From afcd9e5397e3e1bde0392811d0482d76aac391dc Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Wed, 17 Jun 2015 11:14:34 +0530 Subject: add books --- sample_notebooks/vinodkumar/Chapter1.ipynb | 339 +++++++++++++++++++++++++++++ 1 file changed, 339 insertions(+) create mode 100755 sample_notebooks/vinodkumar/Chapter1.ipynb (limited to 'sample_notebooks/vinodkumar') diff --git a/sample_notebooks/vinodkumar/Chapter1.ipynb b/sample_notebooks/vinodkumar/Chapter1.ipynb new file mode 100755 index 00000000..0225739c --- /dev/null +++ b/sample_notebooks/vinodkumar/Chapter1.ipynb @@ -0,0 +1,339 @@ +{ + "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": {} + } + ] +} \ No newline at end of file -- cgit