diff options
| author | Trupti Kini | 2016-01-17 23:30:12 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-01-17 23:30:12 +0600 |
| commit | 2ec8fc25ee45e790f91f2dc0fb342712cc3de4b7 (patch) | |
| tree | fd6440fe7cf46a60449631fa5022acc65a04e81f /Electrical_Network_by_R._Singh/Chapter2_1.ipynb | |
| parent | b68dcdb18f92b07d76db49fc86a98ba287cf15fe (diff) | |
| download | Python-Textbook-Companions-2ec8fc25ee45e790f91f2dc0fb342712cc3de4b7.tar.gz Python-Textbook-Companions-2ec8fc25ee45e790f91f2dc0fb342712cc3de4b7.tar.bz2 Python-Textbook-Companions-2ec8fc25ee45e790f91f2dc0fb342712cc3de4b7.zip | |
Added(A)/Deleted(D) following books
A Electrical_Network_by_R._Singh/Chapter10_1.ipynb
A Electrical_Network_by_R._Singh/Chapter11_1.ipynb
A Electrical_Network_by_R._Singh/Chapter12_1.ipynb
A Electrical_Network_by_R._Singh/Chapter1_1.ipynb
A Electrical_Network_by_R._Singh/Chapter2_1.ipynb
A Electrical_Network_by_R._Singh/Chapter3_1.ipynb
A Electrical_Network_by_R._Singh/Chapter4_1.ipynb
A Electrical_Network_by_R._Singh/Chapter6_1.ipynb
A Electrical_Network_by_R._Singh/Chapter7_1.ipynb
A Electrical_Network_by_R._Singh/Chapter8_1.ipynb
A Electrical_Network_by_R._Singh/screenshots/chapter4_1.png
A Electrical_Network_by_R._Singh/screenshots/chapter6_1.png
A Electrical_Network_by_R._Singh/screenshots/chapter7_1.png
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter10_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter11_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter12_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter13_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter14_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter15_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter16_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter17_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter1_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter2_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter3_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter4_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter5_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter6_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter7_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter8_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/Chapter9_1.ipynb
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter3.png
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter4.png
A Electronic_Circuit_Analysis_And_Design_by_D._A._Neamen/screenshots/chapter5.png
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter1.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter10.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter11.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter12.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter13.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter14.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter15.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter16.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter17.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter18.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter19.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter2.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter20.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter3.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter4.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter5.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter6.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter7.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/Chapter8.ipynb
A Machine_Design_by_T._H._Wentzell,_P._E/screenshots/chapter1.png
A Machine_Design_by_T._H._Wentzell,_P._E/screenshots/chapter2.png
A Machine_Design_by_T._H._Wentzell,_P._E/screenshots/chapter3.png
Diffstat (limited to 'Electrical_Network_by_R._Singh/Chapter2_1.ipynb')
| -rw-r--r-- | Electrical_Network_by_R._Singh/Chapter2_1.ipynb | 1958 |
1 files changed, 1958 insertions, 0 deletions
diff --git a/Electrical_Network_by_R._Singh/Chapter2_1.ipynb b/Electrical_Network_by_R._Singh/Chapter2_1.ipynb new file mode 100644 index 00000000..20e60dfa --- /dev/null +++ b/Electrical_Network_by_R._Singh/Chapter2_1.ipynb @@ -0,0 +1,1958 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b2b5fb16825d4183b23bc2d431db6308dc6c5888398753fd91beb4d61ab4253a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Network Theorem 1"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg2.4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem-1\n",
+ "##pg no.-2.4\n",
+ "##example2.1\n",
+ "print(\"\\nConverting the two delta networks formed by resistors 4.5 Ohm, 3Ohm, and 7.5Ohm into equivalent star networks\");\n",
+ "a=4.5;\n",
+ "b=3.;\n",
+ "c=7.5;\n",
+ "R1= (a*c)/(a+b+c);\n",
+ "R2= (c*b)/(c+b+a);\n",
+ "R3= (a*b)/(a+b+c);\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nR1=R6 = \",R1,\" Ohm\" and \"\\nR2=R5 =\",R2,\" Ohm\" and \"\\nR3=R4 =\" ,R3,\" Ohm\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Converting the two delta networks formed by resistors 4.5 Ohm, 3Ohm, and 7.5Ohm into equivalent star networks\n",
+ "\n",
+ "R1=R6 = 2.25 \n",
+ "R2=R5 = 1.50 \n",
