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diff --git a/Electrical_Network_by_R._Singh/Chapter1.ipynb b/Electrical_Network_by_R._Singh/Chapter1.ipynb new file mode 100644 index 00000000..1f66c550 --- /dev/null +++ b/Electrical_Network_by_R._Singh/Chapter1.ipynb @@ -0,0 +1,634 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d23c488e06827cad82da0b4ae722d620d666ee7bafdcca01ad1272abc13ad886"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "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": {}
+ }
+ ]
+}
\ No newline at end of file |