{
"metadata": {
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter1-Basic Circuit Concepts"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1-pg9"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##page no-1.9\n",
"##example1.1\n",
"print(\"Current through 15Ohm resistor is given by:\");\n",
"print(\"I1=30/15\");\n",
"I1=30/15\n",
"print\"%s %.2f %s \"%(\"current through 15Ohm resistor = \",I1,\" Ampere\")\n",
"print(\"Current through 5Ohm resistor is given by:\")\n",
"print(\"I2=5+2\");\n",
"I2=5+2\n",
"print\"%s %.2f %s \"%(\"current through 5ohm resistor = \",I2,\" Ampere\")\n",
"print(\"R=100-30-5*I2/I1\");\n",
"R=(100-30-5*I2)/I1\n",
"print\"%s %.2f %s \"%(\"R = \",R,\" Ohm\");"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Current through 15Ohm resistor is given by:\n",
"I1=30/15\n",
"current through 15Ohm resistor = 2.00 Ampere \n",
"Current through 5Ohm resistor is given by:\n",
"I2=5+2\n",
"current through 5ohm resistor = 7.00 Ampere \n",
"R=100-30-5*I2/I1\n",
"R = 17.00 Ohm \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2-pg10"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##page no-1.10\n",
"##example1.2\n",
"import math\n",
"import numpy\n",
"print(\"from the given fig:\")\n",
"print(\"I2-I3=13\");\n",
"print(\"-20*I1+8*I2=0\");\n",
"print(\"-12*I1-16*I3=0\");\n",
"##solving these equations in the matrix form\n",
"A=numpy.matrix([[0, 1 ,-1],[-20, 8, 0],[-12 ,0 ,-16]])\n",
"B=numpy.matrix([[13], [0] ,[0]])\n",
"print(\"A=\")\n",
"print[A]\n",
"print(\"B=\")\n",
"print[B]\n",
"X=numpy.dot(numpy.linalg.inv(A),B)\n",
"print(\"X=\")\n",
"print[X]\n",
"print(\"I1 = 4Ampere\")\n",
"print(\"I2 = 10Ampere\")\n",
"print(\"I3 = -3Ampere\")\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"from the given fig:\n",
"I2-I3=13\n",
"-20*I1+8*I2=0\n",
"-12*I1-16*I3=0\n",
"A=\n",
"[matrix([[ 0, 1, -1],\n",
" [-20, 8, 0],\n",
" [-12, 0, -16]])]\n",
"B=\n",
"[matrix([[13],\n",
" [ 0],\n",
" [ 0]])]\n",
"X=\n",
"[matrix([[ 4.],\n",
" [ 10.],\n",
" [ -3.]])]\n",
"I1 = 4Ampere\n",
"I2 = 10Ampere\n",
"I3 = -3Ampere\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3-pg11"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##pg no-1.11\n",
"##example 1.3\n",
"print(\"Iaf=x\")\n",
"print(\"Ife=x-30\")\n",
"print(\"Ied=x+40\")\n",
"print(\"Idc=x-80\")\n",
"print(\"Icb=x-20\")\n",
"print(\"Iba=x-80\")\n",
"print(\"Applying KVL to the closed path AFEDCBA:\")##Applying KVL to the path AFEDCBA\n",
"print(\"x=4.1/0.1\")\n",
"x=4.1/0.1;\n",
"Iaf=x;\n",
"print\"%s %.2f %s \"%(\"\\nIaf = \",Iaf,\" Ampere\");\n",
"Ife=x-30.\n",
"print\"%s %.2f %s \"%(\"\\nIfe = \",Ife,\" Ampere\");\n",
"Ied=x+40.;\n",
"print\"%s %.2f %s \"%(\"\\nIed = \",Ied,\" Ampere\");\n",
"Idc=x-80;\n",
"print\"%s %.2f %s \"%(\"\\nIdc = \",Idc,\" Ampere\");\n",
"Icb=x-20.;\n",
"print\"%s %.2f %s \"%(\"\\nIcb = \",Icb,\" Ampere\");\n",
"Iba=x-80.;\n",
"print\"%s %.2f %s \"%(\"\\nIba = \",Iba,\" Ampere\");"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Iaf=x\n",
