value  of  capacitor  voltage  at  t  =  20s  is  \",round(Vc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    value  of  capacitor  voltage  at  t  =  20s  is   27.07   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 907</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the final value of current, (b) the value of current after 4 ms, \n",
+      "#(c) the value of the voltage across the resistor after 6 ms,\n",
+      "#(d) the value of the voltage across the inductance after 6 ms, and \n",
+      "#(e) the time when the current reaches 15 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.05;#  in  Henry\n",
+      "R  =  5;#  in  Ohm\n",
+      "V  =  110;#  in  VOlts\n",
+      "ti  =  0.004;#  in  sec\n",
+      "tvr  =  0.006;#  in  sec\n",
+      "tvl  =  0.006;#  in  sec\n",
+      "it  =  15;#  in  amperes\n",
+      "\n",
+      " #calculation:\n",
+      " #steady  state  current  i\n",
+      "i  =  V/R\n",
+      " #when  time  t  =  4ms  current  is\n",
+      "i4  =  (V/R)*(1  -  math.e**(-1*ti*R/L))\n",
+      " #resistor  voltage,  VR\n",
+      "VR6  =  V*(1  -  math.e**(-1*tvr*R/L))\n",
+      " #inductor  voltage,  VL\n",
+      "VL6  =  V*(math.e**(-1*tvl*R/L))\n",
+      " #time,  t\n",
+      "ti  =  (-1*L/R)*math.log(1  -  it*R/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  steady  state  current  i  is  \",round(i,2),\"  A\"\n",
+      "print  \"\\n  when  time  t  =  4ms  current  is  is  \",round(i4,2),\"  A\"\n",
+      "print  \"\\n    value  of  resistor  voltage  at  t  =  6ms  is  \",round(VR6,2),\"  V\"\n",
+      "print  \"\\n    value  of  inductor  voltage  at  t  =  6ms  is  \",round(VL6,2),\"  V\"\n",
+      "print  \"\\n    the  time  after  connection  when  the  current  is  15  V  is  \",round(ti,5),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  steady  state  current  i  is   22.0   A\n",
+        "\n",
+        "  when  time  t  =  4ms  current  is  is   7.25   A\n",
+        "\n",
+        "    value  of  resistor  voltage  at  t  =  6ms  is   49.63   V\n",
+        "\n",
+        "    value  of  inductor  voltage  at  t  =  6ms  is   60.37   V\n",
+        "\n",
+        "    the  time  after  connection  when  the  current  is  15  V  is   0.01145   sec"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 909</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine\n",
+      "#(a) the time for the current in the 2 H inductor to fall to 200 mA,\n",
+      "#and (b) the maximum voltage appearing across the resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "i  =  5;#  in  Amperes\n",
+      "L  =  2#  in  Henry\n",
+      "i1  =  0.2;#  in  Amperes\n",
+      "R  =  10;#  in  Ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #time  t\n",
+      "ti  =  (-1*L/R)*math.log(i1/i)\n",
+      " #voltage  across  the  resistor  is  a  maximum  \n",
+      "VRm  =  i*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    time  t  for  the  current  in  the  2  H  inductor  to  fall  to  200  mA  is  \",round(ti,3),\"  sec\"\n",
+      "print  \"\\n    max  voltage  across  the  resistor  is  \",VRm,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    time  t  for  the  current  in  the  2  H  inductor  to  fall  to  200  mA  is   0.644   sec\n",
+        "\n",
+        "    max  voltage  across  the  resistor  is   50   V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 911</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine whether the circuit is over, critical or underdamped. (b) If C D 5 nF, determine the state of damping.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.002#  in  Henry\n",
+      "R  =  1000;#  in  Ohm\n",
+      "C1  =  5E-6;#  in  farad\n",
+      "C2  =  5E-9;#  in  farad\n",
+      "\n",
+      " #calculation:\n",
+      "a  =  (R/(2*L))**2\n",
+      "b  =  1/(L*C1)\n",
+      "if  (a>b):\n",
+      "\ts1  =  \"overdamped\";\n",
+      "elif  (a<b):\n",
+      "\ts1  =  \"underdamped\";\n",
+      "else:\n",
+      "\ts1  =  \"critically  damped\";\n",
+      "c  =  1/(L*C2)\n",
+      "if  (a>c):\n",
+      "\ts2  =  \"overdamped\";\n",
+      "elif  (a<c):\n",
+      "\ts2  =  \"underdamped\";\n",
+      "else:\n",
+      "\ts2  =  \"critically  damped\";\n",
+      "\t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    circuit  is  \",s1\n",
+      "print  \"\\n    if  C  =  5  nF,  circuit  is  \",s2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    circuit  is   overdamped\n",
+        "\n",
+        "    if  C  =  5  nF,  circuit  is   underdamped"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 912</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what value of capacitance will give critical damping ?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.002#  in  Henry\n",
+      "R  =  1000;#  in  Ohm\n",
+      "\n",
+      " #calculation:\n",
+      "a  =  (R/(2*L))**2\n",
+      " #for  critically  damped\n",
+      "C  =  4*L/R**2\n",
+      "\t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    capacitance  C  is  \",C*1E9,\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    capacitance  C  is   8.0 nF"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 913</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the nature of the response and obtain an expression for the current in the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  1.5#  in  Henry\n",
+      "R  =  90;#  in  Ohm\n",
+      "C = 5*1E-6; # in Farad\n",
+      "V = 10; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "a = -1*R/(2*L)\n",
+      "b = (1/(L*C) - (R/(2*L))**2)**0.5\n",
+      "V0 = V\n",
+      "I0 = 0\n",
+      "A = V0\n",
+      "B = (I0 - C*a*V0)/(C*b)\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n  Result  \\n\\n\"\n",
+      "print \"Current, i = e^(\",a,\"t) (\",round((a*C*B - A*C*b),4),\"sin(\",round(b,1),\"t)  + (\",round((-1*a*C*A + B*C*C*b),0),\"cos(\",round(b,1),\"t) Amps.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "Current, i = e^( -30.0 t) ( -0.0183 sin( 363.9 t)  + ( 0.0 cos( 363.9 t) Amps.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_1.ipynb
new file mode 100755
index 00000000..f30323f6
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_1.ipynb
@@ -0,0 +1,435 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 45: Transients and Laplace\n",
+      "transforms</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 903</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the initial value of current flowing, \n",
+      "#(b) the value of current 150 ms after connection, \n",
+      "#(c) the value of capacitor voltage 80 ms after connection, and \n",
+      "#(d) the time after connection when the resistor voltage is 35 V.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  500E-9;#  in  Farad\n",
+      "R  =  100000;#  in  Ohm\n",
+      "V  =  50;#  in  VOlts\n",
+      "ti  =  0.15;#  in  sec\n",
+      "tc  =  0.08;#  in  sec\n",
+      "Vrt  =  35;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #Initial  current,  \n",
+      "i0  =  (V/R)\n",
+      " #when  time  t  =  150ms  current  is\n",
+      "i150  =  (V/R)*math.e**(-1*ti/(R*C))\n",
+      " #capacitor  voltage,  Vc\n",
+      "Vc  =  V*(1  -  math.e**(-1*tc/(R*C)))\n",
+      " #time,  t\n",
+      "tvr  =  -1*R*C*math.log(Vrt/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  initial  value  of  current  flowing  is  \",round(i0*1E3,2),\"mA\"\n",
+      "print  \"\\n  current  flowing  at  t  =  150ms  is  \",round(i150*1E6,2),\"uA\"\n",
+      "print  \"\\n    value  of  capacitor  voltage  at  t  =  80ms  is  \",round(Vc,2),\"  V\"\n",
+      "print  \"\\n    the  time  after  connection  when  the  resistor  voltage  is  35  V  is  \",round(tvr*1E3,2),\"msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  initial  value  of  current  flowing  is   0.5 mA\n",
+        "\n",
+        "  current  flowing  at  t  =  150ms  is   24.89 uA\n",
+        "\n",
+        "    value  of  capacitor  voltage  at  t  =  80ms  is   39.91   V\n",
+        "\n",
+        "    the  time  after  connection  when  the  resistor  voltage  is  35  V  is   17.83 msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 905</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the p.d. across the capacitor after 20 s\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  5E-6;#  in  Farad\n",
+      "R  =  2000000;#  in  Ohm\n",
+      "V  =  200;#  in  VOlts\n",
+      "tc  =  20;#  in  sec\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitor  voltage,  Vc\n",
+      "Vc  =  V*(math.e**(-1*tc/(R*C)))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    value  of  capacitor  voltage  at  t  =  20s  is  \",round(Vc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    value  of  capacitor  voltage  at  t  =  20s  is   27.07   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 907</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the final value of current, (b) the value of current after 4 ms, \n",
+      "#(c) the value of the voltage across the resistor after 6 ms,\n",
+      "#(d) the value of the voltage across the inductance after 6 ms, and \n",
+      "#(e) the time when the current reaches 15 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.05;#  in  Henry\n",
+      "R  =  5;#  in  Ohm\n",
+      "V  =  110;#  in  VOlts\n",
+      "ti  =  0.004;#  in  sec\n",
+      "tvr  =  0.006;#  in  sec\n",
+      "tvl  =  0.006;#  in  sec\n",
+      "it  =  15;#  in  amperes\n",
+      "\n",
+      " #calculation:\n",
+      " #steady  state  current  i\n",
+      "i  =  V/R\n",
+      " #when  time  t  =  4ms  current  is\n",
+      "i4  =  (V/R)*(1  -  math.e**(-1*ti*R/L))\n",
+      " #resistor  voltage,  VR\n",
+      "VR6  =  V*(1  -  math.e**(-1*tvr*R/L))\n",
+      " #inductor  voltage,  VL\n",
+      "VL6  =  V*(math.e**(-1*tvl*R/L))\n",
+      " #time,  t\n",
+      "ti  =  (-1*L/R)*math.log(1  -  it*R/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  steady  state  current  i  is  \",round(i,2),\"  A\"\n",
+      "print  \"\\n  when  time  t  =  4ms  current  is  is  \",round(i4,2),\"  A\"\n",
+      "print  \"\\n    value  of  resistor  voltage  at  t  =  6ms  is  \",round(VR6,2),\"  V\"\n",
+      "print  \"\\n    value  of  inductor  voltage  at  t  =  6ms  is  \",round(VL6,2),\"  V\"\n",
+      "print  \"\\n    the  time  after  connection  when  the  current  is  15  V  is  \",round(ti,5),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  steady  state  current  i  is   22.0   A\n",
+        "\n",
+        "  when  time  t  =  4ms  current  is  is   7.25   A\n",
+        "\n",
+        "    value  of  resistor  voltage  at  t  =  6ms  is   49.63   V\n",
+        "\n",
+        "    value  of  inductor  voltage  at  t  =  6ms  is   60.37   V\n",
+        "\n",
+        "    the  time  after  connection  when  the  current  is  15  V  is   0.01145   sec"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 909</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine\n",
+      "#(a) the time for the current in the 2 H inductor to fall to 200 mA,\n",
+      "#and (b) the maximum voltage appearing across the resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "i  =  5;#  in  Amperes\n",
+      "L  =  2#  in  Henry\n",
+      "i1  =  0.2;#  in  Amperes\n",
+      "R  =  10;#  in  Ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #time  t\n",
+      "ti  =  (-1*L/R)*math.log(i1/i)\n",
+      " #voltage  across  the  resistor  is  a  maximum  \n",
+      "VRm  =  i*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    time  t  for  the  current  in  the  2  H  inductor  to  fall  to  200  mA  is  \",round(ti,3),\"  sec\"\n",
+      "print  \"\\n    max  voltage  across  the  resistor  is  \",VRm,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    time  t  for  the  current  in  the  2  H  inductor  to  fall  to  200  mA  is   0.644   sec\n",
+        "\n",
+        "    max  voltage  across  the  resistor  is   50   V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 911</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine whether the circuit is over, critical or underdamped. (b) If C D 5 nF, determine the state of damping.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.002#  in  Henry\n",
+      "R  =  1000;#  in  Ohm\n",
+      "C1  =  5E-6;#  in  farad\n",
+      "C2  =  5E-9;#  in  farad\n",
+      "\n",
+      " #calculation:\n",
+      "a  =  (R/(2*L))**2\n",
+      "b  =  1/(L*C1)\n",
+      "if  (a>b):\n",
+      "\ts1  =  \"overdamped\";\n",
+      "elif  (a<b):\n",
+      "\ts1  =  \"underdamped\";\n",
+      "else:\n",
+      "\ts1  =  \"critically  damped\";\n",
+      "c  =  1/(L*C2)\n",
+      "if  (a>c):\n",
+      "\ts2  =  \"overdamped\";\n",
+      "elif  (a<c):\n",
+      "\ts2  =  \"underdamped\";\n",
+      "else:\n",
+      "\ts2  =  \"critically  damped\";\n",
+      "\t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    circuit  is  \",s1\n",
+      "print  \"\\n    if  C  =  5  nF,  circuit  is  \",s2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    circuit  is   overdamped\n",
+        "\n",
+        "    if  C  =  5  nF,  circuit  is   underdamped"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 912</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what value of capacitance will give critical damping ?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.002#  in  Henry\n",
+      "R  =  1000;#  in  Ohm\n",
+      "\n",
+      " #calculation:\n",
+      "a  =  (R/(2*L))**2\n",
+      " #for  critically  damped\n",
+      "C  =  4*L/R**2\n",
+      "\t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    capacitance  C  is  \",C*1E9,\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    capacitance  C  is   8.0 nF"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 913</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the nature of the response and obtain an expression for the current in the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  1.5#  in  Henry\n",
+      "R  =  90;#  in  Ohm\n",
+      "C = 5*1E-6; # in Farad\n",
+      "V = 10; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "a = -1*R/(2*L)\n",
+      "b = (1/(L*C) - (R/(2*L))**2)**0.5\n",
+      "V0 = V\n",
+      "I0 = 0\n",
+      "A = V0\n",
+      "B = (I0 - C*a*V0)/(C*b)\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n  Result  \\n\\n\"\n",
+      "print \"Current, i = e^(\",a,\"t) (\",round((a*C*B - A*C*b),4),\"sin(\",round(b,1),\"t)  + (\",round((-1*a*C*A + B*C*C*b),0),\"cos(\",round(b,1),\"t) Amps.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "Current, i = e^( -30.0 t) ( -0.0183 sin( 363.9 t)  + ( 0.0 cos( 363.9 t) Amps.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_2.ipynb
new file mode 100755
index 00000000..f30323f6
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_2.ipynb
@@ -0,0 +1,435 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 45: Transients and Laplace\n",
+      "transforms</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 903</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate (a) the initial value of current flowing, \n",
+      "#(b) the value of current 150 ms after connection, \n",
+      "#(c) the value of capacitor voltage 80 ms after connection, and \n",
+      "#(d) the time after connection when the resistor voltage is 35 V.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  500E-9;#  in  Farad\n",
+      "R  =  100000;#  in  Ohm\n",
+      "V  =  50;#  in  VOlts\n",
+      "ti  =  0.15;#  in  sec\n",
+      "tc  =  0.08;#  in  sec\n",
+      "Vrt  =  35;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #Initial  current,  \n",
+      "i0  =  (V/R)\n",
+      " #when  time  t  =  150ms  current  is\n",
+      "i150  =  (V/R)*math.e**(-1*ti/(R*C))\n",
+      " #capacitor  voltage,  Vc\n",
+      "Vc  =  V*(1  -  math.e**(-1*tc/(R*C)))\n",
+      " #time,  t\n",
+      "tvr  =  -1*R*C*math.log(Vrt/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  initial  value  of  current  flowing  is  \",round(i0*1E3,2),\"mA\"\n",
+      "print  \"\\n  current  flowing  at  t  =  150ms  is  \",round(i150*1E6,2),\"uA\"\n",
+      "print  \"\\n    value  of  capacitor  voltage  at  t  =  80ms  is  \",round(Vc,2),\"  V\"\n",
+      "print  \"\\n    the  time  after  connection  when  the  resistor  voltage  is  35  V  is  \",round(tvr*1E3,2),\"msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  initial  value  of  current  flowing  is   0.5 mA\n",
+        "\n",
+        "  current  flowing  at  t  =  150ms  is   24.89 uA\n",
+        "\n",
+        "    value  of  capacitor  voltage  at  t  =  80ms  is   39.91   V\n",
+        "\n",
+        "    the  time  after  connection  when  the  resistor  voltage  is  35  V  is   17.83 msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 905</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the p.d. across the capacitor after 20 s\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  5E-6;#  in  Farad\n",
+      "R  =  2000000;#  in  Ohm\n",
+      "V  =  200;#  in  VOlts\n",
+      "tc  =  20;#  in  sec\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitor  voltage,  Vc\n",
+      "Vc  =  V*(math.e**(-1*tc/(R*C)))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    value  of  capacitor  voltage  at  t  =  20s  is  \",round(Vc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    value  of  capacitor  voltage  at  t  =  20s  is   27.07   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 907</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine (a) the final value of current, (b) the value of current after 4 ms, \n",
+      "#(c) the value of the voltage across the resistor after 6 ms,\n",
+      "#(d) the value of the voltage across the inductance after 6 ms, and \n",
+      "#(e) the time when the current reaches 15 A.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.05;#  in  Henry\n",
+      "R  =  5;#  in  Ohm\n",
+      "V  =  110;#  in  VOlts\n",
+      "ti  =  0.004;#  in  sec\n",
+      "tvr  =  0.006;#  in  sec\n",
+      "tvl  =  0.006;#  in  sec\n",
+      "it  =  15;#  in  amperes\n",
+      "\n",
+      " #calculation:\n",
+      " #steady  state  current  i\n",
+      "i  =  V/R\n",
+      " #when  time  t  =  4ms  current  is\n",
+      "i4  =  (V/R)*(1  -  math.e**(-1*ti*R/L))\n",
+      " #resistor  voltage,  VR\n",
+      "VR6  =  V*(1  -  math.e**(-1*tvr*R/L))\n",
+      " #inductor  voltage,  VL\n",
+      "VL6  =  V*(math.e**(-1*tvl*R/L))\n",
+      " #time,  t\n",
+      "ti  =  (-1*L/R)*math.log(1  -  it*R/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  steady  state  current  i  is  \",round(i,2),\"  A\"\n",
+      "print  \"\\n  when  time  t  =  4ms  current  is  is  \",round(i4,2),\"  A\"\n",
+      "print  \"\\n    value  of  resistor  voltage  at  t  =  6ms  is  \",round(VR6,2),\"  V\"\n",
+      "print  \"\\n    value  of  inductor  voltage  at  t  =  6ms  is  \",round(VL6,2),\"  V\"\n",
+      "print  \"\\n    the  time  after  connection  when  the  current  is  15  V  is  \",round(ti,5),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  steady  state  current  i  is   22.0   A\n",
+        "\n",
+        "  when  time  t  =  4ms  current  is  is   7.25   A\n",
+        "\n",
+        "    value  of  resistor  voltage  at  t  =  6ms  is   49.63   V\n",
+        "\n",
+        "    value  of  inductor  voltage  at  t  =  6ms  is   60.37   V\n",
+        "\n",
+        "    the  time  after  connection  when  the  current  is  15  V  is   0.01145   sec"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 909</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine\n",
+      "#(a) the time for the current in the 2 H inductor to fall to 200 mA,\n",
+      "#and (b) the maximum voltage appearing across the resistor.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "i  =  5;#  in  Amperes\n",
+      "L  =  2#  in  Henry\n",
+      "i1  =  0.2;#  in  Amperes\n",
+      "R  =  10;#  in  Ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #time  t\n",
+      "ti  =  (-1*L/R)*math.log(i1/i)\n",
+      " #voltage  across  the  resistor  is  a  maximum  \n",
+      "VRm  =  i*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    time  t  for  the  current  in  the  2  H  inductor  to  fall  to  200  mA  is  \",round(ti,3),\"  sec\"\n",
+      "print  \"\\n    max  voltage  across  the  resistor  is  \",VRm,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    time  t  for  the  current  in  the  2  H  inductor  to  fall  to  200  mA  is   0.644   sec\n",
+        "\n",
+        "    max  voltage  across  the  resistor  is   50   V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 911</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#(a) Determine whether the circuit is over, critical or underdamped. (b) If C D 5 nF, determine the state of damping.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.002#  in  Henry\n",
+      "R  =  1000;#  in  Ohm\n",
+      "C1  =  5E-6;#  in  farad\n",
+      "C2  =  5E-9;#  in  farad\n",
+      "\n",
+      " #calculation:\n",
+      "a  =  (R/(2*L))**2\n",
+      "b  =  1/(L*C1)\n",
+      "if  (a>b):\n",
+      "\ts1  =  \"overdamped\";\n",
+      "elif  (a<b):\n",
+      "\ts1  =  \"underdamped\";\n",
+      "else:\n",
+      "\ts1  =  \"critically  damped\";\n",
+      "c  =  1/(L*C2)\n",
+      "if  (a>c):\n",
+      "\ts2  =  \"overdamped\";\n",
+      "elif  (a<c):\n",
+      "\ts2  =  \"underdamped\";\n",
+      "else:\n",
+      "\ts2  =  \"critically  damped\";\n",
+      "\t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    circuit  is  \",s1\n",
+      "print  \"\\n    if  C  =  5  nF,  circuit  is  \",s2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    circuit  is   overdamped\n",
+        "\n",
+        "    if  C  =  5  nF,  circuit  is   underdamped"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 912</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#what value of capacitance will give critical damping ?\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.002#  in  Henry\n",
+      "R  =  1000;#  in  Ohm\n",
+      "\n",
+      " #calculation:\n",
+      "a  =  (R/(2*L))**2\n",
+      " #for  critically  damped\n",
+      "C  =  4*L/R**2\n",
+      "\t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    capacitance  C  is  \",C*1E9,\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    capacitance  C  is   8.0 nF"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 913</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Determine the nature of the response and obtain an expression for the current in the coil.\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  1.5#  in  Henry\n",
+      "R  =  90;#  in  Ohm\n",
+      "C = 5*1E-6; # in Farad\n",
+      "V = 10; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "a = -1*R/(2*L)\n",
+      "b = (1/(L*C) - (R/(2*L))**2)**0.5\n",
+      "V0 = V\n",
+      "I0 = 0\n",
+      "A = V0\n",
+      "B = (I0 - C*a*V0)/(C*b)\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n  Result  \\n\\n\"\n",
+      "print \"Current, i = e^(\",a,\"t) (\",round((a*C*B - A*C*b),4),\"sin(\",round(b,1),\"t)  + (\",round((-1*a*C*A + B*C*C*b),0),\"cos(\",round(b,1),\"t) Amps.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "Current, i = e^( -30.0 t) ( -0.0183 sin( 363.9 t)  + ( 0.0 cos( 363.9 t) Amps.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_3.ipynb
new file mode 100755
index 00000000..4533f303
--- /dev/null
+++ b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint_3.ipynb
@@ -0,0 +1,428 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:76623e2b912c3ee572e7978144ba11bf11add72a64c001ac7cb23291cfabcb07"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h1>Chapter 45: Transients and Laplace\n",
+      "transforms</h1>"
+     ]
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 1, page no. 903</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  500E-9;#  in  Farad\n",
+      "R  =  100000;#  in  Ohm\n",
+      "V  =  50;#  in  VOlts\n",
+      "ti  =  0.15;#  in  sec\n",
+      "tc  =  0.08;#  in  sec\n",
+      "Vrt  =  35;#  in  Volts\n",
+      "\n",
+      " #calculation:\n",
+      " #Initial  current,  \n",
+      "i0  =  (V/R)\n",
+      " #when  time  t  =  150ms  current  is\n",
+      "i150  =  (V/R)*math.e**(-1*ti/(R*C))\n",
+      " #capacitor  voltage,  Vc\n",
+      "Vc  =  V*(1  -  math.e**(-1*tc/(R*C)))\n",
+      " #time,  t\n",
+      "tvr  =  -1*R*C*math.log(Vrt/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  initial  value  of  current  flowing  is  \",round(i0*1E3,2),\"mA\"\n",
+      "print  \"\\n  current  flowing  at  t  =  150ms  is  \",round(i150*1E6,2),\"uA\"\n",
+      "print  \"\\n    value  of  capacitor  voltage  at  t  =  80ms  is  \",round(Vc,2),\"  V\"\n",
+      "print  \"\\n    the  time  after  connection  when  the  resistor  voltage  is  35  V  is  \",round(tvr*1E3,2),\"msec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  initial  value  of  current  flowing  is   0.5 mA\n",
+        "\n",
+        "  current  flowing  at  t  =  150ms  is   24.89 uA\n",
+        "\n",
+        "    value  of  capacitor  voltage  at  t  =  80ms  is   39.91   V\n",
+        "\n",
+        "    the  time  after  connection  when  the  resistor  voltage  is  35  V  is   17.83 msec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 2, page no. 905</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "C  =  5E-6;#  in  Farad\n",
+      "R  =  2000000;#  in  Ohm\n",
+      "V  =  200;#  in  VOlts\n",
+      "tc  =  20;#  in  sec\n",
+      "\n",
+      " #calculation:\n",
+      " #capacitor  voltage,  Vc\n",
+      "Vc  =  V*(math.e**(-1*tc/(R*C)))\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    value  of  capacitor  voltage  at  t  =  20s  is  \",round(Vc,2),\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    value  of  capacitor  voltage  at  t  =  20s  is   27.07   V"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 3, page no. 907</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.05;#  in  Henry\n",
+      "R  =  5;#  in  Ohm\n",
+      "V  =  110;#  in  VOlts\n",
+      "ti  =  0.004;#  in  sec\n",
+      "tvr  =  0.006;#  in  sec\n",
+      "tvl  =  0.006;#  in  sec\n",
+      "it  =  15;#  in  amperes\n",
+      "\n",
+      " #calculation:\n",
+      " #steady  state  current  i\n",
+      "i  =  V/R\n",
+      " #when  time  t  =  4ms  current  is\n",
+      "i4  =  (V/R)*(1  -  math.e**(-1*ti*R/L))\n",
+      " #resistor  voltage,  VR\n",
+      "VR6  =  V*(1  -  math.e**(-1*tvr*R/L))\n",
+      " #inductor  voltage,  VL\n",
+      "VL6  =  V*(math.e**(-1*tvl*R/L))\n",
+      " #time,  t\n",
+      "ti  =  (-1*L/R)*math.log(1  -  it*R/V)\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n  steady  state  current  i  is  \",round(i,2),\"  A\"\n",
+      "print  \"\\n  when  time  t  =  4ms  current  is  is  \",round(i4,2),\"  A\"\n",
+      "print  \"\\n    value  of  resistor  voltage  at  t  =  6ms  is  \",round(VR6,2),\"  V\"\n",
+      "print  \"\\n    value  of  inductor  voltage  at  t  =  6ms  is  \",round(VL6,2),\"  V\"\n",
+      "print  \"\\n    the  time  after  connection  when  the  current  is  15  V  is  \",round(ti,5),\"  sec\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "  steady  state  current  i  is   22.0   A\n",
+        "\n",
+        "  when  time  t  =  4ms  current  is  is   7.25   A\n",
+        "\n",
+        "    value  of  resistor  voltage  at  t  =  6ms  is   49.63   V\n",
+        "\n",
+        "    value  of  inductor  voltage  at  t  =  6ms  is   60.37   V\n",
+        "\n",
+        "    the  time  after  connection  when  the  current  is  15  V  is   0.01145   sec"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 4, page no. 909</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "i  =  5;#  in  Amperes\n",
+      "L  =  2#  in  Henry\n",
+      "i1  =  0.2;#  in  Amperes\n",
+      "R  =  10;#  in  Ohm\n",
+      "\n",
+      " #calculation:\n",
+      " #time  t\n",
+      "ti  =  (-1*L/R)*math.log(i1/i)\n",
+      " #voltage  across  the  resistor  is  a  maximum  \n",
+      "VRm  =  i*R\n",
+      "\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    time  t  for  the  current  in  the  2  H  inductor  to  fall  to  200  mA  is  \",round(ti,3),\"  sec\"\n",
+      "print  \"\\n    max  voltage  across  the  resistor  is  \",VRm,\"  V\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    time  t  for  the  current  in  the  2  H  inductor  to  fall  to  200  mA  is   0.644   sec\n",
+        "\n",
+        "    max  voltage  across  the  resistor  is   50   V"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 5, page no. 911</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.002#  in  Henry\n",
+      "R  =  1000;#  in  Ohm\n",
+      "C1  =  5E-6;#  in  farad\n",
+      "C2  =  5E-9;#  in  farad\n",
+      "\n",
+      " #calculation:\n",
+      "a  =  (R/(2*L))**2\n",
+      "b  =  1/(L*C1)\n",
+      "if  (a>b):\n",
+      "\ts1  =  \"overdamped\";\n",
+      "elif  (a<b):\n",
+      "\ts1  =  \"underdamped\";\n",
+      "else:\n",
+      "\ts1  =  \"critically  damped\";\n",
+      "c  =  1/(L*C2)\n",
+      "if  (a>c):\n",
+      "\ts2  =  \"overdamped\";\n",
+      "elif  (a<c):\n",
+      "\ts2  =  \"underdamped\";\n",
+      "else:\n",
+      "\ts2  =  \"critically  damped\";\n",
+      "\t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    circuit  is  \",s1\n",
+      "print  \"\\n    if  C  =  5  nF,  circuit  is  \",s2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    circuit  is   overdamped\n",
+        "\n",
+        "    if  C  =  5  nF,  circuit  is   underdamped"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 6, page no. 912</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  0.002#  in  Henry\n",
+      "R  =  1000;#  in  Ohm\n",
+      "\n",
+      " #calculation:\n",
+      "a  =  (R/(2*L))**2\n",
+      " #for  critically  damped\n",
+      "C  =  4*L/R**2\n",
+      "\t\n",
+      "\n",
+      "#Results\n",
+      "print  \"\\n\\n  Result  \\n\\n\"\n",
+      "print  \"\\n    capacitance  C  is  \",C*1E9,\"nF\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "\n",
+        "    capacitance  C  is   8.0 nF"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "markdown",
+     "metadata": {},
+     "source": [
+      "<h3>Example 7, page no. 913</h3>"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from __future__ import division\n",
+      "import math\n",
+      "import cmath\n",
+      "#initializing  the  variables:\n",
+      "L  =  1.5#  in  Henry\n",
+      "R  =  90;#  in  Ohm\n",
+      "C = 5*1E-6; # in Farad\n",
+      "V = 10; # in Volts\n",
+      "\n",
+      "#calculation:\n",
+      "a = -1*R/(2*L)\n",
+      "b = (1/(L*C) - (R/(2*L))**2)**0.5\n",
+      "V0 = V\n",
+      "I0 = 0\n",
+      "A = V0\n",
+      "B = (I0 - C*a*V0)/(C*b)\n",
+      "\n",
+      "#Results\n",
+      "print \"\\n\\n  Result  \\n\\n\"\n",
+      "print \"Current, i = e^(\",a,\"t) (\",round((a*C*B - A*C*b),4),\"sin(\",round(b,1),\"t)  + (\",round((-1*a*C*A + B*C*C*b),0),\"cos(\",round(b,1),\"t) Amps.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "  Result  \n",
+        "\n",
+        "\n",
+        "Current, i = e^( -30.0 t) ( -0.0183 sin( 363.9 t)  + ( 0.0 cos( 363.9 t) Amps.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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diff --git a/Electrical_Circuit_Theory_And_Technology/screenshots/find_power_and_work_done.png b/Electrical_Circuit_Theory_And_Technology/screenshots/find_power_and_work_done.png
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diff --git a/Stoichiometry_And_Process_Calculations/README.txt b/Stoichiometry_And_Process_Calculations/README.txt
new file mode 100755
index 00000000..511b05a9
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/README.txt
@@ -0,0 +1,10 @@
+Contributed By: dolar khachariya
+Course: mtech
+College/Institute/Organization: IIT Bombay
+Department/Designation: Electrical Eng. Dept.
