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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 |