From a4206084fd8c2bd696ea4ae4012aa83534979456 Mon Sep 17 00:00:00 2001 From: Jovina Dsouza Date: Tue, 22 Jul 2014 00:00:04 +0530 Subject: adding book --- .../chapter_43-checkpoint.ipynb | 1037 ++++++++++++++++++++ 1 file changed, 1037 insertions(+) create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb') diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb new file mode 100755 index 00000000..618fd99a --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_43-checkpoint.ipynb @@ -0,0 +1,1037 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Chapter 43: Magnetically coupled circuits

" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 1, page no. 842

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the self inductance of coil A, and (b) the mutual inductance.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "Na = 1200; \n", + "Nb = 1000;\n", + "Ia = 0.8;# in amperes\n", + "Phia = 100E-6;# in Wb\n", + "xb = 0.75;\n", + "\n", + " #calculation:\n", + " #self inductance of coil A\n", + "La = Na*Phia/Ia\n", + " #mutual inductance, M\n", + "Phib = xb*Phia\n", + "M = Nb*Phib/Ia\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n self inductance of coil A is \",La,\" H\"\n", + "print \"\\n mutual inductance, M is \",M,\"H\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " self inductance of coil A is 0.15 H\n", + "\n", + " mutual inductance, M is 0.09375 H" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 2, page no. 843

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the e.m.f. induced in the secondary,\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "M = 600E-3;# in Henry\n", + "Ia = 5;# in amperes\n", + "dt = 0.2;# in secs\n", + "\n", + " #calculation:\n", + " #change of current\n", + "dIa = 2*Ia \n", + "dIadt = dIa/dt\n", + " #secondary induced e.m.f., E2\n", + "E2 = -1*M*dIadt\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n secondary induced e.m.f., E2 is \",E2,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " secondary induced e.m.f., E2 is -30.0 V" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 3, page no. 844

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the magnetic coupling coefficient of the pair of coils\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "La = 250E-3;# in Henry\n", + "Lb = 400E-3;# in Henry\n", + "M = 80E-3;# in Henry\n", + "\n", + " #calculation:\n", + " #coupling coefficient,\n", + "k = M/(La*Lb)**0.5\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n coupling coefficient, is \",round(k,3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " coupling coefficient, is 0.253" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 4, page no. 845

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the mutual inductance between the coils, and (b) the coefficient of coupling.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "Lx = 80E-3;# in Henry\n", + "Ly = 60E-3;# in Henry\n", + "Nx = 200;# turns\n", + "Ny = 100;# turns\n", + "Ix = 4;# in Amperes\n", + "Phiy = 0.005;# in Wb\n", + "\n", + "#calculation:\n", + " #mutual inductance, M\n", + "M = Ny*Phiy/(2*Ix)\n", + " #coupling coefficient,\n", + "k = M/(Lx*Ly)**0.5\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n", + "print \"\\n coupling coefficient, is \",round(k,3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " mutual inductance, M is 62.5 mH\n", + "\n", + " coupling coefficient, is 0.902" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 5, page no. 846

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the mutual inductance between the two coils, and (b) the coefficient of coupling.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "La = 40E-3;# in Henry\n", + "Lb = 10E-3;# in Henry\n", + "L = 60E-3;# in Henry\n", + "\n", + " #calculation:\n", + " #mutual inductance, M\n", + "M = (L - La - Lb)/2\n", + " #coupling coefficient,\n", + "k = M/(La*Lb)**0.5\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n mutual inductance, M is \",M*1E3,\"mH\"\n", + "print \"\\n coupling coefficient, is \",k" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " mutual inductance, M is 5.0 mH\n", + "\n", + " coupling coefficient, is 0.25" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 6, page no. 847

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the mutual inductance between the coils and (b) the coefficient of coupling.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "V = 240;# in Volts\n", + "Ra = 5;# in Ohm\n", + "La = 1;# in Henry\n", + "Rb = 10;# in Ohm\n", + "Lb = 5;# in Henry\n", + "I = 8;# in amperes\n", + "dIdt = 15;# in A/sec\n", + "\n", + " #calculation:\n", + " #Kirchhoff\u2019s voltage law\n", + "L = (V - I*(Ra + Rb))/dIdt\n", + " #mutual inductance, M\n", + "M = (L - La - Lb)/2\n", + " #coupling coefficient,\n", + "k = M/(La*Lb)**0.5\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n mutual inductance, M is \",M,\"H\"\n", + "print \"\\n coupling coefficient, is \",round(k,3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " mutual inductance, M is 1.0 H\n", + "\n", + " coupling coefficient, is 0.447" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 7, page no. 848