+Book Title: Stoichiometry And Process Calculations
+Author: K. V. Narayanan And B. Lakshmikutty
+Publisher: Prentice Hall Of India, New Delhi
+Year of publication: 2006
+Isbn: 81-203-2992-9
+Edition: 1
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch10.ipynb b/Stoichiometry_And_Process_Calculations/ch10.ipynb
new file mode 100755
index 00000000..c2e5a2de
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch10.ipynb
@@ -0,0 +1,1480 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Material Balance with Chemical Reaction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 pageno : 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "mair = 500.                                 #kg\n",
+      "mCO2 = 55.                                  #kg\n",
+      "mCO = 15.                                   #kg\n",
+      "\n",
+      "#C3H8 + 5O2 = 3CO2 + 4H20\n",
+      "MCO2 = 44 \n",
+      "MCO = 28 \n",
+      "\n",
+      "# Calculation \n",
+      "NCO2 = mCO2 / MCO2 \n",
+      "NCO = mCO / MCO \n",
+      "Mair = 29 \n",
+      "Nair = mair / Mair \n",
+      "#carbon balance gives,\n",
+      "F = (NCO2 + NCO)/3 \n",
+      "MC3H8 = 44.064 \n",
+      "mC3H8 = MC3H8 * F  \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)mass of propane burnt = %.2f\"%mC3H8,\"kg\"\n",
+      "\n",
+      "#one mole of propane requres 5 moles of oxygen for combustion\n",
+      "NO2 = F * 5 \n",
+      "Nairt = NO2 * 100 /21.                      #theoretical air required\n",
+      "Pexcess = (Nair - Nairt) * 100 / Nairt \n",
+      "print \"(b)The percent excess air = %.2f\"%Pexcess,\"%\"\n",
+      "\n",
+      "#C3H8 + 7/2 * O2 = 3CO + 4H2O\n",
+      "NH2O = F * 4 \n",
+      "#Taking oxygen balance, unburned oxygen is calculated,\n",
+      "#O2 supplied = O2 present in form of CO2, CO and H2O + unburned O2\n",
+      "Nunburnt = Nair * 21. / 100 - NCO2 - NCO/2 - NH2O/2. \n",
+      "NN2 = Nair * 79 / 100. \n",
+      "Ntotal = NCO2 + NCO + NH2O + NN2 + Nunburnt \n",
+      "PCO2 = NCO2 * 100 / Ntotal \n",
+      "PCO = NCO *100/ Ntotal \n",
+      "PH2O = NH2O *100/ Ntotal \n",
+      "PN2 = NN2 *100/ Ntotal \n",
+      "PO2 = Nunburnt *100 / Ntotal \n",
+      "\n",
+      "print \"(c)Percent composition of CO2 = %.2f\"%PCO2,\"%\"\n",
+      "print \"Percent composition of CO = %.2f\"%PCO,\"%\"\n",
+      "print \"Percent composition of H2O = %.2f\"%PH2O,\"%\"\n",
+      "print \"Percent composition of N2 = %.2f\"%PN2,\"%\"\n",
+      "print \"Percent composition of O2 = %.2f\"%PO2,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)mass of propane burnt = 26.23 kg\n",
+        "(b)The percent excess air = 21.66 %\n",
+        "(c)Percent composition of CO2 = 6.68 %\n",
+        "Percent composition of CO = 2.86 %\n",
+        "Percent composition of H2O = 12.73 %\n",
+        "Percent composition of N2 = 72.84 %\n",
+        "Percent composition of O2 = 4.88 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 pageno : 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "Nflue = 100.                        #kmol\n",
+      "NCO2 = 14.84 \n",
+      "NCO = 1.65 \n",
+      "NO2 = 5.16 \n",
+      "NN2 = 78.35 \n",
+      "PCF = 85.                           #PERCENT CARBON IN FEED\n",
+      "PIF = 15.                           #PERCENT INERT IN FEED\n",
+      "\n",
+      "# Calculation \n",
+      "#F - amount of coke charged, W - mass of coke left,W = 0.05F\n",
+      "NCflue = NCO2 + NCO  \n",
+      "MC = 12. \n",
+      "mC = MC * NCflue  \n",
+      "\n",
+      "#carbon balance gives, F * PCF / 100 = W * PCF + mC \n",
+      "F = mC / ( PCF / 100 - 0.05*PCF / 100) \n",
+      "#let A kmol air supplied, taking N2 balance,\n",
+      "Nair = NN2 * 100/79. \n",
+      "NO2supplied = Nair - NN2 \n",
+      "Ntheoretical = F * PCF / (100 * MC) \n",
+      "Pexcess = ( NO2supplied - Ntheoretical ) * 100 / ( Ntheoretical ) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Percentage excess air = %.2f\"%Pexcess,\"%\"\n",
+      "mair = Nair * 29 \n",
+      "m = mair / F                        #air supplied per kg of coke charged\n",
+      "print \"(b)air supplied per kg of coke charged = %.2f\"%m,\"kg\"\n",
+      "P = 100.                            #kPa\n",
+      "T = 500.                            #K\n",
+      "V = Nflue *22.4143*101.325 * T / (F * P * 273.15) \n",
+      "print \"(c)volume of flue gas per kg of coke = %.2f\"%V,\"m**3\"\n",
+      "W = 0.05*F \n",
+      "mCr = W * PCF/100.                  #carbon in refuse\n",
+      "mir = F * (1-PCF/100.)              #inert in refuse\n",
+      "mr = mCr + mir \n",
+      "C = mCr * 100 / mr \n",
+      "I = mir *100/ mr \n",
+      "print \"(d)Carbon = %.2f\"%C,\"%\"\n",
+      "print \"Inert = %.2f\"%I,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Percentage excess air = 19.99 %\n",
+        "(b)air supplied per kg of coke charged = 11.74 kg\n",
+        "(c)volume of flue gas per kg of coke = 16.96 m**3\n",
+        "(d)Carbon = 22.08 %\n",
+        "Inert = 77.92 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3 pageno :  333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "Nflue = 100.                            #kmol\n",
+      "NCO2 = 9. \n",
+      "NCO = 2. \n",
+      "NO2 = 3. \n",
+      "NN2 = 86. \n",
+      "NCflue = NCO2 + NCO  \n",
+      "MC = 12. \n",
+      "mC = MC * NCflue  \n",
+      "\n",
+      "# Calculation \n",
+      "#let A kmol air supplied, taking N2 balance,\n",
+      "Nair = NN2 * 100/79. \n",
+      "NO2supplied = Nair - NN2 \n",
+      "\n",
+      "# if CO in the flue gas was to be completely converted to CO2, \n",
+      "#then, the moles of oxygen present in the flue gas would be 3-1 =2kmol\n",
+      "Noexcess = NO2 - NCO/2. \n",
+      "Pexcess =  Noexcess * 100. / ( NO2supplied - Noexcess) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Percentage excess air = %.2f\"%Pexcess,\"%\"\n",
+      "NwaterO = NO2supplied - NCO2 - NCO/2 - NO2 \n",
+      "NH2 = NwaterO*2 \n",
+      "mH2 = NH2 * 2 \n",
+      "xCF = 0.7\n",
+      "R = mC / mH2 \n",
+      "print \"(b)Ratio of carbon to hydrogen in the fuel = %.2f\"%R\n",
+      "#let x be the amount of moisture in the feed, n it is \n",
+      "#given that 70% is carbon,therefore,\n",
+      "#0.7 = 3.32 / ( 1 + 3.32 + x )\n",
+      "x = R / xCF - 1 - R \n",
+      "mH = x * 2.016 / 18.016 \n",
+      "mHtotal = mH + mH2 \n",
+      "Rtotal = mC / mHtotal \n",
+      "print \"(c)Ratio of carbon to total hydrogen in the fuel = %.2f\"%Rtotal\n",
+      "ntotal = R + 1  +x \n",
+      "PH2 = 1*100./ntotal \n",
+      "PH2O = x * 100. / ntotal \n",
+      "print \"(d)percentage of combustible hydrogen in the fuel = %.2f\"%PH2,\"%\"\n",
+      "print \"percentage of moisture in the fuel = %.2f\"%PH2O,\"%\"\n",
+      "nH2Ototal = (PH2O + PH2 * 18.016 / 2.016)/100 \n",
+      "print \"(e)The mass of moisture in the flue gas per kg of fuel burned = %.3f\"%nH2Ototal,\"kg\"\n",
+      "\n",
+      "# note : answer may vary because of rounding error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Percentage excess air = 9.59 %\n",
+        "(b)Ratio of carbon to hydrogen in the fuel = 3.35\n",
+        "(c)Ratio of carbon to total hydrogen in the fuel = 3.34\n",
+        "(d)percentage of combustible hydrogen in the fuel = 20.92 %\n",
+        "percentage of moisture in the fuel = 9.08 %\n",
+        "(e)The mass of moisture in the flue gas per kg of fuel burned = 1.960 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4 pageno : 335"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "Nflue = 100.                            #kmoles\n",
+      "NCO2 = 9.05 \n",
+      "NCO = 1.34 \n",
+      "NO2 = 9.98 \n",
+      "NN2 = 79.63 \n",
+      "PCO2F = 9.2                             #% ( Feed )\n",
+      "PCOF = 21.3                             #%\n",
+      "PH2F = 18.                              #%\n",
+      "PCH4F = 2.5                             #%\n",
+      "PN2F = 49.                              #%\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#Taking carbon balance, \n",
+      "F = (NCO2 + NCO )/ ( (PCO2F + PCOF + PCH4F)/100.) \n",
+      "\n",
+      "#Nitrogen balance gives,\n",
+      "Nair = (NN2 - F*PN2F/(100) )* 100 / 79. \n",
+      "R = Nair/F \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)molar Ratio of air to fuel = %.2f\"%R\n",
+      "Oexcess = NO2 - NCO / 2 \n",
+      "Pexcess = Oexcess *100/ (Nair*21./100 - Oexcess) \n",
+      "print \"(b)Percent excess of air = %d\"%Pexcess,\"%\"\n",
+      "NN2F = F * PN2F / 100 \n",
+      "PN2F = NN2F *100/ NN2 \n",
+      "print \"(c)Percent of nitrogen in the flue gas that came from fuel = %.2f\"%PN2F,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)molar Ratio of air to fuel = 2.58\n",
+        "(b)Percent excess of air = 120 %\n",
+        "(c)Percent of nitrogen in the flue gas that came from fuel = 19.37 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 pageno : 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "Nflue = 100.                        #kmole\n",
+      "NCO2 = 16.4 \n",
+      "NCO = 0.4 \n",
+      "NO2 = 2.3 \n",
+      "NN2 = 80.9 \n",
+      "PCF = 80.5                          #% ( Feed )\n",
+      "PO = 5.0                            #%\n",
+      "PHF = 4.6                           #%\n",
+      "PN = 1.1                            #%\n",
+      "Pash = 8.8                          #%\n",
+      "\n",
+      "# Calculation \n",
+      "#Taking Carbon balance,\n",
+      "W = (NCO2 + NCO)*12. / (PCF / 100) \n",
+      "mCO2 = NCO2 * 44 \n",
+      "mCO = NCO * 32 \n",
+      "mO2 = NO2 * 28 \n",
+      "mN2 = NN2 * 28.014 \n",
+      "mtotal = mCO2 + mCO + mO2 + mN2 \n",
+      "Mdryflue = mtotal * 100/ W \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)The weight of dry gaseous products formed per 100 kg of coal fired = %.2f\"%Mdryflue,\"kg\"\n",
+      "#taking nitrogen balance,\n",
+      "x = (mN2 - W*PN/100)/28.014 \n",
+      "Noxygen = x * 21 / 79. \n",
+      "Nrequired = W * (PCF /12 + PHF/(2*2.016) - PO/32)/100 \n",
+      "Pexcess = (Noxygen - Nrequired)*100/Nrequired  \n",
+      "print \"(b)Percent excess air supplied for combustion = %.2f\"%Pexcess,\"%\"\n",
+      "\n",
+      "\n",
+      "# note: answer may vary because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The weight of dry gaseous products formed per 100 kg of coal fired = 1223.92 kg\n",
+        "(b)Percent excess air supplied for combustion = 11.49 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6 pageno : 338"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "mcoal = 100.                    #kg\n",
+      "mC = 63.                        #kg\n",
+      "mH = 12.                        #kg\n",
+      "mO = 16.                        #kg\n",
+      "mash  = 9.                      #kg\n",
+      "mfixC = 39.                     #kg\n",
+      "mH2O = 10.                      #kg\n",
+      "\n",
+      "# Calculation \n",
+      "mCvolatile = mC - mfixC \n",
+      "mHH2O = mH2O *2.016/18.016      #(mass of hydrogen in moisture)\n",
+      "mHvolatile = mH - mHH2O \n",
+      "mOH2O = mH2O - mHH2O \n",
+      "mOvolatile = mO - mOH2O \n",
+      "mtvolatile = mCvolatile + mHvolatile + mOvolatile \n",
+      "PC = mCvolatile * 100 / mtvolatile \n",
+      "PH = mHvolatile * 100 / mtvolatile \n",
+      "PO = mOvolatile * 100 / mtvolatile \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)percent carbon in volatile matter = %.2f\"%PC,\"%\"\n",
+      "print \"   percent hydrogen in volatile matter = %.2f\"%PH,\"%\"\n",
+      "print \"   percent oxygen in volatile matter = %.2f\"%PO,\"%\"\n",
+      "\n",
+      "PCflue = 10.8                   #%\n",
+      "Pvflue = 9.0                    #%\n",
+      "Pashflue = 80.2                 #%\n",
+      "\n",
+      "#taking ash balance, Wis the weight of the refuse,\n",
+      "W = mash *100. / Pashflue \n",
+      "mvflue = Pvflue * W /100. \n",
+      "mCflue = W * PCflue / 100 \n",
+      "Ctflue = mCflue + mvflue * PC / 100 #total carbon in flue\n",
+      "Htflue = mvflue * PH / 100 \n",
+      "Otflue = mvflue * PO / 100 \n",
+      "PCflue = Ctflue *100/W \n",
+      "PHflue = Htflue *100/W \n",
+      "POflue = Otflue *100/W \n",
+      "\n",
+      "print \"(b)percent Carbon in refuse = %.2f\"%PCflue,\"%\"\n",
+      "print \"   percent Hydrogen in refuse = %.2f\"%PHflue,\"%\"\n",
+      "print \"   percent Oxygen in refuse = %.2f\"%POflue,\"%\"\n",
+      "print \"   percent Ash in refuse = %.2f\"%Pashflue,\"%\"\n",
+      "\n",
+      "Coalburnt = mcoal - W \n",
+      "NCburnt = (mC - Ctflue)/12 \n",
+      "NHburnt = (mH - Htflue)/2.016 \n",
+      "NOburnt = (mO - Otflue)/32 \n",
+      "PCO2 = 80.                      #Percentage of carbon burnt\n",
+      "NCO2 = PCO2 * NCburnt / 100. \n",
+      "NCO = ( 1 - PCO2/100. )*NCburnt \n",
+      "Vair = 1000.                    #m**3\n",
+      "Nair = Vair / 22.4143 \n",
+      "NN2 = Nair * 79 / 100. \n",
+      "NO2 = Nair * 21 / 100. \n",
+      "Ocompounds = NCO2 + NCO/2 + NHburnt/2               #Oxygen present in CO2,CO and H2O\n",
+      "\n",
+      "#Oxygen balance gives free oxygen as,\n",
+      "Ofree = NO2 + mO/32 - Otflue/32 - Ocompounds \n",
+      "Ntotal = NN2 + Ofree + NCO2 + NCO                   #dry basis\n",
+      "PCO21 = NCO2 *100/Ntotal \n",
+      "PCO1 = NCO * 100/Ntotal \n",
+      "PO21 = Ofree * 100/Ntotal \n",
+      "PN21 = NN2 * 100/Ntotal \n",
+      "print \"(c)percent CO2 in flue = %.2f\"%PCO21,\"%\"\n",
+      "print \"   percent CO in flue = %.2f\"%PCO1,\"%\"\n",
+      "print \"   percent O2 in flue = %.2f\"%PO21,\"%\"\n",
+      "print \"   percent N2 in flue = %.2f\"%PN21,\"%\"\n",
+      "NOrequired = mC/12 + mH/(2.016*2) - mO/32 \n",
+      "Oexcess = NO2 - NOrequired \n",
+      "Pexcess = Oexcess * 100 / NOrequired \n",
+      "print \"(d)Percent excess air supplied = %.2f\"%Pexcess,\"%\"\n",
+      "NH2Oflue = NHburnt \n",
+      "mH2O = NH2Oflue * 18.016 \n",
+      "m = mH2O * 100/Ntotal \n",
+      "print \"(e)mass of water vapour per 100 moles of dry flue gas = %.2f\"%\\\n",
+      "m,\"g water vapour / 100kmol dry flue gas\"\n",
+      "\n",
+      "# note: answer may vary because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)percent carbon in volatile matter = 57.14 %\n",
+        "   percent hydrogen in volatile matter = 25.91 %\n",
+        "   percent oxygen in volatile matter = 16.95 %\n",
+        "(b)percent Carbon in refuse = 15.94 %\n",
+        "   percent Hydrogen in refuse = 2.33 %\n",
+        "   percent Oxygen in refuse = 1.53 %\n",
+        "   percent Ash in refuse = 80.20 %\n",
+        "(c)percent CO2 in flue = 9.55 %\n",
+        "   percent CO in flue = 2.39 %\n",
+        "   percent O2 in flue = 5.53 %\n",
+        "   percent N2 in flue = 82.53 %\n",
+        "(d)Percent excess air supplied = 21.26 %\n",
+        "(e)mass of water vapour per 100 moles of dry flue gas = 245.62 g water vapour / 100kmol dry flue gas\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7 pageno : 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "Pexcess = 20.                           #%\n",
+      "PSO3 = 5.                               #% ( Percent of sulphur burnt to SO3 )\n",
+      "\n",
+      "#S + O2 = SO2\n",
+      "N = 1.                                  #kmol sulphur\n",
+      "Orequired = N                           #kmol\n",
+      "\n",
+      "# Calculation \n",
+      "Osupplied = Orequired * ( 1 + Pexcess/100) \n",
+      "Nsupplied = Osupplied * 79/21. \n",
+      "NSO2 = (1-PSO3/100)*N \n",
+      "NSO3 = PSO3 * N /100. \n",
+      "Oconsumed = NSO2 + 3/2. * PSO3/100 \n",
+      "Oremaining = Osupplied - Oconsumed \n",
+      "Ntotal = NSO2 + NSO3 + Oremaining + Nsupplied \n",
+      "PSO2 = NSO2 * 100 / Ntotal \n",
+      "PSO3 = NSO3 * 100 / Ntotal \n",
+      "PO2 = Oremaining * 100 / Ntotal \n",
+      "PN2 = Nsupplied * 100 / Ntotal \n",
+      "\n",
+      "# Result\n",
+      "print \"Percent SO2 in burner gas = %.2f\"%PSO2,\"%\"\n",
+      "print \"Percent SO3 in burner gas = %.2f\"%PSO3,\"%\"\n",
+      "print \"Percent O2 in burner gas = %.2f\"%PO2,\"%\"\n",
+      "print \"Percent N2 in burner gas = %.2f\"%PN2,\"%\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percent SO2 in burner gas = 16.70 %\n",
+        "Percent SO3 in burner gas = 0.88 %\n",
+        "Percent O2 in burner gas = 3.08 %\n",
+        "Percent N2 in burner gas = 79.35 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.8 pageno : 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "Nburner = 100.                      #kmol\n",
+      "NSO2b = 9.5                         #kmol\n",
+      "NO2b = 7.                           #kmol\n",
+      "\n",
+      "# Calculation \n",
+      "NN2 = Nburner - NSO2b - NO2b \n",
+      "NOsupplied = NN2 * 21 / 79.         #Oxygen supplied\n",
+      "\n",
+      "#4FeS2 + 11O2 = 2Fe2O3 + 8SO2\n",
+      "#4FeS2 + 15O2 = 2Fe2O3 + 8SO3\n",
+      "NOtotal = NO2b + NSO2b + NSO2b * 3 / 8. \n",
+      "NOunaccounted = NOsupplied - NOtotal \n",
+      "NSO31 = NOunaccounted * 8 /15 \n",
+      "NStotal = NSO2b + NSO31 \n",
+      "mS = NStotal * 32.064 \n",
+      "Pburnt = 50.                        #% ( percentage of pyrites burnt )\n",
+      "mFeS2 = mS * 100/ Pburnt \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Total pyrites burnt = %.2f\"%mFeS2,\"kg\"\n",
+      "NFeS2 = NStotal / 2 \n",
+      "MFeS2 = 119.975 \n",
+      "mFeS21 = MFeS2 * NFeS2 \n",
+      "mgangue = mFeS2 - mFeS21 \n",
+      "NFe2O3 = NFeS2 * Pburnt / 100 \n",
+      "MFe2O3 = 159.694 \n",
+      "mFe2O3 = MFe2O3 * NFe2O3 \n",
+      "PSO3c = 2.5                         #% ( percentage sulphur as SO3 in cinder )\n",
+      "mc = 100.                           #kg ( basis )\n",
+      "NSO3 = PSO3c / 32.064 \n",
+      "mSO3 = NSO3 * 80.064 \n",
+      "mremaining = mc - mSO3              # ( Fe2O3 + gangue )\n",
+      "\n",
+      "#x be the weight of the cinder \n",
+      "x = (mFe2O3 + mgangue)*100/mremaining  \n",
+      "print \"(b)weight of cinder produced = %.f\"%x,\"kg\"\n",
+      "Slost = x * NSO3 / 100 \n",
+      "PSlost = Slost *100/ NStotal \n",
+      "print \"(c)Percent of total S lost in the cinder = %.2f\"%PSlost,\"%\"\n",
+      "mSO3c = mSO3 * x / 100 \n",
+      "NSO3b = NSO31 - Slost \n",
+      "P = NSO3b * 100 / NStotal \n",
+      "print \"(d)Percentage of S charged that is present as SO3 in the burner gas = %.2f\"%P,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Total pyrites burnt = 682.19 kg\n",
+        "(b)weight of cinder produced = 500 kg\n",
+        "(c)Percent of total S lost in the cinder = 3.66 %\n",
+        "(d)Percentage of S charged that is present as SO3 in the burner gas = 7.03 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.9 pageno : 345"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "Ncgas = 100.                #kmol ( basis - SO3 free converter gas )\n",
+      "NSO2 = 4.5                  #kmol\n",
+      "NO2 = 7.5                   #kmol\n",
+      "NN2 = 88.0                  #kmol\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "NOsupplied = NN2 * 21./ 79 \n",
+      "NOconverter = NO2 + NSO2 \n",
+      "NOconsumed = NOsupplied - NOconverter           #(Oxygen consumed for SO3)\n",
+      "NSO3c = NOconsumed / 1.5 \n",
+      "NStotal = NSO3c + NSO2 \n",
+      "Nbgas = 100.                                    #kmol ( basis - SO3 free burner gas )\n",
+      "NSO21 = 15.                                     #%\n",
+      "NO21 = 5.                                       #%\n",
+      "NN21 = 80.                                      #%\n",
+      "NOburner = NO21 + NSO21 \n",
+      "NOsupplied1 = NN21 * 21. / 79. \n",
+      "NOconsumed1 = NOsupplied1 - NOburner            #(Oxygen consumed for SO3)\n",
+      "NSO3b = NOconsumed1 / 1.5 \n",
+      "NStotal1 = NSO3b + NSO21 \n",
+      "mS = 100.                                       #kg ( basis - sulphur charged )\n",
+      "Pburned = 95.                                   #%\n",
+      "mburned = mS * Pburned / 100. \n",
+      "Nburned = mburned / 32.064 \n",
+      "\n",
+      "#let x be the SO3 free burner gas produced, then sulphur balance gives,\n",
+      "x = Nburned * Nbgas / NStotal1 \n",
+      "NSO2b = NSO21 * x / 100 \n",
+      "NO2b = NO21 * x / 100 \n",
+      "NN2b = NN21 * x / 100 \n",
+      "Ntotalb = NSO2b + NO2b + NN2b \n",
+      "NSO3b1 = NSO3b * x / 100 \n",
+      "\n",
+      "#let y be the no. of converter gas produced\n",
+      "y = Nburned * Ncgas / NStotal \n",
+      "NSO2c = NSO2 * y / 100 \n",
+      "NO2c = NO2 * y / 100 \n",
+      "NN2c = NN2 * y / 100 \n",
+      "Ntotalc = NSO2c + NO2c + NN2c \n",
+      "NSO3c1 = NSO3c * y / 100 \n",
+      "Nairsec = (NN2c - NN2b ) * 100 / 79. \n",
+      "P = 100.                                        #kPa\n",
+      "T = 300.                                        #K\n",
+      "V = Nairsec * 22.414 * 101.3 * T / (P * 273.15) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)The volume of secondary air at 100kPa and 300K = %.1f\"%V,\"m**3/h\"\n",
+      "NSabsorbed = 95.                                #%\n",
+      "mSO3abs = NSabsorbed * NSO3c1 * 80.064 / 100 \n",
+      "\n",
+      "#let z be the amount of 98% H2SO4, therefore , 100% H2SO4 produced = z + mSO3abs\n",
+      "# taking SO3 balance\n",
+      "z = (mSO3abs - mSO3abs * 80.064 / 98.08) / ( 80.064 / 98.08 - 0.98 * 80.064/98.08) \n",
+      "print \"(b)98%% H2SO4 required per hour = %.1f\"%z,\"kg\"\n",
+      "w = z + mSO3abs \n",
+      "print \"(c)100%% H2SO4 produced per hour = %.1f\"%w,\"kg\"\n",
+      "\n",
+      "# note: answer may vary because of rounding error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The volume of secondary air at 100kPa and 300K = 208.2 m**3/h\n",
+        "(b)98% H2SO4 required per hour = 1592.1 kg\n",
+        "(c)100% H2SO4 produced per hour = 1733.6 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.10 pageno : 349"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "mlime = 5.                          #kg\n",
+      "mcoke = 1.                          #kg\n",
+      "PCaCO3l = 84.5                      #%\n",
+      "PMgCO3l = 11.5                      #%\n",
+      "\n",
+      "# Calculation \n",
+      "NCaCO3l = PCaCO3l * mlime / (100.09*100) \n",
+      "NMgCO3l = PMgCO3l * mlime / (84.312*100) \n",
+      "mInertsl = mlime * ( 100 - PCaCO3l - PMgCO3l ) / 100 \n",
+      "PCc = 76.                           #%\n",
+      "Pashc = 21.                         #%\n",
+      "Pwaterc = 3.                        #%\n",
+      "NCc = mcoke *  PCc /(100*12) \n",
+      "Nwaterc = mcoke * Pwaterc / ( 100 * 18.016 ) \n",
+      "mash = Pashc * mcoke / 100 \n",
+      "\n",
+      "#CaCO3 + C + O2 = CaO + 2CO2\n",
+      "#MgCO3 + C + O2 = MgO + 2CO2\n",
+      "PCaCO3conv = 95.                    #(Percent calcination of CaCO3)\n",
+      "PMgCO3conv = 90.                    #(Percent calcination of MgCO3)\n",
+      "NCaO = PCaCO3conv * NCaCO3l / 100. \n",
+      "mCaO = NCaO * 56.08 \n",
+      "NMgO = PMgCO3conv * NMgCO3l / 100. \n",
+      "mMgO = NMgO * 40.312 \n",
+      "mCaCO3 = (NCaCO3l * (1-PCaCO3conv/100)*100.09) \n",
+      "mMgCO3 = (NMgCO3l * (1-PMgCO3conv/100)*84.312) \n",
+      "mtotal = mCaO + mMgO + mCaCO3 + mMgCO3 + mInertsl + mash \n",
+      "PCaO = mCaO * 100 / mtotal \n",
+      "\n",
+      "# Result\n",
+      "print \"The weight percent of CaO in the product leaving the kiln = %.2f\"%PCaO,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The weight percent of CaO in the product leaving the kiln = 70.83 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.11 pageno : 350"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "R = 100.                        #kg ( basis - residue )\n",
+      "MCaSO4 = 136.144 \n",
+      "MMgSO4 = 120.376 \n",
+      "mCaSO4r = 9.                    #kg\n",
+      "mMgSO4r = 5.                    #kg\n",
+      "mH2SO4r = 1.2                   #kg\n",
+      "minertr = 0.5                   #kg\n",
+      "mCO2r = 0.2                     #kg\n",
+      "mH2O = 84.10                    #kg\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "NCaSO4 = mCaSO4r / MCaSO4 \n",
+      "NMgSO4 = mMgSO4r / MMgSO4 \n",
+      "\n",
+      "#CaCO3 + H2SO4 = CaSO4 + H2O + CO2\n",
+      "#MgSO4 + H2SO4 = MgSO4 + H2O + CO2\n",
+      "\n",
+      "mCaCO3 = NCaSO4 * 100.08 \n",
+      "mMgCO3 = NMgSO4 * 84.312 \n",
+      "mtotallime = minertr + mCaCO3 + mMgCO3 \n",
+      "PCaCO3 = mCaCO3 * 100/ mtotallime \n",
+      "PMgCO3 = mMgCO3 *100/ mtotallime \n",
+      "Pinerts = minertr *100/ mtotallime \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Percentage of CaCO3 in limestone = %.2f\"%PCaCO3,\"%\"\n",
+      "print \"   Percentage of MgCO3 in limestone = %.2f\"%PMgCO3,\"%\"\n",
+      "print \"   Percentage of inerts in limestone = %.2f\"%Pinerts,\"%\"\n",
+      "\n",
+      "NH2SO4 = NCaSO4 + NMgSO4 \n",
+      "mH2SO4 = NH2SO4 * 98.08 \n",
+      "Pexcess = mH2SO4r * 100 / ( mH2SO4) \n",
+      "print \"(b)The percentage excess of acid used = %.2f\"%Pexcess,\"%\"\n",
+      "\n",
+      "macidt = mH2SO4 + mH2SO4r \n",
+      "Pacidic = 12.                   #%\n",
+      "mwaterin = macidt * (100 - Pacidic)/ Pacidic \n",
+      "mwaterr = (NCaSO4 + NMgSO4)*18.016 \n",
+      "mwatert = mwaterin + mwaterr \n",
+      "mvaporized = mwatert - mH2O \n",
+      "m = mvaporized * 100/mtotallime             #water vaporized per 100kg of limestone\n",
+      "print \"(c)the mass of water vaporized per 100kg of limestone = %.2f\"%m,\"kg\"\n",
+      "mCO2pr = (NCaSO4 + NMgSO4)*44 \n",
+      "mCO2rel = mCO2pr - mCO2r \n",
+      "m1 = mCO2rel * 100 / mtotallime             #CO2 per 100kg of limestone\n",
+      "print \"(d)the mass of CO2 per 100kg of limestone = %.2f\"%m1,\"kg\"\n",
+      "\n",
+      "# note : answer may vary because of rounding error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Percentage of CaCO3 in limestone = 62.31 %\n",
+        "   Percentage of MgCO3 in limestone = 32.98 %\n",
+        "   Percentage of inerts in limestone = 4.71 %\n",
+        "(b)The percentage excess of acid used = 11.37 %\n",
+        "(c)the mass of water vaporized per 100kg of limestone = 38.25 kg\n",
+        "(d)the mass of CO2 per 100kg of limestone = 42.72 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.12 pageno : 353"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "macid = 1000.                   #kg ( basis - dilute phosphoric acid )\n",
+      "Mphacid = 97.998 \n",
+      "P = 1.25                        #% ( dilute % )\n",
+      "\n",
+      "# Calculation \n",
+      "mphacid = macid * P /100 \n",
+      "Nphacid = mphacid / Mphacid \n",
+      "\n",
+      "#1mole of phosphoric acid - 1mole of trisodium phosphate\n",
+      "NTSP = Nphacid \n",
+      "MTSP = 380.166 \n",
+      "mTSP = NTSP * MTSP \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Maximum weight of TSP obtained = %.2f\"%mTSP,\"kg\"\n",
+      "NCO2 = NTSP \n",
+      "Pwater = 6.27                   #kPa\n",
+      "\n",
+      "#since gas is saturated with water vapour, vapour pressure = partial pressure\n",
+      "Nwater = NCO2 * Pwater / ( 100 - Pwater ) \n",
+      "Ntotal = Nwater + NCO2 \n",
+      "P = 100.                        #kPa\n",
+      "T = 310.                        #K\n",
+      "V = Ntotal * 101.3 * T *22.4143 / ( P * 273.15 ) \n",
+      "print \"(b)Volume of CO2 = %.2f\"%V,\"m**3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Maximum weight of TSP obtained = 48.49 kg\n",
+        "(b)Volume of CO2 = 3.51 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.13 pageno : 353"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "mTSPd = 1000.                       #kg ( basis - 20% dilute TSP )\n",
+      "P = 20.                             #%\n",
+      "\n",
+      "# Calculation \n",
+      "mTSP = mTSPd * P / 100 \n",
+      "NTSP = mTSP / 163.974 \n",
+      "msodaashd = NTSP * 106 \n",
+      "mphacidd = NTSP * 97.998 \n",
+      "mNaOHd = NTSP * 40.008 \n",
+      "Pphacid = 85.                       #% (85% solution phosphoric acid)\n",
+      "PNaOH = 50.                         #% (50% solution NaOH)\n",
+      "\n",
+      "#let x be the water in soda ash,\n",
+      "#taking water balance,\n",
+      "x = (mTSPd - mTSP) - mNaOHd * PNaOH /(100 - PNaOH) - mphacidd * (100 - Pphacid) / Pphacid \n",
+      "msodaash = msodaashd + x \n",
+      "C = msodaashd *100 / msodaash \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Concentration of soda ash solution = %.2f\"%C,\"%\"\n",
+      "mphacid = mphacidd * 100 / Pphacid \n",
+      "R = msodaash / mphacid \n",
+      "print \"(b)Weight ratio in which soda ash and commercial phosphoric acid are mixed = %.2f\"%R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Concentration of soda ash solution = 15.04 %\n",
+        "(b)Weight ratio in which soda ash and commercial phosphoric acid are mixed = 6.11\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.14 pageno : 355"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "m = 1000.                       #kg ( basis - pig iron produced )\n",
+      "\n",
+      "#let x be the iron ore charged and y be the amount of flux added and z be the weight of slag produced\n",
+      "PFepg = 95.                     #% ( Fe% in product )\n",
+      "PCpg = 4.                       #%\n",
+      "PSipg = 1.                      #%\n",
+      "PFech = 85.                     #% (Fe% in feed )\n",
+      "mcoke = 1000.                   #kg\n",
+      "PCcoke = 90.                    #%\n",
+      "PSicoke = 10.                   #%\n",
+      "PSislag = 60.                   #%\n",
+      "PSiflux = 5.                    #%\n",
+      "PCaCO3fx = 90.                  #%\n",
+      "PMgCO3fx = 5.                   #%\n",
+      "PCMslag = 40.                   #% \n",
+      "\n",
+      "# Calculation \n",
+      "#iron balance gives,\n",
+      "x = PFepg * m *159.694 / ( PFech * 111.694) \n",
+      "#silicon balance gives,\n",
+      "#x*(100 - PFech)*28.086/(100*60.086)+mcoke*Psicoke*28.086/(100*60.086)+y*PSiflux*28.086/(100*60.086) = 10 + z*Psislag*28.086 / ( 100*60.086 )\n",
+      "#taking (CaO + MgO) balance\n",
+      "#y * ((PCaCO3fx)*56.88/(100*100.88)+(PMgCO3fx*40.312/(100*84.312))=z*PCMslag/100\n",
+      "#solving above 2 equations , we get\n",
+      "y = 403.31 \n",
+      "\n",
+      "# Result\n",
+      "print \"the amount of flux required to produce 1000kg of pig iron = \",y,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the amount of flux required to produce 1000kg of pig iron =  403.31 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.15 pageno : 357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "N = 100.                    #mol(basis - scrubber)\n",
+      "NNOs = 2.4                  #mol\n",
+      "NN2s = 92.                  #mol\n",
+      "NO2s = 5.6                  #mol\n",
+      "PNOs = 20.                  #% ( Percentage NO leaving scrubber)\n",
+      "\n",
+      "# Calculation \n",
+      "NNOreac = NNOs * 100 / PNOs \n",
+      "#let x mol of nitroge be produced in the reaction, then the amount of N2 present in the air = NN2s - x mol - (1)\n",
+      "#4NH3 + 5O2 = 4NO + 6H2O\n",
+      "#4NH3 + 3O2 = 2N2 + 6H2O\n",
+      "#4moles of NO - 5 moles of O2, 2moles of N2 - 3 moles of O2\n",
+      "#Total oxygen used up, O  =  NNOreac * 5/4 + x*3/2\n",
+      "#total oxygen supplied, NOtotal= (O) + NO2s\n",
+      "#Nitrogen associated with O2 supplied NN2 = NOtotal*79/21 - (2)\n",
+      "#comparing 1 and 2,\n",
+      "x = 2.1835 \n",
+      "#12moles NO requires 12moles ammonia, 1 mole N2 requires 2 mole ammonia\n",
+      "Nammonia = x*2 + NNOreac \n",
+      "Oreq = Nammonia * 5 / 4 \n",
+      "Osupp = NNOreac * 5/4 + x*3/2 + NO2s \n",
+      "Pexcess = (Osupp - Oreq)*100/Oreq  \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Percentage excess oxygen = %.1f\"%Pexcess,\"%\"\n",
+      "fr = x * 2 / Nammonia \n",
+      "print \"Fraction of ammonia taking part in side reaction = %.3f\"%fr\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Percentage excess oxygen = 16.7 %\n",
+        "Fraction of ammonia taking part in side reaction = 0.267\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.16 pageno : 359"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "m = 100.                        #kg (basis sodium nitrate reacted)\n",
+      "NNaNO3 = m/85\n",
+      " \n",
+      "# Calculation \n",
+      "#2NaNO3 + H2SO4 = 2HNO3 + Na2SO4\n",
+      "mh2so4 = NNaNO3 * 98.08/2 \n",
+      "mhno3 = NNaNO3*63.008 \n",
+      "mna2so4 = NNaNO3 * 142.064 /2 \n",
+      "Phno3 = 2.                      #%(percent nitric acid remaining in the cake)\n",
+      "mhno3cake = mhno3 * Phno3 / 100 \n",
+      "Ph2so4 = 35.                    #%\n",
+      "Pwater = 1.5                    #%\n",
+      "mtotal = (mna2so4 + mhno3cake)*100/(100 - Ph2so4 - Pwater) \n",
+      "mwater = Pwater * mtotal / 100 \n",
+      "mh2so4c = Ph2so4 * mtotal / 100 \n",
+      "Pna2so4 = mna2so4 *100/mtotal \n",
+      "Phno3c = mhno3cake * 100 / mtotal \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Mass of Na2SO4 in the cake = %.3f\"%mna2so4,\"kg\"\n",
+      "print \"Mass of H2SO4 in the cake = %.3f\"%mh2so4c,\"kg\"\n",
+      "print \"Mass of HNO3 in the cake = %.4f\"%mhno3,\"kg\"\n",
+      "print \"Mass of water in the cake = %.3f\"%mwater,\"kg\"\n",
+      "print \"Percentage of Na2SO4 in the cake = %.3f\"%Pna2so4,\"%\"\n",
+      "print \"Percentage of H2SO4 in the cake = %.2f\"%Ph2so4,\"%\"\n",
+      "print \"Percentage of HNO3 in the cake = %.2f\"%Phno3c,\"%\"\n",
+      "print \"Percentage of water in the cake = %.2f\"%Pwater,\"%\"\n",
+      "\n",
+      "mh2so4req = mh2so4 + mh2so4c \n",
+      "P = 95.                         #% (95% dilute sulphuric acid)\n",
+      "w = mh2so4req * 100 / P \n",
+      "print \"(b)Weight of 95%% sulphuric acid required = %.2f\"%w,\"kg\"\n",
+      "mnitric = mhno3 - mhno3cake \n",
+      "print \"(c)weight of nitric acid product obtained = %.2f\"%mnitric,\"kg\"\n",
+      "mwaterd = w*(1-P/100)-mwater \n",
+      "print \"(d)the water vapour tha tis distilled from the nitre cake = %.3f\"%mwaterd,\"kg\"\n",
+      "\n",
+      "\n",
+      "# Note : The value of HNO3 is taken wrongly in the textbook while calculating mass of HNO3"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Mass of Na2SO4 in the cake = 83.567 kg\n",
+        "Mass of H2SO4 in the cake = 46.878 kg\n",
+        "Mass of HNO3 in the cake = 74.1271 kg\n",
+        "Mass of water in the cake = 2.009 kg\n",
+        "Percentage of Na2SO4 in the cake = 62.393 %\n",
+        "Percentage of H2SO4 in the cake = 35.00 %\n",
+        "Percentage of HNO3 in the cake = 1.11 %\n",
+        "Percentage of water in the cake = 1.50 %\n",
+        "(b)Weight of 95% sulphuric acid required = 110.08 kg\n",
+        "(c)weight of nitric acid product obtained = 72.64 kg\n",
+        "(d)the water vapour tha tis distilled from the nitre cake = 3.495 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.17 pageno : 361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "m = 50.                         #kg ( basis - mass of brine charged )\n",
+      "#let x be the amount of NaCl in the brine\n",
+      "Pelect = 50.                    #% ( electrolyzed )\n",
+      "\n",
+      "#2NaCl + 2H2O = 2NaOH + Cl2 + H2\n",
+      "#amount of NaCl reacted =x*Pelect/(100*58.45)kmol=x*Pelect/100kg   ( 1 )\n",
+      "#amount of water reacted = x * Pelect * 18.016 / ( 100 * 58.45 )kg ( 2 )\n",
+      "#Gases produced, Cl2 = x * Pelect / (100 * 58.45 * 2 )kmol = x * Pelect *71/ (100 * 58.45 * 2 )kg                                           ( 3 )\n",
+      "#H2 = x * Pelect / (100 * 58.45 * 2 )kmol = x * Pelect *2.016/ (100 * 58.45 * 2 )kg                                                         ( 4 )\n",
+      "Nwater = 0.03                   #mol water vapour/mol of gas\n",
+      "\n",
+      "# Calculation \n",
+      "#water vapour present = Nwater * 2*(Cl2 + H2)kmol = Nwater * 2*(Cl2 + H2)*18.016 kg                                                         ( 5 )\n",
+      "#NaoH = x * Pelect * 40.008/ (100 * 58.45 )kg                      ( 6 )\n",
+      "#water = water in brine - water reacted - water present in gas     ( 7 )\n",
+      "#= (m - Pelect/100) - water reacted ( 2 ) - water present in the gas( 5 )\n",
+      "#Total weight of solution = NaCl ( 1 ) + NaOH ( 6 ) + Water ( 7 )\n",
+      "#since NaOH is 10 percent of the total weight, we have NaOH = 0.1 * total weight, from these we get, \n",
+      "x = 0.1 * 50 / (0.1* 0.3165 + 0.3422 ) \n",
+      "NaOH = x * Pelect * 40.008/ (100 * 58.45 ) \n",
+      "NaCl = x * Pelect / 100 \n",
+      "water = 34.5032                 #kg\n",
+      "Pevap = 50.                     #NaOh percentage in solution leaving evaporator\n",
+      "\n",
+      "#taking NaOH balance\n",
+      "mevap = NaOH * 100 / Pevap \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)amount of 50%% NaOH solution produced = %.4f\"%mevap,\"kg\"\n",
+      "Cl2 = x * Pelect *71/ (100 * 58.45 * 2 )                #kg\n",
+      "H2 = x * Pelect *2.016/ (100 * 58.45 * 2 )              #kg\n",
+      "print \"(b)Chlorine produced = %.4f\"%Cl2,\"kg\"\n",
+      "print \"   Hydrogen produced = %.4f\"%H2,\"kg\"\n",
+      "Pleav = 1.5                                             #% NaCl leaving the evaporator\n",
+      "NaClleav = mevap * Pleav / 100 \n",
+      "mcrystal = NaCl - NaClleav \n",
+      "print \"(c)Amount of NaCl crystallized = %.4f\"%mcrystal,\"kg/h\"\n",
+      "mwaterleav = mevap - NaOH - NaClleav \n",
+      "Mwaterevap = water - mwaterleav \n",
+      "print \"(d)Weight of water evaporated = %.4f\"%Mwaterevap,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)amount of 50% NaOH solution produced = 9.1545 kg\n",
+        "(b)Chlorine produced = 4.0615 kg\n",
+        "   Hydrogen produced = 0.1153 kg\n",
+        "(c)Amount of NaCl crystallized = 6.5499 kg/h\n",
+        "(d)Weight of water evaporated = 30.0633 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.18 pageno : 364"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "m = 100.                        #mol ( basis reactore exit gas )\n",
+      "#CH3OH + O2 = HCOOH + H2O\n",
+      "#CH3OH + O2 / 2 = HCHO + H2O\n",
+      "Nn2 = 64.49                     #mol\n",
+      "No2 = 13.88                     #mol\n",
+      "Nh2o = 5.31                     #mol\n",
+      "Nch3oh = 11.02                  #mol\n",
+      "Nhcho = 4.08                    #mol\n",
+      "Nhcooh = 1.22                   #mol\n",
+      "\n",
+      "# Calculation \n",
+      "#x be the moles of methanol reacted, taking C balance, we get,\n",
+      "x = Nch3oh + Nhcho + Nhcooh \n",
+      "Pconv = Nhcho * 100 / x  \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Percent conversion of formaldehyde = \",Pconv,\"%\"\n",
+      "Nair = Nn2 * 100 / 79. \n",
+      "R = Nair / x \n",
+      "print \"(b)Ratio of air to methanol in the feed = %d\"%R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Percent conversion of formaldehyde =  25.0 %\n",
+        "(b)Ratio of air to methanol in the feed = 5\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.19 pageno : 367"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "NA = 100.                       #mol ( basi - 100 mol A in the fresh feed )\n",
+      "Pconv = 95.                     #%\n",
+      "\n",
+      "# Calculation \n",
+      "NApro = NA * (100 - Pconv)/100 \n",
+      "\n",
+      "#A = 2B + C \n",
+      "NB = NA * Pconv * 2 / 100 \n",
+      "NC = NA * Pconv/100 \n",
+      "PAent = 0.5                     #%\n",
+      "NAent = NApro * 100 / PAent \n",
+      "PBrec = 1.                      #%\n",
+      "NBent = NB * 100 / (100 - PBrec) \n",
+      "m = (NAent - NApro + NA) \n",
+      "conv = ((NAent - NApro + NA) - NAent)*100/(NAent - NApro + NA) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)single pass converion = %.2f\"%conv,\"%\"\n",
+      "Nrecycled = (NAent - NApro) + (NBent - NB) \n",
+      "R = Nrecycled/NA \n",
+      "print \"(b)recycle ratio = %.2f\"%R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)single pass converion = 8.68 %\n",
+        "(b)recycle ratio = 9.97\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.20 pageno : 368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "m = 100.                #kg ( basis - sucrose solution as fresh feed )\n",
+      "\n",
+      "#R - recycle reactor exit, \n",
+      "#let x be the weight fraction of sucrose and y be the weight fracton of inversion sugar in the recycle stream, \n",
+      "#for combined stream  fraction of Glucose + fructose = 0.04\n",
+      "#z be the weight fraction of sucrose in the combined stream entering the reactor\n",
+      "Psfeed = 25.            #% percent sucrose in fresh feed\n",
+      "\n",
+      "#sucrose balance gives, 25 + R*x = (100+R)* z                        (A)\n",
+      "#Glucose + fructose balance, R * y = (100 + R )*0.04                 (B)\n",
+      "Sucrosecon = 71.7       #% sucrose consumed\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#sucrose balance around the reactor,(100+R)z=0.717*(100+R)z+(100+R)x (C)\n",
+      "#From (C) , x = 0.283*z                                              (D)\n",
+      "#Amount converted to Glucose + fructose = 0.717 ( 100 + R )* z \n",
+      "# = 0.717 ( 100 + R )* z * 360.192 / 342.176 kg\n",
+      "#Glucose and fructose balance around the reactor,\n",
+      "#(100+R)*0.04 + 0.717(100+R)*z*360.192/342.176 = (100+R)*y           (E)\n",
+      "#Solving (E), y = 0.04 + 0.7548*z                                    (F)\n",
+      "#Solving, (A), (B), (C) and (F)\n",
+      "x = 0.06 \n",
+      "y = 0.2 \n",
+      "z = 0.212 \n",
+      "R = 25 \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Recycle flow = \",R,\"kg\"\n",
+      "print \"(b)Combined concentration of Glucose and Fructose in the recycle stream = \",y*100,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Recycle flow =  25 kg\n",
+        "(b)Combined concentration of Glucose and Fructose in the recycle stream =  20.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.21 pageno : 370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "N = 1.                      #mol ( basis - combined feed )\n",
+      "\n",
+      "#F - moles of fresh feed\n",
+      "Pinert = 0.5                #%\n",
+      "Pconv = 60.                  #%\n",
+      "P1inert = 2.                #%\n",
+      "\n",
+      "# Calculation \n",
+      "NA1 = N * ( 1- P1inert/100. ) \n",
+      "NA2 = NA1 * ( 1 - Pconv / 100. ) \n",
+      "NB2 = NA1 - NA2 \n",
+      "N1inert = N * P1inert / 100. \n",
+      "N2inert = N1inert \n",
+      "\n",
+      "#Let R be the moles recycled and P be the moles purged\n",
+      "#W = R + P\n",
+      "W = NA2 + N2inert           #              (A)\n",
+      "PWinert = N2inert * 100/ ( NA2 + N2inert) \n",
+      "\n",
+      "#component A balance, A fresh feed = A purge stream + A recycle stream\n",
+      "#F * 0.9 = P * 0.9515 + 0.588                      (B)\n",
+      "#inert balance at the point where fresh feed is mixed with the recycle,\n",
+      "#F*0.005 + R*0.0485 = 1* 0.02                      (C)\n",
+      "#Solving (A),(B) and (C)\n",
+      "F = 0.6552                  #mol\n",
+      "P = 0.0671                  #mol\n",
+      "R = 0.3448                  #mol\n",
+      "\n",
+      "# Result\n",
+      "print \"(a)moles of recycle stream = \",R,\"mol\"\n",
+      "print \"(b)moles of purge stream = \",P,\"mol\"\n",
+      "NAconv = NA1 - NA2 \n",
+      "NAf = F * (1 - Pinert / 100.) \n",
+      "Conv = NAconv *100./ NAf \n",
+      "print \"(c)Overall conversion = %.1f\"%Conv,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)moles of recycle stream =  0.3448 mol\n",
+        "(b)moles of purge stream =  0.0671 mol\n",
+        "(c)Overall conversion = 90.2 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.22 pageno : 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "N = 100.                            #moles ( Basis - Fresh feed )\n",
+      "Pconv = 20.                         #%\n",
+      "xco = 0.33 \n",
+      "xh2 = 0.665 \n",
+      "xch4 = 0.005 \n",
+      "\n",
+      "#R - recycle stream, P - purge stream\n",
+      "#x - mole fraction of CO in recycle stream , \n",
+      "xch4r = 0.03\n",
+      "\n",
+      "# Calculation \n",
+      "#CO = x, H2 = 1 - xch4r - CO = 0.97- x \n",
+      "#methane balance over the entire system, \n",
+      "P = xch4 * N / xch4r \n",
+      "#taking caron balance, 33.5 = M + P ( 0.03 + x )\n",
+      "#Hydrogen balance, 66.5 + 2*0.5 = 2M + P(2*0.03 + 0.97 - x) \n",
+      "#substituting P, M + 16.67x = 33.0 and 2M - 16.67x = 50.33\n",
+      "M = (33.0 + 50.33)/3 \n",
+      "x = ((xco + xch4)*N - M ) / P - xch4r \n",
+      "\n",
+      "#methanol balance,(xco*N+Rx) * Poncv/100 = M\n",
+      "R = (M*100 / Pconv - (xco*N))/x \n",
+      "\n",
+      "# Result \n",
+      "print \"(a)moles of recycle stream = %.1f\"%R,\"mol\"\n",
+      "print \"(b)moles of purge stream = %.2f\"%P,\"mol\"\n",
+      "H2 = 1 - xch4r - x \n",
+      "print \"(c)CH4 in purge stream = \",xch4r*100,\"%\"\n",
+      "print \"CO in purge stream = \",x*100,\"%\"\n",
+      "print \"hydrogen in purge stream = \",H2*100,\"%\"\n",
+      "print \"(d)Methanol produced = %.2f\"%M,\"mol\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)moles of recycle stream = 337.9 mol\n",
+        "(b)moles of purge stream = 16.67 mol\n",
+        "(c)CH4 in purge stream =  3.0 %\n",
+        "CO in purge stream =  31.34 %\n",
+        "hydrogen in purge stream =  65.66 %\n",
+        "(d)Methanol produced = 27.78 mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch11.ipynb b/Stoichiometry_And_Process_Calculations/ch11.ipynb
new file mode 100755
index 00000000..b8409421
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch11.ipynb
@@ -0,0 +1,1955 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Energy Balance Thermophysics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 pageno : 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "m = 75.                     #kg\n",
+      "g = 9.81                    #m**2/s\n",
+      "d = 10.                     #m\n",
+      "t = 2.5*60                  #s\n",
+      "\n",
+      "# Calculation \n",
+      "f = m*g \n",
+      "w = f * d \n",
+      "P = w / t \n",
+      "\n",
+      "# Result\n",
+      "print \"The work done = \",w,\"Nm\"\n",
+      "print \"Power required = \",P,\"W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The work done =  7357.5 Nm\n",
+        "Power required =  49.05 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2 pageno : 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "PE = 1.5*10**3                  #J\n",
+      "m = 10.                         #kg\n",
+      "g = 9.81                        #m/s**2\n",
+      "v = 50.                         #m/s\n",
+      "\n",
+      "# Calculation \n",
+      "#PE = mgz\n",
+      "z = PE / (m*g) \n",
+      "KE = m* (v**2) / 2 \n",
+      "\n",
+      "# Result\n",
+      "print \"Height of the body from the ground = %.1f\"%z,\"m\"\n",
+      "print \"Kinetic energy of the body = \",KE/1000,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Height of the body from the ground = 15.3 m\n",
+        "Kinetic energy of the body =  12.5 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 pageno : 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Variables \n",
+      "d = 100. /1000                  #m\n",
+      "m = 50.                         #kg\n",
+      "P = 1.01325*10**5.              #Pa\n",
+      "\n",
+      "# Calculation \n",
+      "A = math.pi * (d**2)/4 \n",
+      "Fatm = P * A \n",
+      "Fwt = m * g \n",
+      "Ftotal = Fatm + Fwt \n",
+      "P = Ftotal / A\n",
+      "\n",
+      "# Result \n",
+      "print \"(a)Pressure of the gas = %.4f\"%(P/10**5),\"bar\"\n",
+      "z = 500/1000.                   #m\n",
+      "w = Ftotal * z \n",
+      "print \"(b)Work done by the gas = %.2f\"%w,\"J\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Pressure of the gas = 1.6378 bar\n",
+        "(b)Work done by the gas = 643.15 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4 pageno : 406"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Sgr = 0.879 \n",
+      "F = 5.                      #m**3/h\n",
+      "\n",
+      "# Calculation \n",
+      "D = Sgr * 1000. \n",
+      "m = F * D/3600.             #kg/s\n",
+      "P = 3500.                   #kPa\n",
+      "W = P * m * 1000/ D \n",
+      "\n",
+      "# Result\n",
+      "print \"Power requirement for the pump = %d\"%W,\"W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power requirement for the pump = 4861 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5 pageno : 408"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Variables \n",
+      "\n",
+      "d = 3.                              #m\n",
+      "m = 12500.                          #kg\n",
+      "P = 7000.                           #kPa\n",
+      "\n",
+      "# Calculation \n",
+      "U = 5.3*10**6                       #kJ\n",
+      "Vtank = 4*math.pi*((d/2)**3) / 3 \n",
+      "Vliq = Vtank / 2 \n",
+      "H = U + P * Vliq \n",
+      "\n",
+      "# Result\n",
+      "print \"Specific enthalpy of the fluid in the tank = %.2f\"%(H/m),\"kJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specific enthalpy of the fluid in the tank = 427.96 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6 pageno : 409"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 101.3                       #kPa\n",
+      "SVl = 1.04 * 10**-3             #m**3/kmol\n",
+      "SVg = 1.675                     #m**3/kmol\n",
+      "Q = 1030.                       #kJ\n",
+      "\n",
+      "# Calculation \n",
+      "W = P * 10**3 * (SVg - SVl)/1000 \n",
+      "U = Q - W \n",
+      "H = U +  P * 10**3 * (SVg - SVl)/1000 \n",
+      "\n",
+      "# Result\n",
+      "print \"Change in internal energy = %.2f\"%U,\"kJ/kmol\"\n",
+      "print \"Change in enthalpy = \",H,\"kJ/kmol\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in internal energy = 860.43 kJ/kmol\n",
+        "Change in enthalpy =  1030.0 kJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7 pageno : 409"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#work is done on the system, hence, W is negative\n",
+      "\n",
+      "W = - 2 * 745.7             #J/s\n",
+      "#heat is transferres to the surrounding, hence, heat transferred is negative,\n",
+      "Q = -3000.                  #kJ/h\n",
+      "\n",
+      "# Calculation \n",
+      "U = Q*1000/3600 - W \n",
+      "\n",
+      "# Result\n",
+      "print \"Change in internal energy = %.2f\"%U,\"J/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in internal energy = 658.07 J/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8 pageno : 410"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#Fe(s) + 2HCl(aq) = FeCl2(aq) + H2(g)\n",
+      "MFe = 55.847 \n",
+      "m = 1.                      #kg\n",
+      "\n",
+      "# Calculation \n",
+      "Nfe = m * 10**3/MFe \n",
+      "Nh2 = Nfe                   #(since 1 mole of Fe produces 1 mole of H2)\n",
+      "T = 300.                    #K\n",
+      "R = 8.314 \n",
+      "\n",
+      "#the change in volume is equal to the volume occupied by hydrogen produced\n",
+      "PV = Nh2 * R * T \n",
+      "W = PV \n",
+      "\n",
+      "# Result\n",
+      "print \"Work done = %.2f\"%W,\"J\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work done = 44661.31 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9 pageno : 412"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Cp =1.436 + 2.18*10**-3*T \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# variables \n",
+      "m = 1000./3600                  #kg/s\n",
+      "T1 = 380.                       #K\n",
+      "T2 = 550.                       #K\n",
+      "\n",
+      "# Calculation \n",
+      "def f0(T): \n",
+      "\t return 1.436 + 2.18*10**-3*T\n",
+      "\n",
+      "x =  quad(f0,T1,T2)[0]\n",
+      "Q = m*x \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat load on the heater = %.1f\"%Q,\"kW\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat load on the heater = 115.7 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.10 page no : 413"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Cp = 26.54 + 42.454*10**-3 * T - 14.298 * 10**-6 * T**2 \n",
+      "\n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# variables \n",
+      "T1 = 300.                   #K\n",
+      "T2 = 1000.                  #K\n",
+      "m = 1.                      #kg\n",
+      "N = m/44                    #kmol\n",
+      "\n",
+      "# Calculation \n",
+      "def f3(T): \n",
+      "\t return 26.54 + 42.454*10**-3 * T - 14.298 * 10**-6 * T**2\n",
+      "\n",
+      "x =  quad(f3,T1,T2)[0]\n",
+      "\n",
+      "Q = N*x \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Heat required = %.2f\"%Q,\"kJ\"\n",
+      "\n",
+      "#for temperature in t degree celsius\n",
+      "#Cp = 26.54 + 42.454*10**-3 * (t + 273.15) - 14.298 * 10**-6 * (t + 273.15)**2\n",
+      "#Cp = 37.068 + 34.643 * 10**-3*t - 14.298* 10**-6 * t**2 (kJ/kmolC)\n",
+      "#Cp = 8.854 + 8.274*10**-3*t -3.415*10**-6*t**2 ( Kcal/kmolC)\n",
+      "#For degree Fehreneit scale,replacet by ( t1 - 32)/18, we get\n",
+      "#Cp = 8.7058 + 4.6642 * 10**-3 *t1 - 1.0540 * 10**-6 * t1**2 ( Btu/lbmolF)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Heat required = 755.85 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.11 pageno : 414"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "%pylab inline\n",
+      "\n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "# variables \n",
+      "T = [273, 373, 473, 573, 673, 773, 873, 973, 1073, 1173, 1273] \n",
+      "Cp = [33.6, 35.1, 36 ,36.6, 37, 37.3, 37.5, 37.6, 37.7, 37.8, 37.9] \n",
+      "\n",
+      "# Calculation \n",
+      "plot(T,Cp)\n",
+      "suptitle(\"Determination of enthalpy change \")\n",
+      "xlabel(\"Temperature, K \")\n",
+      "ylabel(\"Heat capacity, kJ/kmol K\")\n",
+      "\n",
+      "H = 36828.              #kJ/kmol\n",
+      "\n",
+      "# Result\n",
+      "print \"Enthalpy change = \",H,\"kJ/kmol\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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VAxD2JXt6CCEqq1JrYX322WeW90TddXNPj/h42dNDCFFxVhPItm3bmD17Nvv2\n7WP27NkcOnSoXGthFRQUEBoaik6nw9vbm+nTp1veW7lyJUFBQQQEBPC3v/2t1PLx8fEEBATg6+vL\n4sWLK1AlUR5vvGH+l5AAFdwrTAghABuvhZWfn4+bmxsGg4Hw8HAWLVpEXl4eq1ev5rPPPsPZ2Zms\nrCxatWpVrNyNGzfw8fFh//79eHh4EBYWxtq1a9HpdMWDlzGQStmwAf72N/OeHrIsuxD1j83HQEpb\nCyslJYUtW7aUey0st9861QsLCzEajbi7u/Piiy8yc+ZMnJ3Nt749eQAcOnQIPz8/y2PCo0aNIi4u\nrkQCERUne3oIIapLmV1Y169fJycnx/Lvxo0blmPlXUjRZDIRHByMh4cHkZGR+Pn5cerUKbZv305w\ncDBhYWEcPHiwRLn09HTat29vea3VaklPT69E9cStZE8PIUR1srofSFU4OTmRnJxMdnY20dHR6PV6\nTCYTOTk5JCcnc+TIEYYPH85PP/1UrLVTW5aLdyTffy97egghqpdNE8hNzZs3JyYmhqSkJNq3b8+w\nYcMACA0NxcXFhYsXL9K2bVvL+VqtlrS0NMvrtLS0Yi2SW82dO9fyc0REBBERETapQ11mMsHEiTBv\nnnmNKyFE/aLX69Hr9dV+3UovZWJNVlYWLi4uNG3alPz8fKKjo5k5cyanT5/m6tWrzJs3j9OnTxMR\nEUF6ejpOTr/3phUUFODj48OBAwdwd3enV69evPHGG4SEhBQPXgbRy2XtWli3Dg4cACerz90JIRyd\nzQfRqyojI4Px48ejlKKgoICxY8cSExND//79mTBhAv6/dcKvX78eJycnMjIymDhxInFxcbi6urJ6\n9Wqio6MxmUzExsaWSB6ifDIz4cUXYdcuSR5CiOpV4RaIj48PAFOnTmXq1Kk2Caq8pAVi3dix0KGD\neWl2IYQAO7ZATp06xeXLlzl06FCVby5sa/t2SEqCt96ydyRCCEdktVNjxYoV/Prrr8WOtW7dmpiY\nGJsFJaouL8+8Fe3rr0PjxvaORgjhiKwmkIsXLxIaGsrIkSOJj4+XLqM64uWXITQUblsHUwghqk25\nxkBMJhM7duxg/fr1fP3114wcOZIJEybQpUuXmoixTDIGUrqTJ82P6544AffcY+9ohBC1TXV9dpbr\nuRwnJyfatm2Lh4cHDRo04Ndff2XkyJHMmDGjygGI6mUywZNPwoIFkjyEELZltQWyfPly3nvvPVq1\nasVf/vJgpL+mAAAdZUlEQVQXhg4dSsOGDVFK4ePjww8//FBTsZYgLZCSZM6HEMKaGnsK68qVK3zy\nySd07NixRACffvpplQMQ1UfmfAghapLVj5lz586VSB6xsbEA+Pr62iYqUSkzZpgXS7zDdvZCCFFt\nrLZAvvvuu2KvjUajzAGphWTOhxCippXZAlm4cCFNmzbl22+/pWnTppZ/rVq1YtCgQTUZo7BC5nwI\nIezB6iD6rFmzeLWWroMhg+hmL7wA586Zl2oXQghrquuzs8wEcurUKXx8fDh69Gip+3PUhsUNJYHI\nnA8hRMXZPIFMnDiRN998k4iIiFITSGJiYpVvXlX1PYGYTNC7Nzz6KEyebO9ohBB1hc0TSF1Q3xPI\n2rXwzjtw8KA8tiuEKL8am4m+YsUKsrOzLa+zs7N57bXXqnxjUTWZmTB7tjmJSPIQQtiD1RZIUFAQ\nx48fL3YsODiY5ORkmwZWHvW5BTJ2LLRvD4sX2zsSIURdU2MtkMLCwmKvb+4waE1BQQGhoaHodDq8\nvb2ZPn06YN7DXKvVotPp0Ol0xMfHl1rey8uLwMBAdDodPXv2LE9d6o3t2+Grr+Af/7B3JEKI+szq\nRMIHHniA0aNHM3HiRJRSvPnmmzzwwANWL+zq6srevXtxc3PDYDAQHh5OYmIiGo2GGTNmWF2IUaPR\noNfradmyZflrUw/cnPOxahXcdZe9oxFC1GdWE8jy5ctZuXIly5YtA6Bfv37l3srWzc0NMLdijEYj\nHh4eAOVuOtXX7qk7efll6NFD9vkQQtifTZ/CMplMhISEcO7cOSZPnsySJUuYN28e7777Lo0aNaJ7\n9+6sWLGi1FbGvffeS4sWLTAYDDzxxBOlJq36NgYicz6EENWhxh7j/f7775k5cyY//PADRUVFlpv/\n+OOP5b5JdnY20dHRvPrqq/j7+9OqVSvAPB5y7tw5PvjggxJlfvnlF9zd3bl06RIDBgxg8eLFREVF\nFQ++HiUQmfMhhKguNbace2xsLIsXL2bGjBkkJCTw3nvvlRhYt6Z58+bExMSQlJRERESE5fiTTz5J\nZGRkqWXc3d0BaNOmDSNGjODIkSMlEgiYk9BNERERxa7vSN56C4xG82ZRQghREXq9Hr1eX+3XLfdj\nvP7+/pw8eRKA0NBQjhw5cscLZ2Vl4eLiQtOmTcnPzyc6OpqZM2fSs2dP2rRpA8DKlStJTEzkk08+\nKVY2Ly8PgMaNG5Obm8ugQYN49tlneeihh4oHX09aIJmZ5iXad+2CwEB7RyOEqOtqrAXSuHFjlFJ0\n7NiRVatW0bZtW7KysqxeOCMjg/Hjx1se+x07diwxMTHExsZy4sQJCgsL6dixI2+//bbl/IkTJxIX\nF0dmZiZDhw5Fo9GQl5fH6NGjSySP+uTmPh+SPIQQtYnVFsjhw4fx9fXl0qVLzJ49m4KCAp577jl6\n9epVUzGWqT60QLZvh0mTzAPo8tiuEKI61PhaWJcvXwagdevWVb5pdXH0BJKXZ+66eu01eWxXCFF9\namwm+oEDB+jatSvdu3ene/fu+Pj4cPDgwSrfWFgncz6EELWZ1RaIj48P77zzjqXL6quvvuLxxx/n\n1KlTNRLgnThyC0TmfAghbKXGWiDNmjUrNt4RFhZG8+bNq3xjUTaTyfy47vz5kjyEELWX1RbIX//6\nVwoLCxk5ciQAW7ZsoWHDhsTGxgL23ZnQUVsgss+HEMKWamwQ/fYdCZVSxV7bc2dCR0wgFy+aB853\n7pTHdoUQtiE7EuKYCWTcONBqZZ8PIYTt1NhEQpPJxKeffsoPP/yAwWCwHP+HbEZR7XbsMHdb/Tbh\nXwghajWrPewTJkxg69atrFq1CqUUmzdv5qeffqqJ2OqV/HzzIomyz4cQoq4o12O8p06dsqyJlZ+f\nz4ABA9izZ09NxVgmR+rCmj0bzp6FTZvsHYkQwtHVWBdWs2bNzCc6O5OZmUmLFi2kBVLNvvvO/OTV\niRP2jkQIIcrPagIZNGgQ165d49lnnyUwMBAnJycef/zxmoitXpA5H0KIuqpCT2Fdv34dg8FAixYt\nbBlTuTlCF9abb8Lbb8ucDyFEzamxmegrVqwgOzsbgCZNmqDRaHjttdeqfGNhnvMxe7a5+0qShxCi\nrin3hlK3Cg4OJjk52aaBlUddb4HInA8hhD3U2CD67dvX3twgSlTNsWOwdy/UgjUphRCiUqx2nDzw\nwAOMHj2aXbt2sXPnTkaPHs0DDzxg9cIFBQWEhoai0+nw9vZm+vTpgHkPc61Wi06nQ6fTER8fX2r5\n+Ph4AgIC8PX1ZbEDfkVftw4mTpQ5H0KIustqF5bBYGDlypXs2rULgH79+jF16lQaNGhg9eL5+fm4\nublhMBgIDw9n0aJF7N27l6ZNmzJjxowyy924cQMfHx/279+Ph4cHYWFhrF27Fp1OVzz4OtqFdeMG\neHrC11+Dl5e9oxFC1Dc11oXl7OzM9OnTLS2IinBzcwPM3WBGoxEPDw8Aq4EfOnQIPz8/PD09ARg1\nahRxcXElEkhd9b//mRdKlOQhhKjLbPrsj8lkIjg4GA8PDyIjI/H19QXg9ddfp1u3bjz66KNcuXKl\nRLn09HTat29vea3VaklPT7dlqDVq/Xp47DF7RyGEEFVjtQVSFU5OTiQnJ5OdnU10dDR6vZ6nnnrK\nshDj3Llzefrpp/nggw+Klbt1uXhr5s6da/k5IiKCiIiI6gjdZjIzYf9+2LjR3pEIIeoLvV6PXq+v\n9utaTSAff/wxjzzyiNVjd9K8eXNiYmJISkoq9gH/5JNPEhkZWeJ8rVZLWlqa5XVaWlqxFsmtbk0g\ndcGHH8LQoTJ4LoSoObd/uZ43b161XNdqF9bChQtLHHv55ZetXjgrK4ucnBzAPJiekJBAQEAAly5d\nspzz3//+Fz8/vxJlQ0NDOXnyJBcuXKCoqIjNmzczcOBAq/es7ZSS7ishhOMoswXy5Zdf8sUXX3Dh\nwgWefvppy8B3Xl5eubqYMjIyGD9+vGXeyNixY4mJiSE2NpYTJ05QWFhIx44defvtty3nT5w4kbi4\nOFxdXVm9ejXR0dGYTCZiY2PtunVudTl2DPLyIDzc3pEIIUTVlfkY7/Hjx/nmm2/4xz/+wYIFCywJ\npHHjxkRGRtKmTZsaDbQ0de0x3mnToE0bkL24hBD2VGNb2hYWFuLi4lLlG9lCXUogMvdDCFFb1Ng8\nkLNnzzJz5kx++OEHioqKLDf/8ccfq3zz+kTmfgghHI3VQfTY2FieeeYZXF1d0ev1TJgwgXHjxtVE\nbA5FBs+FEI6m3Kvx+vv7c/LkScD8lNSRI0dqJMA7qStdWJmZ0K0bpKfL47tCCPursS6sxo0bo5Si\nY8eOrFq1irZt25KVlVXlG9cnMvdDCOGIrLZADh8+jK+vL5cuXWL27NkUFBTw3HPP0atXr5qKsUx1\noQWilHns4/XXoU8fe0cjhBA1+BTWTTk5OTRt2rTKN6xOdSGBHD0KI0fCmTOy66AQonaosS1t9+zZ\nQ5cuXSwLIX733Xc88cQTVb5xfbF+PfzpT5I8hBCOx+rH2jPPPMPu3btp3bo1AH5+fhw8eNDmgTmC\nGzdgwwYYP97ekQghRPWzmkCUUnTo0KHYsYqsllufydwPIYQjs/oUVvv27Tlw4ABg3p1wzZo13Hvv\nvTYPzBHI3A8hhCOzOoiemZnJlClT2LlzJxqNhqioKNasWSNrYVkhcz+EELVVjT+FVRvV5gSydCl8\n9x288469IxFCiOJsPpFw2rRpJW526/+uWLGiyjd3VDf3/Xj9dXtHIoQQtlNmAunevbslYcyZM4f5\n8+dbMpYMot+Z7PshhKgPytWFpdPp+Oabb2oingqprV1Ysu+HEKI2q7GJhJVVUFBAaGgoOp0Ob29v\npk+fXuz9pUuX4uTkxJUrV0ot7+XlRWBgIDqdjp49e9oqzGoncz+EEPWF1cd4K8vV1ZW9e/fi5uaG\nwWAgPDycxMREIiMjSUtLIyEhgY4dO5ZZXqPRoNfradmypa1CtAmZ+yGEqC/KTCBNmjSxjHXk5+cX\nWwdLo9Fw7do1qxd3c3MDzLsaGo1GPDw8AJgxYwZLlizh4YcfvmP52tg9ZY3M/RBC1BdldmFdv36d\nnJwccnJyMBgMlp9zcnLKlTwATCYTwcHBeHh4EBkZia+vL1u3bkWr1RIYGHjHshqNhn79+hEYGMhr\nr71WsVrZSWYm7N8Pw4fbOxIhhLA9m3VhATg5OZGcnEx2djbR0dF88cUXLFq0iB07dljOKauVkZSU\nhLu7O5cuXWLAgAH4+PgQFRVV4ry5c+dafo6IiCAiIqK6q1Fusu+HEKI20uv16PX6ar9ujU0kXLBg\nARqNhpUrV9K4cWMA0tPT8fT05PDhw7i7u5dZdtGiRQA8//zzxY7XpqewZN8PIURdUeufwsrKyiIn\nJwcwj6EkJCSg0+m4ePEiKSkppKSkoNVqOXbsWInkkZeXR15eHgC5ubnEx8fj5+dnq1Crhcz9EELU\nNzZLIBkZGfTp04fg4GB0Oh1RUVHExMQUO+fWCYkZGRmW9zMzMwkLC7OU7du3Lw899JCtQq0Wsu+H\nEKK+kbWwqsGNG+DpCV9/LY/vCiFqv1rfhVWfyNwPIUR9JAmkGsjcDyFEfSRdWFUk+34IIeoa6cKq\nJWTuhxCivpIEUgU39/2Q7ishRH0kCaQKZO6HEKI+kwRSBTL3QwhRn8kgeiXJ3A8hRF0lg+h2JnM/\nhBD1nSSQSpLBcyFEfSddWJUgcz+EEHWZdGHZkcz9EEIISSAVJnM/hBDCTBJIBcncDyGEMJMEUkEy\n90MIIcxkEL0CZO6HEMIR1PpB9IKCAkJDQ9HpdHh7ezN9+vRi7y9duhQnJyeuXLlSavn4+HgCAgLw\n9fVl8eLFtgqzQj7/XOZ+CCHETc62urCrqyt79+7Fzc0Ng8FAeHg4iYmJREZGkpaWRkJCAh07diy1\n7I0bN5g8eTL79+/Hw8ODsLAw+vfvj06ns1W45SKD50II8Tub9uS7ubkBUFhYiNFoxMPDA4AZM2aw\nZMmSMssdOnQIPz8/PD09cXZ2ZtSoUcTFxdkyVKt+/hkOHIDhw+0ahhBC1Bo2TSAmk4ng4GA8PDyI\njIzE19eXrVu3otVqCQwMLLNceno67du3t7zWarWkp6fbMlSrZO6HEEIUZ7MuLAAnJyeSk5PJzs4m\nOjqaL774gkWLFrFjxw7LOaUN5Gg0mnLfY+7cuZafIyIiiIiIqErIpbo592PVqmq/tBBC2Jxer0ev\n11f7dW2aQG5q3rw5MTExHDt2jJSUFIKCggBzS6N79+4cPnwYd3d3y/larZa0tDTL67S0tGItklvd\nmkBs5ehRyM+XuR9CiLrp9i/X8+bNq5br2qwLKysri5ycHADy8/NJSEhAp9Nx8eJFUlJSSElJQavV\ncuzYsWLJAyA0NJSTJ09y4cIFioqK2Lx5MwMHDrRVqFbJ3A8hhCjJZh+JGRkZ9OnTh+DgYHQ6HVFR\nUcTExBQ759auqoyMDMv7rq6urF69mujoaIKCghg2bBghISG2CvWObtyAjRth/Hi73F4IIWotmUho\nxZYt5rGP3bttehshhKgxtX4ioaOQuR9CCFE6aYHcwc8/g6+v7PshhHAs0gKpATL3QwghyiYJpAyy\n74cQQtyZJJAyyNwPIYS4M0kgZZC5H0IIcWcyiF4K2fdDCOHIZBDdhmTfDyGEsE4SSClk8FwIIayT\nLqzbyNwPIYSjky4sG5G5H0IIUT6SQG4hcz+EEKL8JIHcQuZ+CCFE+UkCuYXM/RBCiPKTQfTfyNwP\nIUR9UV2fnTbb0ragoIDevXtjMBjIzc0lJiaGZcuW8eKLL/L5559jNBpp2bIl69ev59577y1R3svL\ni2bNmtGgQQMaNmzI4cOHbRUqYB7/eOcdSR5CCFFuyoby8vKUUkoVFRWp+++/X+3evVvl5ORY3l+x\nYoUaP358qWW9vLxUVlbWHa9v4/DtLjEx0d4h2JQj18+R66aU1K+uq67PTpv29ru5uQFQWFiI0WjE\nw8ODJk2aWN6/fv0699xzz52Smy3Dq/X0er29Q7ApR66fI9cNpH7CzKYJxGQyERwcjIeHB5GRkfj6\n+gIwe/ZsOnTowLvvvsusWbNKLavRaOjXrx+BgYG89tprtgxTCCFEJdg0gTg5OZGcnEx6ejp79+61\nZPVXXnmF8+fP89hjjzF9+vRSyyYlJXHs2DF27drFunXr2Llzpy1DFUIIUUE19hTWggULaNiwYbEW\nx/nz5+nfvz+nTp26Y9lFixYB8Pzzzxc73qVLF86dO1f9wQohhAPr3LkzZ8+erfJ1bPYUVlZWFi4u\nLjRt2pT8/HwSEhKYOXMmqampeP32qNPWrVsJCAgoUTYvLw+Axo0bk5ubS3x8PM8++2yJ86rjFyCE\nEKJybJZAMjIyGD9+PEopCgoKGDt2LDExMQwbNoxz585RVFREp06deOuttyznT5w4kbi4ODIzMxk6\ndCgajYa8vDxGjx7NQw89ZKtQhRBCVEKdnkgohBDCfmrtoh1paWn06dOHgIAAunbtypIlSwC4cuWK\n5ems6Ohorl69aimzaNEifH19CQgIYMeOHfYKvUKMRiM6nY4hQ4YAjlW/q1ev8sgjjxAUFES3bt1I\nSkpyqPrNmTMHb29vfHx8GDFiBHl5eXW2fhMmTMDDw6NYl3Jl6nL06FF0Oh1+fn4888wzNVqHOymt\nfjNmzMDX1xdfX18GDx5MVlaW5T1HqN9NS5cuxcnJiStXrliOVVv9qmU2iQ1kZmaqb7/9VimlVE5O\njrrvvvtUcnKymjp1qlq2bJlSSqlly5app59+Wiml1Ndff6169OihDAaDSk9PV15eXurGjRt2i7+8\nli5dqsaOHauGDBmilFIOVb8RI0aojz76SCmllNFoVNnZ2Q5TvzNnzqhOnTpZYhw5cqR666236mz9\n9u7dq44dO6b8/f0txypSl8LCQqWUUgEBAerYsWNKKaUefvhh9cknn9RwTUpXWv12796tjEajUkqp\nmTNnqr/+9a9KKcepn1JKnT9/XkVHRxebmF2d9au1LRAPDw/8/f0BaNKkCYGBgVy4cIEvvviC2NhY\nAB599FHi4uIAiIuLY/To0TRo0ABPT0/8/PxsvvxJVaWnp/PFF1/wl7/8xTJp0lHql5WVRXJyMmPG\njAHMj3Q3a9bMYerXsmVLGjZsSG5uLgaDgby8PDp06FBn69e7d2/uvvvuYscqUpdDhw5x/vx5TCYT\nOp2uRBl7K61+kZGROP22cuof/vAHLly4ADhO/cDcyrrZe3NTddav1iaQW6WmpnLkyBHCw8O5dOkS\nrVq1AqB169b88ssvAFy4cAGtVmspo9VqSU9Pt0u85TV9+nT++c9/Wv5PDDhM/c6cOUObNm0YOXIk\n/v7+jB8/npycHIepX8uWLXn22Wfp0KED7dq1o0WLFvTr189h6gcV///ihQsXaN++veW4p6dnra/j\nTWvXruXhhx8GHKd+W7duRavVEhgYWOx4ddav1ieQ69evM2LECJYvX06zZs3sHU61+d///oe7uzs6\nnc4hl2wxmUwcOXKEv/3tb5w8eZKWLVuyYMECe4dVbc6dO8d//vMfUlNTycjI4Pr163zwwQf2DktU\nwiuvvIKLiwvjxo2zdyjVJi8vj4ULFzJv3jzLMVt8ztTqBFJUVMTw4cMZN24cf/zjHwFo06YNly9f\nBszfkNzd3QFzFk1LS7OUTU9PL5ZNa5uDBw+ybds2OnXqxJgxY9i9ezexsbEOU7/27dvj6elJaGgo\nACNGjCA5ORl3d3eHqN/hw4fp1asXrVq1wtnZmWHDhnHgwAGH+ftBxf9bK+34rd90a6N3332XuLg4\nPvzwQ8sxR6jfuXPnSE1NJSgoiE6dOpGenk737t25ePFitdav1iYQpRR//vOf8fX1LbbcyaBBgyzf\n9D744AMGDRpkOb5p0yYMBgPp6emcPHmSnj172iX28li4cCFpaWmkpKSwceNGHnjgAd5//32HqV/7\n9u1p3bo1p0+fBmDnzp1069aNgQMHOkT9unTpQlJSEvn5+Sil2LlzJ507d3aYvx9U/L+19u3b4+Tk\nxDfffAPAhx9+aClTG8XHx7NkyRK2bduGq6ur5bgj1C8gIICLFy+SkpJCSkoKWq2WY8eO4eHhUb31\nq95nAarPvn37lEajUUFBQSo4OFgFBwerL7/8UmVlZamoqCgVEBCg+vXrp3799VdLmVdeeUV169ZN\n+fn5qfj4eDtGXzF6vd7yFJYj1S85OVn16NFD+fr6qoEDB6orV644VP3mzJmjunTpory9vdWoUaNU\nfn5+na3f6NGj1T333KMaNmyotFqteueddypVl6+//loFBwcrX19fNW3aNHtUpVS31+/tt99WXbp0\nUR06dLB8vkyePNlyfl2tn4uLi+Xvd6tOnToV2x6juuonEwmFEEJUSq3twhJCCFG7SQIRQghRKZJA\nhBBCVIokECGEEJUiCUQIIUSlSAIRQghRKZJARJ2XlZWFTqdDp9Nxzz33oNVq0el0hISEYDAY7B1e\nMXv27OGrr76qkXt5eXlZlvA+evQo9957L8ePH6+Re4v6wWY7EgpRU1q1amWZPTtv3jyaNm3KjBkz\n7BaPyWQqtkDmrRITE2natClhYWHlvp7RaKRBgwYVjkOj0QBw4sQJHnnkETZv3kxQUFCFryNEWaQF\nIhyOUoqvvvqKsLAwAgMDiYyMtCzVHRERwYwZM/i///s/unXrxpEjRxg+fDidO3dm5syZgHn1Zx8f\nH8aPH4+/vz+DBw8mLy8P4I7XnT59OmFhYSxfvpzPP/+c+++/n4CAAPr06cPPP/9Mamoqb7zxBsuW\nLSMkJIT9+/fz2GOP8d///tcSe5MmTQDQ6/X07t2boUOHEhgYiNFoZOrUqZbNuVasWFGu38V3333H\n0KFD+eCDD+jRo0e1/Y6FAGrvUiZCVMbcuXPVkiVLVPfu3dWlS5eUUkpt3LhRjRs3TimlVEREhHrh\nhReUUkotX75c3XPPPerSpUvqxo0bql27duqXX35RKSkpSqPRqEOHDimllJo4caJauHChKiwsVCEh\nIery5culXvfmhktKKZWdnW35+c0331RTp061xLd06VLLe4899pjasmWL5XWTJk2UUkolJiaqu+66\nS6Wnp1tiffnll5VSShUUFKiQkBB1+vTpO/4uOnbsqFq2bKm+/PLLCv0OhSgv6cISDsfJyYkzZ87Q\nr18/wNwF5OHhYXl/8ODBAPj7++Pv70/r1q0B8wKJFy5coEWLFrRv396y2OGYMWP417/+Rf/+/Tl7\n9ixRUVGlXnfEiBGWn8+ePcuMGTPIysqiqKiIDh06WN5T5Vw9qGfPnnh6egKwY8cOzpw5w5YtWwC4\ndu0aP/74I/fdd1+Z5TUaDf369ePNN9+kf//+ZXarCVFZkkCEw1FKERQUxN69e0t9v1GjRoA50dz8\n+eZrk8kE/D5+cPN6Go3G6nXvuusuy89Tp07lxRdfZNCgQezZs4e5c+eWWubWe5pMJgoLC0u9HsCa\nNWuIjIwsq9qleu2113jyySeZMmUKa9asqVBZIayRryTC4ZhMJs6fP28ZWDcYDPzwww8Vusb58+c5\ncuQIAJs2bSI8PJzAwMA7XvfWlkVBQQFt27YF4L333rMcd3Nzs4yngHnviaNHjwLmrUaLiopKjSc6\nOpo33njDkmxSUlLIz88HwMfHp8x6ODk58dFHH3Hq1CnmzJlT/l+AEOUgCUQ4HGdnZz7++GMmTZpE\ncHAwwcHB7Nmzp8R5Go2mWEvjVl27dmXlypX4+/tz4cIFnnnmGVxcXO543Vuv9dJLLzF06FDuv/9+\nWrVqZXlvyJAhfPTRRwQHB3PgwAEmTZrE9u3b0el0HDx40DKIfvv1nnrqKcv+1UFBQTz++OMYDAbL\nhk+luVm+UaNGbNu2jW3btrF69epy/haFsE6WcxfiNqmpqQwZMoRvv/3W3qFYFRcXR0pKClOnTrV3\nKKIekjEQIUpRVsuktomJibF3CKIekxaIEEKISpExECGEEJUiCUQIIUSlSAIRQghRKZJAhBBCVIok\nECGEEJUiCUQIIUSl/H/CTl7Bxz0qdgAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x27f4550>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enthalpy change =  36828.0 kJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.12 page no : 415"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Cp = 26.586 + 7.582 * 10 **-3 * T - 1.12 * 10**-6 * T**2\n",
+      "\n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# variables \n",
+      "T1 = 500.                   #K\n",
+      "T2 = 1000.                  #K\n",
+      "\n",
+      "# Calculation \n",
+      "def f5(T): \n",
+      "\t return 26.586 + 7.582 * 10**-3 * T - 1.12 * 10**-6 * T**2\n",
+      "\n",
+      "x =  quad(f5,T1,T2)[0]\n",
+      "\n",
+      "Cpm = 1 *x / ( T2 - T1 )  \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Mean molal heat capacity = %.3f\"%Cpm,\"kJ/kmolK\"\n",
+      "V = 500.                    #m**3 \n",
+      "N = V / 22.4143 \n",
+      "Q = N * Cpm * ( T2 - T1 ) \n",
+      "print \"(b)Heat to be supplied = %.3e\"%Q,\"kJ/h\"\n",
+      "T3 = 1500.                  #K\n",
+      "Q1 = Cpm * (T3 - T1) \n",
+      "\n",
+      "def f6(T): \n",
+      "\t return 26.586 + 7.582 * 10 **-3 * T - 1.12 * 10**-6 * T**2\n",
+      "\n",
+      "y =  quad(f6,T1,T3)[0]\n",
+      "\n",
+      "Q2 = y  \n",
+      "Perror = (Q2 - Q1) * 100 / Q2 \n",
+      "print \"(c)Percent error = %.1f\"%Perror,\"%\"\n",
+      "\n",
+      "# note : answer may vary because of rounding error.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Mean molal heat capacity = 31.619 kJ/kmolK\n",
+        "(b)Heat to be supplied = 3.527e+05 kJ/h\n",
+        "(c)Percent error = 4.1 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.13 page no : 416"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# variables \n",
+      "T1 = 1500.                      #K\n",
+      "Tr = 273.                       #K\n",
+      "T2 = 400.                       #K\n",
+      "Cpm1 = 50.                      #kJ/kmol\n",
+      "Cpm2 = 35.                      #kJ/mol\n",
+      "\n",
+      "# Calculation \n",
+      "H = Cpm1 * ( T1 - Tr ) - Cpm2 * ( T2 - Tr ) \n",
+      "\n",
+      "# Result\n",
+      "print \"Enthalpy change = \",H,\"kJ/kmol\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enthalpy change =  56905.0 kJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.14 pageno : 417"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# variables \n",
+      "#CO,  26.586 + 7.582*10**-3*T - 1.12*10**-6*T**2\n",
+      "#CO2, 26.540 + 42.454*10**-3*T - 14.298*10**-6*T**2\n",
+      "#O2, 25.74 + 12.987*10**-3*T - 3.864*10**-6*T**2\n",
+      "#N2, 27.03 + 5.815*10**-3*T - 0.289*10**-6*T**2\n",
+      "#Cpmix = summation ( yi*Cpi ) = summation(yi*ai + yi*bi*T + yi*ci*T**2)\n",
+      "\n",
+      "xco2 = 0.09 \n",
+      "xco = 0.02 \n",
+      "xo2 = 0.07 \n",
+      "xn2 = 0.82 \n",
+      "T1 = 600.                   #K\n",
+      "T2 = 375.                   #K\n",
+      "\n",
+      "# Calculation \n",
+      "sumai = xco * 26.586 +xco2 * 26.540 + xo2 * 25.74 + xn2*27.03 \n",
+      "sumbi = xco * 7.582*10**-3 + xco2*42.454*10**-3+xo2*12.987*10**-3 + xn2*5.815*10**-3 \n",
+      "sumci = -(xco * 1.12*10**-6 + xco2*14.298*10**-6+xo2*3.864*10**-6+xn2*0.289*10**-6) \n",
+      "\n",
+      "def f8(T): \n",
+      "\t return sumai+sumbi*T+sumci*T**2\n",
+      "\n",
+      "H =  quad(f8,T2,T1)[0]\n",
+      "\n",
+      "# Result\n",
+      "print \"Enthalpy change = %.2f\"%H,\"kJ/kmol\"\n",
+      "\n",
+      "# note : Calculating integration using quad may vary answer. Because Python use's its own methodology to calculate quad."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enthalpy change = 7009.12 kJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.15 pageno : 420"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Hna = 26.04                     #J/g-atomK\n",
+      "Hs = 22.6                       #J/g-atomK\n",
+      "Ho = 16.8                       #J/g-atomK\n",
+      "Hh = 9.6                        #J/g-atomK\n",
+      "\n",
+      "# Calculation \n",
+      "Hna2so410h2o = 2*Hna + Hs + 14*Ho  + 20*Hh \n",
+      "Hexp = 592.2                    #J/molK\n",
+      "Deviation = (Hexp - Hna2so410h2o)*100/Hexp \n",
+      "\n",
+      "# Result\n",
+      "print \"Deviation in heat capacity = %.2f\"%Deviation,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Deviation in heat capacity = 15.25 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.16 pageno : 421"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P1 = 75.                    #kPa\n",
+      "T1 = 573.                   #K\n",
+      "Tvap = 365.                 #K\n",
+      "Tbasis = 273.               #K\n",
+      "\n",
+      "#Since, the boiling point of water at 75kPa is 375K, the vapour at 573K is superheated \n",
+      "H1 = 3075.                  #kJ/kg\n",
+      "Cliq = 4.2                  #kJ/kgK\n",
+      "Cvap = 1.97                 # kJ/kg/K\n",
+      "m = 1.                      #kg\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#let assume converting liq. water into superheated stream occurs in 3 steps,\n",
+      "\n",
+      "Hc1 = m*Cliq * ( Tvap - Tbasis ) \n",
+      "\n",
+      "\n",
+      "Hc3 = m*Cvap*(T1 - Tvap) \n",
+      "\n",
+      "#total enthalpy = 3075 = Hc1 + Hc2 + Hc3, therefore\n",
+      "Hc2 = H1 - Hc1 - Hc3 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat of vapourisation = \",Hc2,\"kJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat of vapourisation =  2278.84 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.17 pageno : 422"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.integrate import quad\n",
+      "\n",
+      "# variables \n",
+      "T1 = 250.                           #K\n",
+      "T = 273.15                          #K\n",
+      "T2 = 400.                           #K\n",
+      "Cice = 2.037                        #kJ/kgK\n",
+      "T3 = 373.15                         #K\n",
+      "Cliq = 75.726                       #kJ/kmolK\n",
+      "\n",
+      "#Cp = 30.475 + 9.652*10**-3*T + 1.189*10**-6*T**2\n",
+      "Hfusion = 6012.                     #kJ/kmol\n",
+      "Hvap = 40608.                       #kJ/kmol\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#1 - Heat for raising the temperature of ice, H1\n",
+      "H1 = Cice * (T - T1) \n",
+      "#2 - Latent heat of fusion of ice,  Hf\n",
+      "Hf = Hfusion / 18.016               #kJ\n",
+      "\n",
+      "#3 - Sensible heat of raising the temperature of water, H2\n",
+      "H2 = Cliq * ( T3 - T)/18.016 \n",
+      "#4 - Latent heat of vaporization of water, Hv\n",
+      "Hv = Hvap / 18.016 \n",
+      "#5 - Sensible heat of raising the temperature of water vapou, H3\n",
+      "\n",
+      "def f4(T): \n",
+      "\t return 30.475 + 9.652*10.**-3*T + 1.189*10.**-6*T**2.\n",
+      "\n",
+      "H3 =  51.243883    #quadrature(f4,T3,T2)[0]\n",
+      "\n",
+      "Q = H1 + H2 + H3 + Hf + Hv \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat required = %.1f\"%Q,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat required = 3106.4 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.18 pageno : 423"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "#Cp = 0.16 + 4.78 * (10**-3) * T ( organic liquid )\n",
+      "#Cp = 0.7935 + 1.298 * (10**-4) * T ( CCL4 )\n",
+      "Tb = 349.9                          #K\n",
+      "Hv = 195.                           #kJ/kg\n",
+      "Cp = 0.4693                         #kJ/kgK\n",
+      "\n",
+      "# Calculation \n",
+      "#Let T be the final temperature\n",
+      "#integration(T - 650)(0.16 + 4.78 * (10**-3) * T)dt = integration(295 - T)(0.7935 + 1.298 * (10**-4) * T)dt\n",
+      "# the above equation yields, 2.4549*(10**-3)*T**2 + 0.9535*T - 1353.51 = 0, from this we get\n",
+      "T = 573.3                           #K\n",
+      "\n",
+      "#since this temperature is above boiling point of CCl4,\n",
+      "#heat balance is, integration(T - 650)(0.16 + 4.78 * (10**-3) * T)dt = integration(295 - 349.9)(0.7935 + 1.298 * (10**-4) * T)dt + Hv + integration(349.9 - T)*0.4693*dT\n",
+      "#solving above equation, we get,\n",
+      "T1 = 540.1                          #K\n",
+      "\n",
+      "# Result\n",
+      "print \"equilibrium temperature of the mixture = \",T1,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "equilibrium temperature of the mixture =  540.1 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.19 pageno : 427"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "T1 = 363.                       #K\n",
+      "T2 = 373.                       #K\n",
+      "P1s = 70.11                     #kPa\n",
+      "P2s = 101.3                     #kPa\n",
+      "R = 8.314                       #kJ/kmolK\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "# ln(P2s / P1s) = Hv / R * (1/T1 - 1/T2) \n",
+      "Hv = (math.log(P2s/P1s)*R)/(1/T1 - 1/T2) \n",
+      "Hv1 = Hv / (18) \n",
+      "\n",
+      "# Result\n",
+      "print \"Mean heat of vaporization = %.f\"%Hv1,\"kJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mean heat of vaporization = 2302 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.20 pageno : 427"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "T = 273.15 - 30     \n",
+      "R = 8.314                        #K\n",
+      "\n",
+      "# Calculation \n",
+      "#lnPs = 14.2410 - 2137.72 / (T-26.72)\n",
+      "#dlnPs/dT = Hv / RT2\n",
+      "Hv = 2137.72 * R * T**2 / ( T - 26.72 )**2 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat of vaporization = %.2f\"%Hv,\"kJ/kmol\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat of vaporization = 22432.33 kJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.21 pageno : 427"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Hv1 = 2256.                     #kJ/kg\n",
+      "T1 = 373.                       #K\n",
+      "T2 = 473.                       #K\n",
+      "Tc = 647.                       #K\n",
+      "\n",
+      "# Calculation \n",
+      "Tr1 = T1 / Tc \n",
+      "Tr2 = T2 / Tc \n",
+      "#Hv2 / Hv1 = ((1-Tr2)/(1-Tr1))**0.38\n",
+      "Hv2 = Hv1*(((1-Tr2)/(1-Tr1))**0.38) \n",
+      "\n",
+      "# Result\n",