+ "R3=R4 = 0.90 Ohm \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg2.5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem-1\n",
+ "##pg no.-2.2\n",
+ "##example2.5\n",
+ "##converting delta network to star network\n",
+ "a=10.;\n",
+ "b=10.;\n",
+ "c=10.;\n",
+ "R=(a*b)/(a+b+c);\n",
+ "print(\"\\nConverting the delta formed by three resistors of 10 Ohm into an equivalent star network\");\n",
+ "print'%s %.2f %s'%(\"\\nR1=R2=R3= \",R,\" Ohm\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Converting the delta formed by three resistors of 10 Ohm into an equivalent star network\n",
+ "\n",
+ "R1=R2=R3= 3.33 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg2.7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem-1\n",
+ "##pg no.-2.7\n",
+ "##example2.3\n",
+ "a=4.;\n",
+ "b=3.;\n",
+ "c=6.;\n",
+ "##star to delta conversion\n",
+ "R1=c+a+((a*c)/b);\n",
+ "R2=c+b+((c*b)/a);\n",
+ "R3=a+b+((a*b)/c);\n",
+ "x=1.35;\n",
+ "y=0.9;\n",
+ "RAB=(c*(x+y))/(c+x+y);\n",
+ "print'%s %.2f %s'%(\"\\nR1 = \",R1,\" Ohm\");\n",
+ "print'%s %.2f %s'%(\"\\nR2 = \",R2,\" Ohm\");\n",
+ "print'%s %.2f %s'%(\"\\nR3 = \",R3,\" Ohm\");\n",
+ "print'%s %.2f %s'%(\"\\nThe network can be simplified as, \\nRAB = \",RAB,\" Ohm\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "R1 = 18.00 Ohm\n",
+ "\n",
+ "R2 = 13.50 Ohm\n",
+ "\n",
+ "R3 = 9.00 Ohm\n",
+ "\n",
+ "The network can be simplified as, \n",
+ "RAB = 1.64 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg2.9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem-1\n",
+ "##pg no.-2.9\n",
+ "##example2.5\n",
+ "##converting delta network to star network\n",
+ "a=25.;\n",
+ "b=20.;\n",
+ "c=35.;\n",
+ "R1=(b*c)/(a+b+c);\n",
+ "R2=(a*b)/(a+b+c);\n",
+ "R3=(a*c)/(a+b+c);\n",
+ "print(\"\\nConverting the delta formed by resistors 20 Ohm ,25 Ohm, 35 Ohm into an equivalent star network\");\n",
+ "print'%s %.2f %s'%(\"\\nR1= \",R1,\" Ohm\");\n",
+ "print'%s %.2f %s'%(\"\\nR2= \",R2,\" Ohm\");\n",
+ "print'%s %.2f %s'%(\"\\nR3= \",R3,\" Ohm\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Converting the delta formed by resistors 20 Ohm ,25 Ohm, 35 Ohm into an equivalent star network\n",
+ "\n",
+ "R1= 8.75 Ohm\n",
+ "\n",
+ "R2= 6.25 Ohm\n",
+ "\n",
+ "R3= 10.94 Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg2.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem-1\n",
+ "##pg no.-2.15\n",
+ "##example2.8\n",
+ "a=5;\n",
+ "b=4;\n",
+ "c=3;\n",
+ "##Star to delta conversion\n",
+ "R1=a+b+((a*b)/c);\n",
+ "R2=c+b+((c*b)/a);\n",
+ "R3=a+c+((a*c)/b);\n",
+ "a1=6;\n",
+ "b1=4;\n",
+ "c1=8;\n",
+ "##Satr to delta conversion\n",
+ "R4=a1+b1+((a1*b1)/c1);\n",
+ "R5=c1+b1+((c1*b1)/a1);\n",
+ "R6=a1+c1+((a1*c1)/b1);\n",
+ "x=6.17;\n",
+ "y=9.78;\n",
+ "RAB=(x*y)/(x+y);\n",
+ "print(\"\\nConverting star network formed by 3 Ohm,4 Ohm ,5 Ohm into equivalent delta network \");\n",
+ "print'%s %.2f %s %.2f %s %.2f %s'%(\"\\nR1= \",R1,\" Ohm\" and \" \\nR2= \",R2,\" Ohm\" and \" \\nR3 = \",R3,\" Ohm\");\n",
+ "print(\"\\nSimilarly, converting star network formed by 6 Ohm,4 Ohm ,8 Ohm into equivalent delta network\");\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nR4= \",R4,\" Ohm\" and \" \\nR5= \",R5,\" Ohm \" and \"\\nR6 =\",R6,\" Ohm\");\n",
+ "print'%s %.2f %s'%(\"\\n Simplifying the parallel networks, we get \\nRAB = \",RAB,\" Ohms\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Converting star network formed by 3 Ohm,4 Ohm ,5 Ohm into equivalent delta network \n",
+ "\n",
+ "R1= 15.00 \n",
+ "R2= 9.00 \n",
+ "R3 = 11.00 Ohm\n",
+ "\n",
+ "Similarly, converting star network formed by 6 Ohm,4 Ohm ,8 Ohm into equivalent delta network\n",
+ "\n",
+ "R4= 13.00 \n",
+ "R5= 17.00 \n",
+ "R6 = 26.00 Ohm \n",
+ "\n",
+ " Simplifying the parallel networks, we get \n",
+ "RAB = 3.78 Ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg2.18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.18\n",
+ "##example2.9\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"Applying KVL to mesh 1\");\n",
+ "print(\"10*I1-3*I2-6*I3=0\")##equation 1\n",
+ "print(\"Applying KVL to mesh 2\");\n",
+ "print(\"-3*I1+10*I2=-5\");##equation 2\n",
+ "print(\"Applying KVL to mesh 3\");\n",
+ "print(\"-6*I1+10*I3=25\");##equation 3\n",
+ "print(\"Solving the three equations\");\n",
+ "A=numpy.matrix([[10, -3, -6],[-3 ,10 ,0],[-6, 0, 10]]) ##solving the equations in matrix form\n",
+ "B=numpy.matrix([[10], [-5], [25]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B)\n",
+ "\n",
+ "print[X];\n",
+ "print(\"I1=4.27 A\");\n",
+ "print(\"I2=0.78 A\");\n",
+ "print(\"I3=5.06 A\");\n",
+ "print(\"I5ohm=4.27 A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KVL to mesh 1\n",
+ "10*I1-3*I2-6*I3=0\n",
+ "Applying KVL to mesh 2\n",
+ "-3*I1+10*I2=-5\n",
+ "Applying KVL to mesh 3\n",
+ "-6*I1+10*I3=25\n",
+ "Solving the three equations\n",
+ "[matrix([[ 4.27272727],\n",
+ " [ 0.78181818],\n",
+ " [ 5.06363636]])]\n",
+ "I1=4.27 A\n",
+ "I2=0.78 A\n",
+ "I3=5.06 A\n",
+ "I5ohm=4.27 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg2.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.19\n",
+ "##example 2.10\n",
+ "import math\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "print(\"Applying KVL to mesh 1\");\n",
+ "print(\"7*I1-I2=10\")##equation 1\n",
+ "print(\"Applying KVL to mesh 2\");\n",
+ "print(\"-I1+6*I2-3*I3=0\");##equation 2\n",
+ "print(\"Applying KVL to mesh 3\");\n",
+ "print(\"-3*I2+13*I3=-20\");##equation 3\n",
+ "print(\"Solving the three equations\");\n",
+ "A=numpy.matrix([[7 ,-1, 0],[-1, 6, -3],[0 ,-3, 13]]);##solving the equations in matrix form\n",
+ "B=numpy.matrix([[10], [0], [-20]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"I1=1.34 A\");\n",
+ "print(\"I1=-0.62 A\");\n",
+ "print(\"I3=-1.68 A\");\n",
+ "print(\"I2ohm=1.34 A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KVL to mesh 1\n",
+ "7*I1-I2=10\n",
+ "Applying KVL to mesh 2\n",
+ "-I1+6*I2-3*I3=0\n",
+ "Applying KVL to mesh 3\n",
+ "-3*I2+13*I3=-20\n",
+ "Solving the three equations\n",
+ "[matrix([[ 1.34042553],\n",
+ " [-0.61702128],\n",
+ " [-1.68085106]])]\n",
+ "I1=1.34 A\n",
+ "I1=-0.62 A\n",
+ "I3=-1.68 A\n",
+ "I2ohm=1.34 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg2.20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.20\n",
+ "##example 2.11\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"Applying KVL to mesh 1\");\n",
+ "print(\"3*I1-I2-2*I3=8\");##equation 1\n",
+ "print(\"Applying KVL to mesh 2\");\n",
+ "print(\"-I1+8*I2-3*I3=10\");##equation 2\n",
+ "print(\"Applying KVL to mesh 3\");\n",
+ "print(\"-2*I1-3*I2+10*I3=12\");##equation 3\n",
+ "print(\"Solving the three equations\");\n",
+ "A=numpy.matrix([[3 ,-1 ,-2],[-1, 8 ,-3],[-2 ,-3 ,10]]);##solving the equations in matrix form\n",
+ "B=numpy.matrix([[8], [10] ,[12]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"I1=6.01 A\");\n",
+ "print(\"I1=3.27 A\");\n",
+ "print(\"I3=3.38 A\");\n",
+ "print(\"I5ohm=3.38 A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KVL to mesh 1\n",
+ "3*I1-I2-2*I3=8\n",
+ "Applying KVL to mesh 2\n",
+ "-I1+8*I2-3*I3=10\n",
+ "Applying KVL to mesh 3\n",
+ "-2*I1-3*I2+10*I3=12\n",
+ "Solving the three equations\n",
+ "[matrix([[ 6.01257862],\n",
+ " [ 3.27044025],\n",
+ " [ 3.3836478 ]])]\n",
+ "I1=6.01 A\n",
+ "I1=3.27 A\n",
+ "I3=3.38 A\n",
+ "I5ohm=3.38 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg2.21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.21\n",
+ "##example 2.12\n",
+ "import numpy\n",
+ "print(\"Applying KVL to mesh 1\");\n",
+ "print(\"8*I1-I2-4*I3=4\");##equation 1\n",
+ "print(\"Applying KVL to mesh 2\");\n",
+ "print(\"-I1+8*I2-5*I3=0\");##equation 2\n",
+ "print(\"Applying KVL to mesh 3\");\n",
+ "print(\"-4*I1-5*I2+15*I3=0\");##equation 3\n",
+ "print(\"Solving the three equations\");\n",
+ "A=numpy.matrix([[8 ,-1 ,-4],[-1 ,8 ,-5],[-4 ,-5 ,15]]);##solving the equations in matrix form\n",
+ "B=numpy.matrix([[4] ,[0], [0]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"I1=0.66\");\n",
+ "print(\"I1=0.24 A\");\n",
+ "print(\"I3=0.26 A\");\n",
+ "print(\"current supplied by the battery = 0.66 A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KVL to mesh 1\n",
+ "8*I1-I2-4*I3=4\n",
+ "Applying KVL to mesh 2\n",
+ "-I1+8*I2-5*I3=0\n",
+ "Applying KVL to mesh 3\n",
+ "-4*I1-5*I2+15*I3=0\n",
+ "Solving the three equations\n",
+ "[matrix([[ 0.65857886],\n",
+ " [ 0.24263432],\n",
+ " [ 0.25649913]])]\n",
+ "I1=0.66\n",
+ "I1=0.24 A\n",
+ "I3=0.26 A\n",
+ "current supplied by the battery = 0.66 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg2.22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.22\n",
+ "##example 2.13\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"Applying KVL to mesh 1\");\n",
+ "print(\"V+13*I1-2*I2-5*I3=20\");##mesh equation 1\n",
+ "print(\"Applying KVL to mesh 2\");\n",
+ "print(\"2*I1-6*I2+I3=0\");##mesh equation 2\n",
+ "print(\"Applying KVL to mesh 3\");\n",
+ "print(\"V+5*I1+I2-10*I3=0\");##mesh equation 3\n",
+ "print(\"putting I1=0 in equation 1, 2 and 3 we get\"); \n",
+ "print(\"V-2*I2-5*I3=20\");##equation 1\n",
+ "print(\"-6*I2+I3=0\");##equation 2\n",
+ "print(\"V+I2-10*I3=0\");##equation 3\n",
+ "print(\"Solving the three equations\");\n",
+ "A=numpy.matrix([[1 ,-2 ,-5],[0 ,-6 ,1],[1 ,1 ,-10]]);##solving the equations in matrix form\n",
+ "B=numpy.matrix([[20], [0] ,[0]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"V=43.7 V\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KVL to mesh 1\n",
+ "V+13*I1-2*I2-5*I3=20\n",
+ "Applying KVL to mesh 2\n",
+ "2*I1-6*I2+I3=0\n",
+ "Applying KVL to mesh 3\n",
+ "V+5*I1+I2-10*I3=0\n",
+ "putting I1=0 in equation 1, 2 and 3 we get\n",
+ "V-2*I2-5*I3=20\n",
+ "-6*I2+I3=0\n",
+ "V+I2-10*I3=0\n",
+ "Solving the three equations\n",
+ "[matrix([[ 43.7037037 ],\n",
+ " [ 0.74074074],\n",
+ " [ 4.44444444]])]\n",
+ "V=43.7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg2.22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.13\n",
+ "##example2.14\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"Mesh 1 contains a current source of 6A.Hence, we cannot write KVL equation for Mesh 1.direction of current source and mesh current I1 are same,\");\n",
+ "print(\"I1=6A\");##equation 1\n",
+ "print(\"Applying KVL to mesh 2\");\n",
+ "print(\"18*I2-6*I3=108\");##equation 2\n",
+ "print(\"Applying KVL to mesh 3\");\n",
+ "print(\"6*I2-11*I3=9\");##equation 3\n",
+ "print(\"Solving the three equations\");\n",
+ "A=numpy.matrix([[18, -6],[6 ,-11]])##solving the equations in matrix form\n",
+ "B=numpy.matrix([[108] ,[9]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"I3 = 3A\");\n",
+ "print(\"I2ohm = 3A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mesh 1 contains a current source of 6A.Hence, we cannot write KVL equation for Mesh 1.direction of current source and mesh current I1 are same,\n",
+ "I1=6A\n",
+ "Applying KVL to mesh 2\n",
+ "18*I2-6*I3=108\n",
+ "Applying KVL to mesh 3\n",
+ "6*I2-11*I3=9\n",
+ "Solving the three equations\n",
+ "[matrix([[ 7.],\n",
+ " [ 3.]])]\n",
+ "I3 = 3A\n",
+ "I2ohm = 3A\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg2.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.23\n",
+ "##example2.15\n",
+ "print(\"from the fig,\");\n",
+ "print(\"IA=I1\");##equation 1\n",
+ "print(\"IB=I2\");##equation 2\n",
+ "print(\"Applying Kvl to mesh 1:\");\n",
+ "print(\"5-5*I1-10*IB-10*(I1-I2)-5*IA=0\");\n",
+ "print(\"5-5*I1-10*I2-10*I1+10*I2-5*I1=0\");\n",
+ "print(\"-20*I1=-5\");\n",
+ "I1=5./20.;\n",
+ "print'%s %.2f %s'%(\"I1= \",I1,\" A\");##equation 3\n",
+ "print(\"Applying Kvl to mesh 2:\");\n",
+ "print(\"15*I1-15*I2=10\");##equation 4\n",
+ "print(\"Put I1=0.25 A in equation 4\");\n",
+ "print(\"-6.25=15*I2\");\n",
+ "I2=-6.25/15.;\n",
+ "print'%s %.2f %s '%(\"I2= \",I2,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "from the fig,\n",
+ "IA=I1\n",
+ "IB=I2\n",
+ "Applying Kvl to mesh 1:\n",
+ "5-5*I1-10*IB-10*(I1-I2)-5*IA=0\n",
+ "5-5*I1-10*I2-10*I1+10*I2-5*I1=0\n",
+ "-20*I1=-5\n",
+ "I1= 0.25 A\n",
+ "Applying Kvl to mesh 2:\n",
+ "15*I1-15*I2=10\n",
+ "Put I1=0.25 A in equation 4\n",
+ "-6.25=15*I2\n",
+ "I2= -0.42 A \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg2.24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.25\n",
+ "##example2.17\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"from the fig,\");\n",
+ "print(\"V1=-5*I1\");##equation 1\n",
+ "print(\"V2=2*I2\");##equation 2\n",
+ "print(\"Applying Kvl to mesh 1:\");\n",
+ "print(\"20*I1+3*I2=-5\");##equation 3\n",
+ "print(\"Applying Kvl to mesh 2:\");\n",
+ "print(\"11*I1-3*I2=10\");##equation 4\n",
+ "print(\"Solving equations 3 and 4\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[20, 3],[11, -3]]);\n",
+ "B=numpy.matrix([[-5], [10]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"I1=0.161 A\");\n",
+ "print(\"I2=-2.742 A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "from the fig,\n",
+ "V1=-5*I1\n",
+ "V2=2*I2\n",
+ "Applying Kvl to mesh 1:\n",
+ "20*I1+3*I2=-5\n",
+ "Applying Kvl to mesh 2:\n",
+ "11*I1-3*I2=10\n",
+ "Solving equations 3 and 4\n",
+ "[matrix([[ 0.16129032],\n",
+ " [-2.74193548]])]\n",
+ "I1=0.161 A\n",
+ "I2=-2.742 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18-pg2.25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem-1\n",
+ "##pg no.-2.25\n",
+ "##example2.18\n",
+ "import numpy\n",
+ "print(\"from the fig,\");\n",
+ "print(\"Iy=I1\");##equation 1\n",
+ "print(\"Ix=I1-I2\");##equation 2\n",
+ "print(\"Applying Kvl to mesh 1:\");\n",
+ "print(\"-10*I1+3*I2=5\");##equation 3\n",
+ "print(\"Applying Kvl to mesh 2:\");\n",
+ "print(\"-I1-3*I2=10\");##equation 4\n",
+ "print(\"Solving equations 3 and 4\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[-10 ,3],[-1, -3]]);\n",
+ "B=numpy.matrix([[5] ,[10]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"I1=-1.364 A\");\n",
+ "print(\"I2=-2878 A\");\n",
+ "x=-1.364;\n",
+ "y=-2.878;\n",
+ "Ix=x-y;\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nIy = \",x,\" A\" and \" \\nIx = \",Ix,\" A\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "from the fig,\n",
+ "Iy=I1\n",
+ "Ix=I1-I2\n",
+ "Applying Kvl to mesh 1:\n",
+ "-10*I1+3*I2=5\n",
+ "Applying Kvl to mesh 2:\n",
+ "-I1-3*I2=10\n",
+ "Solving equations 3 and 4\n",
+ "[matrix([[-1.36363636],\n",
+ " [-2.87878788]])]\n",
+ "I1=-1.364 A\n",
+ "I2=-2878 A\n",
+ "\n",
+ "Iy = -1.36 \n",
+ "Ix = 1.51 A \n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg2.26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.26\n",
+ "##example2.19\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"Applying KVL to mesh 1:\");\n",
+ "print(\"11*I1-10*I2=2\");##equation 1\n",
+ "print(\"Writing current equation to supermesh:\")\n",
+ "print(\"I3-I2=4\");##equation 2\n",
+ "print(\"Applying KVL to outer path of supermesh:\");\n",
+ "print(\"2*I1-3*I2-3*I3=0\");##equation 3\n",
+ "print(\"solving these equations we get :\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[11, -10, 0],[0 ,-1 ,1],[2 ,-3, -3]]);\n",
+ "B=numpy.matrix([[2] ,[4] ,[0]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "I1=-2.35\n",
+ "I2=-2.78\n",
+ "I3=1.22\n",
+ "I4=I1-I2;\n",
+ "print'%s %.2f %s'%(\"\\ncurrent through the 10 ohm resistor = I1-I2 = \",I4,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KVL to mesh 1:\n",
+ "11*I1-10*I2=2\n",
+ "Writing current equation to supermesh:\n",
+ "I3-I2=4\n",
+ "Applying KVL to outer path of supermesh:\n",
+ "2*I1-3*I2-3*I3=0\n",
+ "solving these equations we get :\n",
+ "[matrix([[-2.34782609],\n",
+ " [-2.7826087 ],\n",
+ " [ 1.2173913 ]])]\n",
+ "\n",
+ "current through the 10 ohm resistor = I1-I2 = 0.43 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex20-pg2.26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.26\n",
+ "##example2.20\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"writing equation for supermesh,\");\n",
+ "print(\"I1-I3=7\");##equation 1\n",
+ "print(\"Applying Kvl to the outer path of the supermesh:\");\n",
+ "print(\"-I1+4*I2-4*I3 = -7\");##equation 2\n",
+ "print(\"Applying Kvl to mesh 2:\");\n",
+ "print(\"I1-6*I2+3*I3 = 0\");##equation 3\n",
+ "print(\"Solving equations 1 ,2 and 3\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[1 ,0 ,-1],[-1 ,4 ,-4],[1 ,-6 ,3]]);\n",
+ "B=numpy.matrix([[7], [-7], [0]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"I1=9 A\");\n",
+ "print(\"I2=-2.5 A\");\n",
+ "print(\"I3=-2 A\");\n",
+ "x=2.5;\n",
+ "y=2;\n",
+ "z=x-y;\n",
+ "print'%s %.2f %s'%(\"\\nCurrent through the 3-Ohm resistor = I2-I3 =\",z,\" A\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "writing equation for supermesh,\n",
+ "I1-I3=7\n",
+ "Applying Kvl to the outer path of the supermesh:\n",
+ "-I1+4*I2-4*I3 = -7\n",
+ "Applying Kvl to mesh 2:\n",
+ "I1-6*I2+3*I3 = 0\n",
+ "Solving equations 1 ,2 and 3\n",
+ "[matrix([[ 9. ],\n",
+ " [ 2.5],\n",
+ " [ 2. ]])]\n",
+ "I1=9 A\n",
+ "I2=-2.5 A\n",
+ "I3=-2 A\n",
+ "\n",
+ "Current through the 3-Ohm resistor = I2-I3 = 0.50 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex21-pg2.27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.27\n",
+ "##example2.21\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"Applying KVL to mesh 1:\");\n",
+ "print(\"15*I1-10*I2-5*I3=50\");##equation 1\n",
+ "print(\"Writing current equation to supermesh:\")\n",
+ "print(\"I2-I3=2 A\");##equation 2\n",
+ "print(\"Applying KVL to outer path of supermesh:\");\n",
+ "print(\"-15*I1+12*I2+6*I3=0\");##equation 3\n",
+ "print(\"solving these equations we get :\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[15, -10, -5],[0 ,1 ,-1],[-15, 12, 6]]);\n",
+ "B=numpy.matrix([[50] ,[2] ,[0]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "I1=20.\n",
+ "I2=17.33\n",
+ "I3=15.33\n",
+ "I4=I1-I3;\n",
+ "print'%s %.2f %s'%(\"\\ncurrent through the 5 ohm resistor = I1-I3 = \",I4,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KVL to mesh 1:\n",
+ "15*I1-10*I2-5*I3=50\n",
+ "Writing current equation to supermesh:\n",
+ "I2-I3=2 A\n",
+ "Applying KVL to outer path of supermesh:\n",
+ "-15*I1+12*I2+6*I3=0\n",
+ "solving these equations we get :\n",
+ "[matrix([[ 20. ],\n",
+ " [ 17.33333333],\n",
+ " [ 15.33333333]])]\n",
+ "\n",
+ "current through the 5 ohm resistor = I1-I3 = 4.67 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex22-pg2.28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.28\n",
+ "##example2.22\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"from the fig,\");\n",
+ "print(\"I4=40\");##equation 1\n",
+ "print(\"\\nmeshes 2 and 3 form a supermesh. current equation for supermesh,\")\n",
+ "print(\"-I1+2*I2-I3 = 0\");##equation 2\n",
+ "print(\"Applying Kvl to supermesh:\");\n",
+ "print(\"-1/5(I2-I1)-1/20*I2-1/15*I3-1/2(I3-I4)=0\");##equation 3\n",
+ "print(\"applying KVL to mesh 1\");\n",
+ "print(\"-1/10*I1-1/5(I1-I2)-1/6(I1-I4)=6\");##equation 4\n",
+ "print(\"Solving equations 1 ,2 ,3 and 4\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[0 ,0 ,0 ,1],[-1, 2 ,-1 ,0],[0.2, -0.25, -17/30, 0.5],[-7/15, 0.2, 0 ,1/6]]);\n",
+ "B=numpy.matrix([[40], [0] ,[0], [6]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"I1=10 A\");\n",
+ "print(\"I2=-20 A\");\n",
+ "print(\"I3=30 A\");\n",
+ "print(\"I4=40 A\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "from the fig,\n",
+ "I4=40\n",
+ "\n",
+ "meshes 2 and 3 form a supermesh. current equation for supermesh,\n",
+ "-I1+2*I2-I3 = 0\n",
+ "Applying Kvl to supermesh:\n",
+ "-1/5(I2-I1)-1/20*I2-1/15*I3-1/2(I3-I4)=0\n",
+ "applying KVL to mesh 1\n",
+ "-1/10*I1-1/5(I1-I2)-1/6(I1-I4)=6\n",
+ "Solving equations 1 ,2 ,3 and 4\n",
+ "[matrix([[ -4.72636816],\n",
+ " [ 6.3681592 ],\n",
+ " [ 17.46268657],\n",
+ " [ 40. ]])]\n",
+ "I1=10 A\n",
+ "I2=-20 A\n",
+ "I3=30 A\n",
+ "I4=40 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex24-pg2.29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.29\n",
+ "import numpy\n",
+ "##example2.24\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"2*V1-V2 = 2\");##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"3*V2-V1 = 4\");##equation 2\n",
+ "print(\"Solving equations 1 and 2\");##solving equations in matrix form\n",
+ "A=([[2 ,-1],[-1 ,3]]);\n",
+ "B=([[2] ,[4]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"V1= 2 V\");\n",
+ "print(\"V2=-2 V\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "2*V1-V2 = 2\n",
+ "Applying KCL to node 2:\n",
+ "3*V2-V1 = 4\n",
+ "Solving equations 1 and 2\n",
+ "[array([[ 2.],\n",
+ " [ 2.]])]\n",
+ "V1= 2 V\n",
+ "V2=-2 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex25-pg2.30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.30\n",
+ "##example2.25\n",
+ "import numpy\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"8*VA-2*VB = 50\");##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"-3*VA+9*VB = 85\");##equation 2\n",
+ "print(\"Solving equations 1 and 2\");##solving equations in matrix form\n",
+ "A=([[8 ,-2],[-3 ,9]]);\n",
+ "B=([[50], [85]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"VA= 9.39 V\");\n",
+ "print(\"VB= 12.58 V\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "8*VA-2*VB = 50\n",
+ "Applying KCL to node 2:\n",
+ "-3*VA+9*VB = 85\n",
