{ "metadata": { "name": "", "signature": "sha256:232ca964f6cb50454cfbdf0b82acf046a9abb3f3780b179ff1b54c28005cb225" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter3-Network Theorem 2" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1-pg3.2" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.2\n", "##example 3.1\n", "print(\"When 10-V source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I1=10.*(0.87/(1.0+0.87));\n", "print\"%s %.2f %s\"%(\"I1=10*(0.87/(10+0.87))= \",I1,\" A (down)\");\n", "print(\"When 4 A source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I2=2.86*(0.875/(10.+0.875));\n", "print\"%s %.2f %s\"%(\"I2=2.86*(0.875/(10+0.875))= \",I2,\" A (down)\");\n", "print(\"By superposition theorem:\");\n", "I=I1+I2;\n", "print\"%s %.2f %s\"%(\"\\nI=I1+I2=0.8+0.23= \",I,\" A (down)\");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "When 10-V source is acting alone:\n", "By current-division formula :\n", "I1=10*(0.87/(10+0.87))= 4.65 A (down)\n", "When 4 A source is acting alone:\n", "By current-division formula :\n", "I2=2.86*(0.875/(10+0.875))= 0.23 A (down)\n", "By superposition theorem:\n", "\n", "I=I1+I2=0.8+0.23= 4.88 A (down)\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2-pg3.3" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.4\n", "##example 3.2\n", "print(\"When 4-A source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I1=3.33*(3.53/(6.+3.53));\n", "print\"%s %.2f %s\"%(\"I1=3.33*(3.53/(6+3.53)) = \",I1,\" A (down)\");\n", "print(\"When 10-V source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I2=0.833*(3.53/(6.+3.53));\n", "print\"%s %.2f %s\"%(\"I2=0.833*(3.53/(6+3.53))= \",I2,\" A (up)\");\n", "print(\"When 3-A source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I3=3*(3.53/(6.+3.53));\n", "print\"%s %.2f %s\"%(\"I3=3*(3.53/(6+3.53))= \",I3,\" A (down)\");\n", "print(\"By superposition theorem:\");\n", "I=I1-I2+I3;\n", "print\"%s %.2f %s\"%(\"\\nI=I1-I2+I3=1.23-0.31+1.11= \",I,\" A (down)\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "When 4-A source is acting alone:\n", "By current-division formula :\n", "I1=3.33*(3.53/(6+3.53)) = 1.23 A (down)\n", "When 10-V source is acting alone:\n", "By current-division formula :\n", "I2=0.833*(3.53/(6+3.53))= 0.31 A (up)\n", "When 3-A source is acting alone:\n", "By current-division formula :\n", "I3=3*(3.53/(6+3.53))= 1.11 A (down)\n", "By superposition theorem:\n", "\n", "I=I1-I2+I3=1.23-0.31+1.11= 2.04 A (down)\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex3-pg3.5" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.5\n", "##example 3.3\n", "print(\"When 4-A source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I1=4./(2.+1.);\n", "print\"%s %.2f %s\"%(\"I1=4/(2+1) = \",I1,\" A (down)\");\n", "print(\"When 3-A source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I2=3.*(2./(2.+1.));\n", "print\"%s %.2f %s\"%(\"I2=3*(2/(2+1)) = \",I2,\" A (down)\");\n", "print(\"When 1-A source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I3=1.*(2./(2.+1.));\n", "print\"%s %.2f %s\"%(\"I3=1*(2/(2+1)) = \",I3,\" A (down)\");\n", "print(\"By superposition theorem:\");\n", "I=I1+I2+I3;\n", "print\"%s %.2f %s\"%(\"\\nI=I1+I2+I3=1.33+2+0.66= \",I,\" A (down)\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "When 4-A source is acting alone:\n", "By current-division formula :\n", "I1=4/(2+1) = 1.33 A (down)\n", "When 3-A source is acting alone:\n", "By current-division formula :\n", "I2=3*(2/(2+1)) = 2.00 A (down)\n", "When 1-A source is acting alone:\n", "By current-division formula :\n", "I3=1*(2/(2+1)) = 0.67 A (down)\n", "By superposition theorem:\n", "\n", "I=I1+I2+I3=1.33+2+0.66= 4.00 A (down)\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4-pg3.7" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.\n", "##example 3.4\n", "print(\"When 6-V source is acting alone:\");\n", "VAB1=6.;\n", "print\"%s %.2f %s\"%(\"VAB1 = \",VAB1,\" V\");\n", "print(\"When 10-V source is acting alone:\");\n", "print(\"Since the resistor of 5 ohm is shorted,the voltage across it is zero\")\n", "VAB2=10.;\n", "print\"%s %.2f %s\"%(\"VAB2= \",VAB2,\" V\" );\n", "print(\"When 5-A source is acting alone:\");\n", "print(\"Due to short circuit in both the parts\");\n", "VAB3=0.;\n", "print\"%s %.2f %s\"%(\"VAB3 = \",VAB3,\" V\");\n", "print(\"By superposition theorem:\");\n", "VAB=VAB1+VAB2+VAB3;\n", "print\"%s %.2f %s\"%(\"\\nVAB=VAB=VAB1+VAB2+VAB3= \",VAB,\" V\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "When 6-V source is acting alone:\n", "VAB1 = 6.00 V\n", "When 10-V source is acting alone:\n", "Since the resistor of 5 ohm is shorted,the voltage across it is zero\n", "VAB2= 10.00 V\n", "When 5-A source is acting alone:\n", "Due to short circuit in both the parts\n", "VAB3 = 0.00 V\n", "By superposition theorem:\n", "\n", "VAB=VAB=VAB1+VAB2+VAB3= 16.00 V\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex5-pg3.7" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.7\n", "##example 3.5\n", "print(\"When 5-A source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I1=5.*(2./(2.+4.));\n", "print\"%s %.2f %s\"%(\"I1=5*(2/(2+4)) = \",I1,\" A (down)\");\n", "print(\"When 2-A source is acting alone:\");\n", "print(\"By current-division formula :\");\n", "I2=2.*(2./(2.+4.));\n", "print\"%s %.2f %s\"%(\"I2=2*(2/(2+4)) = \",I2,\" A (down)\");\n", "print(\"When 6-V source is acting alone:\");\n", "print(\"Applying KVL to the mesh\");\n", "print(\"-2*I3-6-4*I3=0\");\n", "print(\"I3=-1\");\n", "I3=-1.;\n", "print\"%s %.2f %s\"%(\"I3=-1 A= \",I3,\" A (down)\");\n", "print(\"By superposition theorem:\");\n", "I=I1+I2+I3;\n", "print\"%s %.2f %s\"%(\"\\nI=I1+I2+I3=1.67+0.67-1= \",I,\" A (down)\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "When 5-A source is acting alone:\n", "By current-division formula :\n", "I1=5*(2/(2+4)) = 1.67 A (down)\n", "When 2-A source is acting alone:\n", "By current-division formula :\n", "I2=2*(2/(2+4)) = 0.67 A (down)\n", "When 6-V source is acting alone:\n", "Applying KVL to the mesh\n", "-2*I3-6-4*I3=0\n", "I3=-1\n", "I3=-1 A= -1.00 A (down)\n", "By superposition theorem:\n", "\n", "I=I1+I2+I3=1.67+0.67-1= 1.33 A (down)\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6-pg3.8" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.8\n", "##example 3.6\n", "a=15./38.;\n", "b=10./38.;\n", "x=a+b;\n", "print\"%s %.2f %s\"%(\"\\nApplying KCL at node 1, \\nI1 =\",a,\"\");##When the 15 V source is acting alone\n", "print\"%s %.2f %s\"%(\"\\nApplying KCL at node 1, \\nI1 = \",b,\"\");##When the 10 V source is acting alone\n", "print\"%s %.2f %s\"%(\"\\nBy superposition theorem, \\nI = I1+I2 = \",x,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "Applying KCL at node 1, \n", "I1 = 0.39 \n", "\n", "Applying KCL at node 1, \n", "I1 = 0.26 \n", "\n", "By superposition theorem, \n", "I = I1+I2 = 0.66 A\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex7-pg3.9" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.8\n", "##example 3.7\n", "a=3.;\n", "b=2.;\n", "x=a+b;\n", "print\"%s %.2f %s\"%(\"\\napplying KCL at node 1, \\nIx1 = \",a,\" A\");##when the 30 V source is acting alone\n", "print\"%s %.2f %s\"%(\"\\napplying KCL at the mesh, \\nIx2 = \",b,\" A\");##when the 20 V source is acting alone\n", "print\"%s %.2f %s\"%(\"\\nBy superposition theorem, Ix = Ix1+Ix2 = \",x,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "applying KCL at node 1, \n", "Ix1 = 3.00 A\n", "\n", "applying KCL at the mesh, \n", "Ix2 = 2.00 A\n", "\n", "By superposition theorem, Ix = Ix1+Ix2 = 5.00 A\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex8-pg3.10" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.10\n", "##example 3.8\n", "##when 5 V source is acting alone\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "print(\"Vx+10I1=5\");##equation 1\n", "print(\"Applying KVL to mesh,\");\n", "print(\"4Vx+12I1=5\");##equation 2\n", "A=numpy.matrix([[1, 10],[4 ,12]]);##solving equation in matrix form\n", "B=([[5],[5]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1 = 0.535 A\");\n", "##when the 2 A source is acting alone\n", "print(\"Vx+10I2=0\");##equation 1\n", "print(\"Applying KCL at Node x,\");\n", "print(\"Vx=-10/7\");##equation 2\n", "A1=numpy.matrix([[1, 10],[1 ,0]]);##solving equation in matrix form\n", "B1=numpy.matrix([[0], [-10/7]])\n", "X1=numpy.dot(numpy.linalg.inv(A1),B1);\n", "print[X1];\n", "print(\"I2 = 0.1428 A\");\n", "a=0.535;\n", "b=0.1428;\n", "x=a+b;\n", "print\"%s %.3f %s\"%(\"\\nBy superposition theorem, \\nI = I1+I2 = \",x,\" A \");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Vx+10I1=5\n", "Applying KVL to mesh,\n", "4Vx+12I1=5\n", "[matrix([[-0.35714286],\n", " [ 0.53571429]])]\n", "I1 = 0.535 A\n", "Vx+10I2=0\n", "Applying KCL at Node x,\n", "Vx=-10/7\n", "[matrix([[-2. ],\n", " [ 0.2]])]\n", "I2 = 0.1428 A\n", "\n", "By superposition theorem, \n", "I = I1+I2 = 0.678 A \n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9-pg3.10" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.10\n", "##example 3.9\n", "##when 100 V source is acting alone\n", "import numpy\n", "from numpy import linalg\n", "print(\"Vx-5I1=0\");##equation 1\n", "print(\"Applying KVL to mesh,\");\n", "print(\"10Vx-15I1=-100\");##equation 2\n", "A=numpy.matrix([[1, -5],[10 ,-15]]);##solving equation in matrix form\n", "B=numpy.matrix([[0], [-100]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];##negative because of opposite direction\n", "print(\"I1 = 2.857 A\");\n", "##when the 10 A source is acting alone\n", "print(\"9Vx+10I2=0\");##equation 1\n", "print(\"Applying KCL at Node 1,\");\n", "print(\"Vx=-100/7\");##equation 2\n", "A=numpy.matrix([[9, 10],[1, 0]]);##solving equation in matrix form\n", "B=numpy.matrix([[0] ,[-100/7]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I2 = 12.857 A\");\n", "a=2.857;\n", "b=12.857;\n", "x=a+b;\n", "print\"%s %.2f %s\"%(\"\\nBy superposition theorem, \\nI = I1+I2 = \",x,\" A \");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Vx-5I1=0\n", "Applying KVL to mesh,\n", "10Vx-15I1=-100\n", "[matrix([[-14.28571429],\n", " [ -2.85714286]])]\n", "I1 = 2.857 A\n", "9Vx+10I2=0\n", "Applying KCL at Node 1,\n", "Vx=-100/7\n", "[matrix([[-15. ],\n", " [ 13.5]])]\n", "I2 = 12.857 A\n", "\n", "By superposition theorem, \n", "I = I1+I2 = 15.71 A \n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex10-pg3.11" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.11\n", "##example 3.10\n", "##when 17 V source is acting alone\n", "import numpy\n", "from numpy import linalg\n", "print(\"Vx+2I1=0\");##equation 1\n", "print(\"Applying KVL to mesh,\");\n", "print(\"-5Vx-5I1=17\");##equation 2\n", "A=numpy.matrix([[1, 2],[-5 ,-5]]);##solving equation in matrix form\n", "B=([[0], [17]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1 = 3.4 A\");\n", "##when the 1 A source is acting alone\n", "print(\"4Vx+3I2=0\");##equation 1\n", "print(\"Applying KCL at Node x,\");\n", "print(\"Vx=-6/5\");##equation 2\n", "A=numpy.matrix([[4, 3],[1, 0]]);##solving equation in matrix form\n", "B=numpy.matrix([[0],[-6/5]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I2 = 1.6 A\");\n", "a=3.4;\n", "b=1.6;\n", "x=a+b;\n", "print\"%s %.2f %s\"%(\"\\nBy superposition theorem, \\nI = I1+I2 = \",x,\" A \");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Vx+2I1=0\n", "Applying KVL to mesh,\n", "-5Vx-5I1=17\n", "[matrix([[-6.8],\n", " [ 3.4]])]\n", "I1 = 3.4 A\n", "4Vx+3I2=0\n", "Applying KCL at Node x,\n", "Vx=-6/5\n", "[matrix([[-2. ],\n", " [ 2.66666667]])]\n", "I2 = 1.6 A\n", "\n", "By superposition theorem, \n", "I = I1+I2 = 5.00 A \n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex11-pg3.12" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.12\n", "##example 3.11\n", "##when 5 A source is acting alone\n", "print(\"-V1+4I=0\");##equation 1\n", "print(\"Applying KCL to node 1,\");\n", "print(\"1.25V1-4I=5\");##equation 2\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[-1, 4],[1.25 ,-4]]);##solving equation in matrix form\n", "B=([[0] ,[5]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"V1 = 20 V\");\n", "##when the 20 V source is acting alone\n", "print(\"from the figure,\");\n", "print(\"V2-3I=0\");##equation 1\n", "print(\"Applying KVL to the mesh,\");\n", "print(\"I=-20\");##equation 2\n", "A=([[1 ,-3],[0 ,1]]);##solving equation in matrix form\n", "B=([[0] ,[-20]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"V2 = -60 V\");\n", "a=20.;\n", "b=-60.;\n", "x=a+b;\n", "print\"%s %.2f %s\"%(\"\\nBy superposition theorem, \\n V = V1+V2 =\",x,\" V \");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-V1+4I=0\n", "Applying KCL to node 1,\n", "1.25V1-4I=5\n", "[matrix([[ 20.],\n", " [ 5.]])]\n", "V1 = 20 V\n", "from the figure,\n", "V2-3I=0\n", "Applying KVL to the mesh,\n", "I=-20\n", "[array([[-60.],\n", " [-20.]])]\n", "V2 = -60 V\n", "\n", "By superposition theorem, \n", " V = V1+V2 = -40.00 V \n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex12-pg3.13" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.13\n", "##example 3.12\n", "##when 18 V source is acting alone\n", "print(\"Vx+I1=0\");##equation 1\n", "print(\"Applying KVL to mesh,\");\n", "print(\"3Vx-6I1=-18\");##equation 2\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[1, 1],[3 ,-6]]);##solving equation in matrix form\n", "B=([[0] ,[-18]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1 = 2 A\");\n", "##when the 3 A source is acting alone\n", "print(\"from the figure,\");\n", "print(\"Vx=2 V\");##equation 1\n", "print(\"Applying KCL at node 1,\");\n", "print(\"3Vx-6I2=0\");##equation 2\n", "A=([[1 ,0],[3 ,-6]]);##solving equation in matrix form\n", "B=([[2] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I2 =1 V\");\n", "a=2;\n", "b=1;\n", "x=a+b;\n", "print\"%s %.2f %s\"%(\"\\nBy superposition theorem, \\n I = I1+I2 = \",x,\" A \");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Vx+I1=0\n", "Applying KVL to mesh,\n", "3Vx-6I1=-18\n", "[matrix([[-2.],\n", " [ 2.]])]\n", "I1 = 2 A\n", "from the figure,\n", "Vx=2 V\n", "Applying KCL at node 1,\n", "3Vx-6I2=0\n", "[array([[ 2.],\n", " [ 1.]])]