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the self inductance of each coil, and (b) the mutual inductance.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "k = 0.7;# coefficient of coupling\n", + "L1 = 15E-3;# in Henry\n", + "L2 = 10E-3;# in Henry\n", + "\n", + "#calculation:\n", + " #L1 = La + Lb + 2*k*(La*Lb)**0.5\n", + " #L2 = La + Lb - 2*k*(La*Lb)**0.5\n", + " #self inductance of coils\n", + "a = ((L1 - (L1 + L2)/2)/(2*k))**2\n", + "La1 =((L1 + L2)/2 + (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n", + "La2 =((L1 + L2)/2 - (((L1 + L2)/2)**2 - 4*a)**0.5)/2\n", + "Lb1 = (L1 + L2)/2 - La1\n", + "Lb2 = (L1 + L2)/2 - La2\n", + " #mutual inductance, M\n", + "M = (L1 - L2)/4\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\nself inductance of coils are \",round(La1*1E3,2),\"mH and \",round( Lb1*1E3,2),\"mH\"\n", + "print \"\\n mutual inductance, M is \",round(M*1E3,2),\"mH\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "self inductance of coils are 12.24 mH and 0.26 mH\n", + "\n", + " mutual inductance, M is 1.25 mH" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 8, page no. 850

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the p.d. E2 which appears across the open-circuited secondary winding,\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "E1 = 8;# in Volts\n", + "thetae1 = 0;# in degrees\n", + "w = 2500;# in rad/sec\n", + "R = 15;# in ohm\n", + "L = 5E-3;# in Henry\n", + "M = 0.1E-3;# in Henry\n", + "\n", + " #calculation:\n", + " #voltage\n", + "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n", + " #Impedance of primary\n", + "Z1 = R + 1j*w*L\n", + " #Primary current I1\n", + "I1 = E1/Z1\n", + " #E2\n", + "E2 = 1j*w*M*I1\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\nE2 is \",round(abs(E2),3),\"/_\",round(cmath.phase(complex(E2.real,E2.imag))*180/math.pi,2),\"deg V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "E2 is 0.102 /_ 50.19 deg V" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 9, page no. 850

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the magnitude of the open circuit e.m.f. induced in y.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "Lx = 20E-3;# in Henry\n", + "Ly = 80E-3;# in Henry\n", + "k = 0.75;# coupling coeff.\n", + "Ex = 5;# in Volts\n", + "\n", + " #calculation:\n", + " #mutual inductance\n", + "M = k*(Lx*Ly)**0.5\n", + " #magnitude of the open circuit e.m.f. induced\n", + "Ey = M*Ex/Lx\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n mutual inductance is \",M,\" H\"\n", + "print \"\\n magnitude of the open circuit e.m.f. induced is \",Ey,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " mutual inductance is 0.03 H\n", + "\n", + " magnitude of the open circuit e.m.f. induced is 7.5 V" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 10, page no. 852

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the value of the secondary current I2\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "E1 = 2;# in Volts\n", + "thetae1 = 0;# in degrees\n", + "f = 1000/math.pi;# in Hz\n", + "R1 = 4;# in ohm\n", + "R2 = 16;# in ohm\n", + "R3 = 16;# in ohm\n", + "R4 = 50;# in ohm\n", + "L = 10E-3;# in Henry\n", + "M = 2E-3;# in Henry\n", + "\n", + "#calculation:\n", + "w = 2*math.pi*f\n", + " #R1e is the real part of Z1e\n", + "R1e = R1 + R2 + ((R3 + R4)*(M**2)*(w**2))/((R3 + R4)**2 + (w*L)**2)\n", + " #X1e is the imaginary part of Z1e\n", + "X1e = w*L - (L*(M**2)*(w**3))/((R3 + R4)**2 + (w*L)**2)\n", + "Z1e = R1e + 1j*X1e\n", + "Z2e = R3 + R4 + 1j*w*L\n", + " #primary current, I1\n", + "I1 = E1/Z1e\n", + " #E2\n", + "E2 = 1j*w*M*I1\n", + " #secondary current I2\n", + "I2 = E2/Z2e\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"secondary current I2 is \",round(abs(I2)*1E3,3),\"/_\", round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg mA\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "secondary current I2 is 4.085 /_ 28.55 deg mA\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 11, page no. 853