+ "Solving equations 1 and 2\n",
+ "[array([[ 9.39393939],\n",
+ " [ 12.57575758]])]\n",
+ "VA= 9.39 V\n",
+ "VB= 12.58 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex26-pg2.30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.30\n",
+ "import numpy\n",
+ "##example2.26\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"5*V1-2*V2 = -24\");##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"10*V1-31*V2+6*V3 = 300\");##equation 2\n",
+ "print(\"Applying KCL to node 3:\");\n",
+ "print(\"-4*V2 +9*V3 = 160\");##equation 3\n",
+ "print(\"Solving equations 1,2 and 3\");##solving equations in matrix form\n",
+ "A=([[5, -2 ,0],[10, -31, 6],[0, -4, 9]]);\n",
+ "B=([[-24] ,[300] ,[160]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"V1= -8.77 V\");\n",
+ "print(\"V2= -9.92 V\");\n",
+ "print(\"V3= 13.37 V\");\n",
+ "x=13.37;\n",
+ "y=-9.92;\n",
+ "z=(x-y)/5.;\n",
+ "print'%s %.2f %s'%(\"\\ncurrent through the 5 ohm resistor = V3-V2/5 = \",z,\" A\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "5*V1-2*V2 = -24\n",
+ "Applying KCL to node 2:\n",
+ "10*V1-31*V2+6*V3 = 300\n",
+ "Applying KCL to node 3:\n",
+ "-4*V2 +9*V3 = 160\n",
+ "Solving equations 1,2 and 3\n",
+ "[array([[ -8.76712329],\n",
+ " [ -9.91780822],\n",
+ " [ 13.36986301]])]\n",
+ "V1= -8.77 V\n",
+ "V2= -9.92 V\n",
+ "V3= 13.37 V\n",
+ "\n",
+ "current through the 5 ohm resistor = V3-V2/5 = 4.66 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex27-pg2.31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.31\n",
+ "##example2.27\n",
+ "import numpy\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"50*V1-20*V2 = 2400\");##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"-10*V1+19*V2 = 240\");##equation 2\n",
+ "print(\"Solving equations 1 and 2\");##solving equations in matrix form\n",
+ "A=([[50, -20],[-10, 19]]);\n",
+ "B=([[2400] ,[240]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"V1= 67.2 V\");\n",
+ "print(\"V2=-48 V\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "50*V1-20*V2 = 2400\n",
+ "Applying KCL to node 2:\n",
+ "-10*V1+19*V2 = 240\n",
+ "Solving equations 1 and 2\n",
+ "[array([[ 67.2],\n",
+ " [ 48. ]])]\n",
+ "V1= 67.2 V\n",
+ "V2=-48 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex28-pg2.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.32\n",
+ "##example2.28\n",
+ "import numpy\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"4*VA-2*VB = 5\");##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"-2*VA+3*VB = 4\");##equation 2\n",
+ "print(\"Solving equations 1 and 2\");##solving equations in matrix form\n",
+ "A=([[4, -2],[-2 ,3]]);\n",
+ "B=([[5], [4]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "\n",
+ "print(X);\n",
+ "print(\"VA= 2.88 V\");\n",
+ "print(\"VB= 3.25 V\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "4*VA-2*VB = 5\n",
+ "Applying KCL to node 2:\n",
+ "-2*VA+3*VB = 4\n",
+ "Solving equations 1 and 2\n",
+ "[[ 2.875]\n",
+ " [ 3.25 ]]"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "VA= 2.88 V\n",
+ "VB= 3.25 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex29-pg2.33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import numpy\n",
+ "##Network Theorem 1\n",
+ "##page no-2.33\n",
+ "##example2.29\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"4*V1-2*V2-V3 = -24\");##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"-50*V1+71*V2-20*V3 = 0\");##equation 2\n",
+ "print(\"Applying KCL to node 3:\");\n",
+ "print(\"-5V1-4*V2 +10*V3 = 180\");##equation 3\n",
+ "print(\"Solving equations 1,2 and 3\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[4, -2, -1],[-50 ,71 ,-20],[-5, -4 ,10]]);\n",
+ "B=numpy.matrix([[-24], [0], [180]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"V1= 6.35 V\");\n",
+ "print(\"V2= 11.76 V\");\n",
+ "print(\"V3= 25.88 V\");\n",
+ "x=25.88;\n",
+ "y=11.76;\n",
+ "z=(x-y);\n",
+ "print'%s %.2f %s'%(\"\\ncurrent through the 5 ohm resistor = V3-V2/5 = \",z,\" A\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "4*V1-2*V2-V3 = -24\n",
+ "Applying KCL to node 2:\n",
+ "-50*V1+71*V2-20*V3 = 0\n",
+ "Applying KCL to node 3:\n",
+ "-5V1-4*V2 +10*V3 = 180\n",
+ "Solving equations 1,2 and 3\n",
+ "[matrix([[ 6.35294118],\n",
+ " [ 11.76470588],\n",
+ " [ 25.88235294]])]\n",
+ "V1= 6.35 V\n",
+ "V2= 11.76 V\n",
+ "V3= 25.88 V\n",
+ "\n",
+ "current through the 5 ohm resistor = V3-V2/5 = 14.12 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex30-pg2.34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.34\n",
+ "##example2.30\n",
+ "import numpy\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"8*V1-V2 = 50\");##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"-2*V1+11*V2 = -500\");##equation 2\n",
+ "print(\"Solving equations 1 and 2\");##solving equations in matrix form\n",
+ "A=([[8, -1],[-2, 17]]);\n",
+ "B=([[50] ,[-500]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"V1= 2.61 V\");\n",
+ "print(\"V2=-29.1 V\");\n",
+ "x=2.61;\n",
+ "y=-29.1;\n",
+ "I1=-x/2.;\n",
+ "I2=(x-y)/10.;##current through 10 Ohm resistor\n",
+ "I3=(y+50)/2.;##50 volts is the supply to the circuit\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nI1= \",I1,\" A\" and \"\\nI2= \",I2,\"A\" and \" \\nI3= \",I3,\" A\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "8*V1-V2 = 50\n",