+      "print \"Latent heat of vaporization of water at 473K = %d\"%Hv2,\"kJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Latent heat of vaporization of water at 473K = 1898 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.22 pageno : 428"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "\n",
+      "# Variables \n",
+      "#Cp = a + b*T\n",
+      "\n",
+      "T1 = 293.15                     #K\n",
+      "Cp1 = 131.05                    #J/molK\n",
+      "T2 = 323.                       #K\n",
+      "Cp2 = 138.04                    #J/molK\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#a + 293*b = 131.05\n",
+      "#a + 323*b = 138.04\n",
+      "b = (Cp1 - Cp2)/(T1 - T2) \n",
+      "a = Cp1 - b * T1 \n",
+      "\n",
+      "#Cp = 62.781 + 0.233*T\n",
+      "# Hvb / Tb = 36.63 + 8.31lnTb\n",
+      "Tb = 273.15 + 80.1              #K\n",
+      "\n",
+      "Hvb = (36.63 + 8.31*math.log(Tb)) * Tb \n",
+      "m = 100.                        #kg\n",
+      "\n",
+      "def f7(T): \n",
+      "\t return 62.781 + 0.233*T\n",
+      "\n",
+      "k = quad(f7,T1,Tb)[0]\n",
+      "\n",
+      "H = m*(10.**3) * ( k)/78.048 + m*(10.**3) * Hvb/78.048\n",
+      "\n",
+      "# Result\n",
+      "print \"Heat required = %.3e J\" %H\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat required = 4.928e+07 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.23 pageno : 430"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 10.                 #kPa\n",
+      "T1 = 323.15             #K\n",
+      "T2 = 373.15             #K\n",
+      "T = 358.15              #K\n",
+      "H1 = 2592.6             #kJ/kg\n",
+      "H2 = 2687.5             #kJ/kg\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#H by interpolation,\n",
+      "H = H1 + ((H2 - H1)/(T2 - T1))*(T - T1) \n",
+      "Hl = 697.061            #kJ/kg\n",
+      "Hg = 2762.              #kJ/kg\n",
+      "\n",
+      "#H = x*Hl + ( 1 - x )* Hg\n",
+      "x = (H - Hg)/(Hl - Hg)  \n",
+      "Pmois = x*100 \n",
+      "Psteam = ( 1 - x )*100 \n",
+      "\n",
+      "# Result\n",
+      "print \"Percentage of moisture = %.f\"%Pmois,\"%\"\n",
+      "print \"Percentage of dry saturated steam = %.f\"%Psteam,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage of moisture = 5 %\n",
+        "Percentage of dry saturated steam = 95 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.24 pageno : 430"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 3500.                   #kPa\n",
+      "T = 673.15                  #K\n",
+      "SV = 0.08453                #m**3/kg\n",
+      "Vcondensed = 1/2. \n",
+      "m = 100.                    #kg\n",
+      "\n",
+      "# Calculation \n",
+      "V = m * SV / (m/2) \n",
+      "#m*(Vl+Vg)*Vcondensed = m * SV\n",
+      "#But Vl is negligible,\n",
+      "Vg = m * SV / (m * Vcondensed) \n",
+      "#using steam table\n",
+      "T1 = 459.5                  #K\n",
+      "P1 = 1158.                  #kPa\n",
+      "\n",
+      "#internal energy of superheated steam from steam table\n",
+      "I = 2928.4                  #kJ/kg\n",
+      "U1 = m * I \n",
+      "Ul = 790.                   #kJ/kg\n",
+      "Ug = 2585.9                 #kJ/kg\n",
+      "U2 = m*Vcondensed*Ul + m*(1-Vcondensed)*Ug \n",
+      "Q = U2 - U1 \n",
+      "\n",
+      "# Result\n",
+      "print \"The amount of heat removed fromt he system = \",Q,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of heat removed fromt he system =  -124045.0 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.25 pageno : 433"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 1000.                       #kg/h ( basis  mass of 10% NaOH solution )\n",
+      "Pfeed = 10.                     #%\n",
+      "Ppro = 50.                      #(Percentage NaOH in product)\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#Taking NaOH balance,P being the weight of the product\n",
+      "P = Pfeed * m / Ppro \n",
+      "\n",
+      "#W be the weight of water evaporized\n",
+      "W = m - P \n",
+      "\n",
+      "#step1 - cooling 1000kg/h of 10% solution from 305K to 298K\n",
+      "T1 = 305.                       #K\n",
+      "T2 = 298.                       #K\n",
+      "Cliq = 3.67                     #kJ/kgK\n",
+      "H1 = m*Cliq * (T2 - T1) \n",
+      "\n",
+      "#step2 - separation into pure components\n",
+      "Hsolution = -42.85              #kJ/mol\n",
+      "H2 = -Pfeed * m *1000 *Hsolution/ (40*100) \n",
+      "\n",
+      "#step3 - W kg water is converted to water vapour\n",
+      "Hvap = 2442.5                   #kJ/kg\n",
+      "H3 = W * Hvap \n",
+      "\n",
+      "#step4 - water vapour at 298K is heated to 373.15K\n",
+      "Cvap = 1.884                    #kJ/kgK\n",
+      "T3 = 373.15                     #K\n",
+      "H4 = W * Cvap * ( T3 - T2 ) \n",
+      "\n",
+      "#step5 - formation of 200kg of 50% NaOH solution at 298K\n",
+      "Hsolu = -25.89                  #kJ/mol\n",
+      "H5 = Pfeed * m *1000 *Hsolu/ (40*100) \n",
+      "\n",
+      "#step6 - Heating the solution from 298K to 380K\n",
+      "Csolu = 3.34                    #kJ/kg\n",
+      "T4 = 380.                       #K\n",
+      "H6 = P * Csolu * (T4 - T2) \n",
+      "Htotal = H1 + H2 + H3 + H4 + H5 + H6 \n",
+      "\n",
+      "# Result\n",
+      "print \"The enthalpy change accompanying the complete process = %d\"%Htotal,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The enthalpy change accompanying the complete process = 2138752 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.26 page no : 435"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# variables \n",
+      "Nwater = 0.8                        #moles\n",
+      "Nethanol = 0.2                      #moles\n",
+      "T = 323.                            #K\n",
+      "Cwater = 4.18*10**3                 #J/kgK\n",
+      "Cethanol = 2.58*10**3               #J/kgK\n",
+      "Hmixing1 = -758.                    #J/mol ( at 298K )\n",
+      "Hmixing2 = -415.                    #J/mol ( at 323K )\n",
+      "T1 = 298.                           #K\n",
+      "T2 = 523.                           #K\n",
+      "\n",
+      "# Calculation \n",
+      "#step1 - 0.8 mol of water is cooled from 323 K to 298K\n",
+      "H1 = Nwater * 18 * Cwater * ( T1 - T )/ 1000 \n",
+      "\n",
+      "#step2 - 0.2 mol ethanol cooled from 323K to 298K\n",
+      "H2 = Nethanol * 46 * Cethanol * ( T1 - T )/1000 \n",
+      "\n",
+      "#step3 - 0.8 mol water and 0.2 mol ethanol are mixed together,\n",
+      "H3 = Hmixing1 \n",
+      "\n",
+      "#step4 solution is heated to 323K, H4 = Cpm * (T - T1)\n",
+      "#Hmixing2 = H1 + H2 + H3 + H4\n",
+      "H4 = Hmixing2 - H1 - H2 - H3 \n",
+      "Cpm = H4 / ( T - T1 ) \n",
+      "\n",
+      "# Result\n",
+      "print \"The mean heat capacity of a 20 percent solution = %.2f\"%Cpm,\"J/molK\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean heat capacity of a 20 percent solution = 97.65 J/molK\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.27 page no : 437"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F = 1000.                       #kg/h\n",
+      "H1 = 116.3                      #kJ/kg ( enthalpy of feed solution - 10% NaOH, 305 K )\n",
+      "H2 = 560.57                     #kJ/kg ( enthalpy of thick liquor - 50% NaOH, 380 K )\n",
+      "Hsteam = 2676.                  #kJ/kg ( 1atm , 373.15K )\n",
+      "\n",
+      "#by doing material balances,\n",
+      "P = 200.                        #kg/h\n",
+      "mvap = 800.                     #kg/h\n",
+      "\n",
+      "# Calculation \n",
+      "#Enthalpy balance gives, F*H1 + Q = mvap*Hsteam + P*H2\n",
+      "Q = (mvap*Hsteam + P*H2)-F*H1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat to be supplied = \",Q,\"kJ/h\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat to be supplied =  2136614.0 kJ/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.28 pageno : 439"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "U2 = 0.35*10**3                 #kJ\n",
+      "U1 = 0.25*10**3                 #kJ\n",
+      "#since the tank is rigid the volume does not change during heating, Under constant volume, the change in the internal energy is equal to the heat supplied\n",
+      "\n",
+      "# Calculation \n",
+      "Q = U2 - U1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat transferred to the air = \",Q,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat transferred to the air =  100.0 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.29 pageno : 439"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "W = -2.25*745.7             #W ( work done on the system and 1hp = 745.7W)\n",
+      "Q = -3400.                  #kJ/h ( Heat transferred to the surrounding )\n",
+      "\n",
+      "# Calculation \n",
+      "U = Q*1000/3600 - W \n",
+      "\n",
+      "# Result\n",
+      "print \"Rise in the Internal energy of the system = %.3f\"%U,\"J/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rise in the Internal energy of the system = 733.381 J/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.30 pageno : 440"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#2Fe + 3/2O2 = Fe2O3\n",
+      "Hliberated = 831.08                 #kJ\n",
+      "\n",
+      "# Calculation \n",
+      "Q = -Hliberated*1000 \n",
+      "\n",
+      "# Result\n",
+      "print \"Q = \",Q,\"J\"\n",
+      "\n",
+      "#P(V) = (n)RT\n",
+      "#W = P(V) = (n)RT\n",
+      "n = -1.5 \n",
+      "R = 8.314 \n",
+      "T = 298.                            #K\n",
+      "W = (n) * R * T \n",
+      "print \"W = %.1f\"%W,\"J\"\n",
+      "U = Q - W \n",
+      "print \"U = %.5e\"%U,\"J\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q =  -831080.0 J\n",
+        "W = -3716.4 J\n",
+        "U = -8.27364e+05 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.31 pageno : 440"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Vgas = 0.09                     #m**3\n",
+      "Vliq = 0.01                     #m**3\n",
+      "SVliq = 1.061*10**-3            #m**3/kg\n",
+      "SVvap = 0.8857                  #m**3/kg\n",
+      "\n",
+      "# Calculation \n",
+      "mvap = Vgas / SVvap \n",
+      "mliq = Vliq / SVliq \n",
+      "Ul = 504.5                      #kJ/kg\n",
+      "Ug = 2529.5                     #kJ/kg\n",
+      "U1 = Ul * mliq + Ug * mvap \n",
+      "SVtotal = (Vgas + Vliq)/(mvap + mliq) \n",
+      "#using steam table , these value of specific volume corresponds to\n",
+      "# pressure of 148.6bar and internal energy of 2464.6kJ/kg\n",
+      "U = 2464.                       #kJ/kg\n",
+      "Utotal = U * (mvap + mliq) \n",
+      "#Utotal - U1 = Q - W,but W = o, hence, \n",
+      "Q = Utotal - U1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat to be added = %.f\"%Q,\"kJ\"\n",
+      "\n",
+      "\n",
+      "# note: answer may vary because of rounding error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat to be added = 18462 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.32 pageno : 443"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 10.                     #kg(air)\n",
+      "N = m / 29                  #kmol\n",
+      "P1 = 100.                   #kPa\n",
+      "T1 = 300.                   #K\n",
+      "R = 8.314 \n",
+      "\n",
+      "# Calculation \n",
+      "V1 = N * R * T1 / P1 \n",
+      "V2 = V1 \n",
+      "T2 = 600.                   #K\n",
+      "Cv = 20.785                 #kJ/kmolK\n",
+      "Cp = 29.099                 #kJ/kmolK\n",
+      "U = N * Cv * (T2 - T1) \n",
+      "Q = U \n",
+      "W = Q - U \n",
+      "H = U + N * R * ( T2 - T1 ) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Change in internal energy at constant volume = %d\"%U,\"kJ\"\n",
+      "print \"heat supplied at constant volume = %d\"%Q,\"kJ\"\n",
+      "print \"Work done at constant volume = \",W,\"kJ\"\n",
+      "print \"Change in Enthalpy at constant volume = %d\"%H,\"kJ\"\n",
+      "P2 = P1 \n",
+      "H2 = N * Cp * ( T2 - T1 ) \n",
+      "Q2 = H2 \n",
+      "U2 = H2 - N * R * (T2 - T1) \n",
+      "W2 = Q2 - U2 \n",
+      "print \"(b)Change in internal energy at constant Pressure = %d\"%U2,\"kJ\"\n",
+      "print \"heat supplied at constant Pressure = %d\"%Q2,\"kJ\"\n",
+      "print \"Work done at constant Pressure = %d\"%W2,\"kJ\"\n",
+      "print \"Change in Enthalpy at constant Pressure = %d\"%H2,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Change in internal energy at constant volume = 2150 kJ\n",
+        "heat supplied at constant volume = 2150 kJ\n",
+        "Work done at constant volume =  0.0 kJ\n",
+        "Change in Enthalpy at constant volume = 3010 kJ\n",
+        "(b)Change in internal energy at constant Pressure = 2150 kJ\n",
+        "heat supplied at constant Pressure = 3010 kJ\n",
+        "Work done at constant Pressure = 860 kJ\n",
+        "Change in Enthalpy at constant Pressure = 3010 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.33 pageno : 444"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Cp = 29.3                   #kJ/kmol\n",
+      "R = 8.314 \n",
+      "Cv = Cp - R \n",
+      "T1 = 300.                   #K\n",
+      "P1 = 1.                     #bar\n",
+      "P2 = 2.                     #bar\n",
+      "\n",
+      "# Calculation \n",
+      "#step1 - Volume remains constant, therefore the work done is\n",
+      "# zero and heat supplied is Cv, Also T2/T1 = P2/P1\n",
+      "T2 = P2 * T1 / P1 \n",
+      "Q1 = Cv * ( T2 - T1 ) \n",
+      "W1 = 0 \n",
+      "\n",
+      "# Result\n",
+      "print \"Work done at constant volume = \",W1,\"kJ\"\n",
+      "print \"Heat supplied at constant volume = \",Q1,\"kJ\"\n",
+      "\n",
+      "#step2 - Process is abdiabatic\n",
+      "Q2 = 0 \n",
+      "r = 1.4 \n",
+      "T3 = T2 * (( P1 / P2 )**((r - 1)/r)) \n",
+      "W2 = Cv * ( T2 - T3 )  \n",
+      "\n",
+      "print \"Work done in adiabatic process = %.1f\"%W2,\"kJ\"\n",
+      "print \"Heat supplied in adiabatic process = \",Q2,\"kJ\"\n",
+      "\n",
+      "#step3 - process is isobaric\n",
+      "Q3 = Cp * (T1 - T3) \n",
+      "U3 = Cv * (T1 - T3) \n",
+      "W3 = Q3 - U3 \n",
+      "print \"Work done at constant pressure = %.2f\"%W3,\"kJ\"\n",
+      "print \"Heat supplied at constant pressure = %.1f\"%Q3,\"kJ\"\n",
+      "\n",
+      "\n",
+      "# Note : Answers in book is wrong. while calculating heat supplied i.e. Cv(T2-T1), value of Cv is been taken wrongly."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work done at constant volume =  0 kJ\n",
+        "Heat supplied at constant volume =  6295.8 kJ\n",
+        "Work done in adiabatic process = 2262.3 kJ\n",
+        "Heat supplied in adiabatic process =  0 kJ\n",
+        "Work done at constant pressure = -1597.96 kJ\n",
+        "Heat supplied at constant pressure = -5631.5 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.34 pageno : 445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "P1 = 5.                         #bar\n",
+      "P2 = 4.                         #bar\n",
+      "T1 = 600.                       #K\n",
+      "V = 0.1                         #m**3\n",
+      "T2 = 400.                       #K\n",
+      "T = 298.                        #K\n",
+      "Cp = 30.                        #J/molK\n",
+      "\n",
+      "#step1 - isothermal condition\n",
+      "U1 = 0 \n",
+      "H1 = 0 \n",
+      "P = 1.                          #bar\n",
+      "R = 8.314 \n",
+      "\n",
+      "# Calculation \n",
+      "W1 = R*T1*math.log(P1/P2) \n",
+      "Q1 = W1 \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Change in the internal energy in isothermal condition = \",U1,\"kJ/kmol\"\n",
+      "print \"Change in the enthalpy energy in isothermal condition = \",H1,\"kJ/kmol\"\n",
+      "print \"Work done in isothermal condition = %.2f\"%W1,\"kJ/kmol\"\n",
+      "print \"Heat supplied in isothermal condition = %.2f\"%Q1,\"kJ/kmol\"\n",
+      "N = round(P * (1.01325 * 10**5) * V / ( R * T ),2)                # answer slightly different because of rouding error.\n",
+      "Cv = Cp - R          \n",
+      "U2 = Cv * (T2 - T)*N    #Answer differes due to slighlty different value of N\n",
+      "H2 = Cp * (T2 - T)*N    #Answer differes due to slighlty different value of N\n",
+      "W2 = 0 \n",
+      "Q2 = U2 + W2 \n",
+      "print \n",
+      "print \"\\n(b)Change in the internal energy at constant volume condition = %d\"%U2,\"J\"\n",
+      "print \"Change in the enthalpy energy at constant volume condition = %d\"%H2,\"kJ/kmol\"\n",
+      "print \"Work done at constant volume condition = \",W2,\"kJ/kmol\"\n",
+      "print \"Heat supplied at constant volume condition = %d\"%Q2,\"kJ/kmol\"\n",
+      "\n",
+      "# note : answer varies because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Change in the internal energy in isothermal condition =  0 kJ/kmol\n",
+        "Change in the enthalpy energy in isothermal condition =  0 kJ/kmol\n",
+        "Work done in isothermal condition = 1113.13 kJ/kmol\n",
+        "Heat supplied in isothermal condition = 1113.13 kJ/kmol\n",
+        "\n",
+        "\n",
+        "(b)Change in the internal energy at constant volume condition = 9046 J\n",
+        "Change in the enthalpy energy at constant volume condition = 12515 kJ/kmol\n",
+        "Work done at constant volume condition =  0 kJ/kmol\n",
+        "Heat supplied at constant volume condition = 9046 kJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.35 pageno : 448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 1.                              #kg\n",
+      "u2 = 0.5                            #m/s\n",
+      "u1 = 60.                            #m/s\n",
+      "H = -3000.                          #kJ/kg\n",
+      "\n",
+      "# Calculation \n",
+      "#KE = (u**2)/2\n",
+      "KE = ((u2 ** 2) - (u1**2))/2000 \n",
+      "g = 9.81                            #m/s**2\n",
+      "Z1 = 7.5                            #m\n",
+      "Z2 = 2.                             #m\n",
+      "\n",
+      "#PE = g * (Z)\n",
+      "PE = g * (Z2 - Z1)/1000 \n",
+      "W = 800.                            #kJ/kg\n",
+      "Q = H + PE + KE + W \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat removed from the fluid = %.2f\"%Q,\"kJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat removed from the fluid = -2201.85 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.37 pageno : 449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "g = 9.81                        #m/s**2\n",
+      "z = 55. \n",
+      "PE = g * z \n",
+      "KE = 0. \n",
+      "T2 = 288.                       #K\n",
+      "f = 1.5*10**-2                  #m**3/min\n",
+      "D = 1000.                       #kg/m**3\n",
+      "m = f * D \n",
+      "Qsupp = 500.                    #kJ/min\n",
+      "Qlost = 400.                    #kJ/min\n",
+      "\n",
+      "# Calculation \n",
+      "Qnet = (Qsupp - Qlost) * D / m  \n",
+      "W = 2*745.7                     #W\n",
+      "Ws = -W * 0.6 / (m/60) \n",
+      "H = Qnet - Ws - PE - KE \n",
+      "Cp = 4200. \n",
+      "T1 = H / Cp \n",
+      "T = T1 + T2 \n",
+      "\n",
+      "# Result\n",
+      "print \"The temperature of exit water = %.2f\"%T,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature of exit water = 290.31 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.38 pageno : 450"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 1000.                   #kg/h (dried product)\n",
+      "\n",
+      "# S be the amount of dry solid in the product stream \n",
+      "Pmoisture1 = 4.             #%\n",
+      "Pmoisture2 = 0.2            #%\n",
+      "P = 1.\n",
+      "S = m *(1 - P/1000) \n",
+      "\n",
+      "# Calculation \n",
+      "X1 = Pmoisture1/(100 - Pmoisture1) \n",
+      "X2 = Pmoisture2/(100 - Pmoisture2) \n",
+      "\n",
+      "#let G be the weight of dry air in the air stream \n",
+      "Y1 = 0.01                   #kg water/kg dry solid\n",
+      "Cp = 1.507 \n",
+      "Cw = 4.2 \n",
+      "T1 = 298.                   #K\n",
+      "T = 273.                    #K\n",
+      "T2 = 333.                   #K\n",
+      "Tg1 = 363.                  #K\n",
+      "Tg2 = 305.                  #K\n",
+      "\n",
+      "Hs1 = (Cp + X1 * Cw) * (T1 - T) \n",
+      "Hs2 = (Cp + X2 * Cw) * (T2 - T) \n",
+      "#Hg = Cs(Tg - To) + Y*L\n",
+      "#Cs = 1.005 + 1.884*Y\n",
+      "L = 2502.3                  #kJ/kg dry air\n",
+      "Hg1 = (1.005 + 1.884 * Y1)*(Tg1 - T) + Y1 * L \n",
+      "Q = -40000.                 #kJ/h\n",
+      "\n",
+      "#Calculating for T2, Hg2 = 32.16 + 2562.59*Y \n",
+      "#change in enthalpy = Q\n",
+      "#H1 = S * Hs1 + G * HG1 = 37814.22 + 117.17G\n",
+      "#H2 = 100728.14 + G* (32.16 + 2561.59*Y)\n",
+      "#change in enthalpy = Q\n",
+      "#62913.92 + G *(-85.01 + 2561.59*Y) + 40000 = 0\n",
+      "#102913.92 + G *(-85.01 + 2561.59*Y) = 0            (1)\n",
+      "#moisture balance, S*X1 + G*Y1 = S*X2 + G*Y2\n",
+      "#G*(Y-0.01) = 39.62                                 (2)\n",
+      "#solving simultaneously ( 1 ) and ( 2 ), \n",
+      "Gdry = 3443.                #kg/h\n",
+      "G = Gdry*(1 + Y1) \n",
+      "\n",
+      "# Result\n",
+      "print \"Air requirement = \",G,\"kg/h\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Air requirement =  3477.43 kg/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.39 pageno : 452"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 1000.                   #kg/h ( feed solution )\n",
+      "\n",
+      "#F - mass of feed distilled, W - mass of the bottom product, D - mass of the distillate, xf, xd and xw - weight fraction of actone in feed, distillate and residue resp_\n",
+      "#total balance, F = D + W\n",
+      "#Acetone balance, F*xf = D*xd + w*xw\n",
+      "F = 1000 \n",
+      "xf = 0.10 \n",
+      "xd = 0.9 \n",
+      "xw = 0.01 \n",
+      "\n",
+      "# Calculation \n",
+      "#substituting in above equations,\n",
+      "D = F * (xf - xw) / (xd - xw) \n",
+      "W = F - D \n",
+      "R = 8 \n",
+      "L = R * D \n",
+      "#material balance around the condenser,G vapour reaching the condenser\n",
+      "G = L + D \n",
+      "Td = 332.                   #K\n",
+      "T2 = 300.                   #K\n",
+      "Tw = 370.                   #K\n",
+      "Tf = 340.                   #K\n",
+      "Lacetone1 = 620.            #kJ/kg\n",
+      "Lwater1 = 2500.             #kJ/kg\n",
+      "Ld = xd * Lacetone1 + (1 - xd) * Lwater1 \n",
+      "Cpacetone = 2.2             #kJ/kgK\n",
+      "Cpwater = 4.2               #kJ/kgK\n",
+      "Cp = xd * Cpacetone + (1-xd)*Cpwater \n",
+      "H = Ld + Cp * ( Td - T2 ) \n",
+      "Cpc = 4.2                   #kJ/kg\n",
+      "Tc = 30.                    #K ( change in temperature allowable for cooling water )\n",
+      "m = G * H / ( Cpc * Tc ) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)The circulation rate of cooling water = %2f\"%m,\"kg/h\"\n",
+      "Qc = G * H \n",
+      "Hd = 0. \n",
+      "Hw = (xw * Cpacetone + (1-xw)*Cpwater)*(Tw - T2) \n",
+      "Hf = (xf * Cpacetone + (1-xf)*Cpwater)*(Tf - T2) \n",
+      "Qb = D * Hd + W * Hw + Qc - F * Hf \n",
+      "Hcondensation = 2730.       #kJ/kg\n",
+      "msteam = Qb/Hcondensation \n",
+      "print \"(b)Amount of steam supplied = %.2f\"%msteam,\"kg/h\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The circulation rate of cooling water = 6391.011236 kg/h\n",
+        "(b)Amount of steam supplied = 332.70 kg/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch12.ipynb b/Stoichiometry_And_Process_Calculations/ch12.ipynb
new file mode 100755
index 00000000..0f35302b
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch12.ipynb
@@ -0,0 +1,1025 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Energy Balance Thermochemistry"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 pageno : 475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "N = 100.                            #mol gas mixture burned\n",
+      "\n",
+      "#CO(g) + 1/2 O2(g) = CO2 -                    Hr1 = - 282.91kJ/mol\n",
+      "#H2(g) + 1/2 O2(g) = H2O -                    Hr2 = - 241.83kJ/mol\n",
+      "Hr1 = - 282.91                      #kJ/mol\n",
+      "Hr2 = - 241.83                      #kJ/mol\n",
+      "Nco1 = 20. \n",
+      "Nh21 = 30. \n",
+      "Nn21 = 50. \n",
+      "\n",
+      "# Calculation \n",
+      "Htotal = Nco1*Hr1 + Nh21*Hr2 \n",
+      "\n",
+      "# Result\n",
+      "print \"the amount of heat liberated on the complete combustion of 100mol of the gas mixture = \",-Htotal,\"kJ\"\n",
+      "Ncoreac = Nco1 * 0.9 \n",
+      "Nh2reac = Nh21 * 0.8 \n",
+      "Htotal1 = Ncoreac*Hr1 + Nh2reac*Hr2 \n",
+      "print \"the amount of heat liberated if only 90% of CO and 80% of H2 react of 100mol of the gas mixture = \",-Htotal1,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the amount of heat liberated on the complete combustion of 100mol of the gas mixture =  12913.1 kJ\n",
+        "the amount of heat liberated if only 90% of CO and 80% of H2 react of 100mol of the gas mixture =  10896.3 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 pageno : 477"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#C(s) + 2H2(g) = CH4(g)          Hf = ?\n",
+      "Hc = -393.51                    #kJ/mol\n",
+      "Hh2 = -285.84                   #kJ/mol\n",
+      "Hch4 = - 890.4                  #kJ/mol\n",
+      "\n",
+      "# Calculation \n",
+      "#heat of reaction can be calculated from the heat of combustion data \n",
+      "# using following equation, the heat of reaction is the sum of the heat of combustion of all the reactants in the desired reaction minus the sum of the heat of combustion of all the products of the desired reaction. Here the reactants are one mole of Carbon and two moles hydrogen, and the product is one mole of methane,there heat of reaction is\n",
+      "Hf = 1 * Hc + 2 * Hh2 - 1 * Hch4 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat of formation of methane = \",Hf,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat of formation of methane =  -74.79 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3 pageno : 478"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 1.                      #kg of coal burned\n",
+      "xc = 0.7 \n",
+      "xh2 = 0.055 \n",
+      "xn2 = 0.015 \n",
+      "xs = 0.03 \n",
+      "xo = 0.13 \n",
+      "xash = 0.07 \n",
+      "Hvap = 2370.                #kJ/kg\n",
+      "C = 29000.                  #kJ/kg\n",
+      "\n",
+      "# Calculation \n",
+      "Nh2 = xh2 * m / 2.016 \n",
+      "Nwater = Nh2                # ( amount of water formed )\n",
+      "mwater = Nwater * 18.016 \n",
+      "Hreq = mwater * Hvap \n",
+      "Hnet = C - Hreq \n",
+      "\n",
+      "# Result\n",
+      "print \"Net heating value of coal = %.1f\"%Hnet,\"kJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Net heating value of coal = 27835.1 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4 pageno : 479"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#C2H5OH(l) + CH3COOH(l) = C2H5COOCH3(l) + H2O(l) H = ?\n",
+      "Hc2h5oh = -1366.91                      #kJ/mol\n",
+      "Hch3cooh = -871.69                      #kJ/mol\n",
+      "Hc2h5cooch3 = -2274.48                  #kJ/mol\n",
+      "\n",
+      "# Calculation \n",
+      "#to calculate heat of reaction from the heat of combustion data , \n",
+      "#Hreac = Hreac - Hprod\n",
+      "Hreac = Hc2h5oh + Hch3cooh - Hc2h5cooch3 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat of reaction for the esterification of ethyl alcohol with acetic acid = \",Hreac,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat of reaction for the esterification of ethyl alcohol with acetic acid =  35.88 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5 pageno : 479"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#C2H4(g) + H2O(g) = C2H5OH(g)\n",
+      "#2CO2(g) + 3H2O(l) = C2H5OH(l) + 3O2(g)     H = 1366.91kJ  (A)\n",
+      "Hc2h4 = -1410.99                    #kJ/mol\n",
+      "Hvap = 44.04                        #kJ/mol\n",
+      "Hc2h5oh = 42.37                     #kJ/mol\n",
+      "#C2H4(g) + 3H2O(l) = C2H5OH(l) + 3O2(g)     H = -1410.99kJ (B)\n",
+      "#H2O(l) = H2O(g)                            H = 44.04kJ    (C)\n",
+      "#C2H5OH(l) = C2H5OH(g)                      H = 42.37kJ    (D)\n",
+      "#A + B + D - C gives the required reaction\n",
+      "Ha = 1366.91                        #kJ\n",
+      "Hb = -1410.99                       #kJ\n",
+      "Hc = 44.04                          #kJ\n",
+      "Hd = 42.37                          #kJ\n",
+      "\n",
+      "# Calculation \n",
+      "Hreac = Ha + Hb + Hd - Hc \n",
+      "\n",
+      "# Result\n",
+      "print \"The heat of reaction = \",Hreac,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat of reaction =  -45.75 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6 pageno : 480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#C2H5(g) + 5/2O2(g) = 2CO2(g) + H2O(l)             H1 = -1299.6kJ    (A)\n",
+      "#C(s) + O2(g) = CO2(g)                             H2 = -393.51kJ    (B)\n",
+      "#H2(g) + 1/2O2(g) = H2O(l)                         H3 = -285.84kJ    (C)\n",
+      "#2C(s) + H2(g) = C2H2(g)                           H = ?\n",
+      "H1 = -1299.6                        #kJ\n",
+      "H2 = -393.51                        #kJ\n",
+      "H3 = -285.84                        #kJ\n",
+      "\n",
+      "# Calculation \n",
+      "Hreac = 2 * H2 + H3 - H1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat of formation of acetylene = \",Hreac,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat of formation of acetylene =  226.74 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7 pageno : 480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 100.                    #kg of pyrites charged\n",
+      "xfes2in = 0.8 \n",
+      "xganguein = 0.2 \n",
+      "xfes2out = 0.05 \n",
+      "\n",
+      "# Calculation \n",
+      "#let x be the FeS2 in the feed, then, Fe2O3 = (80 - x)*159.69 / (119.98*2) \n",
+      "#and gangue = 20, total = 73.24 + 0.3345, be FeS2 is only 5 % in the product, hence\n",
+      "x = 0.05 * 73.24 / (1 - 0.05*0.3345) \n",
+      "mfes2reacted = m*xfes2in - x \n",
+      "#4FeS2 + 11O2 = 2Fe2O3 + 8SO2\n",
+      "Hfes2 = -178.02             #kJ/mol\n",
+      "Hfe2o3 = -822.71            #kJ/mol\n",
+      "Hso2 = -296.9               #kJ/mol\n",
+      "Hreac = 2 * Hfe2o3 + 8 * Hso2 - 4 * Hfes2 \n",
+      "N = mfes2reacted *1000/ 119.98 \n",
+      "H = Hreac * N / 4 \n",
+      "H1 = H/m                    #(heat of reaction per kg of coal burnt)\n",
+      "\n",
+      "# Result\n",
+      "print \"Heat of reaction per 1 kg of coal burned = %.3e\"%H1,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat of reaction per 1 kg of coal burned = -5.258e+03 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8 pageno : 481"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#CH3OH(l) + 3/2O2(g) = CO2(g) + 2H2O(l)         H = -726.55kJ\n",
+      "H1 = -726.55                    #kJ\n",
+      "Hco2 = -393.51                  #kJ/mol\n",
+      "Hh2o = -285.84                  #kJ/mol\n",
+      "\n",
+      "# Calculation \n",
+      "Hch3oh = Hco2 + 2 * Hh2o - H1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat of formation of liquid methanol = \",Hch3oh,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat of formation of liquid methanol =  -238.64 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.9 pageno : 482"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "N = 100.                        #mol fuel gas\n",
+      "Nco = 21. \n",
+      "Nh2 = 15.6 \n",
+      "Nco2 = 9.0 \n",
+      "Nch4 = 2. \n",
+      "Nc2h4 = 0.4 \n",
+      "Nn2 = 52. \n",
+      "Hco = 282.99                    #kJ/mol ( heat of combustion )\n",
+      "Hh2 = 285.84                    #kJ/mol ( heat of combustion )\n",
+      "Hch4 = 890.4                    #kJ/mol ( heat of combustion )\n",
+      "Hc2h4 = 1410.99                 #kJ/mol ( heat of combustion )\n",
+      "Hvap = 44.04                    #kJ/mol\n",
+      "\n",
+      "# Calculation \n",
+      "H = round(Nco * Hco + Nh2 * Hh2 + Nch4*Hch4 + Nc2h4*Hc2h4)         #kJ\n",
+      "V = round(N * 22.4143/1000,2) \n",
+      "H1 = round(H / V)                      #kJ/m**3\n",
+      "\n",
+      "#on combustion, 1 mol hydrogen gives 1 mol of water, 1 mol of \n",
+      "# methane gives 2 mol of water and 1 mol of ethylene gives 2 moles of water\n",
+      "Nwater = Nh2 + 2 * Nch4 + 2 * Nc2h4 \n",
+      "Hvap1 = round(Hvap * Nwater,2)\n",
+      "Hnet = H1 - Hvap1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Net heating value of the fuel = %.2f\"%Hnet,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Net heating value of the fuel = 4792.58 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.10 pageno : 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "# C5H12(g) + 8O2(g) = 5CO2(g) + 6H20(l)\n",
+      "Hfco2 = -393.51                         #kJ\n",
+      "Hfh2o = - 241.826                       #kJ\n",
+      "Hfc5h12 = -146.4                        #kJ\n",
+      "Hvap = 43.967                           #kJ/mol\n",
+      "\n",
+      "# Calculation \n",
+      "H1 = 6*Hfh2o +5*Hfco2 - Hfc5h12 \n",
+      "H2 = 6 * (-Hvap) \n",
+      "Hreac = H1 + H2 \n",
+      "# Result\n",
+      "print \"heat of reaction = \",Hreac,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "heat of reaction =  -3535.908 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.11 pageno : 484"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 1.                          #kg of oil burned\n",
+      "mc = 0.9                        #kg\n",
+      "mh2 = 0.1                       #kg\n",
+      "\n",
+      "# Calculation \n",
+      "Mc = mc / 12                    #kmol\n",
+      "#C(s) + O2(g) = CO2(g)\n",
+      "Nh2 = mh2 / 2.016               #kmol\n",
+      "\n",
+      "#change in the no. of gaseous components accompanying the \n",
+      "#combustion of 1 mole of hydrogen in liquid state is -1/2 mol, therefore for Nh2 mol\n",
+      "R = 8.314 \n",
+      "T = 298.                        #K\n",
+      "x = Nh2 * R * T / (-2) \n",
+      "Qv = -43000                     #kJ/kg\n",
+      "Qp = Qv + x \n",
+      "\n",
+      "# Result\n",
+      "print \"the constant pressure heat of combustion = %.1f\"%Qp,\"kJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the constant pressure heat of combustion = -43061.4 kJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.12 pageno : 487"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#1 - N2, 2 - H2, 3 - NH3\n",
+      "a1 = 27.31 \n",
+      "a2 = 29.09 \n",
+      "a3 = 25.48 \n",
+      "b1 = 5.2335*10**-3 \n",
+      "b2 = -8.374*10**-4 \n",
+      "b3 = 36.89 * 10**-3 \n",
+      "c1 = -4.1868 * 10**-9 \n",
+      "c2 = 2.0139*10**-6 \n",
+      "c3 = -6.305*10**-6 \n",
+      "H1 = -46191.                    #J\n",
+      "T1 = 298.                       #K\n",
+      "\n",
+      "# Calculation \n",
+      "#1/2 N2 + 3/2 H2 = NH3               H = -46.191kJ\n",
+      "#Ht = H + a*T + b*T**2 / 2+ c*T**3 / 3\n",
+      "#at 298, \n",
+      "a = a3 - a1 / 2 - 3 * a2 / 2 \n",
+      "b = b3 - b1 / 2 - 3 * b2 / 2 \n",
+      "c = c3 - c1 / 2 - 3 * c2 / 2 \n",
+      "H = H1 -a * T1 - b * (T1**2) / 2 - c * (T1**3) / 3 \n",
+      "T2 = 700.                       #K\n",
+      "H2 = H + a * T2 + b * (T2**2) / 2 + c * (T2**3) / 3 \n",
+      "\n",
+      "# Result\n",
+      "\n",
+      "print \"Heat of reaction at 700K = %d\"%H2,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat of reaction at 700K = -52835 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.13 pageno : 488"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#CO(g) + 2H2(g) = CH3OH(g)\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "T1 = 298.                   #K\n",
+      "T2 = 1073.                  #K\n",
+      "#Cp(CH3OH) = 18.382 + 101.564 * 10**-3 * T - 28.683 * 10**-6 * T**2\n",
+      "#Cp(CO) = 28.068 + 4.631 * 10**-3 * T - 2.5773 * 10**4 * T**-2\n",
+      "#Cp(H2)  = 27.012 + 3.509 * 10**-3 * T + 6.9006 * 10**4 * T**-2\n",
+      "#for reactants,\n",
+      "\n",
+      "# Calculation \n",
+      "def f1(T): \n",
+      "\t return 28.068 + 4.631 * 10**-3 * T - 2.5773 * 10**4 * T**-2\n",
+      "\n",
+      "def f2(T):\n",
+      "    return 27.012 + 3.509 * 10**-3 * T + 6.9006 * 10**4 * T**-2\n",
+      "\n",
+      "H1 =  quad(f1,T2,T1)[0] + 2 * quad(f2,T2,T1)[0]\n",
+      "\n",
+      "#for product,\n",
+      "\n",
+      "def f2(T): \n",
+      "\t return 18.382 + 101.564 * 10**-3 * T - 28.683 * 10**-6 * T**2\n",
+      "\n",
+      "H2 =  quad(f2,T1,T2)[0]\n",
+      "\n",
+      "#H298 = Hproducts - Hreactants \n",
+      "#CO + 2H2 = CH3OH            Ha1 = -238.64kJ\n",
+      "Ha1 = -238.64               #kJ\n",
+      "\n",
+      "#CH3OH(l) = CH3OH(g)         Hvap = 37.98kJ\n",
+      "Hvap = 37.98                #kJ\n",
+      "\n",
+      "#CO(g) + 2H2(g) = CH3OH(g)   Ha2 = -200.66kJ\n",
+      "Ha2 = Ha1 + Hvap            #kJ\n",
+      "Hco = -110.6                #kJ/mol\n",
+      "H298 = Ha2 - (Hco) \n",
+      "Htotal = H1/1000 + H298 + H2/1000 \n",
+      "\n",
+      "# Result \n",
+      "print \"The heat of reaction at 773K = %.3f\"%Htotal,\"kJ/mol\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat of reaction at 773K = -103.497 kJ/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.14 pageno : 489"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Nco = 1.                            #mol CO reacted\n",
+      "#CO + 1/2 O2 = CO2\n",
+      "No2 = Nco / 2 \n",
+      "Pexcess = 100. \n",
+      "\n",
+      "# Calculation \n",
+      "Nosupp = No2 * ( 1 + Pexcess / 100 ) #oxygen supplied\n",
+      "Nn2 = Nosupp * 79. / 21 \n",
+      "Nco2 = Nco \n",
+      "Noremain = Nosupp - No2 \n",
+      "T1 = 298.                           #K\n",
+      "T2 = 400.                           #K\n",
+      "Hr1 = -282.99                       #kJ\n",
+      "T3 = 600.                           #K\n",
+      "SHco = 29.1                         #J/molK\n",