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate (a) the self impedance of the primary circuit, \n", + "#(b) the self impedance of the secondary circuit, \n", + "#(c) the impedance reflected the primary circuit, \n", + "#(d) the effective primary impedance, \n", + "#(e) the primary current, and \n", + "#(f) the secondary current\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "E1 = 50;# in Volts\n", + "thetae1 = 0;# in degrees\n", + "w = 500;# in rad/sec\n", + "R1 = 300;# in ohm\n", + "L1 = 0.2;# in Henry\n", + "L2 = 0.5;# in Henry\n", + "L3 = 0.3;# in Henry\n", + "R2 = 500;# in ohm\n", + "C = 5E-6;# in farad\n", + "M = 0.2;# in Henry\n", + "\n", + "#calculation:\n", + " # Self impedance of primary circuit\n", + "Z1 = R1 + 1j*w*(L1 + L2)\n", + " #Self impedance of secondary circuit,\n", + "Z2 = R2 + 1j*(w*L3 - 1/(w*C))\n", + " #reflected impedance, Zr\n", + "Zr = (w*M)**2/Z2\n", + " #Effective primary impedance,\n", + "Z1e = Z1 + Zr\n", + " #Primary current I1 \n", + "I1 = E1/Z1e\n", + " #Secondary current I2\n", + "E2 = 1j*w*M*I1\n", + "I2 = E2/Z2\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n Self impedance of primary circuit, Z1 is \",Z1.real,\" + (\", Z1.imag,\")i ohm\"\n", + "print \"\\n Self impedance of secondary circuit, Z2 is \",Z2.real,\" + (\", Z2.imag,\")i ohm\"\n", + "print \"\\n reflected impedance, Zr is \",Zr.real,\" +(\", Zr.imag,\")i ohm\"\n", + "print \"\\n Effective primary impedance Z1(eff) is \",Z1e.real,\" +(\",Z1e.imag,\")i ohm\"\n", + "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n", + "print \"\\n secondary current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " Self impedance of primary circuit, Z1 is 300.0 + ( 350.0 )i ohm\n", + "\n", + " Self impedance of secondary circuit, Z2 is 500.0 + ( -250.0 )i ohm\n", + "\n", + " reflected impedance, Zr is 16.0 +( 8.0 )i ohm\n", + "\n", + " Effective primary impedance Z1(eff) is 316.0 +( 358.0 )i ohm\n", + "\n", + " primary current I1 is 0.1 /_ -48.57 deg A\n", + "\n", + " secondary current I2 is 0.02 /_ 68.0 deg A" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 12, page no. 855

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the resonant frequency, \n", + "#(b) the value of capacitor C2, \n", + "#(c) the effective primary impedance, \n", + "#(d) the primary current, (e) the voltage across capacitor C2 and \n", + "#(f) the coefficient of coupling.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "E1 = 20;# in Volts\n", + "thetae1 = 0;# in degrees\n", + "R1 = 15;# in ohm\n", + "C1 = 400E-12;# in farad\n", + "R2 = 30;# in ohm\n", + "L1 = 0.001;# in Henry\n", + "L2 = 0.0002;# in Henry\n", + "R3 = 50;# in ohm\n", + "M = 10E-6;# in Henry\n", + "\n", + "#calculation:\n", + " #voltage\n", + "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n", + " #the resonant frequency, fr \n", + "fr = 1/(2*math.pi*(L1*C1)**0.5)\n", + " #The secondary is also tuned to a resonant frequency\n", + " #capacitance,C2\n", + "C2 = 1/(L2*(2*math.pi*fr)**2)\n", + " #the effective primary impedance Z1eff\n", + "w = 2*math.pi*fr\n", + "Z1e = R1 + R2 + ((w*M)**2)/R3\n", + " #Primary current I1 \n", + "I1 = E1/Z1e\n", + " #Secondary current I2\n", + "E2 = 1j*w*M*I1\n", + "I2 = E2/Z1e\n", + " #voltage across capacitor C2\n", + "Vc2 = I2*(-1*1j/(w*C2))\n", + " #coefficient of coupling, k \n", + "k = M/(L1*L2)**0.5\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n the resonant frequency,fr is \",round(fr/1000,2),\"KHz\"\n", + "print \"\\n capacitance,C2 is \",round(C2*1E9,2),\"nF\"\n", + "print \"\\n Effective primary impedance Z1(eff) is \",round(abs(Z1e),2),\" ohm\"\n", + "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag)),0),\"deg A\"\n", + "print \"\\n voltage across capacitor C2 is \",round(abs(Vc2),2),\"/_\",round(abs(cmath.phase(complex(Vc2.real,Vc2.imag))),0),\"deg V\"\n", + "print \"\\n coefficient of coupling, k is \",round(k,4),\"\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " the resonant frequency,fr is 251.65 KHz\n", + "\n", + " capacitance,C2 is 2.0 nF\n", + "\n", + " Effective primary impedance Z1(eff) is 50.0 ohm\n", + "\n", + " primary current I1 is 0.4 /_ 0.0 deg A\n", + "\n", + " voltage across capacitor C2 is 40.0 /_ 0.0 deg V\n", + "\n", + " coefficient of coupling, k is 0.0224 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 13, page no. 858

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the values of currents I1 and I2\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "E1 = 250;# in Volts\n", + "thetae1 = 0;# in degrees\n", + "R1 = 50j;# in ohm\n", + "R2 = 10;# in ohm\n", + "R3 = 10;# in ohm\n", + "R4 = 50j;# in ohm\n", + "R5 = 50;# in ohm\n", + "M = 10j;# in ohm\n", + "\n", + " #calculation:\n", + " #voltage\n", + "E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n", + " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n", + " #(R1 + R2)*I1 - M*I2 = E1\n", + " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n", + " #-1*M*I1 + ( R3 + R4 + R5)*I2 = 0\n", + " #solving these two\n", + "I2 = E1/((R1 + R2)*(R3 + R4 + R5)/(-1*M) + (-1*M))\n", + "I1 = I2*(R3 + R4 + R5)/(-1*M)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n primary current I1 is \",round(I1.real,2),\" +(\",round( I1.imag,2),\")i A\"\n", + "print \"\\n secondary current I2 is \",round(I2.real,2),\" +(\",round( I2.imag,2),\")i A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " primary current I1 is 0.85 +( -4.77 )i A\n", + "\n", + " secondary current I2 is -0.54 +( 0.31 )i A" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 14, page no. 859

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine for the circuit (a) the mutual inductance M, (b) the primary current I1 and (c) the secondary terminal p.d.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "E1 = 40;# in Volts\n", + "thetae1 = 0;# in degrees\n", + "R1 = 5;# in ohm\n", + "L1 = 0.001;# in Henry\n", + "L2 = 0.006;# in Henry\n", + "R2 = 40;# in ohm\n", + "rzl = 200;# in ohm\n", + "thetazl = -60;# in degrees\n", + "k = 0.70\n", + "f = 20000;# in Hz\n", + "\n", + " #calculation:\n", + "w = 2*math.pi*f\n", + " #voltage\n", + "#E1 = E1*math.cos(thetae1*math.pi/180) + 1j*E1*math.sin(thetae1*math.pi/180)\n", + " #impedance\n", + "ZL = rzl*math.cos(thetazl*math.pi/180) + 1j*rzl*math.sin(thetazl*math.pi/180)\n", + " #mutual inductance, M\n", + "M = k*(L1*L2)**0.5\n", + " #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the primary circuit gives\n", + " #(R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n", + " #Applying Kirchhoff\u00e2\u20ac\u2122s voltage law to the secondary circuit gives\n", + " #-1j*w*M*I1 + ( R2 + ZL + 1j*w*L2)*I2 = 0\n", + " #solving these two\n", + "\n", + "a = R1 + 1j*w*L1\n", + "b = 1j*w*M\n", + "c = R2 + ZL + 1j*w*L2\n", + "I1 = E1/(1*a - (b**2)/c)\n", + "d = -1*cmath.phase(complex(I1.real,I1.imag))\n", + "e = abs(I1)\n", + "I2 = (b/c)*(e*math.cos(d) + 1j*e*math.sin(d))\n", + "pd2 = I2*ZL\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n mutual induction M is \",round(M*1E3,3),\"mH\"\n", + "print \"\\n primary current I1 is \",round(abs(I1),3),\"/_\",round(-1*cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n", + "print \"\\n secondary current I2 is \",round(abs(pd2),1),\"/_\",round(cmath.phase(complex(pd2.real,pd2.imag))*180/math.pi,2),\"deg V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " mutual induction M is 1.715 mH\n", + "\n", + " primary current I1 is 0.724 /_ 65.15 deg A\n", + "\n", + " secondary current I2 is 52.2 /_ 18.7 deg V" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 15, page no. 860

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the generator current I1 and (b) the load current I2.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "E1 = 50;# in Volts\n", + "thetae1 = 0;# in degrees\n", + "r = 5;# in ohm\n", + "R1 = 20;# in ohm\n", + "L1 = 0.2;# in Henry\n", + "L2 = 0.4;# in Henry\n", + "R2 = 25;# in ohm\n", + "RL = 20;# in ohm\n", + "M = 0.1;# in Henry\n", + "f = 75/math.pi;# in Hz\n", + "\n", + "#calculation:\n", + "w = 2*math.pi*f\n", + " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n", + " #(r + R1 + 1j*w*L1)*I1 - 1j*w*M*I2 = E1\n", + " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n", + " #-1*1j*w*M*I1 + ( R2 + RL + 1j*w*L2)*I2 = 0\n", + " #solving these two\n", + "I2 = E1/((r + R1 + 1j*w*L1)*(R2 + RL + 1j*w*L2)/(1j*w*M) + (-1*1j*w*M))\n", + "I1 = I2*(R2 + RL + 1j*w*L2)/(1j*w*M)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n", + "print \"\\n load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " primary current I1 is 1.3 /_ -45.84 deg A\n", + "\n", + " load current I2 is 0.26 /_ -8.97 deg A" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 16, page no. 862

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the source and load currents for (a) the windings as shown , and (b) with one winding reversed\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "E1 = 50;# in Volts\n", + "thetae1 = 0;# in degrees\n", + "r = 5;# in ohm\n", + "R1 = 20;# in ohm\n", + "L1 = 0.2;# in Henry\n", + "R = 8;# in ohm\n", + "L = 0.1;# in Henry\n", + "L2 = 0.4;# in Henry\n", + "R2 = 25;# in ohm\n", + "RL = 20;# in ohm\n", + "M = 0.1;# in Henry\n", + "f = 75/math.pi;# in Hz\n", + "\n", + "#calculation:\n", + "w = 2*math.pi*f\n", + " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n", + " #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1 - (1j*w*M + R + 1j*w*L)*I2 = E1\n", + " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n", + " #-1*(1j*w*M + R + 1j*w*L)*I1 + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2 = 0\n", + " #solving these two\n", + "I2 = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L) + (-1*(1j*w*M + R + 1j*w*L)))\n", + "I1 = I2*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(1j*w*M + R + 1j*w*L)\n", + " #reversing\n", + " #Applying Kirchhoff\u2019s voltage law to the primary circuit gives\n", + " #(r + R1 + 1j*w*L1 + R + 1j*w*L)*I1r - (-1*1j*w*M + R + 1j*w*L)*I2r = E1\n", + " #Applying Kirchhoff\u2019s voltage law to the secondary circuit gives\n", + " #-1*(-1*1j*w*M + R + 1j*w*L)*I1r + (R2 + RL + 1j*w*L2 + R + 1j*w*L)*I2r = 0\n", + " #solving these two\n", + "I2r = E1/((r + R1 + 1j*w*L1 + R + 1j*w*L)*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L) + (-1*(-1*1j*w*M + R + 1j*w*L)))\n", + "I1r = I2r*(R2 + RL + 1j*w*L2 + R + 1j*w*L)/(-1*1j*w*M + R + 1j*w*L)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"primary current I1 is \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real,I1.imag))*180/math.pi,2),\"deg A\"\n", + "print \"load current I2 is \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real,I2.imag))*180/math.pi,2),\"deg A\"\n", + "print \"reversed primary current I1r is \",round(abs(I1r),2),\"/_\",round(cmath.phase(complex(I1r.real,I1r.imag))*180/math.pi,2),\"deg A\"\n", + "print \"reversed load current I2r is \",round(abs(I2r),2),\"/_\",round(cmath.phase(complex(I2r.real,I2r.imag))*180/math.pi,2),\"deg A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "primary current I1 is 1.03 /_ -45.47 deg A\n", + "load current I2 is 0.35 /_ -25.16 deg A\n", + "reversed primary current I1r is 0.89 /_ -54.42 deg A\n", + "reversed load current I2r is 0.08 /_ -109.17 deg A\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit