{ "metadata": { "name": "", "signature": "sha256:573d3c26bc06249cf2f9f4748e6d10f443e0ca37c6357e39541ace7a1534cbe6" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 15 : Mutual Inductance and Transformers" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 15.4 Page No : 306" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "#Example 15.4\")\n", "\n", "# Given\n", "#L1 = 0.1H L2 = 0.2H\")\n", "#i1 = 4A i2 = 10A\")\n", "L1 = 0.1;L2 = 0.2\n", "i1 = 4;i2 = 10;\n", "#The energy stored in coupled coils is\n", "#W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2\")\n", "\n", "#a)\")\n", "M = 0.1;\n", "W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2;\n", "print \"Total Energy in the coils = %3.2fJ\"%(W);\n", "\n", "#b)\")\n", "M = math.sqrt(2)/10;\n", "W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2;\n", "print \"Total Energy in the coils = %3.2fJ\"%(W);\n", "\n", "#c)\")\n", "M = -0.1;\n", "W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2;\n", "print \"Total Energy in the coils = %3.2fJ\"%(W);\n", "\n", "#a)\")\n", "M = -math.sqrt(2)/10;\n", "W = (L1*i1**2)/2+(L2*i2**2)/2+M*i1*i2;\n", "print \"Total Energy in the coils = %3.2fJ\"%(W);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Total Energy in the coils = 14.80J\n", "Total Energy in the coils = 16.46J\n", "Total Energy in the coils = 6.80J\n", "Total Energy in the coils = 5.14J\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 15.7 Page No : 311" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from scipy.linalg import polar\n", "\n", "#Example 15.7\")\n", "\n", "# Given\n", "#N1 = 20 N2 = N3 = 10\")\n", "#I2 = 10(-53.13 deg) I3 = 10(-45 deg)\")\n", "N1 = 20;\n", "N2 = 10;\n", "N3 = 10;\n", "I2mag = 10;\n", "I2ph = -53.13;\n", "I3mag = 10;\n", "I3ph = -45;\n", "#From figure 15.14\n", "#N1*I1-N2*I2-N3*I3 = 0\")\n", "#Solving for I1\n", "Xmag = N2*I2mag \n", "Xph = I2ph\n", "x = Xmag*math.cos((Xph*math.pi)/180);\n", "y = Xmag*math.sin((Xph*math.pi)/180);\n", "z = complex(x,y)\n", "\n", "Ymag = N3*I3mag \n", "Yph = I3ph\n", "x1 = Ymag*math.cos((Yph*math.pi)/180);\n", "y1 = Ymag*math.sin((Yph*math.pi)/180);\n", "z1 = complex(x1,y1)\n", "\n", "I1 = (z+z1)/N1\n", "R,Theta = polar([[I1]]);\n", "R = R[0][0].real\n", "Theta = Theta[0][0].real\n", "\n", "print \"I1 = %3.2f%3.2f deg) A\"%(R,(Theta*180)/math.pi);\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "I1 = 0.66571.52 deg) A\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 15.8 Page No : 316" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from scipy.linalg import polar\n", "#Example 15.8\")\n", "\n", "# Given\n", "#L1 = 0.2H L2 = 0.1H\")\n", "#M = 0.1H R = 10ohm\")\n", "#v1 = 142.3*math.sin(100*t)\")\n", "L1 = 0.2;L2 = 0.1\n", "M = 0.1;R = 10;\n", "v1mag = 142.3;\n", "w = 100;\n", "#Let Input impedance be Z1 and can be calculated as\n", "#From the equations in 15.10\n", "#Z1 = 1j*w*L1+((M*w)**2)/(Z2+1j*w*L2)\")\n", "Z1 = 1j*w*L1+((M*w)**2)/(R+1j*w*L2)\n", "R,Theta = polar([[Z1]])\n", "R = R[0][0].real\n", "Theta = Theta[0][0].real\n", "\n", "#If I1 is the input current\n", "I1mag = v1mag/R\n", "I1ph = -(Theta*180)/math.pi\n", "#In time domain form\n", "print \"i1 = %3.1f*math.sin%d*t%3.1f deg) A)\"%(I1mag,w,I1ph);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i1 = 450.0*math.sin100*t-905.9 deg) A)\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 15.9 Page No : 318" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from sympy import Symbol\n", "\n", "s = Symbol('s')\n", "# Given\n", "#L1 = 0.2H L2 = 0.1H\")\n", "#M = 0.1H R = 10ohm\")\n", "#v1 = u(t) a unit step function\")\n", "L1 = 0.2;\n", "L2 = 0.1\n", "M = 0.1;\n", "R = 10;\n", "v1 = 1;\n", "w = 100;\n", "#Let Input impedance be Z1 and can be calculated as\n", "#From the equations in 15.10\n", "#Z1(s) = L1*s-((M*s)**2)/(R+L2*s)\")\n", "Z1 = L1*s-(((M*s)**2)/(R+L2*s))\n", "#Proper rearranging of co-efficients\n", "Num = Z1/0.01\n", "Den = Z1*100\n", "\n", "print \"Z1(s)\",Num/Den\n", "Y1 = 1./Z1\n", "print \"Y1(s)\",Den/Num\n", "\n", "#As the input is unit step function the value is 1V for t>0\n", "#In exponential form the value is represented as exp(s*t) with s = 0 as the pole of Y1(s)\n", "\n", "#Therefore forced response\n", "k = 1/L1;\n", "print \"Forced response i1,f = %d*t) A)\"%(k);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Z1(s) 1\n", "Y1(s) 1\n", "Forced response i1,f = 5*t) A)\n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }