{
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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-pg1.9"
]
},
{
"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-pg1.10"
]
},
{
"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-pg1.11"
]
},
{
"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-pg1.12"
]
},
{
"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": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5-pg1.12"
]
},
{
"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
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex6-pg1.13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##pg no-1.13\n",
"##example 1.6\n",
"import math\n",
"#calculate the \n",
"I1=20/15.;\n",
"print'%s %.2f %s'%(\"I1=2/5= \",I1,\" Ampere\")\n",
"I2=15./10.;\n",
"print'%s %.2f %s'%(\"\\nI2=4/8= \",I2,\" Ampere\")\n",
"print(\"Voltage between points A and B = VAB = VA-VB\");\n",
"print(\"Writing KVL equations for the path A to B:\");##Writing KVL equations for the path A to B\n",
"print(\"VA - 5*I1 - 5 - 15 + 6*I2 - VB = 0\");\n",
"print(\"VA - VB = 5*1.33 + 5 + 15 + 6*1.5\");\n",
"VAB=(5*1.33)+5.+15.-(6*1.5);\n",
"print'%s %.2f %s'%(\"VAB = \",VAB,\" Volt\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I1=2/5= 1.33 Ampere\n",
"\n",
"I2=4/8= 1.50 Ampere\n",
"Voltage between points A and B = VAB = VA-VB\n",
"Writing KVL equations for the path A to B:\n",
"VA - 5*I1 - 5 - 15 + 6*I2 - VB = 0\n",
"VA - VB = 5*1.33 + 5 + 15 + 6*1.5\n",
"VAB = 17.65 Volt\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7-pg1.13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##page no-1.13\n",
"##example1.7\n",
"import math\n",
"#calculate the \n",
"I1=5./2.;\n",
"print'%s %.2f %s'%(\"I1=2/5= \",I1,\" Ampere\")\n",
"I2=2.;\n",
"print'%s %.2f %s'%(\"\\nI2=4/8= \",I2,\" Ampere\")\n",
"print(\"Potential difference VAB = VA - VB\");\n",
"print(\"Writing KVL equations for path A to B\") ##Writing KVL equations for path A to B\n",
"print(\"VA - 2*I1 + 8 - 5*I2 - VB = 0\");\n",
"print(\"VA - VB = (2*2.5) - 8 5 + (5*2)\");\n",
"VAB=(2.*2.5)-8.+(5.*2.)\n",
"print'%s %.2f %s'%(\"VAB = \",VAB,\" Volt\");"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I1=2/5= 2.50 Ampere\n",
"\n",
"I2=4/8= 2.00 Ampere\n",
"Potential difference VAB = VA - VB\n",
"Writing KVL equations for path A to B\n",
"VA - 2*I1 + 8 - 5*I2 - VB = 0\n",
"VA - VB = (2*2.5) - 8 5 + (5*2)\n",
"VAB = 7.00 Volt\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8-pg1.14"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##page no-1.14\n",
"##example1.8\n",
"import math\n",
"#calculate the \n",
"I1=10./8.;\n",
"print'%s %.2f %s'%(\"I1=2/5= \",I1,\" Ampere\")\n",
"I2=5.;\n",
"print'%s %.2f %s'%(\"\\nI2=4/8= \",I2,\" Ampere\")\n",
"print(\"Applying KVL to the path from A to B\") ##Applying KVL to the path from A to B\n",
"print(\"VA - 3*I1 - 8 + 3*I2 - VB = 0\");\n",
"print(\"VA - VB = 3*1.25 + 8 - 3*5\")\n",
"VAB= (3*1.25)+8.-(3.*5.);\n",
"print'%s %.2f %s'%(\"VAB = \",VAB,\" Volt\");"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I1=2/5= 1.25 Ampere\n",
"\n",
"I2=4/8= 5.00 Ampere\n",
"Applying KVL to the path from A to B\n",
"VA - 3*I1 - 8 + 3*I2 - VB = 0\n",
"VA - VB = 3*1.25 + 8 - 3*5\n",
"VAB = -3.25 Volt\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.12-pg1.17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##page no-1.17\n",
"##example1.12\n",
"print(\"Applying KVL to the circuit :\");\n",
"print(\"50 - 5*I - 1.2*I - 16 = 0\")\n",
"I=(50.-16.)/6.2;\n",
"print'%s %.2f %s'%(\"I= \",I,\" Amp\");\n",
"P=50.*I;\n",
"print'%s %.2f %s'%(\"\\nPower delivered 50 V source = 50 * 5.48= \",P,\" W\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Applying KVL to the circuit :\n",
"50 - 5*I - 1.2*I - 16 = 0\n",
"I= 5.48 Amp\n",
"\n",
"Power delivered 50 V source = 50 * 5.48= 274.19 W\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.13-pg1.18"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##page no-1.18\n",
"##example1.13\n",
"print(\"By Current Division formula ;\");\n",
"I4=4.*(2./(2.+4.));\n",
"print'%s %.2f %s'%(\"I4 = 4 * (2/(2+4)) = \",I4,\" Amp\");"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"By Current Division formula ;\n",
"I4 = 4 * (2/(2+4)) = 1.33 Amp\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.14-pg1.19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##page no-1.19\n",
"##example1.14\n",
"print(\"Applying KVL to the mesh\");\n",
"print(\"15 - 50*I - 50*I - 5*I\");\n",
"I=15./105.;\n",
"print'%s %.2f %s'%(\"I=15/105 = \",I,\" Amp\");\n",
"V=15-(50*0.143);\n",
"print'%s %.2f %s'%(\"\\nVoltage at node 2 = 15 - 50*I = \",V,\" Volt\");"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Applying KVL to the mesh\n",
"15 - 50*I - 50*I - 5*I\n",
"I=15/105 = 0.14 Amp\n",
"\n",
"Voltage at node 2 = 15 - 50*I = 7.85 Volt\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.15-pg1.20"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Basic Circuit Concepts\n",
"##pg no.-1.20\n",
"##example 1.15\n",
"r1=3.;\n",
"r2=2.33;\n",
"r3=6.;\n",
"v1=18.;\n",
"v2=5.985;\n",
"print(\"\\nApplying KCL at the node, \\n(Va-18)/3+(Va-5.985)/2.33+Va/6 = 0\");\n",
"Va=((v1*r2*r3)+(v2*r1*r3))/((r2*r3)+(r1*r3)+(r1*r2));\n",
"print'%s %.2f %s'%(\"\\nSolving the equation,we get, \\nVa = \",Va,\" V\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"Applying KCL at the node, \n",
"(Va-18)/3+(Va-5.985)/2.33+Va/6 = 0\n",
"\n",
"Solving the equation,we get, \n",
"Va = 9.22 V\n"
]
}
],
"prompt_number": 12
}
],
"metadata": {}
}
]
}