{
"metadata": {
"name": "",
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"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": {}
}
]
}