\n", "I2 =1 V\n", "\n", "By superposition theorem, \n", " I = I1+I2 = 3.00 A \n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex13-pg3.14" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.14\n", "##example 3.13\n", "##when 120 V source is acting alone\n", "print(\"Applying KVL to mesh,\");\n", "print(\"Iy1=5.45 A\");\n", "##when the 12 A source is acting alone\n", "print(\"from the figure,\");\n", "print(\"V1+4Iy2=0\");##equation 1\n", "print(\"Applying KCL at node 1,\");\n", "print(\"-V1/8 +9/4Iy2=-12\");##equation 2\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[1, 4],[-1/8 ,9/4]]);##solving equation in matrix form\n", "B=([[0] ,[-12]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"Iy2 =-4.36 A\");\n", "##when 40 V source is acting alone\n", "print(\"Applying KVL to mesh,\");\n", "print(\"Iy3=-1.82 A\");\n", "a=5.45;\n", "b=-4.36;\n", "c=-1.82;\n", "x=a+b+c;\n", "print\"%s %.2f %s\"%(\"\\nBy superposition theorem, \\n I = Iy1+Iy2+Iy3 = \",x,\" A \");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Applying KVL to mesh,\n", "Iy1=5.45 A\n", "from the figure,\n", "V1+4Iy2=0\n", "Applying KCL at node 1,\n", "-V1/8 +9/4Iy2=-12\n", "[matrix([[ 8.],\n", " [-2.]])]\n", "Iy2 =-4.36 A\n", "Applying KVL to mesh,\n", "Iy3=-1.82 A\n", "\n", "By superposition theorem, \n", " I = Iy1+Iy2+Iy3 = -0.73 A \n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex14-pg3.15" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.15\n", "##example 3.14\n", "##when 18 V source is acting alone\n", "print(\"Vx1-31=0\");##equation 1\n", "print(\"Applying KVL to mesh,\");\n", "print(\"-3Vx1-9I=-18\");##equation 2\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[1, -3],[-3 ,-9]]);##solving equation in matrix form\n", "B=([[0] ,[-18]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"Vx1 = 3 V\");\n", "##when the 5 A source is acting alone\n", "print(\"from the figure,\");\n", "print(\"V1+Vx2=0\");##equation 1\n", "print(\"Applying KCL at node 1,\");\n", "print(\"1/2V1-1/2Vx2=5\");##equation 2\n", "A=numpy.matrix([[1, 1],[1/2 ,-1/2]]);##solving equation in matrix form\n", "B=([[0] ,[5]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"Vx2= -5 V\");\n", "##when the 36 V source is acting alone\n", "print(\"from the figure,\");\n", "print(\"Vx3+3I=0\");##equation 1\n", "print(\"Applying KVL to the mesh,\");\n", "print(\"3Vx3-9I=-36\");##equation 2\n", "A=numpy.matrix([[1, 3],[3 ,-9]]);##solving equation in matrix form\n", "B=([[0] ,[-36]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"Vx3= -6 V\");\n", "a=3.;\n", "b=-5.;\n", "c=-6.;\n", "x=a+b+c;\n", "print\"%s %.2f %s\"%(\"\\nBy superposition theorem, \\n Vx = Vx1+Vx2+Vx3 = \",x,\" V \");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Vx1-31=0\n", "Applying KVL to mesh,\n", "-3Vx1-9I=-18\n", "[matrix([[ 3.],\n", " [ 1.]])]\n", "Vx1 = 3 V\n", "from the figure,\n", "V1+Vx2=0\n", "Applying KCL at node 1,\n", "1/2V1-1/2Vx2=5\n", "[matrix([[ 5.],\n", " [-5.]])]\n", "Vx2= -5 V\n", "from the figure,\n", "Vx3+3I=0\n", "Applying KVL to the mesh,\n", "3Vx3-9I=-36\n", "[matrix([[-6.],\n", " [ 2.]])]\n", "Vx3= -6 V\n", "\n", "By superposition theorem, \n", " Vx = Vx1+Vx2+Vx3 = -8.00 V \n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex15-pg3.16" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.16\n", "##example 3.15\n", "a=10.;\n", "b=2.;\n", "c=(5.*a)-(20.*b);\n", "x=20.;\n", "y=30.;\n", "z=5.;\n", "r=z+((x*y)/(x+y));\n", "i=c/(r+c);\n", "##Calculation of Vth(Thevenin's voltage)\n", "print(\"removing the 10 ohm resistor from the circuit\");\n", "print\"%s %.2f %s\"%(\"\\nFor mesh 1, \\nI1 = \",a,\" A\");\n", "print\"%s %.2f %s\"%(\"\\nApplying KVL to mesh 2,, \\nI2 = \",b,\" A\");\n", "print\"%s %.2f %s\"%(\"\\nWriting Vth equation, \\n Vth = \",c,\" V\");\n", "##Calculation of Rth(Thevenin's Resistance)\n", "print(\"replacing the current source of 10 A with an open circuit and voltage source of 100 V with a short circuit,\");\n", "print\"%s %.2f %s\"%(\"\\nRth = \",r,\" Ohm\");\n", "##Calculation of IL(load current)\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "removing the 10 ohm resistor from the circuit\n", "\n", "For mesh 1, \n", "I1 = 10.00 A\n", "\n", "Applying KVL to mesh 2,, \n", "I2 = 2.00 A\n", "\n", "Writing Vth equation, \n", " Vth = 10.00 V\n", "replacing the current source of 10 A with an open circuit and voltage source of 100 V with a short circuit,\n", "\n", "Rth = 17.00 Ohm\n", "\n", "IL = 0.37 A\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex16-pg3.17" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.17\n", "##example 3.16\n", "a=30.;\n", "b=20.;\n", "c=50.;\n", "d=5.;\n", "e=24.;\n", "v=220.;\n", "x=(v/(a+c));\n", "y=(v/(b+d));\n", "z=(20.*y)-(30.*x);\n", "r=((a*c)/(a+c))+((b*d)/(b+d));\n", "i=z/(r+e);\n", "##Calculation the Vth (Thevenin's voltage)\n", "print(\"removing the 24 Ohm resistor from the network\");\n", "print\"%s %.2f %s\"%(\"\\nI1 = \",x,\" A\");\n", "print\"%s %.2f %s\"%(\"\\nI2 = \",y,\" A\");\n", "print\"%s %.2f %s\"%(\"\\nWriting Vth equation, \\n Vth = \",z,\" V\");\n", "##Calculation of Rth (Thevenin's resistance)\n", "print(\"replacing the 220 V source with short circuit\");\n", "print\"%s %.2f %s\"%(\"\\nRth = \",r,\" Ohm\");\n", "##Calculation of IL (load current)\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "removing the 24 Ohm resistor from the network\n", "\n", "I1 = 2.75 A\n", "\n", "I2 = 8.80 A\n", "\n", "Writing Vth equation, \n", " Vth = 93.50 V\n", "replacing the 220 V source with short circuit\n", "\n", "Rth = 22.75 Ohm\n", "\n", "IL = 2.00 A\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex17-pg3.18" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.18\n", "##example 3.17\n", "print(\"removing the 3 Ohm resistor from the network\");\n", "print(\"Applying KVL to mesh 1\");\n", "print(\"11*I1-9*I2=50\");##equation 1\n", "print(\"Applying KVL to mesh 2\");\n", "print(\"-9*I1+18*I2=0\");##equation 2\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[11, -9],[-9 ,18]]);##solving equation in matrix form\n", "B=numpy.matrix([[50] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1=7.69 A\");\n", "print(\"I2=3.85 A\");\n", "##Calculation of Vth (Thevenin's voltage)\n", "a=7.69;\n", "b=3.85;\n", "v=-((5.*b)+(8.*(b-a)));##the B terminal is positive w.r.t A\n", "print\"%s %.2f %s\"%(\"\\nWriting Vth equation, \\n Vth = \",v,\" V\");\n", "##Calculation of Rth (Thevenin's resistance)\n", "x=4.;\n", "y=2.;\n", "z=5.;\n", "##delta into star network\n", "r1=((x*y)/(x+y+z));\n", "r2=((x*z)/(x+y+z));\n", "r3=((z*y)/(x+y+z));\n", "print\"%s %.2f %s %.2f %s %.2f %s \"%(\"\\nR1 = \",r1,\" Ohm\"and \" \\nR2 = \",r2,\" Ohm\" and \"\\nR3 =\",r3,\" Ohm\");\n", "m=1.73;\n", "n=8.91;\n", "r=(r2+(m*n)/(m+n));\n", "print\"%s %.2f %s\"%(\"\\nRth = \",r,\" Ohm\");\n", "##Claculation of IL (Load Current)\n", "i=v/(r+3.);\t\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "removing the 3 Ohm resistor from the network\n", "Applying KVL to mesh 1\n", "11*I1-9*I2=50\n", "Applying KVL to mesh 2\n", "-9*I1+18*I2=0\n", "[matrix([[ 7.69230769],\n", " [ 3.84615385]])]\n", "I1=7.69 A\n", "I2=3.85 A\n", "\n", "Writing Vth equation, \n", " Vth = 11.47 V\n", "\n", "R1 = 0.73 \n", "R2 = 1.82 \n", "R3 = 0.91 Ohm \n", "\n", "Rth = 3.27 Ohm\n", "\n", "IL = 1.83 A\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex18-pg3.21" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.21\n", "##example 3.18\n", "print(\"removing the 20 Ohm resistor from the network\");\n", "print(\"Applying KVL to mesh 1\");\n", "print(\"30*I1-15*I2=-75\");##equation 1\n", "print(\"Applying KVL to mesh 2\");\n", "print(\"-15*I1+20*I2=20\");##equation 2\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[30, -15],[-15 ,20]]);##solving equation in matrix form\n", "B=numpy.matrix([[-75] ,[20]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1=-3.2 A\");\n", "print(\"I2=-1.4 A\");\n", "##Calculation of Vth (Thevenin's voltage)\n", "a=-3.2;\n", "b=-1.4;\n", "v=45.;\n", "v1=45.