+ "Applying KCL to node 2:\n",
+ "-2*V1+11*V2 = -500\n",
+ "Solving equations 1 and 2\n",
+ "[array([[ 2.6119403 ],\n",
+ " [-29.10447761]])]\n",
+ "V1= 2.61 V\n",
+ "V2=-29.1 V\n",
+ "\n",
+ "I1= -1.30 \n",
+ "I2= 3.17 \n",
+ "I3= 10.45 A \n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex31-pg2.34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.34\n",
+ "##example2.31\n",
+ "import numpy\n",
+ "print(\"Applying KCL to node a:\");\n",
+ "print(\"0.5*Va-0.2*Vb = 34.2\");#equation 1\n",
+ "print(\"Applying KCL to node b:\");\n",
+ "print(\"0.1*Va-0.4*Vb = -32.4\");##equation 2\n",
+ "print(\"Solving equations 1 and 2\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[0.5,0.2],[0.1, -0.4]]);\n",
+ "B=numpy.matrix([[34.2], [-32.4]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"Va= 112 V\");\n",
+ "print(\"Vb= 109 V\");\n",
+ "x=112.;\n",
+ "y=109.;\n",
+ "I1=(120.-x)/0.2;\n",
+ "I2=(x-y)/0.3;\n",
+ "I3=(110-y)/0.1;\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nI1=\",I1,\" A \" and \"\\nI2= \",I2,\" A\" and \" \\nI3= \",I3,\" A\");\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node a:\n",
+ "0.5*Va-0.2*Vb = 34.2\n",
+ "Applying KCL to node b:\n",
+ "0.1*Va-0.4*Vb = -32.4\n",
+ "Solving equations 1 and 2\n",
+ "[matrix([[ 32.72727273],\n",
+ " [ 89.18181818]])]\n",
+ "Va= 112 V\n",
+ "Vb= 109 V\n",
+ "\n",
+ "I1= 40.00 \n",
+ "I2= 10.00 \n",
+ "I3= 10.00 A \n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex32-pg2.35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.35\n",
+ "##example2.35\n",
+ "import numpy\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"V1 = 50\");##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"-2*V1+17*V2 = 50\");#equation 2\n",
+ "print(\"Solving equations 1 and 2\");##solving equations in matrix form\n",
+ "A=([[1, 0],[-2, 17]]);\n",
+ "B=([[50] ,[50]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"V1= 50 V\");\n",
+ "print(\"V2= 8.82 V\");\n",
+ "x=8.82;\n",
+ "y=(x/10.);\n",
+ "print'%s %.2f %s'%(\"\\ncurrent in the 10-Ohm resistor =V2/10 =\",y,\" A\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "V1 = 50\n",
+ "Applying KCL to node 2:\n",
+ "-2*V1+17*V2 = 50\n",
+ "Solving equations 1 and 2\n",
+ "[array([[ 50. ],\n",
+ " [ 8.82352941]])]\n",
+ "V1= 50 V\n",
+ "V2= 8.82 V\n",
+ "\n",
+ "current in the 10-Ohm resistor =V2/10 = 0.88 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex33-pg2.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.36\n",
+ "##example2.33\n",
+ "import numpy\n",
+ "print(\"Applying KCL to node a:\");\n",
+ "print(\"6*Va-5*Vb = -20\");##equation 1\n",
+ "print(\"Applying KCL to node b:\");\n",
+ "print(\"-10*Va+17*Vb-5*Vc = 0\");##equation 2\n",
+ "print(\"At node c\");\n",
+ "print(\"Vc = 20\");\n",
+ "print(\"Solving equations 1,2 and 3\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[6 ,-5 ,0],[-10 ,17 ,-5],[0, 0, 1]]);\n",
+ "B=numpy.matrix([[-20], [0] ,[20]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"Va= 3.08 V\");\n",
+ "print(\"Vb= 7.69 V\");\n",
+ "x=3.08;\n",
+ "y=7.69;\n",
+ "z=20.;\n",
+ "Va = x-y;\n",
+ "Vb = y-z;\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nV1 = Va-Vb = \",Va,\"V\" and \"\\nV2 = Vb-Vc = \",Vb,\" V\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node a:\n",
+ "6*Va-5*Vb = -20\n",
+ "Applying KCL to node b:\n",
+ "-10*Va+17*Vb-5*Vc = 0\n",
+ "At node c\n",
+ "Vc = 20\n",
+ "Solving equations 1,2 and 3\n",
+ "[matrix([[ 3.07692308],\n",
+ " [ 7.69230769],\n",
+ " [ 20. ]])]\n",
+ "Va= 3.08 V\n",
+ "Vb= 7.69 V\n",
+ "\n",
+ "V1 = Va-Vb = -4.61 \n",
+ "V2 = Vb-Vc = -12.31 V \n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex34-pg2.37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.37\n",
+ "##example2.334\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"At node A:\");\n",
+ "print(\"VA = 60\");##equation 1\n",
+ "print(\"Applying KCL to node B:\");\n",
+ "print(\"-VA+3*VB-VC = 12\");##equation 2\n",
+ "print(\"Applying KCL to node C:\");\n",
+ "print(\"-2*VA-5*VB+10*VC\");##equation 3\n",
+ "print(\"Solving equations 1,2 and 3\");##solving equations in matrix \n",
+ "A=numpy.matrix([[1 ,0 ,0],[-1 ,3 ,-1],[-2 ,-5 ,10]]);\n",
+ "B=numpy.matrix([[60], [12], [24]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B)\n",
+ "print[X];\n",
+ "print(\"VC= 31.68 V\");\n",
+ "print(\"Voltage across the 100 Ohm resistor = 31.68 V\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "At node A:\n",
+ "VA = 60\n",
+ "Applying KCL to node B:\n",
+ "-VA+3*VB-VC = 12\n",
+ "Applying KCL to node C:\n",
+ "-2*VA-5*VB+10*VC\n",
+ "Solving equations 1,2 and 3\n",
+ "[matrix([[ 60. ],\n",
+ " [ 34.56],\n",
+ " [ 31.68]])]\n",
+ "VC= 31.68 V\n",
+ "Voltage across the 100 Ohm resistor = 31.68 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex35-pg2.38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.38\n",
+ "##example2.35\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"2.5*V1-0.5*V2 = 5\");##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"V1-V2 = 0\");##equation 2\n",
+ "print(\"Solving equations 1 and 2\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[2.5, -0.5],[1 ,-1]]);\n",
+ "B=numpy.matrix([[5] ,[0]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B);\n",
+ "print[X];\n",
+ "print(\"V1= 2.5 V\");\n",
+ "print(\"V2=-2.5 V\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "2.5*V1-0.5*V2 = 5\n",
+ "Applying KCL to node 2:\n",
+ "V1-V2 = 0\n",
+ "Solving equations 1 and 2\n",
+ "[matrix([[ 2.5],\n",
+ " [ 2.5]])]\n",
+ "V1= 2.5 V\n",
+ "V2=-2.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex37-pg2.39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.39\n",
+ "##example2.37\\\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"2*V1+17*V2 = 0\")##equation 1\n",
+ "print(\"Applying KCL to node 2:\");\n",
+ "print(\"V1+6V2 = 0\");##equation 2\n",
+ "print(\"Solving equations 1 and 2\")##solving equations in matrix form\n",
+ "A=numpy.matrix([[2 ,17],[1, 6]]);\n",
+ "B=numpy.matrix([[0], [0]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B)\n",
+ "print(X);\n",
+ "print(\"V1= 0 V\");\n",
+ "print(\"V2= 0 V\");\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node 1:\n",
+ "2*V1+17*V2 = 0\n",
+ "Applying KCL to node 2:\n",
+ "V1+6V2 = 0\n",
+ "Solving equations 1 and 2\n",
+ "[[ 0.]\n",
+ " [ 0.]]\n",
+ "V1= 0 V\n",
+ "V2= 0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex38-pg2.40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.40\n",
+ "##example2.38\n",
+ "import numpy\n",
+ "import math\n",
+ "print(\"Applying KCL to node a:\");\n",
+ "print(\"2*Va-0.5*Vb-0.5*Vc = 5\");##equation 1\n",
+ "print(\"Applying KCL to node b:\");\n",
+ "print(\"-3/2*Va+5/6*Vb+2/3*Vc = -1\");##equation 2\n",
+ "print(\"Applying KCL to node c:\");\n",
+ "print(\"1/2*Va+1/3*Vb-31/30*Vc = -1\");##equation 3\n",
+ "print(\"Solving equations 1,2 and 3\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[2, -0.5, -0.5],[-3/2, 5/6, 2/3],[0.5 ,1/3 ,-31/30 ]]);\n",
+ "B=numpy.matrix([[5], [-1], [0]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B)\n",
+ "print[X];\n",
+ "\n",
+ "print(\"Va= 4.303 V\");\n",
+ "print(\"Vb= 3.87 V\");\n",
+ "print(\"Vc= 3.33 V\");\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KCL to node a:\n",
+ "2*Va-0.5*Vb-0.5*Vc = 5\n",
+ "Applying KCL to node b:\n",
+ "-3/2*Va+5/6*Vb+2/3*Vc = -1\n",
+ "Applying KCL to node c:\n",
+ "1/2*Va+1/3*Vb-31/30*Vc = -1\n",
+ "Solving equations 1,2 and 3\n",
+ "[matrix([[ 0.5 ],\n",
+ " [-8.125],\n",
+ " [ 0.125]])]"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Va= 4.303 V\n",
+ "Vb= 3.87 V\n",
+ "Vc= 3.33 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex39-pg2.41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.41\n",
+ "##example2.39\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"from the figure\");\n",
+ "print(\"V4= 40 V\");##equation 1\n",
+ "print(\"nodes 2 and 3 form suoernode:\");\n",
+ "print(\"V1-2*V2+V3 = 0\");##equation 2\n",
+ "print(\"Applying KCL to node 1:\");\n",
+ "print(\"7/15*V1-1/5*V2 = 2/3\");##equation 3\n",
+ "print(\"Applying KCL to supernode :\");\n",
+ "print(\"-23/30*V1 +83/60*V3 = 20\");##equation 4\n",
+ "print(\"Solving equations 1,2,3 and 4\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[0, 0 ,0 ,1],[1 ,-2 ,1 ,0],[7/15 ,-1/5, 0, 0],[-23/30, 83/60, 0 ,0]]);\n",
+ "B=numpy.matrix([[40] ,[0], [2/3], [20]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B)\n",
+ "print[X];\n",
+ "print(\"V1= 10 V\");\n",
+ "print(\"V2= 20 V\");\n",
+ "print(\"V3= 30 V\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "from the figure\n",
+ "V4= 40 V\n",
+ "nodes 2 and 3 form suoernode:\n",
+ "V1-2*V2+V3 = 0\n",
+ "Applying KCL to node 1:\n",
+ "7/15*V1-1/5*V2 = 2/3\n",
+ "Applying KCL to supernode :\n",
+ "-23/30*V1 +83/60*V3 = 20\n",
+ "Solving equations 1,2,3 and 4\n",
+ "[matrix([[-20.],\n",
+ " [ 0.],\n",
+ " [ 20.],\n",
+ " [ 40.]])]\n",
+ "V1= 10 V\n",
+ "V2= 20 V\n",
+ "V3= 30 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex40-pg2.42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Network Theorem 1\n",
+ "##page no-2.42\n",
+ "##example2.40\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"selecting central node as reference node\");\n",
+ "print(\"V1= -12 V\");##equation 1\n",
+ "print(\"Applying KCL at node 1:\");\n",
+ "print(\"-2*V1+2.5*V2-0.5V3 = 14\");##equation 2\n",
+ "print(\"nodes 3 and 4 form a supernode\");\n",
+ "print(\"0.2*V1+V3-1.2*V4 = 0\");##equation 3\n",
+ "print(\"Applying KCL to supernode :\");\n",
+ "print(\"0.1*V1-V2+0.5*V3+1.4*V4 = 0\");##equation 4\n",
+ "print(\"Solving equations 1,2,3 and 4\");##solving equations in matrix form\n",
+ "A=numpy.matrix([[1, 0, 0 ,0],[-2 ,2.5 ,-0.5 ,0],[0.2 ,0 ,1, -1.2],[0.1 ,-1 ,0.5, 1.4]]);\n",
+ "B=numpy.matrix([[-12], [14] ,[0],[0]])\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B)\n",
+ "print[X];\n",
+ "print(\"V1= -12 V\");\n",
+ "print(\"V2= -4 V\");\n",
+ "print(\"V3= 0\");\n",
+ "print(\"V4= -2 V\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "selecting central node as reference node\n",
+ "V1= -12 V\n",
+ "Applying KCL at node 1:\n",
+ "-2*V1+2.5*V2-0.5V3 = 14\n",
+ "nodes 3 and 4 form a supernode\n",
+ "0.2*V1+V3-1.2*V4 = 0\n",
+ "Applying KCL to supernode :\n",
+ "0.1*V1-V2+0.5*V3+1.4*V4 = 0\n",
+ "Solving equations 1,2,3 and 4\n",
+ "[matrix([[ -1.20000000e+01],\n",
+ " [ -4.00000000e+00],\n",
+ " [ 4.44089210e-16],\n",
+ " [ -2.00000000e+00]])]\n",
+ "V1= -12 V\n",
+ "V2= -4 V\n",
+ "V3= 0\n",
+ "V4= -2 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |