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+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:cd92732b89aba9dec0c4f42de79fa75bac0a4c551a65b1e08ee3d7196bca88ee"

+ },

+ "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-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": {}

+  }

+ ]

+}
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