-10.*(a-b);\n", "print\"%s %.2f %s\"%(\"\\nWriting Vth equation, \\n Vth = \",v1,\" V\");\n", "##Calculation of Rth (Thevenin's resistance)\n", "x=10.;\n", "y=5.;\n", "z=5.;\n", "##delta into star network\n", "r1=((x*y)/(x+y+z));\n", "r2=((x*z)/(x+y+z));\n", "r3=((z*y)/(x+y+z));\n", "print\"%s %.2f %s %.2f %s %.2f %s\"%(\"\\nR1 = \",r1,\" Ohm\" and \" \\nR2 = \",r2,\" Ohm \" and \"\\nR3 = \",r3,\" Ohm\");\n", "m=16.25;\n", "r=((m*r1)/(m+r1))+r1;\n", "print\"%s %.2f %s\"%(\"\\nRth = \",r,\" Ohm\");\n", "##Claculation of IL (Load Current)\n", "i=v1/(r+20.);\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "removing the 20 Ohm resistor from the network\n", "Applying KVL to mesh 1\n", "30*I1-15*I2=-75\n", "Applying KVL to mesh 2\n", "-15*I1+20*I2=20\n", "[matrix([[-3.2],\n", " [-1.4]])]\n", "I1=-3.2 A\n", "I2=-1.4 A\n", "\n", "Writing Vth equation, \n", " Vth = 63.00 V\n", "\n", "R1 = 2.50 \n", "R2 = 2.50 \n", "R3 = 1.25 Ohm\n", "\n", "Rth = 4.67 Ohm\n", "\n", "IL = 2.55 A\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex19-pg3.22" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.22\n", "##example 3.19\n", "print(\"removing the 3 Ohm resistor from the network\");\n", "print(\"Applying KVL to mesh 1\");\n", "print(\"I1=6\");##equation 1\n", "print(\"Applying KVL to mesh 2\");\n", "print(\"-12*I1+18*I2=42\");##equation 2\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix ([[1, 0],[-12 ,18]]);##solving equation in matrix form\n", "B=numpy.matrix ([[6] ,[42]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I2= 6.33 A\");\n", "##Calculation of Vth (Thevenin's voltage)\n", "a=6.33;\n", "v=6.*a;\n", "print\"%s %.2f %s\"%(\"\\nWriting Vth equation, \\n Vth = \",v,\" V\");\n", "##Calculation of Rth (Thevenin's resistance)\n", "print(\"replacing the voltage source with short circuit and current source by open circuit\");\n", "x=6.;\n", "y=12.;\n", "r=(x*y)/(x+y);\n", "print\"%s %.2f %s\"%(\"\\nRth = \",r,\" Ohm\");\n", "##Calculation of IL (load current)\n", "i=v/(r+3.);\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "removing the 3 Ohm resistor from the network\n", "Applying KVL to mesh 1\n", "I1=6\n", "Applying KVL to mesh 2\n", "-12*I1+18*I2=42\n", "[matrix([[ 6. ],\n", " [ 6.33333333]])]\n", "I2= 6.33 A\n", "\n", "Writing Vth equation, \n", " Vth = 37.98 V\n", "replacing the voltage source with short circuit and current source by open circuit\n", "\n", "Rth = 4.00 Ohm\n", "\n", "IL = 5.43 A\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20-pg3.23" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.23\n", "##example 3.20\n", "print(\"removing the 30 Ohm resistor from the network\");\n", "print(\"Applying KVL to supermesh \");\n", "print(\"-I1+I2=13\");##equation 1\n", "print(\"15*I1+100*I2=150\");##equation 2\n", "##Calculation of Vth (Thevenin's voltage)\n", "a=3.;\n", "v=(40.*a)-50.;\n", "print\"%s %.2f %s\"%(\"\\nWriting Vth equation, \\n Vth = \",v,\" V\");\n", "##Calculation of Rth (Thevenin's resistance)\n", "print(\"replacing the voltage source with short circuit and current source by open circuit\");\n", "r=(75.*40.)/(75.+40.);\n", "print\"%s %.2f %s\"%(\"\\nRth = \",r,\" Ohm\");\n", "##Calculation of IL (load current)\n", "i=v/(r+30.);\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "removing the 30 Ohm resistor from the network\n", "Applying KVL to supermesh \n", "-I1+I2=13\n", "15*I1+100*I2=150\n", "\n", "Writing Vth equation, \n", " Vth = 70.00 V\n", "replacing the voltage source with short circuit and current source by open circuit\n", "\n", "Rth = 26.09 Ohm\n", "\n", "IL = 1.25 A\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex21-pg3.24" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.25\n", "##example 3.21\n", "##Calculation of Vth\n", "v=100.;\n", "r=20.;\n", "x=v/r;\n", "print(\"Removing the 20 Ohm resistor from the network\");\n", "print\"%s %.2f %s\"%(\"\\nVth = \",v,\" V \");\n", "##calculation of Rth\n", "print(\"replacing the voltage source with short circuit and current source by open circuit\");\n", "print(\"Rth = 0\");\n", "##calculation of IL\n", "print\"%s %.2f %s\"%(\"\\nIL = \",x,\" A\");\n" ], "language": "python", "metadata": {}, "outputs": [] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex22-pg3.25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.25\n", "##example 3.22\n", "print(\"removing the 10 Ohm resistor from the network\");\n", "print(\"Applying KVL to mesh 1\");\n", "print(\"4*I1-I2=-25\");##equation 1\n", "print(\"Applying KVL to mesh 2\");\n", "print(\"-I1+4*I2=10\");##equation 2\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[4, -1],[-1 ,4]]);##solving equation in matrix form\n", "B=([[-25] ,[10]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1=-6 A\");\n", "print(\"I2=1 A\");\n", "##Calculation of Vth (Thevenin's voltage)\n", "a=-6.;\n", "b=1.;\n", "v=-((2.*a)+(2.*b));##the terminal B is positive w.r.t A\n", "print\"%s %.2f %s\"%(\"\\nWriting Vth equation, \\n Vth = \",v,\" V\");\n", "##Calculation of Rth (Thevenin's resistance)\n", "x=2.;\n", "y=2.;\n", "z=1.;\n", "##star into delta network\n", "r1=x+y+((x*y)/z);\n", "r2=x+z+((x*z)/y);\n", "r3=z+y+((z*y)/x);\n", "print\"%s %.2f %s %.2f %s %.2f %s \"%(\"\\nR1 = \",r1,\" Ohm\" and \" \\nR2 = \",r2,\" Ohm\" and \"\\nR3 = \",r3,\" Ohm\");\n", "##Claculation of IL (Load Current)\n", "r=1.33;\n", "i=v/(r+v);\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "removing the 10 Ohm resistor from the network\n", "Applying KVL to mesh 1\n", "4*I1-I2=-25\n", "Applying KVL to mesh 2\n", "-I1+4*I2=10\n", "[matrix([[-6.],\n", " [ 1.]])]\n", "I1=-6 A\n", "I2=1 A\n", "\n", "Writing Vth equation, \n", " Vth = 10.00 V\n", "\n", "R1 = 8.00 \n", "R2 = 4.00 \n", "R3 = 4.00 Ohm \n", "\n", "IL = 0.88 A\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex23-pg3.28" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 2\n", "##pg no 3.28\n", "##example 3.23\n", "print(\"removing the 1 Ohm resistor from the network\");\n", "print(\"writing current equation for meshes 1 & 2 \");\n", "print(\"I1= -3 A\");##equation 1\n", "print(\"I2=1 A\");##equation 2\n", "##Calculation of Vth (Thevenin's voltage)\n", "a=-3.;\n", "b=1.;\n", "r=2.;\n", "v=4.-2.