"Ife=x-30\n",
"Ied=x+40\n",
"Idc=x-80\n",
"Icb=x-20\n",
"Iba=x-80\n",
"Applying KVL to the closed path AFEDCBA:\n",
"x=4.1/0.1\n",
"\n",
"Iaf = 41.00 Ampere \n",
"\n",
"Ife = 11.00 Ampere \n",
"\n",
"Ied = 81.00 Ampere \n",
"\n",
"Idc = -39.00 Ampere \n",
"\n",
"Icb = 21.00 Ampere \n",
"\n",
"Iba = -39.00 Ampere \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4-pg12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##pg no- 1.12\n",
"##example 1.4\n",
"import math\n",
"import numpy\n",
"print(\"Applying KVL to the closed path OBAO\");##Applying KVL to the closed path OBAO\n",
"print(\"3*x-3*y=2\");\n",
"print(\"Applying KVL to the closed path ABCA\");##Applying KVL to the closed path ABCA\n",
"print(\"9*x+12*y=4\");\n",
"a=numpy.matrix([[3, -3],[9, 12]]);\n",
"b=([[2] ,[4]])\n",
"print(\"a=\")\n",
"print[a]\n",
"print(\"b=\")\n",
"print[b]\n",
"X=numpy.dot(numpy.linalg.inv(a),b)\n",
"\n",
"print(X)\n",
"print(\"x=0.5714286 Ampere\");\n",
"print(\"y=-0.095238 Ampere\");\n",
"print(\"Ioa=0.57A\")\n",
"print(\"Iob=1-0.57\")\n",
"Iob=1-0.57;\n",
"print\"%s %.2f %s \"%(\"\\nIob = \",Iob,\" A\");\n",
"print(\"Iab = 0.095\");\n",
"Iac=0.57-0.095;\n",
"print\"%s %.2f %s \"%(\"\\nIac =\",Iac,\" A\");\n",
"print(\"Iab=1-0.57 + 0.095\")\n",
"Iab=1-0.57 + 0.095;\n",
"print\"%s %.2f %s \"%(\"\\nIob = \",Iab,\" A\") "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Applying KVL to the closed path OBAO\n",
"3*x-3*y=2\n",
"Applying KVL to the closed path ABCA\n",
"9*x+12*y=4\n",
"a=\n",
"[matrix([[ 3, -3],\n",
" [ 9, 12]])]\n",
"b=\n",
"[[[2], [4]]]\n",
"[[ 0.57142857]\n",
" [-0.0952381 ]]\n",
"x=0.5714286 Ampere\n",
"y=-0.095238 Ampere\n",
"Ioa=0.57A\n",
"Iob=1-0.57\n",
"\n",
"Iob = 0.43 A \n",
"Iab = 0.095\n",
"\n",
"Iac = 0.47 A \n",
"Iab=1-0.57 + 0.095\n",
"\n",
"Iob = 0.53 A \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5-pg12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##pg no-1.12\n",
"##example 1.5\n",
"I1=2./5.;\n",
"print\"%s %.2f %s \"%(\"I1=2/5= \",I1,\" Ampere\")\n",
"I2=4./8.;\n",
"print\"%s %.2f %s \"%(\"\\nI2=4/8= \",I2,\" Ampere\")\n",
"print(\"\\nPotential difference between points x and y = Vxy = Vx-Vy\")\n",
"print(\"\\nWriting KVL equations for the path x to y\")##Writing KVL equation from x to y\n",
"print(\"\\nVs+3*I1+4-3*I2-Vy=0\")\n",
"print(\"\\nVs+3*(0.4) + 4- 3*(0.5) -Vy = 0\")\n",
"print(\"\\nVs+3*I1+4-3*I2-Vy = 0\")\n",
"print(\"\\nVx-Vy = -3.7\")\n",
"print(\"\\nVxy = -3.7V\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I1=2/5= 0.40 Ampere \n",
"\n",
"I2=4/8= 0.50 Ampere \n",
"\n",
"Potential difference between points x and y = Vxy = Vx-Vy\n",
"\n",
"Writing KVL equations for the path x to y\n",
"\n",
"Vs+3*I1+4-3*I2-Vy=0\n",
"\n",
"Vs+3*(0.4) + 4- 3*(0.5) -Vy = 0\n",
"\n",
"Vs+3*I1+4-3*I2-Vy = 0\n",
"\n",
"Vx-Vy = -3.7\n",
"\n",
"Vxy = -3.7V\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}