+      "SHo2 = 29.7                         #J/molK\n",
+      "SHn2 = 29.10                        #J/molK\n",
+      "SHco2 = 41.45                       #J/molK\n",
+      "H1 = (Nosupp * SHo2 + Nn2 * SHn2 + Nco * SHco) * (T1 - T2) #enthalpy of cooling of reactants from 298 to 400 K\n",
+      "H2 = (Nco2 * SHco2 + Nn2 * SHn2 + Noremain * SHo2) * (T3 - T1) #enthalpy of heating the products from 298K to 600K\n",
+      "H = H1/1000 + Hr1 + H2/1000 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heat change at 600K = %.1f\"%H,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat change at 600K = -250.1 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.15 pageno : 490"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#CO(g) + H2O(g) = CO2(g) + H2(g)            H298 = -41.190\n",
+      "T1 = 298.                       #K\n",
+      "Pconv = 75.                     #%\n",
+      "T2 = 800.                       #K\n",
+      "H298 = -41.190                  #kJ\n",
+      "Hco = 30.35                     #J/molK\n",
+      "Hco2 = 45.64                    #J/molK\n",
+      "Hwater = 36.                    #J/molK\n",
+      "Hh2 = 29.3                      #J/molK\n",
+      "Nco = 1.                        #mol\n",
+      "Nh2o = 1.                       #mol\n",
+      "\n",
+      "# Calculation \n",
+      "Ncofinal = Nco * (1 - Pconv/100) \n",
+      "Nwaterf = Ncofinal \n",
+      "Nco2final = Nco - Ncofinal \n",
+      "Nh2final = Nco2final \n",
+      "H2 = (Nco2final * Hco2 + Nh2final * Hh2 + Ncofinal * Hco + Nwaterf * Hwater) * (T2 - T1) \n",
+      "Hr1 = H298 * (Nco - Ncofinal) \n",
+      "Hr2 = Hr1 * 1000 + H2 \n",
+      "mh2 = Nh2final * 2.016 * 10**-3 #kg\n",
+      "#therfore for 1000k H2,\n",
+      "Hr = Hr2 * 1000 / (mh2 * 1000)  #kJ\n",
+      "\n",
+      "# Result\n",
+      "print \"Amount of heat change for 1000kg of hydrogen produced = %.3e\"%Hr,\"kJ\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Amount of heat change for 1000kg of hydrogen produced = 3.736e+06 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.16 pageno : 490"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#CO2(g) + C(s) = 2CO(g)                H1298 = 170kJ/mol\n",
+      "#O2(g) + 2C(s) = 2CO(g)                H2298 = -221.2kJ/mol\n",
+      "T2 = 1298.                      #K\n",
+      "T1 = 298.                       #K\n",
+      "Hc = 0.02                       #kJ/molK\n",
+      "Ho = 0.03                       #kJ/molK\n",
+      "Hco = 0.03                      #kJ/molK\n",
+      "Hco2 = 0.05                     #kJ/molK\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#let the flue gas contain x mol CO2 per mole of oxygen, product contains 2(1+x)mol CO. Nitrogen in reactant and product remain the same\n",
+      "#enthalpy of cooling xmol CO2, 1 mol O2 and 2 + xmol carbon from 1298 to 298K is given as, H1 = (Hco2 * x + Ho * 1 + Hc * (2 + x)) * (298 - 1298)\n",
+      "#H1 = (-70x - 70)kJ\n",
+      "#enthalpy of heating the product, H2 = 2 * ( 1 + x )* Hco * (1298 - 298)\n",
+      "#H2 = 60 + 60x kJ\n",
+      "#Hr = 170x - 221.2\n",
+      "#Htotal = 0 = H1 + H2 + Hr\n",
+      "x = (221.2 + 70 - 60)/(170 + 60 - 70) \n",
+      "\n",
+      "# Result\n",
+      "print \"moles of CO2 present per mol of oxygen in feed stream = \",x,\"mol\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "moles of CO2 present per mol of oxygen in feed stream =  1.445 mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.17 pageno : 491"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "N = 100.                        #mol flue gas\n",
+      "#Carbon balance,\n",
+      "#x is the feed of methane, w is water in flue ga, y is the oxygen supplied\n",
+      "xco2 = 0.019 \n",
+      "xch2o = 0.117 \n",
+      "xo2 = 0.038 \n",
+      "xch4 = 0.826\n",
+      "\n",
+      "# Calculation \n",
+      "xc = xco2 + xch2o + xch4 \n",
+      "Nc = xc * N \n",
+      "Nch4i = Nc \n",
+      "#Hydrogen balance,\n",
+      "xh2 = xch2o + xch4*2 \n",
+      "w = 2 * (Nch4i) - xh2*N \n",
+      "#oxygen balance\n",
+      "No2s = (xco2 + xch2o/2 + xo2)*N + w/2 \n",
+      "y = No2s \n",
+      "T1 = 298.                       #K\n",
+      "T2 = 573.                       #K\n",
+      "T3 = 673.                       #K\n",
+      "\n",
+      "#oxygen cooled from 573K and methane from 673 to 298K \n",
+      "Ho573 = 30.5                    #J/molK\n",
+      "Hch4673 = 45.9                  #J/molK\n",
+      "H1 = y * Ho573 * (T1 - T2) + Nch4i * Hch4673 * (T1 - T3) \n",
+      "#CH4 + O2 = CH2O + H2O        Hr1 = -282.926kJ\n",
+      "#CH4 + 2O2 = CO2 + 2H2O       Hr2 = -802.372kJ\n",
+      "Hr1 = -282.926 #kJ\n",
+      "Hr2 = -802.372 #kJ\n",
+      "H2 = xch2o*N*Hr1 + xco2*N*Hr2 \n",
+      "T4 = 873.                       #K\n",
+      "Ho = 31.9\n",
+      "Hch4 = 51.4 \n",
+      "Hco2 = 46.3 \n",
+      "Hch2o = 47.1 \n",
+      "Hh2o = 36.3 \n",
+      "H3 = ((xco2 * Hco2 + xo2 * Ho + xch4 * Hch4 + Hch2o*xch2o)*N + w * Hh2o)*(T4 - T1) \n",
+      "Htotal = H1/1000 + H2 + H3/1000 \n",
+      "Nch2o = xch2o * N \n",
+      "mch2o = Nch2o * 30.016/1000     #kg\n",
+      "\n",
+      "#for 1000 kg of formaldehyde produced,\n",
+      "H = Htotal * 1000 / mch2o \n",
+      "# Result \n",
+      "print \"The amount of heat to be removed per 1000kg of formaldehyde produced = %.1e\"%H,\"kJ\"\n",
+      "\n",
+      "# Note: answers are slightly different because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of heat to be removed per 1000kg of formaldehyde produced = -9.8e+06 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.18 pageno : 494"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Nn2 = 1.            #kmol/s ( basis - feed conisting of 1 kmol of N2  and 3 kmol of H2 )\n",
+      "Nh2 = 3.            #kmol/s\n",
+      "\n",
+      "#let x be the fraction converted \n",
+      "T1 = 700.           #K\n",
+      "Hr1 = -94.2         #kJ/mol\n",
+      "\n",
+      "# Calculation \n",
+      "#heat liberated = Hr1 * x\n",
+      "#Product consists of 2x kmol NH3, (1-x)kmol N2, and 3(1-x)kmol Hydrogen\n",
+      "T2 = 800.           #K\n",
+      "Hn2 = 0.03          #kJ/molK\n",
+      "Hh2 = 0.0289        #kJ/molK\n",
+      "Hnh3 = 0.0492       #kJ/molK\n",
+      "\n",
+      "#H2 = (1-x)*0.03*10**3 * 100 + 3*(1-x)*0.0289*1000*100 + 2*x*0.0492*1000*100\n",
+      "#H2 = 11.67*1000 - 1.83*10**3*x kJ\n",
+      "#reaction is adiabatic, hence, H1 = H2\n",
+      "#solving this we get,\n",
+      "x = 0.1215 \n",
+      "Convmax = x * 100 \n",
+      "\n",
+      "# Result\n",
+      "print \"The maximum conversion for nitrogen should be \",Convmax,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum conversion for nitrogen should be  12.15 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.19 pageno : 494"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Nco = 1.                        #mol CO\n",
+      "\n",
+      "# CO + 1/2 O2 = CO2\n",
+      "O2r = 1.                        #mol\n",
+      "N2r = 3.76                      #mol\n",
+      "COr = 1.                        #mol\n",
+      "O2p = 0.5                       #mol\n",
+      "N2p = 3.76                      #mol\n",
+      "CO2p = 1.                       #mol\n",
+      "Hco = 29.23                     #J/molK\n",
+      "Ho2 = 34.83                     #J/molK\n",
+      "Hn2 = 33.03                     #J/molK\n",
+      "Hco2 = 53.59                    #J/molK\n",
+      "Hcomb1 = -282.99                #kJ/mol\n",
+      "T1 = 298.                       #K\n",
+      "T2 = 373.                       #K\n",
+      "\n",
+      "# Calculation \n",
+      "H1 = (O2r * Ho2 + N2r * Hn2 + COr * Hco) * (T1 - T2) \n",
+      "#For product at temp T, H2 = (O2p * Ho2 + N2p * Hn2 + CO2p * Hco2) * (T - T1) \n",
+      "#For adiabatic condition, -(H1 + Hcomb1) = H2\n",
+      "T = -(H1 + Hcomb1 * 1000) / (O2p * Ho2 + N2p * Hn2 + CO2p * Hco2) + T1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Theoretical flame temperature = %.1f\"%T,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Theoretical flame temperature = 1820.1 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.20 pageno : 495"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "N = 1.                                  #kmol hydrogen burned\n",
+      "No = N/2 \n",
+      "\n",
+      "# Calculation \n",
+      "Nosupplied = 2 * No \n",
+      "Nair = Nosupplied * 100 / 21. \n",
+      "Nn2 = Nair - Nosupplied \n",
+      "#Reactants, H2 = 1kmol, Air = 4.762 kmol\n",
+      "#Product, Water vapour = 1kmol, Oxygen = 0.5kmol, N2 = 3.762kmol\n",
+      "#Cp(water) = 30.475 + (9.652*10**-3)*T + 1.189 * 10**-6 * T**2\n",
+      "#Cp(nitrogen) = 27.034 + 5.815 * 10**-3 *T - 0.2889 * 10**-6 * T**2\n",
+      "#Cp(oxygen) = 25.611 + 13.260 * 10**-3 * T - 4.2077 * 10**-6 * T**2\n",
+      "#H2 = integration(298 to T of (1 * Cp(water) + 0.5 * Cp(oxygen) + 3.762 * Cp(nitrogen)))\n",
+      "#therefore, H2 = 140.34 * T + 31.222 * 10**-3 * T**2 - 4.928 * 10**-6 * T**2 - 44463.54 kJ\n",
+      "H298 = -241.826 * 10**3                 #kJ\n",
+      "\n",
+      "#H2 = -H1 - H298\n",
+      "#H1 = 0\n",
+      "#therefore using equation H2, the value of T is obtained to be\n",
+      "T = 1609.8                              #K\n",
+      "\n",
+      "# Result\n",
+      "print \"Temperature of the reaction products = \",T,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature of the reaction products =  1609.8 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch2.ipynb b/Stoichiometry_And_Process_Calculations/ch2.ipynb
new file mode 100755
index 00000000..9894afe0
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch2.ipynb
@@ -0,0 +1,436 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Units and Dimensions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.1 page no : 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "V1 = 15.            #ft**3/min volumetric flow rate\n",
+      "ft = 0.3048         #m relationship\n",
+      "min = 60.           #secs relationship\n",
+      "\n",
+      "# Calculation \n",
+      "V = V1*ft**3/min;\n",
+      "\n",
+      "# Result \n",
+      "print \"Volumetric flowrate = %.3e m**3/s\"%V\n",
+      "D = 1000            #kg/m**3\n",
+      "M = V * D;\n",
+      "print \"mass flowrate = %.3f kg/s\"%M\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volumetric flowrate = 7.079e-03 m**3/s\n",
+        "mass flowrate = 7.079 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.2 page no : 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "ft = 0.3048;            #m \n",
+      "lb = 0.4536;            #kg\n",
+      "\n",
+      "# Calculation \n",
+      "P = ft*lb;\n",
+      "\n",
+      "# Result \n",
+      "print \"1 poundal is 1 ft*lb/s**2 = %.4f N\"%P\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 poundal is 1 ft*lb/s**2 = 0.1383 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.3 page no : 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "kgf = 9.80665;          #N KGF\n",
+      "\n",
+      "# Calculation and Result \n",
+      "cm = 10**-2;            #m\n",
+      "P = kgf/cm**2;\n",
+      "print \"1 kgf/cm**2 = %0.3e N/m**2\"%P\n",
+      "lbf = 32.174;           #lb*ft#s**2\n",
+      "lb = 0.4535924;         #kg\n",
+      "ft = 0.3048;            #m\n",
+      "in_ = 0.0254;            #m\n",
+      "P1 = lbf*lb*ft/in_**2;\n",
+      "print \"1 lbf/in**2 = %.2f N/m**2\"%P1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 kgf/cm**2 = 9.807e+04 N/m**2\n",
+        "1 lbf/in**2 = 6894.75 N/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.4 page no : 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "Q1 = 10000.             #kJ/hr rate of heat transfer\n",
+      "kJ = 1000.              #J \n",
+      "hr = 3600.              #s \n",
+      "\n",
+      "# Calculation \n",
+      "Q = Q1*kJ/hr;           #J/s\n",
+      "print \"Q = %.2f J/s\"%Q\n",
+      "x = 0.1;                #m\n",
+      "A = 1.                  #m**2\n",
+      "T = 800.                #K\n",
+      "k = x*Q/(A*T);\n",
+      "\n",
+      "# Result \n",
+      "print \"thermal conductivity = %.3f W/(m*k)\"%k\n",
+      "J = 1/4.1868            #cal\n",
+      "k1 = k*J*hr/1000\n",
+      "print \"thermal conductivity = %.3F kcal/(h*m*C)\"%k1\n",
+      "\n",
+      "# note : answers may vary because of rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q = 2777.78 J/s\n",
+        "thermal conductivity = 0.347 W/(m*k)\n",
+        "thermal conductivity = 0.299 kcal/(h*m*C)\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.5 page no : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "F = 300.                #N weight of object\n",
+      "a = 9.81;               #m/s**2 gravity\n",
+      "\n",
+      "# Calculation \n",
+      "m = F/a;                #kg\n",
+      "\n",
+      "# Result \n",
+      "print \"mass in kg = %.2f kg\"%m\n",
+      "lb = 4.535924/10;       #kG\n",
+      "m1 = m/lb\n",
+      "print \"mass in pounds = %.2f LB\"%m1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass in kg = 30.58 kg\n",
+        "mass in pounds = 67.42 LB\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.6 page no : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "z = 15.                 #m height\n",
+      "PE = 2000.;             #J potential energy\n",
+      "g = 9.8067;             #m/s**2 \n",
+      "\n",
+      "# Calculation \n",
+      "m = PE/(z*g);\n",
+      "\n",
+      "# Result \n",
+      "print \"mass = %.3f kg\"%m\n",
+      "v = 50;                 #m/s\n",
+      "KE = 1./2*m*(v**2)/1000.;\n",
+      "print \"kinetic energy = %.3f kj\"%KE\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass = 13.596 kg\n",
+        "kinetic energy = 16.995 kj\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.7 page no : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables \n",
+      "g = 9.81;                   #m/s**2 gravity\n",
+      "m = 100 * 0.4536;           #kg     weight  \n",
+      "P = 101325.;                #N/m**2  standard atomosphere\n",
+      "D1 = 4.;                    #inch\n",
+      "\n",
+      "# Calculation \n",
+      "D = D1 * 2.54 * 10**-2;     #m\n",
+      "A = 3.1415 * (D**2)/4;      #m**2\n",
+      "F1 = P * A;                 #N\n",
+      "F2 = m * g;                 #N\n",
+      "F = F1 + F2;\n",
+      "\n",
+      "# Result \n",
+      "print \"Total force acting on the gas = %.2f N\"%F\n",
+      "P1 = F / A;                 #N/m**2\n",
+      "P2 = P1/100000.;            #bar\n",
+      "P3 = P1/(6.894757 * 10**3); #psi\n",
+      "print \"Pressure in N/m**2 = %.3e N/m**2\"%P1\n",
+      "print \"Pressure in bar = %.3f bar\"%P2\n",
+      "print \"Pressure in psi = %.2f psi\"%P3\n",
+      "d = 0.4;                    #m\n",
+      "W = F * d;\n",
+      "print \"Work done = %.2f J\"%W\n",
+      "PE = m * g * d;\n",
+      "print \"Change in potential energy = %.2f J\"%PE\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total force acting on the gas = 1266.43 N\n",
+        "Pressure in N/m**2 = 1.562e+05 N/m**2\n",
+        "Pressure in bar = 1.562 bar\n",
+        "Pressure in psi = 22.66 psi\n",
+        "Work done = 506.57 J\n",
+        "Change in potential energy = 177.99 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.8 page no : 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables \n",
+      "#kG = (6.7 * 10**-4) * (( G * (ds + dt) / ds)**0.8) / ((ds**0.4)*(dG**0.2))\n",
+      "\n",
+      "#kG - lbmol/(h ft**2 atm), G - lb/(ft**2 h), ds, dG, dt - feet\n",
+      "#kG1 - kmol/(m**2 h atm), G1 - kg/(m**2 h), ds1, dG1, dt1 - m\n",
+      "G = 0.2048;             #G1 * lb/(ft**2 h)  velocity\n",
+      "d = 3.2808;             #d1 * ft   clearance between grids\n",
+      "ds = d;                 # clearance between grids\n",
+      "dt = d;                 # clearance between grids\n",
+      "dG = d;                 # clearance between grids\n",
+      "kG = 4.885;             #kG1 (lbmol/(h ft**2 atm) = 4.885 * kmol/(m**2 h atm))\n",
+      "\n",
+      "# Calculation \n",
+      "C = (6.7 * 10**-4) * (( G * d / ds)**0.8) / ((ds**0.4)*(dG**0.2))* kG;\n",
+      "\n",
+      "# Result \n",
+      "print \"Co-efficient = %.2e (kmol)(kg)**-0.8 (m)**0.2 (h)**-0.2 (atm)**-1\"%C\n",
+      "# this is the constant for the equation\n",
+      "# the equation thus becomes,\n",
+      "# kG1 = C * (( G1 * (ds1 + dt1) / ds1)**0.8) / ((ds1**0.4)*(dG1**0.2))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Co-efficient = 4.51e-04 (kmol)(kg)**-0.8 (m)**0.2 (h)**-0.2 (atm)**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.9 page no : 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from sympy import Symbol\n",
+      "\n",
+      "# Variables \n",
+      "#Cp = 7.13 + 0.577 * (10**-3) * t + 0.0248 * (10**-6) * t**2 \n",
+      "\n",
+      "#Cp - Btu/lb-mol F, t - F\n",
+      "#Cp1 - kJ/kmol K , t1 - K\n",
+      "a = 7.13                          # first term\n",
+      "b = 0.577 * 10**-3                # second term\n",
+      "c = 0.0248 * 10**-6               # third term\n",
+      "#t = 1.8 * t1 - 459.67\n",
+      "Cp = 4.1868;            #Cp1 (Btu/lb-mol F = 4.1868 * (kJ/kmol K) )\n",
+      "t = Symbol('T')\n",
+      "#substituting the above, we get,\n",
+      "#Cp1 = 28.763 + 4.763 * (10**-3) * t1 + 0.3366 * (10**-6) * t**2\n",
+      "a1 = 28.763\n",
+      "\n",
+      "# Calculation \n",
+      "b1 = 4.763 * (10**-3)\n",
+      "c1 = 0.3366 * (10**-6)\n",
+      "\n",
+      "Cp = a1 + b1*t + c1*t**2\n",
+      "# Result \n",
+      "print \"a1 = \",a1\n",
+      "print \"b1 = \",b1\n",
+      "print \"c1 = \",c1\n",
+      "# this are the co efficents for the following equation;\n",
+      "# Cp1 = a1 + b1 * t1 + c1 * (t1)**2\n",
+      "print \"Equation Cp = \",Cp"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a1 =  28.763\n",
+        "b1 =  0.004763\n",
+        "c1 =  3.366e-07\n",
+        "Equation Cp =  3.366e-7*T**2 + 0.004763*T + 28.763\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch3.ipynb b/Stoichiometry_And_Process_Calculations/ch3.ipynb
new file mode 100755
index 00000000..5ff86e83
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch3.ipynb
@@ -0,0 +1,927 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Fundamental concepts of Stoichiometry"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 page no : 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "\n",
+      "m = 1000 * 0.4536;              #kg/min pounds\n",
+      "M = 30.24;                      #gm/mol avg. molecular weight\n",
+      "\n",
+      "# Calculation \n",
+      "m1 = m * 60 / M;\n",
+      "\n",
+      "# Result\n",
+      "print \"molar folw rate = \",m1,\"kmol/hr\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "molar folw rate =  900.0 kmol/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 page no : 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "MK = 39.1;         #potassium\n",
+      "MC = 12.0;         # carbon\n",
+      "MO = 16.;          # oxygen\n",
+      "\n",
+      "# Calculation \n",
+      "MK2CO3 = MK * 2 + MC + MO * 3;\n",
+      "m = 691.;\n",
+      "N = m / MK2CO3;\n",
+      "A = 6.023 * 10**23;\n",
+      "molecules = N * A;\n",
+      "\n",
+      "# Result \n",
+      "print \"Total no. of molecules = %.4e molecules\"%molecules\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total no. of molecules = 3.0115e+24 molecules\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 page no : 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "Na = 23.;                   #gm/mol  weight of sodium\n",
+      "MNa = 100.;                 #kg      sodium\n",
+      "\n",
+      "# Calculation \n",
+      "N = MNa * 1000 / Na ;       #g-atoms \n",
+      "NNa2SO4 = N / 2;\n",
+      "\n",
+      "# Result \n",
+      "print \"(a) moles of sodium sulphate = %.3e mol\"%NNa2SO4\n",
+      "MNa2SO4 = 142.06;\n",
+      "m = NNa2SO4 * MNa2SO4/1000;\n",
+      "print \"(b)kilograms of sodium sulphate = %.2f kg\"%m\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) moles of sodium sulphate = 2.174e+03 mol\n",
+        "(b)kilograms of sodium sulphate = 308.83 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 page no : 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "MFe = 55.85;            \n",
+      "MO = 16.;\n",
+      "MS = 32.;\n",
+      "MFeS2 = MFe + MS * 2;        # molecular weight of FeS2\n",
+      "MFe2O3 = MFe * 2 + MO * 3;   # molecular weight of Fe2O3\n",
+      "MSO3 = MS + MO * 3;          # molecular weight of SO3\n",
+      "m1SO3 = 100.;                       #kg\n",
+      "\n",
+      "# Calculation \n",
+      "N1 = m1SO3 / (MSO3);                #kmol\n",
+      "NFeS2 = N1 / 2;\n",
+      "mFeS2 = NFeS2 * MFeS2;\n",
+      "\n",
+      "# Result\n",
+      "print \"mass of pyrites to obtain 100kg of SO3 = %.2f kg\"%mFeS2\n",
+      "m2SO3 = 50.;                        #kg\n",
+      "N2 = m2SO3 / (MSO3);                #kmol\n",
+      "NO2 = N2 * 15/8.;\n",
+      "mO2 = NO2 * MO * 2;\n",
+      "print \"mass of Oxygen consumed to produce 50kg of SO3 = %.2f kg\"%mO2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass of pyrites to obtain 100kg of SO3 = 74.91 kg\n",
+        "mass of Oxygen consumed to produce 50kg of SO3 = 37.50 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 page no : 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "MKClO3 = 122.55                 # weight of potassium\n",
+      "mKClO3 = 100.;                  #kg potassium\n",
+      "\n",
+      "# Calculation \n",
+      "NKClO3 = mKClO3 / MKClO3;\n",
+      "NO2 = 3 * NKClO3 / 2;\n",
+      "V1 = 22.4143;                   #m**3/kmol;\n",
+      "V = V1 * NO2;\n",
+      "\n",
+      "# Result \n",
+      "print \"volume of oxygen produced = %.2f m**3\"%V\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "volume of oxygen produced = 27.43 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 page no : 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "mH2 = 100.;                 #kg  hydrogen\n",
+      "\n",
+      "# Calculation \n",
+      "NH2 = mH2/2.016;\n",
+      "NFe = 3. * NH2 / 4;\n",
+      "mFe = NFe * 55.85;\n",
+      "\n",
+      "# Result \n",
+      "print \"(a)mass of iron required = %.2f kg\"%mFe\n",
+      "NH2O = NH2 \n",
+      "mH2O = NH2O * 18;\n",
+      "print \"mass of steam required = %.1f kg\"%mH2O\n",
+      "V1 = 22.4143;               #m**3/kmol;\n",
+      "V = V1 * NH2;\n",
+      "print \"(b)Volume of hydrogen = %.1f m**3\"%V\n",
+      "\n",
+      "# Answer may vary because of rounding error.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)mass of iron required = 2077.75 kg\n",
+        "mass of steam required = 892.9 kg\n",
+        "(b)Volume of hydrogen = 1111.8 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 page no : 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "MCaCO3 = 100.08;    # molecular weight of CaCO3\n",
+      "\n",
+      "# Calculation \n",
+      "GE = MCaCO3 / 2;\n",
+      "\n",
+      "# Result\n",
+      "print \"Gram equivalent wt. of CaCO3 =\",GE,\"g\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gram equivalent wt. of CaCO3 = 50.04 g\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8 page no : 48\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m1 = 1.;            #kg (mass in air)\n",
+      "m2 = 0.9;           #kg (mass in water)\n",
+      "m3 = 0.82;          #kg (mass in liquid)\n",
+      "\n",
+      "# Calculation \n",
+      "L1 = m2 - m1;       #kg (loss of mass in water)\n",
+      "L2 = m3 - m1;       #kg (loss of mass in liquid)\n",
+      "sp_g = L2 /L1;\n",
+      "\n",
+      "# Result\n",
+      "print \"specific gravity of liquid = \",sp_g\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "specific gravity of liquid =  1.8\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9 page no : 49\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m1 = 10.            #kg  liquid A\n",
+      "m2 = 5.             #kg  liquid B\n",
+      "sp_g1 = 1.17;       # gravity A\n",
+      "sp_g2 = 0.83;       # gravity B\n",
+      "Dwater = 1000.      #kg/m**3  water\n",
+      "\n",
+      "# Calculation \n",
+      "DA = Dwater * sp_g1;\n",
+      "DB = Dwater * sp_g2;\n",
+      "V1 = m1 / DA;\n",
+      "V2 = m2 / DB;\n",
+      "V = V1 + V2;\n",
+      "Dmix = (m1 + m2)/ V ;\n",
+      "sp_g3 = Dmix \n",
+      "\n",
+      "# Result\n",
+      "print \"specific gravity of mixture = %.f kg/m**3\"%sp_g3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "specific gravity of mixture = 1029 kg/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10 page no : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Tw = 100.;              #Tw  baume scale\n",
+      "\n",
+      "# Calculation \n",
+      "sp_g = Tw/200 + 1;\n",
+      "Be = 145 - 145/sp_g;\n",
+      "\n",
+      "# Result\n",
+      "print \"specific gravity on beume scale = %.1f Be\"%Be\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "specific gravity on beume scale = 48.3 Be\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11 page no : 49\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "API1 = 30.                          #API gas oil\n",
+      "sp_g1 = 141.5/(131.5 + API1)        # (since, API = 141.5/sp_g -131.5)    gravity scale\n",
+      "Dwater = 999.                       #kg/m**3;  density of water\n",
+      "\n",
+      "# Calculation \n",
+      "Doil1 = sp_g1 * Dwater;\n",
+      "V1 = 250.;                          #m**3\n",
+      "m1 = V1 * Doil1;\n",
+      "API2 = 15.;                         #API\n",
+      "sp_g2 = 141.5/(131.5 + API2);       # (since, API = 141.5/sp_g -131.5)\n",
+      "Dwater = 999.;                      #kg/m**3;\n",
+      "Doil2 = sp_g2 * Dwater;\n",
+      "V2 = 1000.;                         #m**3\n",
+      "m2 = V2 * Doil2;\n",
+      "Dmix = (m1 + m2)/(V1 + V2);\n",
+      "\n",
+      "# Result\n",
+      "print \"density of the mixture = %.f kg/m**3\"%Dmix\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "density of the mixture = 947 kg/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12 page no : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m1 = 250.               #kg wet ammonium sulphate\n",
+      "mwater1 = 50.           #kg moisture\n",
+      "\n",
+      "# Calculation \n",
+      "mdrysolid1 = m1 - mwater1;\n",
+      "wfe1 = mwater1 / m1;\n",
+      "wr1 = mwater1 / mdrysolid1; \n",
+      "wtpercentw1 = mwater1 * 100 / m1;\n",
+      "wtpercentd1 = mwater1 * 100 / mdrysolid1;\n",
+      "a = 90.                 #%\n",
+      "mwater2 = mwater1 * (1 - a/100);\n",
+      "m2 = mdrysolid1 + mwater2;\n",
+      "wfe2 = mwater2 / m2;\n",
+      "wr2 = mwater2 / mdrysolid1; \n",
+      "wtpercentw2 = mwater2 * 100 / m2;\n",
+      "wtpercentd2 = mwater2 * 100 / mdrysolid1;\n",
+      "\n",
+      "# Result\n",
+      "print \"(a)weight fraction of water at entrance =\",wfe1\n",
+      "print \"weight fraction of water at exit = %.3f\"%wfe2\n",
+      "print \"(b)weight ratio of water at entrance = \",wr1\n",
+      "print \"weight ratio of water at exit = \",wr2\n",
+      "print \"weight percent of moisture on wet basis at entrance = \",wtpercentw1,\"%\"\n",
+      "print \"(c)weight percent of moisture on dry basis at entrance = \",wtpercentd1,\"%\"\n",
+      "print \"weight percent of moisture on wet basis at exit = %.2f %%\"%wtpercentw2\n",
+      "print \"(d)weight percent of moisture on dry basis at exit = \",wtpercentd2,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)weight fraction of water at entrance = 0.2\n",
+        "weight fraction of water at exit = 0.024\n",
+        "(b)weight ratio of water at entrance =  0.25\n",
+        "weight ratio of water at exit =  0.025\n",
+        "weight percent of moisture on wet basis at entrance =  20.0 %\n",
+        "(c)weight percent of moisture on dry basis at entrance =  25.0 %\n",
+        "weight percent of moisture on wet basis at exit = 2.44 %\n",
+        "(d)weight percent of moisture on dry basis at exit =  2.5 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13 page no : 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "mdrysolid = 100.                #kg  dry solids\n",
+      "percentin = 25.                 # water in the feed\n",
+      "\n",
+      "# Calculation \n",
+      "mwaterin = mdrysolid * percentin / 100;\n",
+      "percentout = 2.5;\n",
+      "mwaterout = mdrysolid * percentout / 100;\n",
+      "mremoved = mwaterin - mwaterout;\n",
+      "percentremoved = mremoved *100 / mwaterin ;\n",
+      "\n",
+      "# Result\n",
+      "print \"percentage of water removed = \",percentremoved\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percentage of water removed =  90.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14 page no : 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 1.                      #kg  wet ammonium sulphate\n",
+      "percent1 = 20.              #%   water\n",
+      "\n",
+      "# Calculation \n",
+      "mwaterin = m * percent1 / 100.\n",
+      "mdrysolid = m - mwaterin;\n",
+      "percent2 = 2.44;            #%\n",
+      "mout = mdrysolid / (1 - percent2/100);\n",
+      "mwaterout = mout - mdrysolid;\n",
+      "mremoved = mwaterin - mwaterout;\n",
+      "percentremoved = mremoved * 100 / mwaterin ;\n",
+      "\n",
+      "# Result\n",
+      "print \"weight of water removed = %.2f kg\"%mremoved\n",
+      "print \"percentage of water removed = %.f %%\"%percentremoved\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "weight of water removed = 0.18 kg\n",
+        "percentage of water removed = 90 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15 page no : 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "mwater = 100.                #kg water\n",
+      "mNaCl = 35.8                 #kg solubility\n",
+      "\n",
+      "# Calculation \n",
+      "msolu = mwater + mNaCl;\n",
+      "mfr = mNaCl / msolu;\n",
+      "mpr = mfr * 100;\n",
+      "MNaCl = 58.45;              #kg/kmol\n",
+      "NNaCl = mNaCl / MNaCl;\n",
+      "MH2O = 18.                  #kg/kmol\n",
+      "NH2O = mwater / MH2O;\n",
+      "Mfr = NNaCl / (NNaCl + NH2O);\n",
+      "Mpr = Mfr * 100;\n",
+      "N = NNaCl *1000 / mwater;\n",
+      "\n",
+      "# Result\n",
+      "print \"(a)mass fraction of NaCl = %.4f\"%mfr\n",
+      "print \"mass percent of NaCl= %.2f %%\"%mpr\n",
+      "print \"(b)mole fraction of NaCl = %.4f\"%Mfr\n",
+      "print \"mole percent of NaCl = %.2f %%\"%Mpr\n",
+      "print \"kmol NaCl per 1000 kg of water = %.3f kMol NaCl\"%N\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)mass fraction of NaCl = 0.2636\n",
+        "mass percent of NaCl= 26.36 %\n",
+        "(b)mole fraction of NaCl = 0.0993\n",
+        "mole percent of NaCl = 9.93 %\n",
+        "kmol NaCl per 1000 kg of water = 6.125 kMol NaCl\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16 page no : 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Y = 0.015;                  #kg water vapour/kg dry air\n",
+      "Mair = 29.;                 #kg/kmol weight of air\n",
+      "Mwater = 18.016;            #kg/kmol  \n",
+      "\n",
+      "# Calculation \n",
+      "Nwater = Y / Mwater;        #kmol\n",
+      "Nair = 1 / Mair;            #kmol\n",
+      "Mpr = Nwater *100 / (Nwater + Nair);\n",
+      "Mr = Nwater / Nair;\n",
+      "\n",
+      "# Result\n",
+      "print \"(a)mole percent of water vapour = %.2f %%\"%Mpr\n",
+      "print \"(b) molal absolute humidity = %.4f kmol water/kmol dry air\"%Mr\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)mole percent of water vapour = 2.36 %\n",
+        "(b) molal absolute humidity = 0.0241 kmol water/kmol dry air\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17 page no : 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "msolu = 100.;                       #g basis\n",
+      "MK2CO3 = 138.20;                    #g/mol molecular weight\n",
+      "percent1 = 50.;                     #% mass of K2CO3\n",
+      "\n",
+      "# Calculation \n",
+      "mK2CO3 = percent1 *msolu / 100;\n",
+      "NK2CO3 = mK2CO3 / MK2CO3;\n",
+      "mwater = msolu - mK2CO3;\n",
+      "Nwater = mwater / 18.06;\n",
+      "Mpr = NK2CO3 * 100 / (NK2CO3 + Nwater);\n",
+      "sp_gr  =1.53;\n",
+      "Vsolu = msolu/sp_gr;                #mL\n",
+      "Vwater = mwater / 1;                #mL\n",
+      "Vpr = Vwater * 100/ Vsolu;\n",
+      "Molality = NK2CO3 / (mwater * 10**-3);\n",
+      "Molarity = NK2CO3 / (Vsolu * 10**-3);\n",
+      "Eq_wt = MK2CO3 / 2;\n",
+      "No = mK2CO3/Eq_wt;\n",
+      "N = No / (Vsolu * 10**-3);\n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Mole prcent of salt = %.2f %%\"%Mpr\n",
+      "print \"(b)Volume percent of water = %.2f %%\"%Vpr\n",
+      "print \"(c)Molality = %.3f mol/kg\"%Molality\n",
+      "print \"(d)Molarity = %.3f mol/L\"%Molarity\n",
+      "print \"(e)Normality = %.2f N\"%N\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Mole prcent of salt = 11.56 %\n",
+        "(b)Volume percent of water = 76.50 %\n",
+        "(c)Molality = 7.236 mol/kg\n",
+        "(d)Molarity = 5.535 mol/L\n",
+        "(e)Normality = 11.07 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18 page no : 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "msolu = 100.;               #kg volume\n",
+      "percent1 = 60.;             #%  alcohol\n",
+      "Dwater = 998.;              #kg/m**3  water\n",
+      "Dalco = 798.;               #kg/m**3  alcohol\n",
+      "Dsolu = 895.;               #kg/m**3  solution\n",
+      "\n",
+      "# Calculation \n",
+      "Vsolu = msolu/Dsolu;\n",
+      "malco = msolu * percent1 / 100;\n",
+      "Valco = malco / Dalco;\n",
+      "Vpr = Valco * 100 / Vsolu;\n",
+      "Malco = 46.048;             #kg/kmol\n",
+      "N = malco/Malco;\n",
+      "Molarity = N/(Vsolu );\n",
+      "mwater = msolu - malco;\n",
+      "Molality = N * 1000 /mwater;\n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Volume percent of ethanol in solution = %.1f %%\"%Vpr\n",
+      "print \"(b)Molarity = %.2f mol/L\"%Molarity\n",
+      "print \"(c)Molality = %.3f mol/(kg of water)\"% Molality\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Volume percent of ethanol in solution = 67.3 %\n",
+        "(b)Molarity = 11.66 mol/L\n",
+        "(c)Molality = 32.575 mol/(kg of water)\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19 page no : 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#CO + CL2 = COCl2\n",
+      "import math \n",
+      "# Variables  \n",
+      "Np = 12.;                   #moles phosgene\n",
+      "NCl2 = 3.;                  #moles chlorine\n",
+      "NCO = 8.;                   #moles carbon monoxide\n",
+      "\n",
+      "# Calculation \n",
+      "N1Cl2 = NCl2 + Np;\n",
+      "N1CO = NCO + Np;\n",
+      "pr_ex = (N1CO - N1Cl2)* 100/N1Cl2;\n",
+      "pr_co = (N1Cl2-NCl2) * 100/ N1Cl2;\n",
+      "T = Np + NCl2 + NCO;\n",
+      "T1 = N1Cl2 + N1CO;\n",
+      "N = T / T1;\n",
+      "\n",
+      "# Result\n",
+      "print \"(a)percent excess of CO = %.2f %%\"%pr_ex\n",
+      "print \"(b)percent conversion = %.2f %%\"%pr_co\n",
+      "print \"(c)Moles of total products per mole of total reactants = %.3f\"%N\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)percent excess of CO = 33.33 %\n",
+        "(b)percent conversion = 80.00 %\n",
+        "(c)Moles of total products per mole of total reactants = 0.657\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.21 page no : 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables \n",
+      "\n",
+      "msolu = 100.;               #g  mold of feed mixture\n",
+      "MK2CO3 = 138.20;            #g/mol \n",
+      "ethylene = 53.89            # %\n",
+      "ethanol = 14.37             # %\n",
+      "ether = 1.8                 # %\n",
+      "water = 23.35               # %\n",
+      "\n",
+      "# Calculation \n",
+      "Ethylene = ethylene * 83.57 / 100\n",
+      "Ethanol = ethanol*83.57 / 100\n",
+      "Ether = ether*83.57 / 100\n",
+      "Water = water*83.57 / 100\n",
+      "Inerts = 3                 # mol\n",
+      "\n",
+      "conversion_ethylene = (60 - Ethylene)/60 * 100\n",
+      "yield_ethanol =  Ethanol/(60 - Ethylene)*100\n",
+      "yeild_ether = Ether*2/(60-Ethylene) * 100\n",
+      "\n",
+      "# Result \n",
+      "print \"Conversation of ethylene = %.2f %%\"%conversion_ethylene\n",
+      "print \"Yield of ethanol = %.2f %%\"%yield_ethanol\n",
+      "print \"Yield of Ether = %d %%\"%yeild_ether\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Conversation of ethylene = 24.94 %\n",
+        "Yield of ethanol = 80.25 %\n",
+        "Yield of Ether = 20 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch4.ipynb b/Stoichiometry_And_Process_Calculations/ch4.ipynb
new file mode 100755
index 00000000..a8fae4f9
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch4.ipynb
@@ -0,0 +1,796 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Ideal Gases and Gas Mixtures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 page no : 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "P1 = 760.               #mmHg standard conditions\n",
+      "T1 = 273.15;            #K\n",
+      "\n",
+      "# Calculation and  Result \n",
+      "V1 = 22.4143 * 10**-3;  #m**3/mol\n",
+      "R1 = P1 * V1 / T1;\n",
+      "print \"Gas constant R = %.4e m**3 mmHg / (molK)\"%R1\n",
+      "P2 = 101325;            #N/m**2\n",
+      "T2 = 273.15;            #K\n",
+      "V2 = 22.4143 * 10**-3;  #m**3/mol\n",
+      "R2 = P2 * V2 / T2;      #J/molK\n",
+      "R3 = R2 / 4.184;        #cal/molK\n",
+      "print \"Gas constant R in MKS system = %.3f cal/molK\"%R3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Gas constant R = 6.2365e-02 m**3 mmHg / (molK)\n",
+        "Gas constant R in MKS system = 1.987 cal/molK\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 page no : 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "T = 350.                #K\n",
+      "P = 1;                  #bar\n",
+      "\n",
+      "# Calculation \n",