*(a-b);\n", "print\"%s %.2f %s\"%(\"\\nWriting Vth equation, \\n Vth = \",v,\" V\");\n", "##Calculation of Rth (Thevenin's resistance)\n", "print(\"replacing the voltage source with short circuit and current source by open circuit\");\n", "print(\"Rth = 2 Ohm\");\n", "##Calculation of IL (load current)\n", "i=v/(r+1.);\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "removing the 1 Ohm resistor from the network\n", "writing current equation for meshes 1 & 2 \n", "I1= -3 A\n", "I2=1 A\n", "\n", "Writing Vth equation, \n", " Vth = 12.00 V\n", "replacing the voltage source with short circuit and current source by open circuit\n", "Rth = 2 Ohm\n", "\n", "IL = 4.00 A\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex24-pg3.29" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.29\n", "##example3.24\n", "##calculation of Isc (short-circuit current)\n", "print(\"Applying KVL to mesh 1:\");\n", "print(\"I1=2\");##equation 1\n", "print(\"Writing current equation to supermesh:\");##meshes 2 & 3 will form a supermesh \n", "print(\"I3-I2=4\");##equation 2\n", "print(\"Applying KVL to supermesh:\");\n", "print(\"-5I2-15I3=0\");##equation 3\n", "print(\"solving these equations we get :\");##solving equations in matrix form\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=([[1, 0 ,0],[0 ,-1, 1],[0 ,-5, -15]]);\n", "B=([[2], [4] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1 = 2 A\");\n", "print(\"I2 = -3 A\");\n", "print(\"I3 = 1 A\");\n", "a=2.;\n", "b=-3.;\n", "x=a-b;\n", "print\"%s %.2f %s\"%(\"\\nIsc = \",x,\" A\");\n", "##calculation of Rn (norton's resistance)\n", "print(\"replacing the voltage source with short circuit and current source by open circuit\");\n", "c=1.;\n", "m=15.;\n", "y=(c*(m+x))/(c+m+x);\n", "print\"%s %.2f %s\"%(\"\\nRn = \",y,\" Ohm\");\n", "##calculation of IL (load current)\n", "z=10.;\n", "i=x*(y/(z+y));\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Applying KVL to mesh 1:\n", "I1=2\n", "Writing current equation to supermesh:\n", "I3-I2=4\n", "Applying KVL to supermesh:\n", "-5I2-15I3=0\n", "solving these equations we get :\n", "[array([[ 2.],\n", " [-3.],\n", " [ 1.]])]\n", "I1 = 2 A\n", "I2 = -3 A\n", "I3 = 1 A\n", "\n", "Isc = 5.00 A\n", "replacing the voltage source with short circuit and current source by open circuit\n", "\n", "Rn = 0.95 Ohm\n", "\n", "IL = 0.43 A\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex25-pg3.30" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.30\n", "##example3.25\n", "##calculation of Isc (short-circuit current)\n", "print(\"Applying KVL to mesh 1:\");\n", "print(\"7*I1-2*I2=20\");##equation 1\n", "print(\"Applying KVL to mesh 2,\"); \n", "print(\"-2*I1+10*I2=-12\");##equation 2\n", "print(\"solving these equations we get :\");##solving equations in matrix form\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[7, -2],[-2 ,10]]);##solving equation in matrix form\n", "B=([[20] ,[-12]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I2 = -0.67 A\");\n", "a=-0.67;\n", "print\"%s %.2f %s\"%(\"\\nIsc = I2 = \",a,\" A\");\n", "##calculation of Rn (norton's resistance)\n", "print(\"replacing the voltage source with short circuit \");\n", "b=5.;\n", "c=2.;\n", "d=8.;\n", "y=((b*c)/(b+c))+d;\n", "print\"%s %.2f %s\"%(\"\\nRn = \",y,\" Ohm\");\n", "##calculation of IL (load current)\n", "z=10.;\n", "i=-a*(y/(10.+y));\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Applying KVL to mesh 1:\n", "7*I1-2*I2=20\n", "Applying KVL to mesh 2,\n", "-2*I1+10*I2=-12\n", "solving these equations we get :\n", "[matrix([[ 2.66666667],\n", " [-0.66666667]])]\n", "I2 = -0.67 A\n", "\n", "Isc = I2 = -0.67 A\n", "replacing the voltage source with short circuit \n", "\n", "Rn = 9.43 Ohm\n", "\n", "IL = 0.33 A\n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex26-pg3.31" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.31\n", "##example3.26\n", "##calculation of Isc (short-circuit current)\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*I2-3*I3=20\");##equation 3\n", "print(\"solving these equations we get :\");##solving equations in matrix form\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=([[7, -1 ,0],[-1 ,6, -3],[0 ,3, -3]]);\n", "B=([[10], [0] ,[20]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1 = -13.17 A\");\n", "a=13.17;\n", "print\"%s %.2f %s\"%(\"\\nIsc = \",a,\" A\");\n", "##calculation of Rn (norton's resistance)\n", "print(\"replacing the voltage source with short circuit \");\n", "c=1.;\n", "b=6.;\n", "x=(c*b)/(c+b);\n", "y=x+2.;\n", "z=(y*3.)/(y+3.);\n", "print\"%s %.2f %s\"%(\"\\nRn = \",z,\" Ohm\");\n", "##calculation of IL (load current)\n", "n=10.;\n", "i=a*(z/(z+n));\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" 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*I2-3*I3=20\n", "solving these equations we get :\n", "[array([[ 0.5 ],\n", " [ -6.5 ],\n", " [-13.16666667]])]\n", "I1 = -13.17 A\n", "\n", "Isc = 13.17 A\n", "replacing the voltage source with short circuit \n", "\n", "Rn = 1.46 Ohm\n", "\n", "IL = 1.68 A\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex27-pg3.32" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.32\n", "##example3.27\n", "##calculation of Isc (short-circuit current)\n", "print(\"Applying KVL to mesh 1:\");\n", "print(\"20*I1-20*I2=10\");##equation 1\n", "print(\"Applying KVL to mesh 2:\"); \n", "print(\"-20*I1+60*I2-20*I3=40\");##equation 2\n", "print(\"Applying KVL to mesh 3:\");\n", "print(\"-20*I2+50*I3=-100\");##equation 3\n", "print(\"solving these equations we get :\");##solving equations in matrix form\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=([[20, -20 ,0],[-20 ,60, -20],[0 ,-20, -50]]);\n", "B=([[10], [40] ,[-100]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1 = 0.81A\");\n", "a=0.81;\n", "print\"%s %.2f %s\"%(\"\\nIsc = \",a,\" A\");\n", "##calculation of Rn (norton's resistance)\n", "print(\"replacing the voltage source with short circuit \");\n", "c=20.;\n", "b=30.;\n", "x=(c*b)/(c+b);\n", "y=x+c;\n", "z=(y*c)/(y+c);\n", "print\"%s %.2f %s\"%(\"\\nRn = \",z,\" Ohm\");\n", "##calculation of IL (load current)\n", "n=10.;\n", "i=a*(z/(z+n));\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Applying KVL to mesh 1:\n", "20*I1-20*I2=10\n", "Applying KVL to mesh 2:\n", "-20*I1+60*I2-20*I3=40\n", "Applying KVL to mesh 3:\n", "-20*I2+50*I3=-100\n", "solving these equations we get :\n", "[array([[ 2.375],\n", " [ 1.875],\n", " [ 1.25 ]])]\n", "I1 = 0.81A\n", "\n", "Isc = 0.81 A\n", "replacing the voltage source with short circuit \n", "\n", "Rn = 12.31 Ohm\n", "\n", "IL = 0.45 A\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex28-pg3.33" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.33\n", "##example3.28\n", "##calculation of Isc (short-circuit current)\n", "print(\"Applying KVL to mesh 1:\");\n", "print(\"90*I1-60*I2=120\");##equation 1\n", "print(\"Applying KVL to mesh 2:\"); \n", "print(\"-60*I1+100*I2-30*I3=40\");##equation 2\n", "print(\"Applying KVL to mesh 3:\");\n", "print(\"30*I2-30*I3=-10\");##equation 3\n", "print(\"solving these equations we get :\");##solving equations in matrix form\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=([[90, -60 ,0],[-60 ,100, -30],[0 ,30, -30]]);\n", "B=([[102], [40] ,[-10]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I3 = 4.67A\");\n", "a=4.67;\n", "print\"%s %.2f %s\"%(\"\\nIsc = \",a,\" A\");\n", "##calculation of Rn (norton's resistance)\n", "print(\"replacing the voltage source with short circuit \");\n", "c=30.;\n", "b=60.;\n", "x=(c*b)/(c+b);\n", "y=x+10.;\n", "z=(y*c)/(y+c);\n", "print\"%s %.2f %s\"%(\"\\nRn = \",z,\" Ohm\");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Applying KVL to mesh 1:\n", "90*I1-60*I2=120\n", "Applying KVL to mesh 2:\n", "-60*I1+100*I2-30*I3=40\n", "Applying KVL to mesh 3:\n", "30*I2-30*I3=-10\n", "solving these equations we get :\n", "[array([[ 3.75555556],\n", " [ 3.93333333],\n", " [ 4.26666667]])]\n", "I3 = 4.67A\n", "\n", "Isc = 4.67 A\n", "replacing the voltage source with short circuit \n", "\n", "Rn = 15.00 Ohm\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex29-pg3.34" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.34\n", "##example3.29\n", "##calculation of Isc (short-circuit current)\n", "print(\"Writing current equation for supermesh :\");\n", "print(\"I2-I1=2\");##equation 1\n", "print(\"Applying KVL to supermesh ,\"); \n", "print(\"12*I1= 55\");##equation 2\n", "print(\"solving these equations we get :\");##solving equations in matrix form\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[-1, 1],[12 ,0]]);##solving equation in matrix form\n", "B=([[2] ,[55]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1 = 4.58 A\");\n", "print(\"I2 = 6.58 A\");\n", "a=6.58;\n", "print\"%s %.2f %s\"%(\"\\nIsc = I2 = \",a,\" A\");\n", "##calculation of Rn (norton's resistance)\n", "print(\"replacing the voltage source with short circuit and current source with open circuit \");\n", "b=12.;\n", "c=4.;\n", "y=((b*c)/(b+c));\n", "print\"%s %.2f %s\"%(\"\\nRn = \",y,\" Ohm\");\n", "##calculation of IL (load current)\n", "z=8.;\n", "i=a*(y/(z+y));\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Writing current equation for supermesh :\n", "I2-I1=2\n", "Applying KVL to supermesh ,\n", "12*I1= 55\n", "solving these equations we get :\n", "[matrix([[ 4.58333333],\n", " [ 6.58333333]])]\n", "I1 = 4.58 A\n", "I2 = 6.58 A\n", "\n", "Isc = I2 = 6.58 A\n", "replacing the voltage source with short circuit and current source with open circuit \n", "\n", "Rn = 3.00 Ohm\n", "\n", "IL = 1.79 A\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex30-pg3.35" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.35\n", "##example3.30\n", "##calculation of Isc (short-circuit current)\n", "print(\"Applying KVL to mesh 1:\");\n", "print(\"5*I1-2*I2=-2\");##equation 1\n", "print(\"Applying KVL to mesh 2:\"); \n", "print(\"4*I2-2*I3=-1\");##equation 2\n", "print(\"Applying KVL to mesh 3:\");\n", "print(\"-2*I1-2*I2+4*I3=0\");##equation 3\n", "print(\"solving these equations we get :\");##solving equations in matrix form\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "A=([[5, -2 ,0],[0 ,4, -2],[-2,-2, 4]]);\n", "B=([[-2], [-1] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "print[X];\n", "print(\"I1 = -0.64A\");\n", "print(\"I2 = -0.55A\");\n", "print(\"I3 = -0.59A\");\n", "a=-0.64;\n", "b=-0.55;\n", "c=-0.59;\n", "print\"%s %.2f %s\"%(\"\\nIsc = I3 = \",a,\" A\");\n", "##calculation of Rn (norton's resistance)\n", "print(\"replacing the voltage source with short circuit \");\n", "z=2.2;\n", "print\"%s %.2f %s\"%(\"\\nRn = \",z,\" Ohm\");\n", "##calculation of IL (load current)\n", "n=1.;\n", "i=-c*(z/(z+n));\n", "print\"%s %.2f %s\"%(\"\\nIL = \",i,\" A\");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Applying KVL to mesh 1:\n", "5*I1-2*I2=-2\n", "Applying KVL to mesh 2:\n", "4*I2-2*I3=-1\n", "Applying KVL to mesh 3:\n", "-2*I1-2*I2+4*I3=0\n", "solving these equations we get :\n", "[array([[-0.61538462],\n", " [-0.53846154],\n", " [-0.57692308]])]\n", "I1 = -0.64A\n", "I2 = -0.55A\n", "I3 = -0.59A\n", "\n", "Isc = I3 = -0.64 A\n", "replacing the voltage source with short circuit \n", "\n", "Rn = 2.20 Ohm\n", "\n", "IL = 0.41 A\n" ] } ], "prompt_number": 29 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex31-pg3.37" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.39\n", "##example3.31\n", "##calculation of Vth (Thevenin's voltage)\n", "a=0.25;\n", "import math\n", "import numpy\n", "from numpy import linalg\n", "v=(10.*a)+(8.*a);\n", "print(\"Writing Vth equation,\");\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Isc (short-circuit current)\n", "print(\"Applying KVL to mesh 1:\");\n", "print(\"4*I1-2*I2 = 1\");##equation 1\n", "print(\"Applying KVL to mesh 2:\"); \n", "print(\"-18*I1-11*I2=0\");##equation 2\n", "A=numpy.matrix([[4, -2],[8 ,-11]]);##solving equation in matrix form\n", "B=([[1] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "print(\"I2 = 2.25 A\");\n", "a=2.25;\n", "print'%s %.2f %s'%(\"\\nIsc = I2 = \",a,\" A\");\n", "##Calculation of Rth\n", "x=v/a;\n", "print'%s %.2f %s'%(\"\\nRth = \",x,\" Ohm\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Writing Vth equation,\n", "\n", "Vth = 4.50 V\n", "Applying KVL to mesh 1:\n", "4*I1-2*I2 = 1\n", "Applying KVL to mesh 2:\n", "-18*I1-11*I2=0\n", "[matrix([[ 0.39285714],\n", " [ 0.28571429]])]\n", "I2 = 2.25 A\n", "\n", "Isc = I2 = 2.25 A\n", "\n", "Rth = 2.00 Ohm\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex33-pg3.39" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.39\n", "##example3.33\n", "##calculation of Vth (Thevenin's voltage)\n", "a=0.25;\n", "import numpy\n", "from numpy import linalg\n", "v=(10.*a)+(8.*a);\n", "print(\"Writing Vth equation,\");\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Isc (short-circuit current)\n", "print(\"Applying KVL to mesh 1:\");\n", "print(\"4*I1-2*I2 = 1\");##equation 1\n", "print(\"Applying KVL to mesh 2:\"); \n", "print(\"-18*I1-11*I2=0\");##equation 2\n", "A=numpy.matrix([[4, -2],[8 ,-11]]);##solving equation in matrix form\n", "B=([[1] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "print(\"I2 = 2.25 A\");\n", "a=2.25;\n", "print'%s %.2f %s'%(\"\\nIsc = I2 = \",a,\" A\");\n", "##Calculation of Rth\n", "x=v/a;\n", "print'%s %.2f %s'%(\"\\nRth = \",x,\" Ohm\");\n", "\n", "\n", "\n", "\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[4, -2],[8 ,-11]]);##solving equation in matrix form\n", "B=([[1] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Writing Vth equation,\n", "\n", "Vth = 4.50 V\n", "Applying KVL to mesh 1:\n", "4*I1-2*I2 = 1\n", "Applying KVL to mesh 2:\n", "-18*I1-11*I2=0\n", "[matrix([[ 0.39285714],\n", " [ 0.28571429]])]\n", "I2 = 2.25 A\n", "\n", "Isc = I2 = 2.25 A\n", "\n", "Rth = 2.00 Ohm\n", "[matrix([[ 0.39285714],\n", " [ 0.28571429]])]\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex41-pg3.47" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.47\n", "##example3.41\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "print(\"I2-I1=4\");##equation 1\n", "print(\"Applying KVL at the outerpath:\"); \n", "print(\"-6*I1-5*I2=2\");##equation 2\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[-1, 1],[-6 ,-5]]);##solving equation in matrix form\n", "B=([[4] ,[2]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "print(\"I1 = -2 A\");\n", "print(\"I2 = 2 A\");\n", "print(\"Writing Vth equation,\");\n", "a=-2;\n", "v=8-a;\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the voltage source with short circuit and current source by an open circuit \");\n", "x=(v*1.)/(v+1.);\n", "print'%s %.2f %s'%(\"\\nRth = \",x,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",x,\" Ohm\");\n", "##calculation of Pmax\n", "m=(v**2)/(4.*x);\n", "print'%s %.2f %s'%(\"\\nPmax = \",m,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "I2-I1=4\n", "Applying KVL at the outerpath:\n", "-6*I1-5*I2=2\n", "[matrix([[-2.],\n", " [ 2.]])]\n", "I1 = -2 A\n", "I2 = 2 A\n", "Writing Vth equation,\n", "\n", "Vth = 10.00 V\n", "replacing the voltage source with short circuit and current source by an open circuit \n", "\n", "Rth = 0.91 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 0.91 Ohm\n", "\n", "Pmax = 27.50 W\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex42-pg3.48" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.48\n", "##example3.42\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "print(\"I1=50\");##equation 1\n", "print(\"Applying KVL to mesh 2:\"); \n", "print(\"5*I1-10*I2=0\");##equation 2\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[1, 0],[5 ,-10]]);##solving equation in matrix form\n", "B=([[50] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "\n", "print(\"I2 = 25 A\");\n", "print(\"Writing Vth equation,\");\n", "a=25;\n", "v=3*a;\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the current source of 50 A by an open circuit \");\n", "x=7.;\n", "y=3.;\n", "m=(x*y)/(x+y);\n", "print'%s %.2f %s'%(\"\\nRth = \",m,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",m,\" Ohm\");\n", "##calculation of Pmax\n", "n=(v**2)/(4.*m);\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "I1=50\n", "Applying KVL to mesh 2:\n", "5*I1-10*I2=0\n", "[matrix([[ 50.],\n", " [ 25.]])]\n", "I2 = 25 A\n", "Writing Vth equation,\n", "\n", "Vth = 75.00 V\n", "replacing the current source of 50 A by an open circuit \n", "\n", "Rth = 2.10 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 2.10 Ohm\n", "\n", "Pmax = 669.64 W\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex43-pg3.49" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.49\n", "##example3.43\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "print(\"Writing the current equation for the supermesh\");\n", "print(\"I2-I1=6\");##equation 1\n", "print(\"Applying KVL to the supermesh :\"); \n", "print(\"5*I1+2*I2=10\");##equation 2\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[-1, 1],[5 ,1]]);##solving equation in matrix form\n", "B=([[6] ,[10]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "\n", "print(\"I1 = -0.29 A\");\n", "print(\"I2 = 5.71 A\");\n", "print(\"Writing Vth equation,\");\n", "a=5.71;\n", "v=2*a;\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the current source of 50 A by an open circuit \");\n", "x=5.;\n", "y=2.;\n", "m=((x*y)/(x+y))+3.+4.;\n", "print'%s %.2f %s'%(\"\\nRth = \",m,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",m,\" Ohm\");\n", "##calculation of Pmax\n", "n=(v**2)/(4.*m);\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "Writing the current equation for the supermesh\n", "I2-I1=6\n", "Applying KVL to the supermesh :\n", "5*I1+2*I2=10\n", "[matrix([[ 0.66666667],\n", " [ 6.66666667]])]\n", "I1 = -0.29 A\n", "I2 = 5.71 A\n", "Writing Vth equation,\n", "\n", "Vth = 11.42 V\n", "replacing the current source of 50 A by an open circuit \n", "\n", "Rth = 8.43 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 8.43 Ohm\n", "\n", "Pmax = 3.87 W\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex44-pg3.50" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.50\n", "##example3.44\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "print(\"Applying KVL to mesh 1\");\n", "print(\"15*I1-5*I2=120\");##equation 1\n", "print(\"Applying KVL to the mesh 2:\"); \n", "print(\"I2=-6\");##equation 2\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[15, -5],[0 ,1]]);##solving equation in matrix form\n", "B=([[120] ,[-6]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "\n", "print(\"I1 = 6 A\");\n", "print(\"Writing Vth equation,\");\n", "a=6.;\n", "v=120.-(10.*a);\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the current source of 50 A by an open circuit \");\n", "x=10.;\n", "y=5.;\n", "m=((x*y)/(x+y));\n", "print'%s %.2f %s'%(\"\\nRth = \",m,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",m,\" Ohm\");\n", "##calculation of Pmax\n", "n=(v**2)/(4.*m);\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "Applying KVL to mesh 1\n", "15*I1-5*I2=120\n", "Applying KVL to the mesh 2:\n", "I2=-6\n", "[matrix([[ 6.],\n", " [-6.]])]\n", "I1 = 6 A\n", "Writing Vth equation,\n", "\n", "Vth = 60.00 V\n", "replacing the current source of 50 A by an open circuit \n", "\n", "Rth = 3.33 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 3.33 Ohm\n", "\n", "Pmax = 270.00 W\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex45-pg3.45" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.51\n", "##example3.45\n", "\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "print(\"I1=3 A\");##equation 1\n", "print(\"Applying KVL to the mesh 2:\"); \n", "print(\"-25*I1+41*I2=0\");##equation 2\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[1, 0],[-25 ,41]]);##solving equation in matrix form\n", "B=([[3] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "\n", "print(\"I2 = 1.83 A\");\n", "print(\"Writing Vth equation,\");\n", "a=1.83;\n", "v=-20.+(10.*a)+(6.*a);\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the current source of 50 A by an open circuit \");\n", "x=25.;\n", "y=16.;\n", "m=((x*y)/(x+y));\n", "print'%s %.2f %s'%(\"\\nRth = \",m,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",m,\" Ohm\");\n", "##calculation of Pmax\n", "n=(v**2)/(4.*m);\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "I1=3 A\n", "Applying KVL to the mesh 2:\n", "-25*I1+41*I2=0\n", "[matrix([[ 3. ],\n", " [ 1.82926829]])]\n", "I2 = 1.83 A\n", "Writing Vth equation,\n", "\n", "Vth = 9.28 V\n", "replacing the current source of 50 A by an open circuit \n", "\n", "Rth = 9.76 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 9.76 Ohm\n", "\n", "Pmax = 2.21 W\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex46-pg3.52" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.52\n", "##example3.46\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "print(\"I2-I1=2\");##equation 1\n", "print(\"I2=-3 A\");##equation 2\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[-1, 1],[0 ,1]]);##solving equation in matrix form\n", "B=([[2] ,[-3]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "\n", "print(\"I1 = -5 A\");\n", "print(\"Writing Vth equation,\");\n", "a=-5.;\n", "b=-3.;\n", "v=8.-(2.*a)-b-6.;\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the voltage source with short circuit and current source by an open circuit \");\n", "m=5.;\n", "print'%s %.2f %s'%(\"\\nRth = \",m,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",m,\" Ohm\");\n", "##calculation of Pmax\n", "n=(v**2)/(4.*m);\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "I2-I1=2\n", "I2=-3 A\n", "[matrix([[-5.],\n", " [-3.]])]\n", "I1 = -5 A\n", "Writing Vth equation,\n", "\n", "Vth = 15.00 V\n", "replacing the voltage source with short circuit and current source by an open circuit \n", "\n", "Rth = 5.00 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 5.00 Ohm\n", "\n", "Pmax = 11.25 W\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex47-pg3.53" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.52\n", "##example3.46\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "print(\"By star-delta transformation\");\n", "a=5.;\n", "b=20.;\n", "c=9.;\n", "v=100.;\n", "i=v/(a+a+b+c+c);\n", "print(\"Writing Vth equation,\");\n", "vth=v-(14.*i);\n", "print'%s %.2f %s'%(\"\\nVth = \",vth,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the voltage source with short circuit \");\n", "m=23.92;\n", "print'%s %.2f %s'%(\"\\nRth = \",m,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",m,\" Ohm\");\n", "##calculation of Pmax\n", "n=(vth**2)/(4.*m);\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "By star-delta transformation\n", "Writing Vth equation,\n", "\n", "Vth = 70.83 V\n", "replacing the voltage source with short circuit \n", "\n", "Rth = 23.92 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 23.92 Ohm\n", "\n", "Pmax = 52.44 W\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex48-pg3.55" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.55\n", "##example3.48\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "print(\"Applying KVL to the mesh 1:\"); \n", "print(\"35*I1-30*I2=60\");##equation 1\n", "print(\"Applying KVL to the mesh 2:\"); \n", "print(\"I2=2\");##equation 2\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[35, -30],[60 ,2]]);##solving equation in matrix form\n", "B=([[60] ,[2]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "print(\"I1 = 3.43 A\");\n", "print(\"Writing Vth equation,\");\n", "a=3.43;\n", "b=2.;\n", "v=20.*(a-b)+20.;\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the voltage source with short circuit and current source by an open circuit \");\n", "x=15.;\n", "y=20.;\n", "m=((x*y)/(x+y));\n", "print'%s %.2f %s'%(\"\\nRth = \",m,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",m,\" Ohm\");\n", "##calculation of Pmax\n", "n=(v**2)/(4.*m);\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "Applying KVL to the mesh 1:\n", "35*I1-30*I2=60\n", "Applying KVL to the mesh 2:\n", "I2=2\n", "[matrix([[ 0.09625668],\n", " [-1.88770053]])]\n", "I1 = 3.43 A\n", "Writing Vth equation,\n", "\n", "Vth = 48.60 V\n", "replacing the voltage source with short circuit and current source by an open circuit \n", "\n", "Rth = 8.57 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 8.57 Ohm\n", "\n", "Pmax = 68.89 W\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex49-pg3.56" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.56\n", "##example3.49\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "x=100.;\n", "a=10.;\n", "b=20.;\n", "c=30.;\n", "d=40.;\n", "i1=x/(a+c);\n", "i2=x/(b+d);\n", "print'%s %.2f %s'%(\"\\nI1 = \",i1,\" A\");\n", "print'%s %.2f %s'%(\"\\ni2 = \",i2,\" A\");\n", "print(\"Writing Vth equation,\");\n", "x=2.5;\n", "y=1.66;\n", "v=(20.*y)-(10.*x);\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the voltage source of 100V with short circuit \");\n", "m=((a*c)/(a+c))+((b*d)/(b+d));\n", "print'%s %.2f %s'%(\"\\nRth = \",m,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",m,\" Ohm\");\n", "##calculation of Pmax\n", "n=(v**2)/(4.*m);\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "\n", "I1 = 2.50 A\n", "\n", "i2 = 1.67 A\n", "Writing Vth equation,\n", "\n", "Vth = 8.20 V\n", "replacing the voltage source of 100V with short circuit \n", "\n", "Rth = 20.83 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 20.83 Ohm\n", "\n", "Pmax = 0.81 W\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex50-pg3.57" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.57\n", "##example3.50\n", "##calculation of Vth\n", "print(\"Removing the variable resistor RL from the network:\");\n", "print(\"Applying KVL to the mesh 1:\"); \n", "print(\"9*I1-3*I2=72\");##equation 1\n", "print(\"Applying KVL to the mesh 2:\"); \n", "print(\"-3*I1+9*I2=0\");##equation 2\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[9, -3],[-3 ,9]]);##solving equation in matrix form\n", "B=([[72] ,[0]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "print(\"I1 = 9 A\");\n", "print(\"I2 = 3 A\");\n", "print(\"Writing Vth equation,\");\n", "a=9.;\n", "b=3.;\n", "v=(6.*a)+(2.*b);\n", "print'%s %.2f %s'%(\"\\nVth = \",v,\" V\");\n", "##calculation of Rth\n", "print(\"replacing the voltage source with short circuit and current source by an open circuit \");\n", "x=6.;\n", "y=2.;\n", "z=4.;\n", "m=((x*b)/(x+b))+2.;\n", "l=((m*z)/(m+z));\n", "print'%s %.2f %s'%(\"\\nRth = \",l,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",l,\" Ohm\");\n", "##calculation of Pmax\n", "n=(v**2)/(4.*l);\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Removing the variable resistor RL from the network:\n", "Applying KVL to the mesh 1:\n", "9*I1-3*I2=72\n", "Applying KVL to the mesh 2:\n", "-3*I1+9*I2=0\n", "[matrix([[ 9.],\n", " [ 3.]])]\n", "I1 = 9 A\n", "I2 = 3 A\n", "Writing Vth equation,\n", "\n", "Vth = 60.00 V\n", "replacing the voltage source with short circuit and current source by an open circuit \n", "\n", "Rth = 2.00 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 2.00 Ohm\n", "\n", "Pmax = 450.00 W\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex51-pg3.58" ] }, { "cell_type": "code", "collapsed": false, "input": [ "##Network Theorem 1\n", "##page no-3.58\n", "##example3.51\n", "##Calculation of Vth\n", "print(\"from the figure\");\n", "print(\"Vth=4*I\");\n", "print(\"Applying KVL to the mesh\");\n", "print(\"0.5*Vth-8*I=-12\");\n", "import numpy\n", "from numpy import linalg\n", "A=numpy.matrix([[1, -4],[0.5 ,-8]]);##solving equation in matrix form\n", "B=([[0] ,[-12]])\n", "X=numpy.dot(numpy.linalg.inv(A),B);\n", "\n", "print[X];\n", "\n", "print(\"Vth=8 V\");\n", "##Calculation of Isc\n", "v=8.;\n", "i=12./4.;\n", "print'%s %.2f %s'%(\"\\nIsc = \",i,\" A\");\n", "##Calculation of Rth\n", "r=v/i;\n", "print'%s %.2f %s'%(\"\\nRth = Vth/Isc = \",r,\" Ohm\");\n", "##calculation of RL\n", "print(\"For maximum power transfer\");\n", "print'%s %.2f %s'%(\"\\nRth = RL = \",r,\" Ohm\");\n", "##calculation of Pmax\n", "x=v/(2.*r);\n", "print'%s %.2f %s'%(\"\\nIL = \",x,\" A\");\n", "n=(x**2)*r;\n", "print'%s %.2f %s'%(\"\\nPmax = \",n,\" W\");" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "from the figure\n", "Vth=4*I\n", "Applying KVL to the mesh\n", "0.5*Vth-8*I=-12\n", "[matrix([[ 8.],\n", " [ 2.]])]\n", "Vth=8 V\n", "\n", "Isc = 3.00 A\n", "\n", "Rth = Vth/Isc = 2.67 Ohm\n", "For maximum power transfer\n", "\n", "Rth = RL = 2.67 Ohm\n", "\n", "IL = 1.50 A\n", "\n", "Pmax = 6.00 W\n" ] } ], "prompt_number": 14 } ], "metadata": {} } ] }