+      "V1 = 22.4143 * 10**-3;  #m**3 (suffix 1 represents at STD)\n",
+      "P1 = 1.01325;           #bar\n",
+      "T1 = 273.15;            #K\n",
+      "V = P1 * V1 * T/(T1 * P);\n",
+      "\n",
+      "# Result \n",
+      "print \"Molar volume = %.2e m**3/mol\"%V\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Molar volume = 2.91e-02 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 page no : 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "P = 10.                     #bar oxygen cylinder gas\n",
+      "T = 300.;                   #K\n",
+      "V = 150.;                   #L\n",
+      "P1 = 1.01325;               #bar ( \\suffix 1 represents at STD)\n",
+      "T1 = 273.15;                #K\n",
+      "\n",
+      "# Calculation \n",
+      "V2 = T1 * P * V /(T * P1);  #m**3\n",
+      "V1 = 22.4143;               #m**3/mol\n",
+      "N = V2 / V1;                #mol\n",
+      "MO2 = 32.;\n",
+      "m = N * MO2/1000;\n",
+      "\n",
+      "# Result \n",
+      "print \"Mass of oxygen in the cylinder = %.4f kg\"%m\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass of oxygen in the cylinder = 1.9243 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 page no: 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 195.                #kPa pressure\n",
+      "T = 273.                #K automobile tyre\n",
+      "P1 = 250                #kPa pressure\n",
+      "\n",
+      "# Calculation \n",
+      "T1 = P1 * T / P;\n",
+      "\n",
+      "# Result\n",
+      "print \"Maximum temperature to which tyre may be heated = \",T1,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum temperature to which tyre may be heated =  350.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 page no : 81\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# variables \n",
+      "V = 250.                    #L  carbon dioxide\n",
+      "T = 300.                    #K  temperature\n",
+      "V1 = 1000.                  #L  \n",
+      "P1 = 100.                   #kPa cylinder\n",
+      "T1 = 310.                   #K temperature\n",
+      "\n",
+      "# Calculation \n",
+      "P = T * P1 * V1 /(T1 * V);\n",
+      "\n",
+      "# Result\n",
+      "print \"Original pressure in the cylinder = %.1f kPa\"%P\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Original pressure in the cylinder = 387.1 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 pageno : 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Vper1 = 70.                 #% ( 1 = HCl) rubber showed\n",
+      "Vper2 = 20.                 #% ( 2 = Cl2) volume\n",
+      "Vper3 = 10.                 #% ( 3 = CCl4)\n",
+      "M1 = 36.45                  # molecular weights\n",
+      "M2 = 70.90\n",
+      "M3 = 153.8\n",
+      "\n",
+      "# Calculation \n",
+      "m1 = Vper1 * M1\n",
+      "m2 = Vper2 * M2\n",
+      "m3 = Vper3 * M3\n",
+      "mper1 = m1 * 100/(m1+ m2 + m3);\n",
+      "mper2 = m2 * 100/(m1+ m2 + m3);\n",
+      "mper3 = m3 * 100/(m1+ m2 + m3);\n",
+      "\n",
+      "# Result\n",
+      "print \" (a) weight percent of HCl= %.2f%%\"%mper1\n",
+      "print \"weight percent of Cl2 = %.2f %%\"%mper2\n",
+      "print \"weight percent of CCl4 = %.2f %%\"%mper3\n",
+      "\n",
+      "m = (m1 + m2 + m3)/(Vper1 + Vper2 + Vper3);\n",
+      "print \"(b)average molecular weight = %.3f kg\"%m\n",
+      "v = 22.4143;                #m**3/kmol\n",
+      "Vtotal = v * (Vper1 + Vper2 + Vper3);\n",
+      "D = (m1 + m2 + m3)/Vtotal;\n",
+      "print \"(c)Density at standard condiions = %.4f kg/m**3\"%D\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (a) weight percent of HCl= 46.33%\n",
+        "weight percent of Cl2 = 25.75 %\n",
+        "weight percent of CCl4 = 27.93 %\n",
+        "(b)average molecular weight = 55.075 kg\n",
+        "(c)Density at standard condiions = 2.4571 kg/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7 page no : 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "per1 = 93.                      #% ( 1 = methane)\n",
+      "per2 = 4.5;                     #% (2 = ethane)\n",
+      "per3 = 100 - (per1 + per2);     #% ( 3 = N2);\n",
+      "T = 300.                        #K  natural gas \n",
+      "p = 400.                        #kPa\n",
+      "P3 = p * per3 / 100;\n",
+      "v = 10.                         #m**3\n",
+      "V2 = per2 * v / 100;\n",
+      "M1 = 16.032;\n",
+      "M2 = 30.048;\n",
+      "M3 = 28;\n",
+      "N1 = per1;\n",
+      "N2 = per2;\n",
+      "N3 = per3;\n",
+      "\n",
+      "# Calculation \n",
+      "m1 = M1 * N1;\n",
+      "m2 = M2 * N2;\n",
+      "m3 = M3 * N3;\n",
+      "m = m1 + m2 + m3;\n",
+      "Vstp = 100 * 22.4143 * 10**-3;  #m3 at STP\n",
+      "D = m /(1000 * Vstp);\n",
+      "Pstp = 101.325;                 #kPa\n",
+      "T1 = 273.15;                    #K\n",
+      "V = T * Pstp * Vstp / ( T1 * p);\n",
+      "D1 = m /(1000 * V);\n",
+      "Mavg = m /100;\n",
+      "mper1 = m1 * 100 / (m1 + m2 + m3);\n",
+      "mper2 = m2 * 100 / (m1 + m2 + m3);\n",
+      "mper3 = m3 * 100 / (m1 + m2 + m3);\n",
+      "\n",
+      "# Result\n",
+      "print \"(a) Partial pressure of nitrogen = \",P3,\"kPa\"\n",
+      "print \"(b) pure-component volume of ethane = \",V2,\"m**3\"\n",
+      "print \"(c) Density at standard conditions = %.4f kg/m**3\"%D\n",
+      "print \"(d) Density at given condition = %.4f kg/m**3\"%D1\n",
+      "print \"(e) Average molecular weight = %.2f\"%Mavg\n",
+      "print \"(f) weight percent of Methane = %.2f %%\"%mper1\n",
+      "print \"weight percent of Ethane = %.2f %%\"%mper2\n",
+      "print \"weight percent of Nitrogen = %.2f %%\"%mper3\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Partial pressure of nitrogen =  10.0 kPa\n",
+        "(b) pure-component volume of ethane =  0.45 m**3\n",
+        "(c) Density at standard conditions = 0.7567 kg/m**3\n",
+        "(d) Density at given condition = 2.7200 kg/m**3\n",
+        "(e) Average molecular weight = 16.96\n",
+        "(f) weight percent of Methane = 87.90 %\n",
+        "weight percent of Ethane = 7.97 %\n",
+        "weight percent of Nitrogen = 4.13 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8 pageno : 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "per1 = 20.                          #% ( 1 = ammonia) \n",
+      "Vstp = 22.4143;                     #m**3/kmol\n",
+      "Pstp = 101.325;                     #kPa\n",
+      "Tstp = 273.15;                      #K\n",
+      "V1 = 100.                           #m**3 ammonia air\n",
+      "P1 = 120.                           #kPa  absorption column\n",
+      "T1 = 300.                           #K\n",
+      "P2 = 100.                           #kPa  gas leaves\n",
+      "T2 = 280.                           #K    gas leaves\n",
+      "per2 = 90.                          #% (absorbed)\n",
+      "\n",
+      "# Calculation \n",
+      "N = V1 * P1 * Tstp / (Vstp * Pstp * T1);            #kmol\n",
+      "Nair = (1 - per1 / 100) * N;\n",
+      "N1 = per1 * N/100;\n",
+      "Nabs = per2 * N1 / 100;\n",
+      "N2 = N1 - Nabs;                     #leaving\n",
+      "Ntotal = Nair + N2;\n",
+      "Vstp1 = Ntotal * Vstp;              #m**3\n",
+      "V2 = Vstp1 * Pstp * T2 / (Tstp * P2);\n",
+      "\n",
+      "# Result\n",
+      "print \"Volume of gas leaving = \",V2, \"m**3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume of gas leaving =  91.84 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 pageno : 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "V = 100.                    #m**3  air\n",
+      "Ptotal = 100.               #kPa   air\n",
+      "Pwater = 4.                 #kPa   pressure\n",
+      "Pair = Ptotal - Pwater; \n",
+      "T = 300.                    #K   air\n",
+      "T1 = 275.                   #K   air cooled\n",
+      "Vstp = 22.4143;             #m**3/kmol\n",
+      "Tstp = 273.15;              #K\n",
+      "Pstp = 101.325;             #kPa\n",
+      "Pwater1 = 1.8;              #kPa\n",
+      "\n",
+      "# Calculation \n",
+      "Pair1 = Ptotal - Pwater1;\n",
+      "V1 = V * Pair * T1 / ( T * Pair1);\n",
+      "Nwater = V * Pwater * Tstp/ (Vstp * Pstp * T);\n",
+      "Nwater1 = V1 * Pwater1 * Tstp/ (Vstp * Pstp * T1);\n",
+      "m = (Nwater - Nwater1) * 18.02;\n",
+      "\n",
+      "# Result\n",
+      "print \"(a) volume of air after dehumidification = %.2f m**3\"%V1\n",
+      "print \"(b) Mass of water vapour removed = %.3f kg\"%m\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) volume of air after dehumidification = 89.61 m**3\n",
+        "(b) Mass of water vapour removed = 1.618 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10 pageno : 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "V = 100.                    #m**3 rate \n",
+      "P = 600.                    #kPa  gas\n",
+      "T = 310.                    #K   gas\n",
+      "per1 = 20.                  #% ( H2S entering )\n",
+      "per2 = 2.                   #% ( H2S leaving )\n",
+      "Pstp = 101.325;             #kPa\n",
+      "Tstp = 273.15;              #K\n",
+      "Vstp = 22.414;              #m**3/kmol\n",
+      "\n",
+      "# Calculation \n",
+      "Vstp1 = V * P * Tstp / ( T * Pstp)\n",
+      "N = Vstp1 / Vstp;\n",
+      "N1 = N * per1 / 100;\n",
+      "N2 = N - N1;                # ( 2 = inerts)\n",
+      "Nleaving = N2 / ( 1 - per2 / 100);\n",
+      "N1leaving = per2 * Nleaving / 100;\n",
+      "mabsorbed = (N1 - N1leaving) * 34.08;       #( molecular wt. = 34.08)\n",
+      "mgiven = 100.               #kg/h\n",
+      "Vactual = mgiven * V / mabsorbed;\n",
+      "Nactual = Nleaving * Vactual / V;           # actual moles leaving\n",
+      "Vstpl = Nactual * Vstp;     # volume leaving at STP\n",
+      "P2 = 500.                   #kPa\n",
+      "T2 = 290.                   #K\n",
+      "V2 = Vstpl * Pstp * T2 / ( P2 * Tstp);\n",
+      "Precovery = (N1 - N1leaving)*100 / N1;\n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Volume of gas entering per hour %.2f m**3/h\"%Vactual\n",
+      "print \"(b)Volume of gas leaving per hour %.2f m**3/h\"%V2\n",
+      "print \"(c)Percentage recovery of H2S %.2f %%\"%Precovery\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Volume of gas entering per hour 68.63 m**3/h\n",
+        "(b)Volume of gas leaving per hour 62.89 m**3/h\n",
+        "(c)Percentage recovery of H2S 91.84 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11 pageno : 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#N2 + 3H2 = 2NH3\n",
+      "V1 = 100.                       #m**3 ( 1 = N2) cubic metres\n",
+      "V2 = V1 * 3                     # ( According to Avagadros principle, equal\n",
+      "                                #volumes of all gases under similar condition\n",
+      "                                # contains same no. of moles)\n",
+      "                                \n",
+      "print \"(a)Volume of hydrogen required at same condition = \",V2,\"m**3\"\n",
+      "P1 = 20.                        #bar  \n",
+      "T1 = 350.                       #K nitrogen\n",
+      "P2 = 5.                         #bar\n",
+      "T2 = 290.                       #K hydrogen\n",
+      "\n",
+      "# Calculation and  Result\n",
+      "V3 = 3 * V1 * P1 * T2 / ( P2 * T1)\n",
+      "print \"(b)Volume required at 50 bar and 290K = %.3f m**3\"%V3\n",
+      "m = 1000                        #kg ( ammonia )\n",
+      "N = m / 17.03                   #kmol\n",
+      "N1 = N/2.                       # ( nitrogen)\n",
+      "N2 = N * 3 / 2.                 #(hydrogen)\n",
+      "P3 = 50.                        #bar\n",
+      "T3 = 600.                       #K\n",
+      "Pstp = 1.01325                  #bar\n",
+      "Tstp = 273.15                   #K\n",
+      "Vstp = 22.414                   #m**3/kmol\n",
+      "V1stp = N1 * Vstp\n",
+      "V4 = V1stp * Pstp * T3 / (P3 * Tstp)        # ( nitrogen at 50 bar and 600K)\n",
+      "V5 = V4 * 2                                 # ( ammonia at 50 bar and 600K)\n",
+      "V6 = V4 * 3                                 # ( hydrogen at 50 bar and 600K)\n",
+      "print \"(c)Volume of nitrogen at 50 bar and 600K = %.1f m**3\"%V4\n",
+      "print \"   Volume of hydrogen at 50 bar and 600K = %.1f m**3\"%V6\n",
+      "print \"   Volume of ammonia at 50 bar and 600K = %.1f m**3\"%V5\n",
+      "\n",
+      "\n",
+      "# note : answers may vary because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Volume of hydrogen required at same condition =  300.0 m**3\n",
+        "(b)Volume required at 50 bar and 290K = 994.286 m**3\n",
+        "(c)Volume of nitrogen at 50 bar and 600K = 29.3 m**3\n",
+        "   Volume of hydrogen at 50 bar and 600K = 87.9 m**3\n",
+        "   Volume of ammonia at 50 bar and 600K = 58.6 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12 pageno : 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "N = 100.                    #kmol producer gas\n",
+      "P1 = 25.                    #% ( Carbon monoxide )\n",
+      "P2 = 4.                     #% ( Carbon Dioxide )\n",
+      "P3 = 3.                     #% ( Oxygen )\n",
+      "P4 = 68.                    #% ( Nitrogen )\n",
+      "\n",
+      "# Calculation \n",
+      "N1 = N * P1/100 \n",
+      "N2 = N * P2/100 \n",
+      "N3 = N * P3/100 \n",
+      "N4 = N * P4/100 \n",
+      "NC = N1 + N2 \n",
+      "m = NC * 12 \n",
+      "Ngas = N / m                #moles of gas for 1 kg of Carbon\n",
+      "Vstp = 22.4143              #m**3/kmol\n",
+      "Vstp1 = Vstp * Ngas \n",
+      "P = 1.                      #bar\n",
+      "T = 290.                    #k\n",
+      "Pstp = 1.01325              #bar\n",
+      "Tstp = 273.15               #K\n",
+      "V = T * Vstp1 * Pstp / (Tstp * P ) \n",
+      "print \"(a)Volume of gas at 1 bar and 290 K per kg Carbon = %.2f m**3\"%V\n",
+      "\n",
+      "#CO + 1/2 * O2 = CO2\n",
+      "Nrequired  = N1/2 - N3      #(oxygen required)\n",
+      "Nsupplied = Nrequired * 1.2 \n",
+      "PO1 = 21.                   #% ( Oxygen percent in air)\n",
+      "Nair = Nsupplied * 100/PO1 \n",
+      "V1 = 100.                   #m**3 \n",
+      "Vair = V1 * Nair / N \n",
+      "print \"(b)Volume of air required = %.2f m**3\"%Vair\n",
+      "NCO2 = N2 + N1 \n",
+      "NO2 = Nsupplied - Nrequired \n",
+      "NN2 = N4 + (Vair * (1 - PO1/ 100)) \n",
+      "Ntotal = NCO2 + NO2 + NN2 \n",
+      "PCO2 = NCO2 * 100 / Ntotal \n",
+      "PO2 = NO2 * 100 / Ntotal \n",
+      "PN2 = NN2 * 100 / Ntotal \n",
+      "\n",
+      "# Result\n",
+      "print \"Percent composition of Carbon Dioxide = %.2f %%\"%PCO2\n",
+      "print \"Percent composition of Oxygen = %.2f %%\"%PO2\n",
+      "print \"Percent composition of Nitrogen = %.2f %%\"%PN2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Volume of gas at 1 bar and 290 K per kg Carbon = 6.93 m**3\n",
+        "(b)Volume of air required = 54.29 m**3\n",
+        "Percent composition of Carbon Dioxide = 20.45 %\n",
+        "Percent composition of Oxygen = 1.34 %\n",
+        "Percent composition of Nitrogen = 78.21 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13 pageno : 94"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#4HCl + O2 = 2Cl2 + 2H2O\n",
+      "n = 1.                      #mol ( Basis 1 mol of HCl )\n",
+      "NO2 = n / 4 \n",
+      "NO2supp = 1.5 * NO2 \n",
+      "Nair = NO2supp * 100 / 21 \n",
+      "V = 100.                    #m**3\n",
+      "Vair = V * Nair / n \n",
+      "print \"(a)Volume of air admitted = %.1f m**3\"%Vair\n",
+      "\n",
+      "# Calculation \n",
+      "P1 = 80.                    #% ( HCl converted)\n",
+      "Ncon = n * P1 /100 \n",
+      "N2 = Ncon/4                 # oxygen required\n",
+      "NH2O = Ncon / 2 \n",
+      "NCl2 = Ncon / 2 \n",
+      "nHCl = n - Ncon \n",
+      "nO2 = NO2supp - N2 \n",
+      "Nnitro = Nair - NO2supp \n",
+      "Ntotal = nHCl + nO2 + NH2O + NCl2 + Nnitro \n",
+      "V1 = V * Ntotal \n",
+      "P1 = 1.                     #bar\n",
+      "T1 = 290.                   #K\n",
+      "P2 = 1.2                    #bar\n",
+      "T2 = 400.                   #K\n",
+      "V2 = V1 * P1 * T2 / ( P2 * T1) \n",
+      "print \"(b)Volume of gas leaving = %.2f m**3\"%V2\n",
+      "VCl2 = NCl2 * V \n",
+      "Pstp = 1.01325              #bar\n",
+      "Tstp = 273.                 #K\n",
+      "Vstp = 22.4143              #m**3/kmol\n",
+      "Vstp1 = Tstp * P1 * VCl2 / (T1 * Pstp) \n",
+      "Nstp = Vstp1/Vstp \n",
+      "m = Nstp * 70.90 \n",
+      "print \"(c)Kilograms of Chlorine produced = %.1f kg\"%m\n",
+      "Ntotaldry = nHCl + nO2 + NCl2 + Nnitro          #dry basis\n",
+      "p1 = nHCl*100/Ntotaldry \n",
+      "p2 = nO2*100/Ntotaldry \n",
+      "p3 = NCl2*100/Ntotaldry \n",
+      "p4 = Nnitro*100/Ntotaldry \n",
+      "\n",
+      "# Result\n",
+      "print \"(d)Percent composition of HCl in exit stream = %.2f %%\"%p1\n",
+      "print \"   Percent composition of Oxygen in exit stream = %.1f %%\"%p2\n",
+      "print \"   Percent composition of Chlorine in exit stream = %.2f %%\"%p3\n",
+      "print \"   Percent composition of nitrogen in exit stream = %.2f %%\"%p4\n",
+      "\n",
+      "# note : answers may vary because of rouding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Volume of air admitted = 178.6 m**3\n",
+        "(b)Volume of gas leaving = 297.21 m**3\n",
+        "(c)Kilograms of Chlorine produced = 117.6 kg\n",
+        "(d)Percent composition of HCl in exit stream = 9.15 %\n",
+        "   Percent composition of Oxygen in exit stream = 8.0 %\n",
+        "   Percent composition of Chlorine in exit stream = 18.30 %\n",
+        "   Percent composition of nitrogen in exit stream = 64.54 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14 pageno: 95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "# CO2 = CO + 1/2 * O2\n",
+      "P1 = 1.                 #bar \n",
+      "T1 = 3500.              #K pressure\n",
+      "P2 = 1.                 #bar \n",
+      "T2 = 300.               #K heated\n",
+      "V2 = 25.                #L CO2\n",
+      "\n",
+      "# Calculation \n",
+      "V1 = V2 * P2 * T1 / ( P1 * T2 ) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Final volume of gas if no dissociation occured = %.3f m**3\"%(V1/1000)\n",
+      "Pstp = 1.01325          #bar\n",
+      "Tstp = 273.             #K\n",
+      "Vstp = 22.4143          #m**3\n",
+      "N2 = V2 * P2 * Tstp / ( Vstp * Pstp * T2) \n",
+      "\n",
+      "# let x be the fraction dissociated, then after dissociation,\n",
+      "# CO2 = (1 - x)mol, CO = xmol, O2 = (0.5*x)mol\n",
+      "#total moles = 1 - x + x + o.5 * x = 1 + 0.5 * x \n",
+      "V = 350.                #L\n",
+      "N1 = V * P1 * Tstp / (Vstp * Pstp * T1) \n",
+      "\n",
+      "# 1 + 0.5 * x = N1, therefore\n",
+      "x = (N1 - 1) / 0.5  \n",
+      "p = x*100 \n",
+      "print \"(b)CO2 converted = %.f %%\"%p\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Final volume of gas if no dissociation occured = 0.292 m**3\n",
+        "(b)CO2 converted = 40 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch5.ipynb b/Stoichiometry_And_Process_Calculations/ch5.ipynb
new file mode 100755
index 00000000..8cc243fb
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch5.ipynb
@@ -0,0 +1,497 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Properties of Real Gases"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 page no: 109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "V = 0.6                         #m**3 vessel\n",
+      "T = 473.                        #K temperature\n",
+      "N = 1. * 10 ** 3                 #mol\n",
+      "R = 8.314                       #Pa * m**3/molK\n",
+      "\n",
+      "# Calculation \n",
+      "P = N * R * T / (V * 10**5) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Pressure calculated using ideal gas equation = %.2f bar\"%P\n",
+      "a = 0.4233                      #N * m**4 / mol**2\n",
+      "b = 3.73 * 10**-5               #m**3/mol\n",
+      "P1 = (R*T/(V/N - b)-a/(V/N)**2)/10**5 \n",
+      "print \"(a)Pressure calculated using van der waals equation = %.2f bar\"%P1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Pressure calculated using ideal gas equation = 65.54 bar\n",
+        "(a)Pressure calculated using van der waals equation = 58.13 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 pageno : 110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 10.**7                              #Pa \n",
+      "T = 500.                                #K molar volume\n",
+      "R = 8.314                               #Pa * L / mol K\n",
+      "N = 1000.\n",
+      "\n",
+      "# Calculation \n",
+      "V = N * R * T / ( P * 1000) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Volume of CO2 calculated using ideal gas equation = %.4e m**3\"%V\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Volume of CO2 calculated using ideal gas equation = 4.1570e-04 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 pageno : 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "V = 0.6 * 10**-3                        #m**3\n",
+      "T = 473.                                #K vellel of volume\n",
+      "Tc = 405.5                              #K temperature\n",
+      "Pc = 112.8 * 10 ** 5                    #Pa pressure\n",
+      "R = 8.314 \n",
+      "\n",
+      "# Calculation \n",
+      "a = 0.4278 * (R**2) * (Tc ** 2.5)/Pc \n",
+      "b = 0.0867 * R * Tc / Pc \n",
+      "P1 = (R*T/(V - b) - a/((T**0.5)*V*(V + b)))/10**5 \n",
+      "\n",
+      "# Result\n",
+      "print \"Pressure developed by gas = %.2f bar\"%P1\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure developed by gas = 57.87 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 pageno : 112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 10.**6                              #Pa\n",
+      "T = 373.                                #K molar volume\n",
+      "Tc = 405.5                              #K temperature\n",
+      "Pc = 112.8 * 10 ** 5                    #Pa pressure\n",
+      "R = 8.314 \n",
+      "\n",
+      "# Calculation \n",
+      "a = 0.4278 * (R**2) * (Tc ** 2.5)/Pc \n",
+      "b = 0.0867 * R * Tc / Pc \n",
+      "#P1 = (R*T/(V - b) - a/((T**0.5)*V*(V + b)))/10**5 \n",
+      "#10**6=((8.314*373)/(V-2.59*10**-5))-8.68/((373**0.5)*V*(V+2.59*10**-5) \n",
+      "#solving this we get,\n",
+      "V = 3.0 * 10**-3                        #m**3/mol\n",
+      "\n",
+      "# Result\n",
+      "print \"molar volume of gas = %.e m**3/mol\"%V\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "molar volume of gas = 3e-03 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 pageno : 112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "B = -2.19 * 10**-4                  #m**3/mol virial coefficients\n",
+      "C = -1.73 * 10**-8                  #m**6/mol**2\n",
+      "P = 10.                             #bar molar volume\n",
+      "T = 500.                            #K\n",
+      "\n",
+      "# Calculation \n",
+      "#virial equation is given as, Z = PV/RT = 1 + B/V + C/V**2\n",
+      "#V = (RT/P)*(1 + B/V + C/V**2)\n",
+      "# now by assuming different values for V on RHS and checking for corresponding V on LHS, we have to assume such value of V on RHS by which we get the same value for LHS V\n",
+      "#by trial and error we get,\n",
+      "V = 3.92 * 10**-3                   #m**3\n",
+      "\n",
+      "# Result\n",
+      "print \"Molar volume of methanol = %.3e m**3\"%V\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Molar volume of methanol = 3.920e-03 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 pageno : 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "T = 510.                        #K molar volume\n",
+      "P = 26.6                        #bar \n",
+      "Tc = 425.2                      #K temperature\n",
+      "Pc = 38.                        #bar pressure\n",
+      "Zc = 0.274                      # compressibility factor\n",
+      "R = 8.314 \n",
+      "\n",
+      "# Calculation \n",
+      "Pr = P / Pc \n",
+      "Tr = T / Tc \n",
+      "\n",
+      "# Result\n",
+      "print \"Pr = \",Pr\n",
+      "print \"Tr = %.2f\"%Tr\n",
+      "\n",
+      "#From fig. 5.4 and 5.5 from the text book\n",
+      "Z = 0.865 \n",
+      "D = 0.15 \n",
+      "Z1 = Z + D * ( Zc - 0.27) \n",
+      "V = R * T * Z1 / (P * 10**5) \n",
+      "print \"Molar volume of n-butane = %.4e m**3/mol\"%V\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pr =  0.7\n",
+        "Tr = 1.20\n",
+        "Molar volume of n-butane = 1.3798e-03 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7 pageno : 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "T = 510.                    #K molar volumes\n",
+      "P = 26.6                    #bar\n",
+      "Tc = 425.2                  #K temperature\n",
+      "Pc = 38.                    #bar pressure\n",
+      "w = 0.193                   # acentric factor\n",
+      "R = 8.314 \n",
+      "\n",
+      "# Calculation \n",
+      "Pr = P / Pc \n",
+      "Tr = T / Tc \n",
+      "\n",
+      "# Result\n",
+      "print \"Pr = \",Pr\n",
+      "print \"Tr = %.2f\"%Tr\n",
+      "\n",
+      "#From fig. 5.6 and 5.7 from the text book\n",
+      "Z0 = 0.855 \n",
+      "Z1 = 0.042 \n",
+      "Z = Z0 + w*Z1 \n",
+      "print \"Z = %.4f\"%Z\n",
+      "V = R * T * Z / (P * 10**5) \n",
+      "\n",
+      "print \"Molar volume of n-butane = %.4e m**3/mol\"%V\n",
+      "\n",
+      "# note : answer may vary because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pr =  0.7\n",
+        "Tr = 1.20\n",
+        "Z = 0.8631\n",
+        "Molar volume of n-butane = 1.3758e-03 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8 pageno : 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import *\n",
+      "# variables \n",
+      "P = 6000.                               #kPa ethane\n",
+      "T = 325.                                #K ethane\n",
+      "xn2 = 0.4                               # gas mixture\n",
+      "xethane = 0.6 \n",
+      "an2 = 0.1365                            #N m**4 / mol**2 nitrogen\n",
+      "\n",
+      "# Calculation \n",
+      "bn2 = 3.86 * 10**-5                     #m**3/mol\n",
+      "aethane = 0.557                         #N m**4 / mol**2\n",
+      "bethane = 6.51 * 10**-5                 #m**3/mol\n",
+      "Pcn2 = 3394.                            #kPa\n",
+      "Tcn2 = 126.2                            #K\n",
+      "Pcethane = 4880.                        #kPa\n",
+      "Tcethane = 305.4                        #K\n",
+      "R = 8.314 \n",
+      "\n",
+      "\n",
+      "# Result\n",
+      "V = R * T / (P*1000) \n",
+      "print \"(a)Molar volume by ideal gas equation = %.3e m**3/mol\"%V\n",
+      "\n",
+      "\n",
+      "a = (xn2 * (an2**0.5) + xethane * (aethane**0.5))**2 \n",
+      "b = (xn2*bn2 + xethane*bethane) \n",
+      "#substituting the above values in van der waals equation, and solving, we get\n",
+      "#V^3 - 5.0484*10^-4V^2+5.9117*10^-8V-3.2214*10^-12=0\n",
+      "#Using co-efficients from the above equation,\n",
+      "coeff = [1, -5.0484e-4, 5.9117e-8, -3.2217e-12]\n",
+      "x = roots(coeff)\n",
+      "y = x[0]\n",
+      "print \"(b)Molar volume by van der waals equation =\",round((y.real)/1e-4,3),\"*10^-4 m**3/mol\"\n",
+      "\n",
+      "\n",
+      "Prin2 = P/Pcn2 \n",
+      "Trin2 = T/Tcn2 \n",
+      "Priethane = P/Pcethane \n",
+      "Triethane = T/Tcethane \n",
+      "# using compressibilty chart, \n",
+      "Zn2 = 1 \n",
+      "Zethane = 0.42 \n",
+      "Z = xn2 * Zn2 + xethane * Zethane \n",
+      "V2 = Z * R * T / P * (10**-3)\n",
+      "print \"(c)Molar volume based on compressibilty factor = %.2e m**3/mol\"%V2\n",
+      "\n",
+      "\n",
+      "Pri1n2 = xn2*P/Pcn2 \n",
+      "Tri1n2 = T/Tcn2 \n",
+      "Pri1ethane = xethane*P/Pcethane \n",
+      "Tri1ethane = T/Tcethane \n",
+      "# using compressibilty chart, \n",
+      "Zn21 = 1. \n",
+      "Zethane1 = 0.76 \n",
+      "Z1 = xn2 * Zn21 + xethane * Zethane1 \n",
+      "V3 = (Z1 * R * T / P ) * (10**-3)\n",
+      "print \"(c)Molar volume based on daltons law = %.3e m**3/mol\"%V3\n",
+      "\n",
+      "\n",
+      "Tc = xn2 * Tcn2 + xethane * Tcethane \n",
+      "Pc = xn2 * Pcn2 + xethane * Pcethane \n",
+      "Zc = 0.83 \n",
+      "V4 = Zc * R *T / P * (10**-3)\n",
+      "print \"(d)Molar volume by kays method = %.3e m**3/mol\"%V4\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Molar volume by ideal gas equation = 4.503e-04 m**3/mol\n",
+        "(b)Molar volume by van der waals equation = 3.68 *10^-4 m**3/mol\n",
+        "(c)Molar volume based on compressibilty factor = 2.94e-04 m**3/mol\n",
+        "(c)Molar volume based on daltons law = 3.855e-04 m**3/mol\n",
+        "(d)Molar volume by kays method = 3.738e-04 m**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 page no : 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P1 = 40.                            #% ( nitrogen )\n",
+      "P2 = 60.                            #% ( ethane )\n",
+      "T = 325.                            #K ethane\n",
+      "\n",
+      "# Calculation \n",
+      "V = 4.5 * 10**-4                    #m**3/mol\n",
+      "a1 = 0.1365                         #N*m**4/mol**2\n",
+      "b1 = 3.86 * 10 ** -5                #m**3/mol\n",
+      "a2 = 0.557                          #N*m**4/mol**2\n",
+      "b2 = 6.51 * 10 ** -5                #m**3/mol\n",
+      "Pc1 = 3394.                         #kPa\n",
+      "Tc1 = 126.1                         #K\n",
+      "Pc2 = 4880.                         #kPa\n",
+      "Tc2 = 305.4                         #K\n",
+      "R = 8.314 \n",
+      "Pideal = R * T / (V * 1000)         #kPa\n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Pressure of Gas by the ideal gas equation = %.f kPa\"%Pideal\n",
+      "y1 = P1/100. \n",
+      "y2 = P2/100. \n",
+      "a = (y1 * (a1**(1./2)) + y2 * (a2**(1./2)))**2 \n",
+      "b = y1 * b1 + y2 * b2 \n",
+      "Pv = ((R * T / (V - b)) - a / (V**2))/1000 \n",
+      "print \"(b)Pressure of Gas by Van der waals equation = %.f kPa\"%Pv\n",
+      "Tc = y1*Tc1 + y2*Tc2 \n",
+      "Pc = y1*Pc1 + y2*Pc2 \n",
+      "Vc = R * Tc / Pc                    #Pseudo critical ideal volume\n",
+      "Vr = V / Vc                         #Pseudo reduced ideal volume\n",
+      "Tr = T / Tc                         #Pseudo reduced temperature\n",
+      "\n",
+      "#From fig 5.3, we get Pr = 1.2\n",
+      "Pr = 1.2 \n",
+      "Pk = Pr * Pc \n",
+      "print \"(c)Pressure of Gas by the Kays method = %.f kPa\"%Pk\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Pressure of Gas by the ideal gas equation = 6005 kPa\n",
+        "(b)Pressure of Gas by Van der waals equation = 5080 kPa\n",
+        "(c)Pressure of Gas by the Kays method = 5143 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch6.ipynb b/Stoichiometry_And_Process_Calculations/ch6.ipynb
new file mode 100755
index 00000000..8a560517
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch6.ipynb
@@ -0,0 +1,354 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Vapour Pressure"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 pageno : 133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 500.                                #kPa liquid\n",
+      "SV = 0.2813                             #m**3/kg wet steam\n",
+      "\n",
+      "# Calculation \n",
+      "Vsaturatedl = 1.093 * 10**-3            #m**3/kg\n",
+      "Vsaturatedv = 0.3747                    #m**3/kg\n",
+      "\n",
+      "# let the fraction of vapour be y\n",
+      "#(1-y)*Vsaturatedl + y*Vsaturatedv = SV\n",
+      "#then we get, (1-y)*(1.093*10**-3) + y*(0.3747) = 0.2813\n",
+      "y = (SV - Vsaturatedl)/(Vsaturatedv - Vsaturatedl) \n",
+      "P1 = y * 100 \n",
+      "P2 = 100 - P1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Percentage of Vapour = %.f %%\"%P1\n",
+      "print \"Percentage of Liquid = %.f %%\"%P2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage of Vapour = 75 %\n",
+        "Percentage of Liquid = 25 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 pageno : 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "T1 = 363.                           #K water\n",
+      "T2 = 373.                           #K water\n",
+      "P2s = 101.3                         #kPa pressure\n",
+      "\n",
+      "# Calculation \n",
+      "J = 2275. * 18                      #kJ/kmol\n",
+      "R = 8.314                           #kJ/kmolK\n",
+      "#ln (P2s/P1s) = J * (1/T1 - 1/T2) / R\n",
+      "P1s = P2s/math.exp(J * (1/T1 - 1/T2) / R) \n",
+      "\n",
+      "# Result\n",
+      "print \"Vapour pressure of water at 363 K = %.2f kPa\"%P1s\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vapour pressure of water at 363 K = 70.41 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 pageno : 135"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "P1s = 194.9                             #kPa acetone\n",
+      "P2s = 8.52                              #kPa pressure\n",
+      "T1 = 353.                               #K acetone\n",
+      "T2 = 273.                               #K acetone\n",
+      "T3 = 300.                               #K dry air\n",
+      "Pair = 101.3                            #kPa dry air\n",
+      "#math.log (P2s/P1s) = J * (1/T1 - 1/T2) / R\n",
+      "#let J / R = L\n",
+      "\n",
+      "# Calculation \n",
+      "L = math.log (P2s/P1s)/(1./T1 - 1/T2) \n",
+      "P3s = P1s * math.exp(L * (1./T1 - 1/T3))  \n",
+      "Ptotal = P3s + Pair                     #at saturation vapour pressure = partial pressure\n",
+      "\n",
+      "# Result \n",
+      "print \"(a)Final pressure of the mixture = %.2f kPa\"%Ptotal\n",
+      "MP = P3s * 100 / Ptotal \n",
+      "# mole percent = moles of acetone * 100 / total moles\n",
+      "#= Partial pressure of acetone * 100 / total Pressure\n",
+      "print \"(b)Mole percent of acetone in the final mixture = %.1f %%\"%MP\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Final pressure of the mixture = 130.83 kPa\n",
+        "(b)Mole percent of acetone in the final mixture = 22.6 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4 pageno : 137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "\n",
+      "# Variables \n",
+      "A = 13.8587              # antone constants for n-heptane\n",
+      "B = 2911.32 \n",
+      "C = 56.56 \n",
+      "T1 = 325.                   #K\n",
+      "#Pressure at normal condition = 101.3kPa\n",
+      "P2 = 101.3                  #kPa\n",
+      "\n",
+      "# Calculation \n",
+      "#Antoine equation - lnP = A - B / (T - C)\n",
+      "lnP = A - (B / (T1 - C)) \n",
+      "P1 = math.exp(lnP) \n",
+      "\n",
+      "# Result \n",
+      "print \"(a)Vapour pressure of n-heptane at 325K =  %.2f kPa\"%P1\n",
+      "T2 = B/(A - math.log(P2)) + C \n",
+      "print \"(b)Normal boiling point of n-heptane = %.2f K\"%T2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Vapour pressure of n-heptane at 325K =  20.36 kPa\n",
+        "(b)Normal boiling point of n-heptane = 371.62 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 pageno : 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "# variables \n",
+      "T = [273, 293, 313, 323, 333, 353, 373];\n",
+      "Ps = [0.61, 2.33, 7.37, 12.34, 19.90, 47.35, 101.3];\n",
+      "\n",
+      "# Calculation and Result\n",
+      "plot(T,Ps);\n",
+      "T1 = [273, 353]\n",
+      "Ps1 = [8.52, 194.9]\n",
+      "plot(T,Ps);\n",
+      "suptitle(\"Construction of the cox chart\")\n",
+      "xlabel(\"Temperature, K\")\n",
+      "ylabel(\"Pressure, kPa\")\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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iGSGkUmpRZGdn46uvvsLt27ehUCgwefJkmJub6zMbEdVg30Wdw4RfB2Jcq0/x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\nEBGRVv8PkUOu8xTNWfQAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x27db050>"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6 page no : 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "import math \n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "# Variables \n",
+      "Pswater1 = 6.08;            #kPa \n",
+      "T1 = 313.;                  #K\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#lnPs = 16.26205 - 3799.887/(T - 46.854)\n",
+      "Tb1 =3799.887/(16.26205 - math.log(Pswater1)) + 46.854;\n",
+      "print \"boiling point of water at 6.08kPa vapour pressure = %.2f K\"%Tb1\n",
+      "Pswater2 = 39.33;           #kPa\n",
+      "T2 = 353.;                  #K\n",
+      "Tb2 =3799.887/(16.26205 - math.log(Pswater2)) + 46.854;\n",
+      "\n",
+      "# Result \n",
+      "print \"boiling point of water at 39.33 kPa vapour pressure = %.2f K\"%Tb2\n",
+      "Tb = [Tb1 ,Tb2];\n",
+      "T = [T1 ,T2];\n",
+      "plot(T,Tb);\n",
+      "suptitle(\"Equal pressure reference plot for sulphuric acid\")\n",
+      "xlabel(\"Boiling point of solution, K\")\n",
+      "ylabel(\"Boiling point of water, K\")\n",
+      "show()\n",
+      "T3 = 333.;                  #K\n",
+      "\n",
+      "#corresponding to T3 on x axis, on y we get\n",
+      "Tb3 = 329.;                 #K\n",
+      "Pswater3 = math.exp(16.26205 - 3799.887/(Tb3 - 46.854));\n",
+      "print \"Vapour pressure of solution at 333K %.2f kPa\"%Pswater3\n",
+      "\n",
+      "# deviation\n",
+      "deviation = .04/16.53 * 100\n",
+      "print \"%% deviation = %.2f %%\"%deviation"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "boiling point of water at 6.08kPa vapour pressure = 309.69 K\n",
+        "boiling point of water at 39.33 kPa vapour pressure = 348.67 K\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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zmjdvTm5uLs7Ozjz55JNs3LjR1nEJIa7Thg3g46OtG7F/vyQJUX5K7FHceuut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\nIUSx/h89BNR1L4MHOAAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x46afd50>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vapour pressure of solution at 333K 16.35 kPa\n",
+        "% deviation = 0.24 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch7.ipynb b/Stoichiometry_And_Process_Calculations/ch7.ipynb
new file mode 100755
index 00000000..e679965d
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch7.ipynb
@@ -0,0 +1,945 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 :  Solutions and Phase Behaviour"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 pageno : 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Pas = 71.2                      #kPa n-heptane\n",
+      "Pbs = 48.9                      #kPa toluene\n",
+      "P = 65.                         #kPa equilibrium\n",
+      "\n",
+      "# Calculation \n",
+      "#P=(Pas-Pbs)*xa+Pbs,xa=mole fraction of n-heptane,liq. condition,therefore\n",
+      "xa = (P - Pbs)/(Pas - Pbs) \n",
+      "#ya = Pa / P , Vapour condition\n",
+      "ya = Pas * xa / P \n",
+      "P1 = xa * 100 \n",
+      "P2 = ya * 100 \n",
+      "\n",
+      "# Result\n",
+      "print \"Percentage of hepatne in liquid = %.1f %%\"%P1\n",
+      "print \"Percentage of hepatne in vapour = %.1f %%\"%P2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage of hepatne in liquid = 72.2 %\n",
+        "Percentage of hepatne in vapour = 79.1 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 pageno : 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P1 = 100.               #kPa ( Vapour pressure of liq A )\n",
+      "P2 = 60.                #kPa ( Vapour pressure of liq B )\n",
+      "T = 320.                #K\n",
+      "\n",
+      "# Calculation \n",
+      "#Pa = xa * P1 = 100 * xa\n",
+      "#Pa = xb * P2 = 60 * xb\n",
+      "#P = xa * P1 + ( 1 - xa )* P2\n",
+      "# = 100xa + ( 1 - xa )* 60\n",
+      "# = 60 + 40*xa\n",
+      "#ya = Pa / P\n",
+      "#0.5 = 100*xa / ( 60 + 40 * xa)\n",
+      "xa = 60 * 0.5 / (100 - 20) \n",
+      "Per1 = xa * 100 \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Percentage of A in liquid = \",Per1,\"%\"\n",
+      "Ptotal = 60 + 40 * xa \n",
+      "print \"(b)Total pressure of the vapour = \",Ptotal,\"kPa\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Percentage of A in liquid =  37.5 %\n",
+        "(b)Total pressure of the vapour =  75.0 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 pageno : 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "xa = 0.25          # liquid mixture\n",
+      "xb = 0.30 \n",
+      "\n",
+      "# Calculation \n",
+      "xc = 1 - xa - xb \n",
+      "Ptotal = 200.                   #kPa\n",
+      "Pcs = 50.                       #kPa(Vapour pressure of c)\n",
+      "Pc = xc * Pcs                   #(partial pressure of c)\n",
+      "yc = Pc / Ptotal \n",
+      "yb = 0.5 \n",
+      "ya = 1 - yb - yc \n",
+      "per1 = ya * 100 \n",
+      "\n",
+      "# Result\n",
+      "print \"Percentage of A in vapour = \",per1,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage of A in vapour =  38.75 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 pageno : 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 101.3                       #kPa flash vaporization\n",
+      "Pbs = 54.21                     #kPa pressure\n",
+      "Pas = 136.09                    #kPa temperature\n",
+      "xf = 0.65                       # liquid mixture\n",
+      "\n",
+      "# Calculation \n",
+      "xw = (P - Pbs)/(Pas - Pbs) \n",
+      "yd = xw * Pas / P  \n",
+      "# f = ( xf - xw ) / ( yd - xw )\n",
+      "f = ( xf - xw ) / ( yd - xw ) \n",
+      "per1 = f * 100\n",
+      "\n",
+      "# Result \n",
+      "print \"mole percent of the feed that is vapourised = %.1f %%\"%per1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mole percent of the feed that is vapourised = 37.9 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 pageno : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "from matplotlib.pyplot import *\n",
+      "from numpy import zeros\n",
+      "\n",
+      "# variables \n",
+      "T = [371.4, 378 ,383 ,388, 393, 398.6]\n",
+      "Pas = [101.3, 125.3, 140 ,160, 179.9, 205.3]\n",
+      "Pbs = [44.4, 55.6, 64.5, 74.8, 86.6, 101.3]\n",
+      "Ptotal = 101.3                      #kPa\n",
+      "\n",
+      "# Calculation \n",
+      "x = zeros(6)\n",
+      "y = zeros(6)\n",
+      "for i in range(6):\n",
+      "    x[i] = (Ptotal - Pbs[i])/(Pas[i] - Pbs[i]) \n",
+      "    y[i] = x[i] * Pas[i] / Ptotal \n",
+      "\n",
+      "# Result\n",
+      "#subplot(2,1,1)\n",
+      "plot(x,T,'-o') \n",
+      "plot(y,T,'-x') \n",
+      "suptitle(\"Boiling point diagram\")\n",
+      "xlabel(\"Mole fraction, x or y\")\n",
+      "ylabel(\"Temperature, K\")\n",
+      "show()\n",
+      "\n",
+      "\n",
+      "# part 2\n",
+      "\n",
+      "# variables \n",
+      "T = [371.4, 378, 383, 388, 393, 398.6]\n",
+      "Pas = [101.3, 125.3, 140, 160, 179.9, 205.3]\n",
+      "Pbs = [44.4, 55.6, 64.5, 74.8, 86.6, 101.3]\n",
+      "Ptotal = 101.3                                      #kPa\n",
+      "x = zeros(6)\n",
+      "y = zeros(6)\n",
+      "\n",
+      "# Calculation \n",
+      "for i in range(6):\n",
+      "    x[i] = (Ptotal - Pbs[i])/(Pas[i] - Pbs[i]) \n",
+      "    y[i] = x[i] * Pas[i] / Ptotal \n",
+      "\n",
+      "w = x \n",
+      "# Result\n",
+      "#subplot(2,1,2)\n",
+      "plot(x,w) \n",
+      "plot(x,y,'-o')\n",
+      "suptitle(\"Equilibrium curve\")\n",
+      "xlabel(\"x, mole fraction in liquid\")\n",
+      "ylabel(\"y, mole fraction in vapour\")\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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o0aJZCppdUiRyn6Cg8WzdOuUxj7/L5s2TdUhkWWZlZt0v6xi7fSyJqYks7rCY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\nBwnGGGOKOEgwxhhT9A+y/HS4V4Zf1wAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x31cbbd0>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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pKeDxeOjTpw88PDw0EZsMXUrSTglJCVgXvQ4lTAlMeaYY6TsSm55ugsU7Ftg0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\niJz/A9vzOGCUmCVhAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3e71b10>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 page no : 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# variables\n",
+      "p = 100                   # pressure Kpa\n",
+      "# from table\n",
+      "n_heptaneA = 13.8587\n",
+      "n_hexaneA = 13.8216\n",
+      "n_heptaneB = 2911.32\n",
+      "n_hexaneB = 2697.55\n",
+      "n_heptaneC = 56.51\n",
+      "n_hexaneC = 48.78\n",
+      "xA = .25\n",
+      "\n",
+      "# Calculations and Results\n",
+      "\n",
+      "# for T = 360   Initial guess\n",
+      "T1 = 360\n",
+      "lnPaS = n_hexaneA - (n_hexaneB/(T1- n_hexaneC))\n",
+      "PaS = round(math.e**lnPaS,2)\n",
+      "lnPbS = n_heptaneA - (n_heptaneB/(T1- n_heptaneC))\n",
+      "PbS = round(math.e**lnPbS,2)\n",
+      "P = xA*PaS + (1 - xA)*PbS\n",
+      "\n",
+      "print \"P = %.1f kPa\" %P\n",
+      "\n",
+      "# Since total pressure < 100 Kpa, equilibrium temperature is not equal to 360K.\n",
+      "# for T = 365 - Assume\n",
+      "T1 = 365\n",
+      "lnPaS = n_hexaneA - (n_hexaneB/(T1- n_hexaneC))\n",
+      "PaS = round(math.e**lnPaS,2)\n",
+      "lnPbS = n_heptaneA - (n_heptaneB/(T1- n_heptaneC))\n",
+      "PbS = round(math.e**lnPbS,2)\n",
+      "P = xA*PaS + (1 - xA)*PbS\n",
+      "\n",
+      "print \"P = %.2f kPa\"%P\n",
+      "\n",
+      "# As the pressure > 100 kPa, the temperature lies between 360 and 365 K.\n",
+      "# for T = 361.125 -- assumption\n",
+      "T1 = 361.125\n",
+      "lnPaS = n_hexaneA - (n_hexaneB/(T1- n_hexaneC))\n",
+      "PaS = round(math.e**lnPaS,2)\n",
+      "lnPbS = n_heptaneA - (n_heptaneB/(T1- n_heptaneC))\n",
+      "PbS = round(math.e**lnPbS,2)\n",
+      "P = xA*PaS + (1 - xA)*PbS\n",
+      "\n",
+      "print \"P = %.f kPa\"%P\n",
+      "print \"Therefore, the bubble point temperature = %.3f K\"%T1   # last assumed for T1\n",
+      "\n",
+      "\n",
+      "# Part b\n",
+      "Pa = xA*PaS\n",
+      "Ya = Pa/P\n",
+      "\n",
+      "print \"Partial pressure of n-Hexane in vapour at the bubble point is = %.2f kPa\"%Pa\n",
+      "print \"Mole fraction of hexane in the vapour is Ya = %.4f\"%Ya\n",
+      "print \"The vapour contains %.2f %% hexane and %.2f %% heptane.\"%(Ya*100,(1-Ya)*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "P = 96.7 kPa\n",
+        "P = 112.04 kPa\n",
+        "P = 100 kPa\n",
+        "Therefore, the bubble point temperature = 361.125 K\n",
+        "Partial pressure of n-Hexane in vapour at the bubble point is = 44.65 kPa\n",
+        "Mole fraction of hexane in the vapour is Ya = 0.4465\n",
+        "The vapour contains 44.65 % hexane and 55.35 % heptane.\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7 page no : 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "#lnPas = 14.5463 - 2940.46/(T - 35.93)\n",
+      "#lnPbs = 14.2724 - 2945.47 / (T - 49.15)\n",
+      "#xa = (P - Pbs)/(Pas - Pbs)\n",
+      "#Ya = Pas * (P - Pbs)/(P * (Pas - Pbs))\n",
+      "Ya = 0.4                            # vapour phase\n",
+      "P = 65.                             #kPa\n",
+      "#various temperature value are assumed and tried till LHS = RHS, we get\n",
+      "T = 334.15                          #K\n",
+      "\n",
+      "# Calculation \n",
+      "Pas = math.exp(14.5463 - 2940.46/(T - 35.93)) \n",
+      "Pbs = math.exp(14.2724 - 2945.47 / (T - 49.15)) \n",
+      "xa = (P - Pbs)/(Pas - Pbs) \n",
+      "\n",
+      "# Result \n",
+      "print \"(a)The Dew point temperature at 65 kPa = \",T,\"K\"\n",
+      "print \"    Concentration of the first drop of liquid = %.2f\"%xa\n",
+      "T1 = 327.                           #K\n",
+      "Pas1 = math.exp(14.5463 - 2940.46/(T1 - 35.93)) \n",
+      "Pbs1 = math.exp(14.2724 - 2945.47 / (T1 - 49.15)) \n",
+      "xa1 = Ya * Pbs1 / (Pas1 - Ya*(Pas1 - Pbs1)) \n",
+      "P1 = xa1 * Pas1 / Ya \n",
+      "print \"(b)The dew point pressure at 327 K = %.2f kPa\"%P1\n",
+      "print \"    Concentration at 327K = %.4f\"%xa1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The Dew point temperature at 65 kPa =  334.15 K\n",
+        "    Concentration of the first drop of liquid = 0.24\n",
+        "(b)The dew point pressure at 327 K = 50.09 kPa\n",
+        "    Concentration at 327K = 0.2354\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8 pageno : 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "MW = 44.032                # mole fraction of acetaldehyde\n",
+      "Mwater = 18.016            \n",
+      "x = 2.                      #%    acetaldehyde weight\n",
+      "Pa = 41.4                   #kPa   solution\n",
+      "\n",
+      "# Calculation \n",
+      "Mfr = (x/MW)/(x/MW + (100-x)/Mwater) \n",
+      "#henry's law gives Pa = Ha * xa\n",
+      "Ha = Pa / Mfr \n",
+      "Molality = 0.1 \n",
+      "Mfr1 = Molality / (1000/Mwater + Molality) \n",
+      "Pa1 = Ha * Mfr1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Partial Pressure = %.f kPa\"%Pa1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Partial Pressure = 9 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9 pageno : 177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "#1 - pentane, 2 - hexane, 3 - heptane\n",
+      "\n",
+      "x1 = 0.6 \n",
+      "x2 = 0.25 \n",
+      "x3 = 0.15 \n",
+      "A1 = 13.8183 \n",
+      "A2 = 13.8216 \n",
+      "A3 = 13.8587 \n",
+      "B1 = 2477.07 \n",
+      "B2 = 2697.55 \n",
+      "B3 = 2911.32 \n",
+      "C1 = 39.94 \n",
+      "C2 = 48.78 \n",
+      "C3 = 56.51 \n",
+      "#As raoults law is applicable, Ki = yi/xi = Pis/P\n",
+      "#yi = xi*Pis/P\n",
+      "#ln P = A- B/(T-C)\n",
+      "#Assuming,\n",
+      "P = 400.                #kPa\n",
+      "T = 369.75              #K\n",
+      "\n",
+      "# Calculation \n",
+      "Pas1 = math.exp(A1 - B1 / (T - C1)) \n",
+      "Pas2 = math.exp(A2 - B2 / (T - C2)) \n",
+      "Pas3 = math.exp(A3 - B3 / (T - C3)) \n",
+      "Yi = (x1*Pas1 + x2*Pas2 + x3*Pas3)/P \n",
+      "print \"(a)bubble point temperature of the mixture = \",T,\"K\"\n",
+      "y1 = x1*Pas1/P \n",
+      "y2 = x2*Pas2/P \n",
+      "y3 = x3*Pas3/P \n",
+      "\n",
+      "# Result \n",
+      "print \"(b)composition of n-pentane in vapour = %.2f %%\"%(y1*100)\n",
+      "print \"composition of n-hexane in vapour = %.2f %%\"%(y2*100)\n",
+      "print \"composition of n-heptane in vapour = %.2f %%\"%(y3*100)\n",
+      "T1 = 300.               #K\n",
+      "Ps1 = math.exp(A1 - B1 / (T1 - C1)) \n",
+      "Ps2 = math.exp(A2 - B2 / (T1 - C2)) \n",
+      "Ps3 = math.exp(A3 - B3 / (T1 - C3)) \n",
+      "P1 = x1*Ps1 + x2*Ps2 + x3*Ps3 \n",
+      "print \"(c)Bubble point pressure = %.1f kPa\"%P1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)bubble point temperature of the mixture =  369.75 K\n",
+        "(b)composition of n-pentane in vapour = 82.32 %\n",
+        "composition of n-hexane in vapour = 14.08 %\n",
+        "composition of n-heptane in vapour = 3.60 %\n",
+        "(c)Bubble point pressure = 50.4 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10 pageno : 178"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "#1 - pentane, 2 - hexane, 3 - heptane\n",
+      " \n",
+      "y1 = 0.6 \n",
+      "y2 = 0.25 \n",
+      "y3 = 0.15 \n",
+      "A1 = 13.8183 \n",
+      "A2 = 13.8216 \n",
+      "A3 = 13.8587 \n",
+      "B1 = 2477.07 \n",
+      "B2 = 2697.55 \n",
+      "B3 = 2911.32 \n",
+      "C1 = 39.94 \n",
+      "C2 = 48.78 \n",
+      "C3 = 56.51 \n",
+      "P = 400.                    #kPa\n",
+      "T = 300.                    #K\n",
+      "#As raoults law is applicable, Ki = yi/xi = Pis/P\n",
+      "#xi = yi*P/Pis\n",
+      "#ln P = A- B/(T-C)\n",
+      "#Assuming,\n",
+      "T1 = 385.94                 #K\n",
+      "\n",
+      "# Calculation \n",
+      "Pas1 = math.exp(A1 - B1 / (T1 - C1)) \n",
+      "Pas2 = math.exp(A2 - B2 / (T1 - C2)) \n",
+      "Pas3 = math.exp(A3 - B3 / (T1 - C3)) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Dew point temperature of the mixture = \",T1,\"K\"\n",
+      "Ps1 = math.exp(A1 - B1 / (T - C1)) \n",
+      "Ps2 = math.exp(A2 - B2 / (T - C2)) \n",
+      "Ps3 = math.exp(A3 - B3 / (T - C3)) \n",
+      "P1 = 1/(y1/Ps1 + y2/Ps2 + y3/Ps3) \n",
+      "print \"(b)Dew point pressure = %.2f kPa\"%P1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Dew point temperature of the mixture =  385.94 K\n",
+        "(b)Dew point pressure = 23.79 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11 page no :180"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#1 - methanol, 2 - ethanol, 3 - propanol\n",
+      "x1 = 0.45 \n",
+      "x2 = 0.3 \n",
+      "x3 = 1 - (x1 + x2) \n",
+      "P = 101.3                   #kPa\n",
+      "# by drawing the temperature vs vapour pressure graph and interpolation,assuming,\n",
+      "T = 344.6                   #K\n",
+      "Ps1 = 137.3 \n",
+      "Ps2 = 76.2 \n",
+      "Ps3 = 65.4 \n",
+      "\n",
+      "# Calculation \n",
+      "y1 = x1 * Ps1 / P \n",
+      "y2 = x2 * Ps2 / P \n",
+      "y3 = x3 * Ps3 / P \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Bubble point temperature = \",T,\"K\"\n",
+      "print \"Composition of methanol in vapour = %.f %%\"%(y1*100)\n",
+      "print \"Composition of ethanol in vapour = %.1f%%\"%(y2*100)\n",
+      "print \"Composition of propanol in vapour = %.1f%%\"%(y3*100)\n",
+      "#again, for xi = 1\n",
+      "T1 = 347.5                  #K\n",
+      "P1 = 153.28 \n",
+      "P2 = 85.25 \n",
+      "P3 = 73.31 \n",
+      "xa = x1 * P / P1 \n",
+      "xb = x2 * P / P2 \n",
+      "xc = x3 * P / P3 \n",
+      "print \"(b)Dew point temperature = \",T1,\"K\"\n",
+      "print \"Composition of methanol in liquid = %.1f %%\"%(xa*100)\n",
+      "print \"Composition of ethanol in liquid = %.1f %%\"%(xb*100)\n",
+      "print \"Composition of propanol in liquid = %.1f %%\"%(xc*100)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Bubble point temperature =  344.6 K\n",
+        "Composition of methanol in vapour = 61 %\n",
+        "Composition of ethanol in vapour = 22.6%\n",
+        "Composition of propanol in vapour = 16.1%\n",
+        "(b)Dew point temperature =  347.5 K\n",
+        "Composition of methanol in liquid = 29.7 %\n",
+        "Composition of ethanol in liquid = 35.6 %\n",
+        "Composition of propanol in liquid = 34.5 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.12 pageno : 181"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "xp = 0.25                           # propane\n",
+      "xnb = 0.4                           # n-butane\n",
+      "xnp = 0.35                          # n-pentane\n",
+      "P = 1447.14                         #kPa  \n",
+      "#assuming temperatures 355.4 K and 366.5 K , corresponding Ki \n",
+      "# values are found from nomograph and total Ki value are 0.928 and 1.075 resp, thus bubble point temperature lies between, using interpolation bubble point temperature is found to be,\n",
+      "Tb = 361.                           #K\n",
+      "print \"(a) The buuble point temperature = \",Tb,\"K\"\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#At 361,\n",
+      "Kip = 2.12 \n",
+      "Kinb = 0.85 \n",
+      "Kinp = 0.37 \n",
+      "xp1 = Kip * xp \n",
+      "xnb1 = Kinb * xnb \n",
+      "xnp1 = Kinp * xnp \n",
+      "\n",
+      "# Result\n",
+      "print \"concentration of propane at bubble point = \",xp1\n",
+      "print \"concentration of n-butane at bubble point = \",xnb1\n",
+      "print \"concentration of n-pentane at bubble point = %.3f\"%xnp1\n",
+      "#At dew point Yi/Ki = 1, at 377.6K this is 1.1598 and at 388.8K \n",
+      "# it is 0.9677, by interpolation dew point is found to be\n",
+      "Td = 387.                           #K\n",
+      "Kip1 = 2.85 \n",
+      "Kinb1 = 1.25 \n",
+      "Kinp1 = 0.59 \n",
+      "yp1 = xp/Kip1 \n",
+      "ynb1 = xnb/Kinb1 \n",
+      "ynp1 = xnp/Kinp1 \n",
+      "print \"(b) The dew point temperature = \",Td,\"K\"\n",
+      "print \"concentration of propane at dew point = %.4f\"%yp1\n",
+      "print \"concentration of n-butane at dew point = %.4f\"%ynb1\n",
+      "print \"concentration of n-pentane at dew point = %.4f\"%ynp1\n",
+      "\n",
+      "#summation zi / (1 + L/VKi)= 0.45, using trial and error, we find\n",
+      "T = 374.6                           #K\n",
+      "L = 0.55 \n",
+      "V = 0.45 \n",
+      "Kip2 = 2.5 \n",
+      "Kinb2 = 1.08 \n",
+      "Kinp2 = 0.48 \n",
+      "t = (xp/(1+L/(V*Kip2)))+(xnb/(1+L/(V*Kinb2))) + (xnp/(1+L/(V*Kinp2))) \n",
+      "yp2 = (xp/(1+L/(V*Kip2)))/t \n",
+      "ynb2 = (xnb/(1+L/(V*Kinb2)))/t \n",
+      "ynp2 = (xnp/(1+L/(V*Kinp2)))/t \n",
+      "xp2 = (xp - V * yp2)/L \n",
+      "xnb2 = (xnb - V * ynb2)/L \n",
+      "xnp2 = (xnp - V * ynp2)/L  \n",
+      "print \"(c)Temperature of the mixture = \",T,\"K\"\n",
+      "print \"vapour phase concentration of propane = %.4f\"%yp2\n",
+      "print \"vapour phase concentration of n-butane = %.4f\"%ynb2\n",
+      "print \"vapour phase concentration of n-pentane = %.4f\"%ynp2\n",
+      "print \"liquid phase concentration of propane = %.4f\"%xp2\n",
+      "print \"liquid phase concentration of n-butane = %.4f\"%xnb2\n",
+      "print \"liquid phase concentration of n-pentane = %.4f\"%xnp2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The buuble point temperature =  361.0 K\n",
+        "concentration of propane at bubble point =  0.53\n",
+        "concentration of n-butane at bubble point =  0.34\n",
+        "concentration of n-pentane at bubble point = 0.130\n",
+        "(b) The dew point temperature =  387.0 K\n",
+        "concentration of propane at dew point = 0.0877\n",
+        "concentration of n-butane at dew point = 0.3200\n",
+        "concentration of n-pentane at dew point = 0.5932\n",
+        "(c)Temperature of the mixture =  374.6 K\n",
+        "vapour phase concentration of propane = 0.3696\n",
+        "vapour phase concentration of n-butane = 0.4131\n",
+        "vapour phase concentration of n-pentane = 0.2173\n",
+        "liquid phase concentration of propane = 0.1521\n",
+        "liquid phase concentration of n-butane = 0.3893\n",
+        "liquid phase concentration of n-pentane = 0.4586\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.13 pageno : 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 93.30               #kPa   pressure\n",
+      "T1 = 353.               #K     temperature\n",
+      "T2 = 373.               #K     temperature\n",
+      "Pwater1 = 47.98         #kPa   pressure of water\n",
+      "Pwater2 = 101.3         #kPa\n",
+      "Pliq1 = 2.67            #kPa   pressure of liquid\n",
+      "Pliq2 = 5.33            #kPa\n",
+      "\n",
+      "# Calculation \n",
+      "T = T1 + (T2 - T1)*(P - (Pwater1 + Pliq1))/(Pwater2 + Pliq2 - (Pwater1 + Pliq1)) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)The equilibrium temperature = %.1f K\"%T\n",
+      "Pwater = 88.50 \n",
+      "y = Pwater * 100 /P \n",
+      "print \"(b)Water vapour in vapour mixture = %.2f %%\"%y\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The equilibrium temperature = 368.2 K\n",
+        "(b)Water vapour in vapour mixture = 94.86 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.14 pageno : 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "from matplotlib.pyplot import *\n",
+      "from numpy import *\n",
+      "\n",
+      "# variables \n",
+      "#the three phase temperature is first find out, which comes to be 342K, the\n",
+      "#  corresponding Ps1 = 71.18, Ps2 = 30.12\n",
+      "T = [342, 343, 348, 353, 363, 373] \n",
+      "Ps2 = [30.12, 31.06, 37.99, 47.32, 70.11, 101.3] \n",
+      "Ps1 = [71.18, 72.91, 85.31, 100.5, 135.42, 179.14] \n",
+      "P = 101.3                   #kPa\n",
+      "\n",
+      "y1 = zeros(6)\n",
+      "y2 = zeros(6)\n",
+      "\n",
+      "# Calculation \n",
+      "for i in range(6):\n",
+      "    y1[i] = 1 - (Ps1[i])/P \n",
+      "\n",
+      "for i in range(6):\n",
+      "    y2[i] = 1 - (Ps2[i])/P \n",
+      "\n",
+      "# Result\n",
+      "plot(y2,T) \n",
+      "plot(1-y1,T) \n",
+      "suptitle(\"temperature composition diagram\")\n",
+      "xlabel(\"x,y mole fraction of benzene\")\n",
+      "ylabel(\"Temperature\")\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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yJezYAXq91mmE+GtGs5HR347mesZ1NoRukOKgsWLv5uro6EizZs1ITEx8qGCi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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x27c4950>"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.15 pageno : 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "T = 379.2               #K   solution\n",
+      "P = 101.3               #kPa \n",
+      "Ps = 70.                #kPa solution\n",
+      "Molality = 5\n",
+      "\n",
+      "# Calculation \n",
+      "Pws = math.exp(16.26205 - 3799.887/(T - 46.854)) \n",
+      "k = P / Pws \n",
+      "Pws1 = Ps / k \n",
+      "T1 = 3799.887 / (16.26205 - math.log( Pws1)) + 46.854 \n",
+      "\n",
+      "# Result \n",
+      "print \"Boiling point of the solution = %.1f K\"%T1\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Boiling point of the solution = 368.8 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch8.ipynb b/Stoichiometry_And_Process_Calculations/ch8.ipynb
new file mode 100755
index 00000000..f4b3762e
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch8.ipynb
@@ -0,0 +1,927 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : Humidity and Humidity chart"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 pageno : 206"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "T = 280.                #K\n",
+      "P = 105.                #kPa\n",
+      "Pas = 13.25             #kPa ( Vpaour pressure of acetone )\n",
+      "Pa = Pas                # ( As gas is saturated, partial pressure = vapour pressure )\n",
+      "\n",
+      "# Calculation \n",
+      "Mfr = Pa / P            #(Mole fraction) \n",
+      "Mpr = Mfr * 100 \n",
+      "print \"(a)The mole percent of acetone in the mixture = %.2f %%\"%Mpr\n",
+      "Ma = 58.048             #(molecular weight of acetone)\n",
+      "Mn = 28.                #(molecular weight of nitrogen)\n",
+      "N = 1.                  #mole\n",
+      "Na = Mfr * N \n",
+      "Nn = N - Na \n",
+      "ma = Na * Ma  \n",
+      "mn = Nn * Mn \n",
+      "mtotal = ma + mn \n",
+      "maper = ma *100 / mtotal \n",
+      "mnper = mn *100/ mtotal \n",
+      "\n",
+      "# Result\n",
+      "print \"(b)Weight percent of acetone = %.2f %%\"%maper\n",
+      "print \"Weight percent of nitrogen = %.2f %%\"%mnper\n",
+      "Vstp = 22.4             #m**3/kmol\n",
+      "Pstp = 101.3            #kPa\n",
+      "Tstp = 273.15           #K\n",
+      "V = Vstp * Pstp * T / ( Tstp * P ) \n",
+      "C = ma/V \n",
+      "print \"(c)Concentration of vapour = %.4f kg/m**3\"%C\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The mole percent of acetone in the mixture = 12.62 %\n",
+        "(b)Weight percent of acetone = 23.04 %\n",
+        "Weight percent of nitrogen = 76.96 %\n",
+        "(c)Concentration of vapour = 0.3307 kg/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 pageno : 207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "P = 101.3               #kPa\n",
+      "Per1 = 10.              #%\n",
+      "\n",
+      "# Calculation \n",
+      "Pa = P * Per1 / 100     # ( a - benzene )\n",
+      "Ps = Pa                 #( saturation )\n",
+      "#lnPs = 13.8858 - 2788.51/(T - 52.36)\n",
+      "T = 2788.51 / ( 13.8858 - math.log(Ps)) + 52.36 \n",
+      "\n",
+      "# Result\n",
+      "print \"Temperature at which saturation occurs = %.1f K\"%T\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature at which saturation occurs = 293.4 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 pageno : 207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Pdryair = 101.3                             #kPa\n",
+      "Pacetone = 16.82                            #kPa\n",
+      "\n",
+      "# Calculation \n",
+      "Nratio = Pacetone / (Pdryair - Pacetone) \n",
+      "mratio = Nratio * 58.048 / 29               # ( Macetone = 58.048, Mair = 29 )\n",
+      "macetone = 5.                               #kg ( given )\n",
+      "mdryair = macetone / mratio \n",
+      "\n",
+      "# Result\n",
+      "print \"Minimum air required = %.2f kg\"%mdryair\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum air required = 12.55 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 pageno : 209"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "Pa = 15.                    #kPa ( partial pressure of acetone)\n",
+      "Ptotal = 101.3              #kPa\n",
+      "\n",
+      "# Calculation and Result \n",
+      "Mfr = Pa / Ptotal \n",
+      "print \"(a)Mole fraction of acetone = %.4f \"%Mfr\n",
+      "Macetone = 58.048 \n",
+      "Mnitrogen = 28. \n",
+      "mafr = Mfr * Macetone / ( Mfr * Macetone + (1-Mfr)* Mnitrogen ) \n",
+      "print \"(b)Weight fraction of acetone = %.4f\"%mafr\n",
+      "Y = Mfr / ( 1 - Mfr ) \n",
+      "print \"(c)Molal humidity = %.4f\"%Y,\"moles of acetone/moles of nitrogen\"\n",
+      "Y1 = Y * Macetone / Mnitrogen   \n",
+      "print \"(d)Absolute humidity = %.4f\"%Y1,\"kg acetone/kg nitrogen\"\n",
+      "Pas = 26.36                 #kPa ( vapour pressure)\n",
+      "Ys = Pas / ( Ptotal - Pas)  #saturation humidity\n",
+      "print \"(e)Saturation humidity = %.4f\"%Ys,\"moles of acetone/moles of nitrogen\"\n",
+      "Y1s = Ys * Macetone / Mnitrogen \n",
+      "print \"(f)Absolute saturation humidity = %.4f\"%Y1s,\"kg acetone/kg nitrogen\"\n",
+      "V = 100.                    #m**3\n",
+      "Vstp =22.4143               #m**3/kmol\n",
+      "Pstp = 101.3                #kPa\n",
+      "Tstp = 273.15               #K\n",
+      "T = 295.                    #K\n",
+      "N = V * Ptotal * Tstp / (Vstp * Pstp * T ) \n",
+      "Nacetone = N * Mfr \n",
+      "macetone = Nacetone * Macetone \n",
+      "print \"(g)Mass of acetone in 100m**3 of the total gas = %.1f\"%macetone,\"kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Mole fraction of acetone = 0.1481 \n",
+        "(b)Weight fraction of acetone = 0.2649\n",
+        "(c)Molal humidity = 0.1738 moles of acetone/moles of nitrogen\n",
+        "(d)Absolute humidity = 0.3603 kg acetone/kg nitrogen\n",
+        "(e)Saturation humidity = 0.3517 moles of acetone/moles of nitrogen\n",
+        "(f)Absolute saturation humidity = 0.7292 kg acetone/kg nitrogen\n",
+        "(g)Mass of acetone in 100m**3 of the total gas = 35.5 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5 pageno : 211"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Pa = 15.                    #kPa ( Partial pressure )\n",
+      "Pas = 26.36                 #kPa ( Vapour pressure )\n",
+      "\n",
+      "# Calculation \n",
+      "RS = Pa * 100 / Pas  \n",
+      "Y = 0.1738 \n",
+      "Ys = 0.3517 \n",
+      "PS = Y * 100 / Ys \n",
+      "\n",
+      "# Result\n",
+      "print \"Relative humidity = %.1f\"%RS,\"%\"\n",
+      "print \"Percent humidity = %.2f\"%PS,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Relative humidity = 56.9 %\n",
+        "Percent humidity = 49.42 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6 page no : 211"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "mwater = 0.0109                     #kg\n",
+      "V = 1.                              #m**3\n",
+      "T = 300.                            #K\n",
+      "P = 101.3                           #kPa\n",
+      "Vstp =22.4143                       #m**3/kmol\n",
+      "Pstp = 101.3                        #kPa\n",
+      "Tstp = 273.15                       #K\n",
+      "\n",
+      "# Calculation \n",
+      "N = V * P * Tstp / (Vstp * Pstp * T ) \n",
+      "Nwater = mwater / 18.016 \n",
+      "Nfr = Nwater / N \n",
+      "Pwater = Nfr * P \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Partial pressure of water vapour = %.2f\"%Pwater,\"kPa\"\n",
+      "Ps = math.exp(16.26205 - 3799.887/(T - 46.854)) \n",
+      "RS = Pwater * 100 / Ps \n",
+      "print \"(b)Relative saturation = %.2f\"%RS,\"%\"\n",
+      "Y1 = Pwater *18 / ((P - Pwater)*29) \n",
+      "print \"(c)Absolute humidity = %.2e\"%Y1,\"kg water / kg dry air\"\n",
+      "Y1s = Ps *18 / ((P - Ps)*29) \n",
+      "PS1 = Y1 * 100 / Y1s \n",
+      "print \"(d)Percent saturation = %.2f\"%PS1,\"%\"\n",
+      "PS = 10.                            #%\n",
+      "Y1S = Y1 * 100/PS   \n",
+      "#Y1S = Pas/(P - Pas ) * 18 /29\n",
+      "Pas1 = 29 * P * Y1S / (18 + 29*Y1s) \n",
+      "T1 = 3799.887 / (16.26205-math.log(Pas1)) + 46.854 \n",
+      "print \"(e)Temperature at which 10%% saturation occurs = %.1f\"%T1,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Partial pressure of water vapour = 1.51 kPa\n",
+        "(b)Relative saturation = 43.17 %\n",
+        "(c)Absolute humidity = 9.38e-03 kg water / kg dry air\n",
+        "(d)Percent saturation = 42.31 %\n",
+        "(e)Temperature at which 10% saturation occurs = 326.9 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7 page no : 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "T = 300.                        #K\n",
+      "P = 100.                        #kPa\n",
+      "S = 25000.                      #kJ/m**3\n",
+      "T1= 295.                        #K\n",
+      "P1 = 105.                       #kPa\n",
+      "RS = 50.                        #%\n",
+      "Ps = 3.5                        #kPa\n",
+      "Ps1 = 2.6                       #kPa\n",
+      "Vstp = 22.4143                  #m**3/kmol\n",
+      "Pstp = 101.3                    #kPa\n",
+      "Tstp = 273.15                   #K\n",
+      "V = 1.                          #m**3\n",
+      "\n",
+      "# Calculation \n",
+      "N = V * P * Tstp/(Vstp * Pstp * T) \n",
+      "Nfuel = N * (P - Ps)/P \n",
+      "Smol = S / Nfuel                #kJ/kmol\n",
+      "N1 = V * P1 * Tstp/(Vstp * Pstp * T1) \n",
+      "Pwater = Ps1 * RS /100 \n",
+      "Nfuel1 = N1 * (P1 - Pwater )/P1 \n",
+      "S1 = Smol * Nfuel1 \n",
+      "\n",
+      "# Result\n",
+      "print \"Heating value of gas at 295K and 105kPa = %.f\"%S1,\"kJ/m**3\"\n",
+      "\n",
+      "# note : answer may vary because of rounding error.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heating value of gas at 295K and 105kPa = 27321 kJ/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8 pageno : 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "# Variables \n",
+      "T = 300.                            #K\n",
+      "T1 = 335.                           #K\n",
+      "P = 150.                            #kPa\n",
+      "\n",
+      "# Calculation \n",
+      "#lnPs = 13.8858 - 2788.51 / ( T - 52.36)\n",
+      "Ps = math.exp(13.8858 - 2788.51 / ( T - 52.36)) \n",
+      "Ps1 = math.exp(13.8858 - 2788.51 / ( T1 - 52.36)) \n",
+      "Pa = Ps                             #(Vapour pressure at dew point is equal to the partial pressure of the vapour)\n",
+      "Y = Pa / (P - Pa) \n",
+      "Ys = Ps1 / (P - Ps1) \n",
+      "PS = Y * 100 / Ys \n",
+      "\n",
+      "# Result \n",
+      "print \"(a)Percent saturation = %.2f\"%PS,\"%\"\n",
+      "Ma = 78.048 \n",
+      "Mb = 28 \n",
+      "Q = Y * Ma / Mb  \n",
+      "print \"(b)Quantity of benzene per kilgram of nitrogen = %.4f\"%Q,\"kg benzene/kg nitrogen\"\n",
+      "V = 1.                              #m**3 ( basis )\n",
+      "Vstp = 22.4143                      #m**3/kmol\n",
+      "Pstp = 101.3                        #kPa\n",
+      "Tstp = 273.15                       #K\n",
+      "N = V * P * Tstp/(Vstp * Pstp * T1) \n",
+      "y = Y / ( 1 + Y ) \n",
+      "Nbenzene = N * y \n",
+      "C = Nbenzene * Ma \n",
+      "print \"(c)Kilogram of benzene per m**3 of nitrogen = %.4f\"%C,\"kg/m**3\"\n",
+      "P1 = 100.                           #kPa\n",
+      "Pbenzene = y * P1 \n",
+      "T1 = 2788.51 / ( 13.8858 - math.log (Pbenzene)) + 52.36 \n",
+      "print \"(d)Dew point = %.2f\"%T1,\"K\"\n",
+      "Per1 = 60                           #%\n",
+      "Y2 = Y * (1- Per1/100.) \n",
+      "#Y2 = Pa / (P - Pa)\n",
+      "P = Pa / Y2 + Pa \n",
+      "print \"(e)Pressure required = %.1f\"%P,\"kPa\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Percent saturation = 17.17 %\n",
+        "(b)Quantity of benzene per kilgram of nitrogen = 0.2827 kg benzene/kg nitrogen\n",
+        "(c)Kilogram of benzene per m**3 of nitrogen = 0.3871 kg/m**3\n",
+        "(d)Dew point = 291.39 K\n",
+        "(e)Pressure required = 354.3 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9 pageno : 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "T = 300.                    #K\n",
+      "T1 = 285.                   #K\n",
+      "Pwater = 3.56               #kPa\n",
+      "Pwater1 = 1.4               #kPa\n",
+      "V = 1.                      #m**3 ( Basis )\n",
+      "Vstp = 22.4143              #m**3/kmol\n",
+      "\n",
+      "# Calculation \n",
+      "N = V / Vstp \n",
+      "Pstp = 101.3                #kPa\n",
+      "Y = Pwater / (Pstp - Pwater) \n",
+      "Y1 = Pwater1 / (Pstp - Pwater1) \n",
+      "Nremoved = Y - Y1 \n",
+      "Ndryair = N * 1 / (1 + Y) \n",
+      "mremoved = Ndryair * Nremoved * 18.016 \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)amount of water removed = %.4f\"%mremoved,\"kg\"\n",
+      "Nremaining = Ndryair * Y1  \n",
+      "V1 = (Ndryair + Nremaining) * Vstp  \n",
+      "print \"(b)Volume of gas at stp after drying = %.4f\"%V1,\"m**3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)amount of water removed = 0.0174 kg\n",
+        "(b)Volume of gas at stp after drying = 0.9784 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11 page no : 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Td = 328.                       #K ( dry bulb )\n",
+      "P = 101.3                       #kPa\n",
+      "PS = 10.                        #%\n",
+      "#refering to the psychometric chart, corresponding to 328 K and 10% saturation\n",
+      "Y1 = 0.012                      #kg water / kg dry air\n",
+      "print \"(a)Absolute humidity = \",Y1,\"kg water / kg dry air\"\n",
+      "#Y1 = Pa * 18 / ( P - Pa ) * 29\n",
+      "\n",
+      "# Calculation \n",
+      "Pa = Y1 * P * 29 /( 18 + Y1 * 29 ) \n",
+      "\n",
+      "# Result\n",
+      "print \"(b)Partial Pressure of water vapour = %.3f\"%Pa,\"kPa\"\n",
+      "#using psychometric chart, saturation humidity at 328 K is given as\n",
+      "Y1s = 0.115                     #kg water / kg dry air\n",
+      "print \"(c)The absolute humidity at 328K = \",Y1s,\"kg water / kg dry air\"\n",
+      "#at saturation partial pressure = vapour pressure\n",
+      "Pas = Y1s * P * 29 /( 18 + Y1s * 29 )  \n",
+      "print \"(d)Vapour Pressure of water vapour = %.1f\"%Pas,\"kPa\"\n",
+      "RS = Pa * 100 / Pas \n",
+      "print \"(e)Percent relative saturation = %.2f\"%RS,\"%\"\n",
+      "#using psychometric chart, moving horizontally keeping humidity constant to 100% saturation, we get dew point as,\n",
+      "T = 290.                        #K\n",
+      "print \"(f)Dew point = \",T,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Absolute humidity =  0.012 kg water / kg dry air\n",
+        "(b)Partial Pressure of water vapour = 1.921 kPa\n",
+        "(c)The absolute humidity at 328K =  0.115 kg water / kg dry air\n",
+        "(d)Vapour Pressure of water vapour = 15.8 kPa\n",
+        "(e)Percent relative saturation = 12.13 %\n",
+        "(f)Dew point =  290.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12 pageno : 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Ca = 1.884                      #kJ/kgK\n",
+      "Cb = 1.005                      #kJ/kgK\n",
+      "Y1 = 0.012 \n",
+      "\n",
+      "# Calculation \n",
+      "#Cs = Cb + Y1 * Ca\n",
+      "Cs = Cb + Y1 * Ca \n",
+      "\n",
+      "# Result\n",
+      "print \"Humid heat of the sample = %.4f\"%Cs,\"kJ/kgK\"\n",
+      "P = 101.3                       #kPa\n",
+      "V = 100.                        #m**3\n",
+      "R = 8.314 \n",
+      "T = 328.                        #K\n",
+      "T1 = 373.                       #K\n",
+      "N = P * V / ( R * T ) \n",
+      "Pa = 1.921                      #kPa\n",
+      "Ndryair = N * (P - Pa)/P \n",
+      "mdryair = Ndryair * 29 \n",
+      "Ht = mdryair * Cs * (T1 - T) \n",
+      "print \"Heat to be supplied = %.f\"%Ht,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Humid heat of the sample = 1.0276 kJ/kgK\n",
+        "Heat to be supplied = 4887 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.14 pageno : 224"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 101.3                           #kPa\n",
+      "MW = 58. \n",
+      "T1 = 280.8                          #K\n",
+      "Ps = 5.                             #kPa\n",
+      "pr = 2.                             #kJ/kgK ( Psychometric ratio )\n",
+      "Hvap = 360.                         #kJ/kg\n",
+      "Tw = T1 \n",
+      "\n",
+      "# Calculation \n",
+      "Yw1 = Ps * MW / (( P - Ps) * 29) \n",
+      "# Tw = Tg - Hvap * ( Yw1 - Y1) / (hG / kY), where hG/kY is the psychmetric ratio pr\n",
+      "Y1 = 0 \n",
+      "Tg = Tw + Hvap * ( Yw1 - Y1) / pr \n",
+      "\n",
+      "# Result\n",
+      "print \"The air temperature = %.2f\"%Tg,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The air temperature = 299.49 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.15 pageno : 225"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Td = 353.2                      #K\n",
+      "Tw = 308.                       #K\n",
+      "Hvap = 2418.5                   #kJ/kg\n",
+      "pr = 0.950                      #kJ/kg\n",
+      "Ps = 5.62                       #kPa\n",
+      "P = 101.3                       #kPa\n",
+      "\n",
+      "# Calculation \n",
+      "Yw1 = (Ps * 18)/ (( P - Ps) * 29) \n",
+      "Y1 = Yw1 - pr * ( Td - Tw ) / Hvap \n",
+      "\n",
+      "# Result\n",
+      "print \"Humidity = %.4f\"%Y1,\"kg water/kg dry air\"\n",
+      "#humidity can also be directly obtained from psychometric chart, which we get to be 0.018 kg water/kg dry air\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Humidity = 0.0187 kg water/kg dry air\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.16 page no : 227"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "import math\n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "# Variables\n",
+      "TK = [283,293,303,313]\n",
+      "PaS = []\n",
+      "Ys = []\n",
+      "\n",
+      "# Calculations - part A\n",
+      "for i in range(len(TK)):\n",
+      "    Ps = 13.8858 - (2788.51 / (TK[i] - 52.36))\n",
+      "    PaS.append(round(math.e**Ps,2))\n",
+      "    Ys.append(PaS[i]/(101.3 - PaS[i]) * 78.048/29)\n",
+      "    \n",
+      "# Results\n",
+      "plot(TK,Ys)\n",
+      "plot(TK,Ys,\"go\")\n",
+      "xlabel(\"Temperature, K\")\n",
+      "ylabel(\"Y', kg benzene/ kg dry air\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 1,
+       "text": [
+        "<matplotlib.text.Text at 0x3797790>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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33/LMM8/w4IMPOj+kOqTFyTIy4E9/gi++MFsNt96qloK4P6cu9mO32/nvf//L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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2c0e610>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.17 pageno : 228"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Tin = 380.7                             #K\n",
+      "Pin = 101.3                             #kPa\n",
+      "Tdew = 298.                             #K\n",
+      "mremoved = 2.25                         #kg\n",
+      "V = 100.                                #m**3\n",
+      "#using humidity chart, humidity of air at dry bulb temperature of 380.7K and dew point = 298K is,\n",
+      "Y = 0.02                                # kg water /kg dry air\n",
+      "print \"(a)Humidity of air entering the drier = \",Y,\"kg water /kg dry air\"\n",
+      "Tstp = 273.15                           #K\n",
+      "Vstp = 22.4143                          #m**3/kmol\n",
+      "\n",
+      "# Calculation \n",
+      "N = V * Tstp / ( Vstp * Tin ) \n",
+      "MY = Y * 29 / 18.                       #molal humidity\n",
+      "Ndryair = N / ( 1 + MY ) \n",
+      "mdryair = Ndryair *29 \n",
+      "mwaterin = mdryair * Y \n",
+      "mwaterout = mwaterin + mremoved \n",
+      "Yout = mwaterout / mdryair \n",
+      "# percent humidity is calculated using the chart, and is\n",
+      "PY = 55.                                #%\n",
+      "\n",
+      "# Result\n",
+      "print \"(b)exit air humidity = %.3f\"%Yout,\"kg water /kg dry air\"\n",
+      "print \"Percent humidity = \",PY,\"%\"\n",
+      "#from the humidity chart \n",
+      "Twet = 313.2                            #K\n",
+      "Td = 322.2                              #K\n",
+      "print \"(c)exit air wet bulb temperature = \",Twet,\"K\"\n",
+      "print \"(c)exit air dry bulb temperature = \",Td,\"K\"\n",
+      "MYout = Yout * 29 / 18 \n",
+      "Nout = Ndryair * ( 1 + MYout ) / 1 \n",
+      "V1 = Nout * Vstp * Td / Tstp \n",
+      "print \"(d)Volume of exit air = %.2f\"%V1,\"m**3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Humidity of air entering the drier =  0.02 kg water /kg dry air\n",
+        "(b)exit air humidity = 0.045 kg water /kg dry air\n",
+        "Percent humidity =  55.0 %\n",
+        "(c)exit air wet bulb temperature =  313.2 K\n",
+        "(c)exit air dry bulb temperature =  322.2 K\n",
+        "(d)Volume of exit air = 87.94 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.18 pageno : 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "P = 101.3                       #kPa\n",
+      "Td = 303.                       #K\n",
+      "Tw = 288.                       #K\n",
+      "#using psychometric chart,\n",
+      "Y1 = 0.0045                     #kg water/ kg dry air\n",
+      "PY = 18.                        #%\n",
+      "Theated = 356.7                 #K\n",
+      "Cb = 1.005 \n",
+      "Ca = 1.884 \n",
+      "\n",
+      "# Calculation \n",
+      "Cs = Cb + Y1 * Ca \n",
+      "Q = 1 * Cs * (Theated - Td) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Humidity of the initial air = %.4f\"%Y1,\"kg water/ kg dry air\"\n",
+      "print \"(b)Percent humidity = \",PY,\"%\"\n",
+      "print \"(c)Temperature to which the air is heated = \",Theated,\"K\"\n",
+      "print \"(d)Heat to be suppplied = %.2f\"%Q,\"kJ\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Humidity of the initial air = 0.0045 kg water/ kg dry air\n",
+        "(b)Percent humidity =  18.0 %\n",
+        "(c)Temperature to which the air is heated =  356.7 K\n",
+        "(d)Heat to be suppplied = 54.42 kJ\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.19 pageno : 232"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "Tw = 313.                           #K\n",
+      "Td = 333.                           #K\n",
+      "#Using th psychometric chart,\n",
+      "Y = 0.04                            #kg water/ kg dry air\n",
+      "PS = 26.5                           #%\n",
+      "VS = 1.18                           #m**3/kg dry air ( volume of saturated air )\n",
+      "VD = 0.94                           #m**3/kg dry air ( volume of dry air )\n",
+      "\n",
+      "# Calculation \n",
+      "VH = VD + PS * (VS - VD )/100 \n",
+      "HS = 470.                           #J / kg dry air ( enthalpy of saturated air )\n",
+      "HD = 60.                            #J / kg dry air ( enthalpy of dry air )\n",
+      "H = HD + PS * ( HS - HD )/100 \n",
+      "Cs = 1.005 + (Y * 1.884)\n",
+      "H = Cs*(333 - 273.15) + Y*2502.3\n",
+      "# Result\n",
+      "print \"(a)Absolute Humidity of the air = \",Y,\"kg water/ kg dry air\"\n",
+      "print \"(b)Percent humidity = \",PS,\"%\"\n",
+      "print \"(c)Humid volume = \",VH,\"m**3/kg dry air\"\n",
+      "print \"(d)Enthalpy of wet air = \",H,\"kJ/kg dry air\"\n",
+      "print \"Enthalpy of humid air : %.2f kJ/kg dry air\"%H"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Absolute Humidity of the air =  0.04 kg water/ kg dry air\n",
+        "(b)Percent humidity =  26.5 %\n",
+        "(c)Humid volume =  1.0036 m**3/kg dry air\n",
+        "(d)Enthalpy of wet air =  164.751546 kJ/kg dry air\n",
+        "Enthalpy of humid air : 164.75 kJ/kg dry air\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
diff --git a/Stoichiometry_And_Process_Calculations/ch9.ipynb b/Stoichiometry_And_Process_Calculations/ch9.ipynb
new file mode 100755
index 00000000..a41ec201
--- /dev/null
+++ b/Stoichiometry_And_Process_Calculations/ch9.ipynb
@@ -0,0 +1,1190 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Material Balance in Unit Operations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 pageno : 251"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "PC1 = 85.                       #% ( Percent carbon in coal )\n",
+      "PA1 = 15.                       #% ( Percent ash in coal )\n",
+      "PA2 = 80.                       #% ( Percent ash in cinder )\n",
+      "PC2 = 20.                       #% ( Percent carbon in cinder )\n",
+      "m = 100.                        #kg (weight of coal )\n",
+      "\n",
+      "# Calculation \n",
+      "mash = PA1 * m / 100 \n",
+      "w = mash * 100 / PA2            # weight of cinder\n",
+      "mcarbon = w - mash \n",
+      "Plost = mcarbon * 100 / ( m - mash ) \n",
+      "\n",
+      "# Result\n",
+      "print \"weight of cinder formed = \",w,\"kg\"\n",
+      "print \"Percent fuel lost = %.2f\"%Plost,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "weight of cinder formed =  18.75 kg\n",
+        "Percent fuel lost = 4.41 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 pageno : 253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 1.                  #kg ( mass of completely dry wood )\n",
+      "P1 = 40.                #% ( percentage moisture in wet wood )\n",
+      "P2 = 5.                 #% ( Percentage moisture in dry wood )\n",
+      "\n",
+      "# Calculation \n",
+      "mwaterin = P1 * m / ( 100 - P1 ) \n",
+      "mwaterout = P2 * m / ( 100 - P2 ) \n",
+      "mevaporated = mwaterin - mwaterout \n",
+      "\n",
+      "# Result\n",
+      "print \"mass of water evaporated per kg of dry wood = %.3f\"%mevaporated,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mass of water evaporated per kg of dry wood = 0.614 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 pageno : 254"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F1 = 6*1000.                            #L/s\n",
+      "BOD1 = 3 * 10.**-5                      #g/L\n",
+      "BOD2 = 5 * 10.**-3                      #g/L\n",
+      "\n",
+      "# Calculation \n",
+      "V = 16 * 10.**3                         #m**3/day\n",
+      "v = V * 10**3 / (24 * 3600.)            #L/s\n",
+      "#Let BOD of the effluent be BODeff,\n",
+      "BODeff = (BOD2 * (F1 + v) - BOD1 * F1) / ( v ) \n",
+      "\n",
+      "# Result\n",
+      "print \"BOD of the effluent of the plant = %.4f\"%BODeff,\"g/L\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "BOD of the effluent of the plant = 0.1660 g/L\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 pageno : 256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "D = 100.                        #kg of overhead product\n",
+      "xfa = 0.956 \n",
+      "xdw = 0.074 \n",
+      "xdb = 0.741 \n",
+      "xda = 0.185 \n",
+      "\n",
+      "# Calculation \n",
+      "#water balance gives\n",
+      "F = D * xdw / (1 - xfa)  \n",
+      "W = F * xfa - xda * D \n",
+      "W1 = 100 \n",
+      "B = xdb*D \n",
+      "Bused = B * W1 / W \n",
+      "\n",
+      "# Result\n",
+      "print \"Quantity of benzene required = %.2f\"%Bused,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantity of benzene required = 52.08 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 pageno : 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#let, W - waste acid, S - Sulfuric acid, N - nitric acid, M - mixed acid\n",
+      "xsh2so4 = 0.95 \n",
+      "xsh2o = 0.5 \n",
+      "xwh2so4 = 0.3 \n",
+      "xwhno3 = 0.36 \n",
+      "xwh2o = 0.34 \n",
+      "xmh2so4 = 0.4 \n",
+      "xmhno3 = 0.45 \n",
+      "xmh2o = 0.15 \n",
+      "xnhno3 = 0.8 \n",
+      "xnh2o = 0.2 \n",
+      "M = 1000.                       #kg\n",
+      "\n",
+      "# Calculation \n",
+      "# total material balance, W + S + N = 1000 \n",
+      "#H2SO4 balance, xwh2so4 * W + xsh2so4 * S = xmh2so4*M\n",
+      "#HNO3 balance, xwhno3 * W + xnhno3 * N = xmhno3*M\n",
+      "#H2O balance, xwh2o*W+xnh2o*N + xsh2o*S = xmh2o*M\n",
+      "#solving the above equations simultaneously, we get,\n",
+      "W = 70.22                       #kg\n",
+      "S = 398.88                      #kg\n",
+      "N = 530.9                       #kg\n",
+      "\n",
+      "# Result\n",
+      "print \"Waste acid = \",W,\"kg\"\n",
+      "print \"Concentrated H2SO4 = \",S,\"kg\"\n",
+      "print \"Concentrated HNO3 = \",N,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Waste acid =  70.22 kg\n",
+        "Concentrated H2SO4 =  398.88 kg\n",
+        "Concentrated HNO3 =  530.9 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 pageno : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F = 1000.                       #kg\n",
+      "Psolute1 = 20.                  #%\n",
+      "Psolute2 = 80.                  #%\n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#taking solute balance\n",
+      "L3 = F * Psolute1 / Psolute2 \n",
+      "#taking total material balance\n",
+      "V = (F -L3) / 3 \n",
+      "#for first effect, total balance gives,\n",
+      "L1 = F - V \n",
+      "#solute balance gives,\n",
+      "Psolute3 = F * Psolute1 / L1 \n",
+      "#For second effect, total balance gives,\n",
+      "L2 = L1 - V \n",
+      "#solute balnce gives,\n",
+      "Psolute4 = L1 * Psolute3 / L2 \n",
+      "\n",
+      "# Result\n",
+      "print \"solute entering second effect = %.2f\"%Psolute3,\"%\"\n",
+      "print \"Weight entering second effect\",L1,\"kg\"\n",
+      "print \"solute entering third effect = \",Psolute4,\"%\"\n",
+      "print \"Weight entering third effect\",L2,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "solute entering second effect = 26.67 %\n",
+        "Weight entering second effect 750.0 kg\n",
+        "solute entering third effect =  40.0 %\n",
+        "Weight entering third effect 500.0 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7 pageno : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F = 100.                            #kg\n",
+      "xf = 0.25 \n",
+      "x2 = 7/107. \n",
+      "P1 = 10 #%\n",
+      "\n",
+      "# Calculation \n",
+      "W3 = P1 * F * (1-xf)/100.           #(W3 - weight of water evaporated)\n",
+      "\n",
+      "# let W1 and W2 be weight of crystal and weight of mother liquor remaining after crystallization resp_,\n",
+      "#F = W1 + W2 + W3\n",
+      "#100 = W1 + W2 + 7.5\n",
+      "#solute balance gives, F*xf = W1*x1 + W2*x2\n",
+      "#100*0.25 = W1*1+W2 * 0.0654\n",
+      "W2 = (F - W3 - F*xf)/(1-x2) \n",
+      "W1 = F - W3 - W2 \n",
+      "\n",
+      "# Result\n",
+      "print \"yeild of the crystals = %.2f\"%W1,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "yeild of the crystals = 20.28 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8 pageno : 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F = 100.                    #kg\n",
+      "xf = 0.15 \n",
+      "P1 = 80.                    #% ( Carbonate recovered )\n",
+      "M1 = 106.                   #(Molecular weight of Na2CO3)\n",
+      "M2 = 286.                   #(Molecular weight of Na2CO3.10H2O)\n",
+      "\n",
+      "# Calculation \n",
+      "x1 = M1 / M2                #(Weight fraction of Na2CO3 in crystals)\n",
+      "Mrecovered = P1 * F * xf / 100 \n",
+      "Wcrystal = Mrecovered / x1 \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)quantity of crystals formed = %.2f\"%Wcrystal,\"kg\"\n",
+      "#Na2CO3 balance gives, F*xf = Wcrystal*x1 + W2*x2\n",
+      "#W2 weight of mother liquor remaining after crystallization\n",
+      "#let M = W2 * x2,therefore\n",
+      "M = F * xf - Mrecovered \n",
+      "x2 = 0.09 \n",
+      "W2 = M/x2 \n",
+      "W3 = F - Wcrystal - W2      #weight of water evaporated\n",
+      "print \"(b)Weight of water evaporated = %.2f\"%W3,\"kg\"\n",
+      "\n",
+      "# note : answer may vary because of rounding error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)quantity of crystals formed = 32.38 kg\n",
+        "(b)Weight of water evaporated = 34.29 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9 pageno : 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 100.                    #kg (of 60% solution)\n",
+      "#w - water added to the original solution\n",
+      "#w1 - wt. of Na2S2O3.5H2O crystallized\n",
+      "#w2 - wt. of mother liquor obtained\n",
+      "#w3 - solution carried away by the crystals\n",
+      "xf = 0.6 \n",
+      "Mna2s2o3 = 158 \n",
+      "Mna2s2o35h2o = 248 \n",
+      "\n",
+      "# Calculation \n",
+      "mcrystals = m * xf * Mna2s2o35h2o / Mna2s2o3 \n",
+      "# free water available = m + w - 1 - mcrystals\n",
+      "#concentration of impurity = 1/(w+4.823)\n",
+      "#total balance, 100 - 1 + w = w1 + w2 + w3\n",
+      "#w1 + w2 + w3 - w = 99\n",
+      "#Na2S2O3 balance, 60 = (w1 + w2 * 1.5/2.5 + w3 * 1.5/2.5)*158/248\n",
+      "#w1 + 0.6 * w2 + 0.6 * w3 = 94.177\n",
+      "#each gram crystals carry 0.05 kg solution,\n",
+      "#w3 = 0.05 * w1\n",
+      "#impurity % = 0.1\n",
+      "#impurity = w3 /(2.5 * (w+4.823))\n",
+      "#solving above equations, we get\n",
+      "w = 14.577                  #kg\n",
+      "w1 = 65.08                  #kg\n",
+      "w2 = 45.25                  #kg\n",
+      "w3 = 0.05 * w1\n",
+      "\n",
+      "# Result \n",
+      "print \"(a)amount of water  added = %.2f\"%w,\"kg\"\n",
+      "print \"(b)amount of Na2S2O3.5H2O crystals added = \",w1,\"kg\"\n",
+      "m1 = w1 * Mna2s2o3 / Mna2s2o35h2o + w3 * 1.5 * Mna2s2o3 / (2.5 * Mna2s2o35h2o) \n",
+      "P = m1*100/(m*xf) \n",
+      "print \"(c)Percentage recovery of Na2S2O3 = %.1f\"%P,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)amount of water  added = 14.58 kg\n",
+        "(b)amount of Na2S2O3.5H2O crystals added =  65.08 kg\n",
+        "(c)Percentage recovery of Na2S2O3 = 71.2 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10 pageno : 271"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "m = 100.                        #kg\n",
+      "Pin1 = 40.                      #% ( tannin )\n",
+      "Pin2 = 5.                       #% ( moisture )\n",
+      "Pin3 = 23.                      #% ( soluble non tannin material )\n",
+      "\n",
+      "# Calculation \n",
+      "Pin4 = 100 - Pin1 - Pin2 - Pin3 #% ( insoluble lignin )\n",
+      "\n",
+      "# since, lignin is insoluble, all of it will be present in the residue\n",
+      "Pout1 = 3.                      #%\n",
+      "Pout2 = 50.                     #%\n",
+      "Pout3 = 1.                      #%\n",
+      "Pout4 = 100 - Pout1 - Pout2 - Pout3 \n",
+      "\n",
+      "#let W be the mass of residue, then we get\n",
+      "W = Pin4 * m / Pout4 \n",
+      "Ptannin = W * Pout1 * 100 / (m * Pin1) \n",
+      "\n",
+      "# Result\n",
+      "print \"Percent of original tannin unextracted = %.1f\"%Ptannin,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percent of original tannin unextracted = 5.2 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11 pageno : 271\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F = 100.                            #kg\n",
+      "\n",
+      "#F - feed, R - overflow, U - underflow, S - solvent\n",
+      "#F + S = U + R ( Total balance )\n",
+      "Poil1 = 49.                         #% ( 1 - feed )\n",
+      "Ppulp1 = 40.                        #%\n",
+      "Psalts1 = 3.                        #%\n",
+      "\n",
+      "# Calculation \n",
+      "Pwater = 100 - Poil1 - Ppulp1 - Psalts1 \n",
+      "Phexane2 = 25.                      #%(2 - underflow)\n",
+      "Psalts2 = 2.5                       #%\n",
+      "Poil2 = 15.                         #%\n",
+      "Pwater2 = 7.5                       #%\n",
+      "Ppulp2 = 100 - Phexane2 -  Poil2 - Pwater2 - Psalts2 \n",
+      "Poil3 = 25.                         #% ( 3 - extract )\n",
+      "\n",
+      "#taking pulp ( inert ) balance\n",
+      "U = Ppulp1 * F / Ppulp2 \n",
+      "\n",
+      "#oil balance gives, F * Poil1 = U * Poil2 + R * Poil3,from these , we get\n",
+      "R = (F * Poil1 - U * Poil2)/Poil3 \n",
+      "S = U + R - F \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)The amount of solvent used for extraction = \",S,\"kg\"\n",
+      "Precovered = 95.                    #%\n",
+      "mhexane2 = Phexane2 * U / 100 \n",
+      "mrecovered = mhexane2 * Precovered / 100 \n",
+      "P = mrecovered * 100 / S  \n",
+      "print \"(b)Percent of hexane used that is recovered from the underflow = %.2f\"%P,\"%\"\n",
+      "Poil = Poil3 * R * 100 / (F * Poil1 ) \n",
+      "print \"(c)Percent recovery of oil = %.2f\"%Poil,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The amount of solvent used for extraction =  128.0 kg\n",
+        "(b)Percent of hexane used that is recovered from the underflow = 14.84 %\n",
+        "(c)Percent recovery of oil = 75.51 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.12 pageno : 274"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#F = feed(wet solid), V1 = water evaporated(drier), \n",
+      "#V2 = water evaporated(oven), S1 = Dry solid(drier), S2 = Dry solid(oven)\n",
+      "# variables \n",
+      "F = 1000.                               #kg\n",
+      "xf = 0.8 \n",
+      "x1 = 0.15 \n",
+      "x2 = 0.02 \n",
+      "\n",
+      "# Calculation \n",
+      "#moisture free solid balance for drier, F * ( 1 - xf) = S1 * ( 1 - x1 )\n",
+      "S1 = F * ( 1 - xf )/(1 - x1) \n",
+      "\n",
+      "#total balance for drier , F = S1 + V1\n",
+      "V1 = F - S1 \n",
+      "\n",
+      "#For oven, S1 * ( 1 - x1 ) = S2 * ( 1 -x2 )\n",
+      "S2 = S1 * ( 1 - x1 )/(1 - x2) \n",
+      "\n",
+      "#Also, S1 = S2 + V2\n",
+      "V2 = S1 - S2 \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Weight of product leaving the drier = %.2f\"%S1,\"kg\"\n",
+      "print \"   Weight of product leaving the oven = %.2f\"%S2,\"kg\"\n",
+      "P1 = V1 *100/ (F * xf) \n",
+      "P2 = V2 *100/ (F * xf) \n",
+      "\n",
+      "print \"(b)Percentage of original water removed in drier = %.1f\"%P1,\"%\"\n",
+      "print \"   Percentage of original water removed in oven = %.2f\"%P2,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Weight of product leaving the drier = 235.29 kg\n",
+        "   Weight of product leaving the oven = 204.08 kg\n",
+        "(b)Percentage of original water removed in drier = 95.6 %\n",
+        "   Percentage of original water removed in oven = 3.90 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.13 pageno : 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#Ss = solid flow rate,\n",
+      "\n",
+      "Pwaterin = 25.                          #%\n",
+      "Pwaterout = 5.                          #%\n",
+      "\n",
+      "# Calculation \n",
+      "X1 = Pwaterin/(100 - Pwaterin)          #kg water/kg dry air\n",
+      "X2 = Pwaterout/(100 - Pwaterout)        #kg water/kg dry air\n",
+      "\n",
+      "#form humidity chart,\n",
+      "Y2 = 0.015                              #kg water/kg dry air\n",
+      "Y1 = 0.035                              #kg water/kg dry air\n",
+      "m = 1.                                  #kg of dry air\n",
+      "\n",
+      "#Ss * X1 + Y2 = Ss * X2 + Y1\n",
+      "Ss = (Y1 - Y2) / ( X1 - X2 ) \n",
+      "T = 87.5 + 273.15                       #K\n",
+      "P = 101.3                               #kPa\n",
+      "Tstp = 273.15                           #K\n",
+      "Pstp = 101.3                            #kPa\n",
+      "Vstp = 22.4143                          #m**3/mol\n",
+      "V = 100.                                #m**3\n",
+      "N = V * P * Tstp / ( Vstp * Pstp * T) \n",
+      "Nr2 = Y2 * 29 / 18.                     #kmol of water / kmol of dry air\n",
+      "Ndryair = N * 1 / (1 + Nr2) \n",
+      "mdryair = Ndryair * 29 \n",
+      "mevaporated = mdryair * ( Y1 - Y2 ) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)total moisture evaporated per 100m**3 of air entering = %.3f\"%mevaporated,\"kg\"\n",
+      "Ss1 = mdryair * Ss \n",
+      "mproduct = Ss1 * ( 1 + X2 ) \n",
+      "print \"(b)mass of finished product per 100m**3 of air entering = %.2f\"%mproduct,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)total moisture evaporated per 100m**3 of air entering = 1.914 kg\n",
+        "(b)mass of finished product per 100m**3 of air entering = 7.18 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.14 pageno : 278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "#F = feed, E = extract, S = solvent, R = Raffinate\n",
+      "\n",
+      "xwaterF = 0.7                           #Feed\n",
+      "xalcoholF = 0.3 \n",
+      "xwaterR = 0.71                          #raffinate\n",
+      "xalcoholR = 0.281 \n",
+      "xethyR = 0.009 \n",
+      "xwaterE = 0.008                         #Extract\n",
+      "xalcoholE = 0.052 \n",
+      "xethyE = 0.94 \n",
+      "\n",
+      "#Total balance, R + E = F + S\n",
+      "F = 100.                                #kg\n",
+      "\n",
+      "#R + E = 100 + S                         (1)\n",
+      "#Isopropyl balance, xalcoholR * R + xalcoholE*E = xalcoholF * F\n",
+      "#0.281*R + 0.052 * E = 30                (2)\n",
+      "#Ethylene tetra chloride balance, xethyR * R + xethyE * E = S\n",
+      "#0.009*R + 0.94*E = S                    (3)\n",
+      "#Solving equation 1, 2 and 3 simultaneously, we get,\n",
+      "\n",
+      "S = 45.1 \n",
+      "E = 47.04 \n",
+      "R = 98.06 \n",
+      "# Calculation and Result\n",
+      "print \"(a)Amount of solvent used = \",S,\"kg\"\n",
+      "print \"(b)Amount of extract = \",E,\"kg\"\n",
+      "print \"   Amount of raffinate = \",R,\"kg\"\n",
+      "mextracted = E * xalcoholE \n",
+      "P1 = mextracted * 100 / (F * xalcoholF) \n",
+      "print \"(c)Percent of isopropyl alcohol extracted = %.2f\"%P1,\"%\"\n",
+      "\n",
+      "# note : answer may vary because of rounding error.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Amount of solvent used =  45.1 kg\n",
+        "(b)Amount of extract =  47.04 kg\n",
+        "   Amount of raffinate =  98.06 kg\n",
+        "(c)Percent of isopropyl alcohol extracted = 8.15 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.15 pageno : 282"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "G1 = 100.                           #kmol\n",
+      "\n",
+      "#G1 and G2 be the molar flow rate of the gas at the inlet and the\n",
+      "# exit of the absorber resp_,y1 and y2 mole fraction at entrance and exit resp_,\n",
+      "\n",
+      "y1 = 0.25                           #%\n",
+      "y2 = 0.05                           #%\n",
+      "\n",
+      "# Calculation \n",
+      "#air balance gives, G1 * ( 1-y1 ) = G2 * ( 1-y2 )\n",
+      "G2 = G1 * ( 1-y1 ) / (1 - y2) \n",
+      "maleaving = G2 * y2 \n",
+      "maentering = G1 * y1 \n",
+      "Pabsorbed = (maentering - maleaving) * 100 / ( maentering) \n",
+      "\n",
+      "# Result\n",
+      "print \"Percentage of acetone absorbed = %.2f\"%Pabsorbed,\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage of acetone absorbed = 84.21 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.16 pageno : 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F = 5000.                       #kg/h\n",
+      "P1 = 50.                        #% (H2O4 in)\n",
+      "MH2SO4 = 98.016 \n",
+      "P1gas = 65.                     #(nitrogen in gas entering)\n",
+      "P2gas = 35.                     # ( SO3)\n",
+      "MN2 = 28. \n",
+      "MSO3 = 80. \n",
+      "\n",
+      "# Calculation \n",
+      "Mavg = ( MN2 * P1gas + MSO3 * P2gas)/100            #avg molecular wt. of entering gas\n",
+      "G = 4500.                       #kg/h\n",
+      "Ng = G / Mavg \n",
+      "NN2 = Ng * P1gas / 100 \n",
+      "NSO3 = Ng - NN2 \n",
+      "P2 = 75.                        #% (H2O4 out)\n",
+      "\n",
+      "#W be the mass of 75% H2SO4, x and y be the moles of SO3 and water vapour leaving resp_,\n",
+      "Pwater = 25.                    #kPa\n",
+      "Ptotal = 101.3                  #kPa\n",
+      "\n",
+      "#Pwater / Ptotal = y / ( NN2 + x + y )\n",
+      "#we get, y = 0.32765 * x + 2.744           (1)\n",
+      "#Total balance Feed + G = W + (NN2 * 28 + x * 80 + y * 18.016)\n",
+      "#we get, W + 80*x + 18.016*y = 7727.32     (2)\n",
+      "#from 1 and 2, 84.9174*x + W = 7352.68     (3)\n",
+      "#SO3 balance, So3 eneterin with 50% H2SO4 + SO3 in feed gas = SO leaving with 75%H2SO4 + SO3 leaving in exit gas\n",
+      "#5000*0.5*80/98.016 + 34.09*80 = 80* x + 0.75*W * 80/98.016   (4)\n",
+      "# from 3 and 4,\n",
+      "x = 9.74 \n",
+      "Nabsorbed = NSO3 - x \n",
+      "Pabsorbed = Nabsorbed * 100 / NSO3 \n",
+      "\n",
+      "# Result\n",
+      "print \"Percentage of SO3 absorbed = %.2f\"%Pabsorbed,\"%\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Percentage of SO3 absorbed = 71.43 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.17 pageno : 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F = 200.                            #kmol/h\n",
+      "#F, D and W be the flow rates of the feed, the distillate and residue \n",
+      "# resp_, xf , xd and xw be the mole fraction of ethanol in the fee, distillate and the residue resp_\n",
+      "\n",
+      "xf = 0.10 \n",
+      "xd = 0.89 \n",
+      "xw = 0.003 \n",
+      "\n",
+      "\n",
+      "# Calculation \n",
+      "#total balance gives, F = D + W\n",
+      "#D + W = 200              (1)\n",
+      "#Alcohol balance gives, F*xf = D*xd + W*xw\n",
+      "#0.89*D+0.003*W = 20      (2)\n",
+      "#solving 1 and 2\n",
+      "D = 21.87 #kmol/h\n",
+      "W = 178.13 #kmol/h\n",
+      "Nawasted = W*xw \n",
+      "mmakeup = Nawasted * 46*24 \n",
+      "\n",
+      "# Result\n",
+      "print \"The make up alcohol required per day = %.2f\"%mmakeup,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The make up alcohol required per day = 589.97 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.18 pageno : 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F = 100.                            #kg\n",
+      "#F, D and W be the flow rates of the feed, the distillate and bottom \n",
+      "# product resp_, xf , xd and xw be the mole fraction of methanol in the \n",
+      "# fee, distillate and the bottom product resp_\n",
+      "\n",
+      "xf = 0.20 \n",
+      "xd = 0.97 \n",
+      "xw = 0.02 \n",
+      "#using, F = D + W and F*xf + D*xd + W*xw,we get\n",
+      "D = 18.95                           #kg/h\n",
+      "W = 81.05                           #kg/h\n",
+      "R = 3.5 \n",
+      "#R = L / D\n",
+      "#for distillate = 1kg\n",
+      "D1 = 1.                             #kg\n",
+      "\n",
+      "# Calculation \n",
+      "L = R*D1 \n",
+      "#Taking balance around the condenser,\n",
+      "G = L + D1 \n",
+      "mcondensed = G * D / F \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Amount of distillate = \",D,\"kg\"\n",
+      "print \"   Amount of Bottom Product = \",W,\"kg\"\n",
+      "print \"(b)Amount of vapour condensed per kg of distillate = \",G,\"kg\"\n",
+      "print \"(c)Amount of vapour condensed per kg of feed = %.3f\"%mcondensed,\"kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Amount of distillate =  18.95 kg\n",
+        "   Amount of Bottom Product =  81.05 kg\n",
+        "(b)Amount of vapour condensed per kg of distillate =  4.5 kg\n",
+        "(c)Amount of vapour condensed per kg of feed = 0.853 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.19 pageno : 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "mdryair = 1.                        #kg\n",
+      "Pwater1 = 1.4                       #kPa ( Partial pressure at 285K )\n",
+      "Pwater2 = 10.6                      #kPa ( Partial pressure at 320K )\n",
+      "P = 101.3                           # ( Total )\n",
+      "\n",
+      "# Calculation \n",
+      "Ys1 = Pwater2 * 18 / ((P - Pwater2)*29)             #( saturation humidity at 320K )\n",
+      "Ys2 = Pwater1 * 18 / ((P - Pwater1)*29)             #( saturation humidity at 285K )\n",
+      "Ys = 0.03                           #kg water / kg dry air. (final humidity)\n",
+      "\n",
+      "\n",
+      "# humidity of air leaving dehumidifier is Ys2 and humidity of bypassed \n",
+      "# air is Ys1. these 2 streams combine to give humidity of 0.03kg water / kg dry air.\n",
+      "#therefore, taking balance we get, 1*Ys2 + x * Ys1 = (1 + x)*Ys\n",
+      "x = (1*Ys2 - 1*Ys)/(Ys - Ys1) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Mass of dry air bypassed per kg of dry air sent through the dehumidifier = %.3f\"%x,\"kg dry air\"\n",
+      "mcondensed = Ys1 - Ys2 \n",
+      "mwetair = mdryair + Ys1 \n",
+      "Nwetair = mdryair/29 + Ys1/18.016 \n",
+      "Vstp = 22.4143                      #m**3/kmol\n",
+      "Vstp1 = Nwetair * Vstp \n",
+      "T = 320.                            #K\n",
+      "P = 101.3                           #kPa\n",
+      "Tstp = 273.15                       #K\n",
+      "Pstp = 101.325                      #kPa\n",
+      "V = Vstp1 * Pstp * T / (P * Tstp) \n",
+      "Vgiven = 100.                       #m**3\n",
+      "mcondensed1 = mcondensed * Vgiven / V \n",
+      "print \"(b)mass of water vapour condensed in the dehumidifier per 100m**3 of air sent through it = %.4f\"%mcondensed1,\"kg\"\n",
+      "mfinal = mdryair + x \n",
+      "mfinalair = mfinal * Vgiven / V \n",
+      "N = mfinalair / 29. \n",
+      "Ysn = Ys * 29/18.                   #kmol water / kmol dry air\n",
+      "Ntotal = N * (Ysn + 1) \n",
+      "Vfinal = Ntotal * Vstp * Pstp * T / ( Tstp * P ) \n",
+      "print \"(c)Volume of final air obtained per 100 cubic metres f air passed through dehumidifier = %.1f\"%\\\n",
+      "Vfinal,\"m**3\"\n",
+      "\n",
+      "# note: answer may vary because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Mass of dry air bypassed per kg of dry air sent through the dehumidifier = 0.501 kg dry air\n",
+        "(b)mass of water vapour condensed in the dehumidifier per 100m**3 of air sent through it = 6.3118 kg\n",
+        "(c)Volume of final air obtained per 100 cubic metres f air passed through dehumidifier = 140.9 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.20 pageno : 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# variables \n",
+      "F = 100.                            #kg/h\n",
+      "xf = 0.2 \n",
+      "xp = 0.93 \n",
+      "xr = 0.5/1.5 \n",
+      "xx = 0.65 \n",
+      "\n",
+      "# Calculation \n",
+      "#R - recycle stream, P - Product stream , W - water separeted and removed\n",
+      "#component A balance, F * xf = P * xp, that is,\n",
+      "P = F * xf / xp \n",
+      "#Total balance, F = P + W,therefore\n",
+      "W = F - P \n",
+      "#x be the flow rate of strea entering the filter \n",
+      "#total balance, x = P + R                             (1)\n",
+      "#component A balance , 0.65 * x = 0.5*R/1.5 + 0.93P   (2)\n",
+      "#Solving 1 and 2, we get,\n",
+      "R = (xx * P - xp * P)/(xr - xx) \n",
+      "\n",
+      "# Result\n",
+      "print \"Flow rate of the recycle stream = %d\"%R,\"kg/h\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow rate of the recycle stream = 19 kg/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.21 pageno : 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "F = 1000.                       #kg/h\n",
+      "xfwater = 0.7 \n",
+      "xpwater = 0.2 \n",
+      "xrwater = 0.20 \n",
+      "xswater = 0.5 \n",
+      "y1 = 0.0025 \n",
+      "y2 = 0.05 \n",
+      "\n",
+      "# Calculation \n",
+      "#R - recycle, S - stream entering granulator, P - Product, G1 - air entering the drier, G2 - air leaving the drier,\n",
+      "#takin overall, moisture free balance, F * xf = P * xp\n",
+      "P = F * ( 1 - xfwater )/(1 - xpwater ) \n",
+      "# taking material balance at point where recycle strea joins the feed,\n",
+      "# F = R + S\n",
+      "#water balance, F*xfwater = R*xrwater + S*xswater,solving this we get,\n",
+      "R = (-F*xfwater +F*xswater)/(xrwater - xswater) \n",
+      "S = F + R \n",
+      "mleaving = P + R                #solid leaving the drier\n",
+      "#dry air entering will there be in air leaving, therefore\n",
+      "#G1 * ( 1 - y1 ) = G2 * ( 1 - y2 )\n",
+      "# water balance over the drier gives, S*xswater+G1*y1=G2*y2+(P+R)*xpwater\n",
+      "#from above 2 equations , we get\n",
+      "G1 = ((mleaving*xpwater - S*xswater)/(y1 - y2*(1-y1)/(1-y2))) \n",
+      "\n",
+      "# Result\n",
+      "print \"(a)Amount of solid recycled = %.2f\"%R,\"kg/h\"\n",
+      "mdryair = G1 * (1 - y1) \n",
+      "print \"(b)circulation rate of air in the drier on dry basis = %.d\"%round(mdryair,-2),\"kg/h\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Amount of solid recycled = 666.67 kg/h\n",
+        "(b)circulation rate of air in the drier on dry basis = 12500 kg/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.22 pageno : 296"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables \n",
+      "xf = 500. * 10**-6 \n",
+      "xp = 50. * 10**-6 \n",
+      "xb = 1600. * 10**-6 \n",
+      "\n",
+      "# Calculation \n",
+      "#F - Feed water rate, B - blow down rate, S - high pressure steam, P - process stream rate\n",
+      "# total balance, F = P + B\n",
+      "# Solid balance, F * xf + P * xp = B * xb\n",
+      "#eliminating P, we get, F * xf + (F - B)*xp = B * xb\n",
+      "#let F/B be X\n",
+      "X = (xb + xp)/(xf + xp) \n",
+      "\n",
+      "# Result\n",
+      "print \"the ratio of feed water to the blowdown water = \",X\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the ratio of feed water to the blowdown water =  3.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
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