From efb9ead5d9758d5d0bed7a22069320b14f972e40 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Fri, 25 Jul 2014 12:37:07 +0530 Subject: adding books --- .../chapter_40-checkpoint_3.ipynb | 1319 ++++++++++++++++++++ 1 file changed, 1319 insertions(+) create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb') diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb new file mode 100755 index 00000000..181b437b --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint_3.ipynb @@ -0,0 +1,1319 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b1ccfc3b90d3e80d897264262435b8c5eeae5b8f515ec94674405489033f24fc" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Chapter 40: Field theory

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

Example 1, page no. 725

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 2.8;\n", + "l = 1;# in m\n", + "\n", + "#calculation: \n", + " #From Figure 40.9\n", + "m = 16;# number of parallel squares measured along each equipotential\n", + "n = 6;# the number of series squares measured along each line of force\n", + "C = e0*er*l*m/n\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " capacitance is 66.08 pFarad." + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 2, page no. 725

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 3.4;\n", + "l = 100;# in m\n", + "\n", + " #calculation: \n", + " #From Figure 40.10\n", + "m = 13;# number of parallel squares measured along each equipotential\n", + "n = 4;# the number of series squares measured along each line of force\n", + "C = e0*er*l*m/n\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n capacitance is \",round(C*1E9,2),\"nFarad.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " capacitance is 9.78 nFarad." + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 3, page no. 726

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 2.7;\n", + "ri = 0.0005;# in m\n", + "ro = 0.006;# in m\n", + "\n", + " #calculation: \n", + " #capacitance C\n", + "C = 2*math.pi*e0*er/(math.log(ro/ri))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n capacitance is \",round(C*1E12,2),\"pFarad.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " capacitance is 60.42 pFarad." + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 4, page no. 727

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "C = 80E-12;# in Farads\n", + "e0 = 8.85E-12; \n", + "er = 3.5;\n", + "d0 = 0.008;# in m\n", + "\n", + " #calculation: \n", + " #internal diameter\n", + "di = d0*(math.e**(2*math.pi*e0*er/C))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n internal diameter is \",round(di,2),\" m.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " internal diameter is 0.09 m." + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 5, page no. 728

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 3.5;\n", + "di = 0.08;# in m\n", + "d0 = 0.032;# in m\n", + "r = 0.03;# in m\n", + "V = 40000;# in Volts\n", + "\n", + "#calculation: \n", + " #capacitance C\n", + "C = 2*math.pi*e0*er/(math.log(di/d0))\n", + " #dielectric stress at radius r,\n", + "E = V/(r*math.log(di/d0))\n", + " #maximum dielectric stress,\n", + "Emax = V/((d0/2)*(math.log((di/d0))))\n", + " #minimum dielectric stress,\n", + "Emin = V/((di/2)*(math.log((di/d0))))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n capacitance is \",round(C*1E12,2),\"pF/km\"\n", + "print \"\\n dielectric stress at radius r is \",round(E,2),\"V/m\"\n", + "print \"\\n maximum dielectric stress, is \",round(Emax,2),\"V/m minimum dielectric stress \",round( Emin,2),\"V/m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " capacitance is 212.4 pF/km\n", + "\n", + " dielectric stress at radius r is 1455142.22 V/m\n", + "\n", + " maximum dielectric stress, is 2728391.67 V/m minimum dielectric stress 1091356.67 V/m" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 6, page no. 729

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 3.5;\n", + "V = 60000;# in Volts\n", + "f = 50;# in Hz\n", + "Em = 10E6;# in V/m\n", + "\n", + "\n", + "#calculation: \n", + " #core radius, a\n", + "a = V/Em\n", + " #internal sheath radius,\n", + "b = a*math.e**1\n", + " #capacitance\n", + "C = 2*math.pi*e0*er/(math.log(b/a))\n", + " #Charging current\n", + "I = V*2*math.pi*f*C\n", + " #charging current per kilometre\n", + "Ipkm = I*1000\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n core radius is \",round(a*1000,2),\"mm and internal sheath radius \",round(b*1000,1),\"mm\"\n", + "print \"\\n capacitance is \",round(C*1E12,0),\"pF/m\"\n", + "print \"\\n the charging current per kilometre \",round(Ipkm,2),\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " core radius is 6.0 mm and internal sheath radius 16.3 mm\n", + "\n", + " capacitance is 195.0 pF/m\n", + "\n", + " the charging current per kilometre 3.67 A" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 7, page no. 730

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 2.5;\n", + "di = 0.08;# in m\n", + "d0 = 0.025;# in m\n", + "r = 1000;# in m\n", + "V = 132000;# in Volts\n", + "f = 50;# in Hz\n", + "de = 3.5E-3;# rad.\n", + "\n", + " #calculation:\n", + " #core radius, a\n", + "a = d0/2\n", + " #internal sheath radius,\n", + "b = di/2\n", + " #capacitance\n", + "C = 2*math.pi*e0*er*1E3/(math.log(b/a))\n", + " #Charging current\n", + "I = V*2*math.pi*f*C\n", + " #power loss\n", + "P = (2*math.pi*f*C*math.tan(de))*V**2\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)capacitance for a 1 km length is \",round(C*1E6,2),\"uF\"\n", + "print \"\\n (b)the charging current \",round(I,2),\"A/km\"\n", + "print \"\\n (c)power loss \",round(P,2),\" W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)capacitance for a 1 km length is 0.12 uF\n", + "\n", + " (b)the charging current 4.96 A/km\n", + "\n", + " (c)power loss 2289.78 W" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 8, page no. 732

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 3.2;\n", + "di = 0.06;# in m\n", + "d0 = 0.020;# in m\n", + "\n", + " #calculation:\n", + " #core radius, a\n", + "a = d0/2\n", + " #internal sheath radius,\n", + "b = di/2\n", + " #capacitance\n", + "C = 2*math.pi*e0*er/(math.log(b/a))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n capacitance per m of length is \",round(C*1E9,2),\"nF\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " capacitance per m of length is 0.16 nF" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 9, page no. 736

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 1;\n", + "D = 0.05;# in m\n", + "d = 0.005;# in m\n", + "l = 200;# in m\n", + "\n", + " #calculation:\n", + " #capacitance\n", + "C = math.pi*e0*er/(math.log(D/(d/2)))\n", + " #capacitance of a 200 m length\n", + "C200 = C*l\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n capacitance of a 200 m length is \",round(C200*1E6,5),\"uF\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " capacitance of a 200 m length is 0.00186 uF" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 10, page no. 736

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 1;\n", + "D = 1.2;# in m\n", + "r = 0.004;# in m\n", + "f = 50;# in Hz\n", + "V = 15000;# in Volts\n", + "l = 1000;# in m\n", + "\n", + " #calculation:\n", + " #capacitance\n", + "C = math.pi*e0*er/(math.log(D/r))\n", + " #capacitance of a 1 km length\n", + "Cpkm = C*l\n", + " #Charge Q\n", + "Q = Cpkm*V\n", + " #Charging current\n", + "I = V*2*math.pi*f*Cpkm\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n capacitance per 1km length is \",round(Cpkm*1E9,2),\"nF\"\n", + "print \"\\n Charge Q is \",round(Q*1E6,2),\"uC\"\n", + "print \"\\n Charging current is \",round(I,2),\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " capacitance per 1km length is 4.87 nF\n", + "\n", + " Charge Q is 73.12 uC\n", + "\n", + " Charging current is 0.02 A" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 11, page no. 737

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 1;\n", + "I = 0.015;# in Amperes\n", + "d = 1.25;# in m\n", + "r = 800;# in m\n", + "f = 50;# in Hz\n", + "V = 10000;# in Volts\n", + "\n", + " #calculation:\n", + " #capacitance\n", + "C = I/(2*math.pi*f*V)\n", + " #required maximum value of capacitance\n", + "Cmax = C/r\n", + " #maximum diameter of each conductor\n", + "D = 2*d/(math.e**(math.pi*e0*er/Cmax))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n required maximum value of capacitance is \",round(Cmax*1E12,2),\"pF/m\"\n", + "print \"\\nthe maximum diameter of each conductor is \",round(D,2),\" m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " required maximum value of capacitance is 5.97 pF/m\n", + "\n", + "the maximum diameter of each conductor is 0.02 m" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 12, page no. 739

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 1;\n", + "C = 10E-9;# in Farad\n", + "V = 1000;# in Volts\n", + "t = 10E-6;# in sec\n", + "\n", + " #calculation:\n", + " #energy stored,Wf\n", + "Wf = C*V*V/2\n", + " #average power developed\n", + "Pav = Wf/t\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n the energy stored is \",Wf,\"J\"\n", + "print \"\\nthe average power developed is \",Pav,\" W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " the energy stored is 0.005 J\n", + "\n", + "the average power developed is 500.0 W" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 13, page no. 739

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 1;\n", + "Q = 5E-3;# in Coulomb\n", + "W = 0.625;# in Joules\n", + "\n", + " #calculation:\n", + " #voltage across the plates\n", + "V = 2*W/Q\n", + " #Capacitance C\n", + "C = Q/V\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n voltage across the plates is \",V,\" V\"\n", + "print \"\\n Capacitance C is \",C*1E6,\"uF\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " voltage across the plates is 250.0 V\n", + "\n", + " Capacitance C is 20.0 uF" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 14, page no. 740

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 10;\n", + "C = 0.01E-6;# in Farad\n", + "E = 10E6;# in V/m\n", + "V = 2500;# in Volts\n", + "\n", + " #calculation:\n", + " #thickness of ceramic dielectric,\n", + "d = V/E\n", + " #cross-sectional area of plate\n", + "A = C*d/(e0*er)\n", + " #Maximum energy stored,\n", + "W = C*V*V/2\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n thickness of ceramic dielectric is \",d*1000,\"mm\"\n", + "print \"\\n cross-sectional area of plate, is \",round(A,2),\"m2\"\n", + "print \"\\n Maximum energy stored is \",round(W,3),\" J\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " thickness of ceramic dielectric is 0.25 mm\n", + "\n", + " cross-sectional area of plate, is 0.03 m2\n", + "\n", + " Maximum energy stored is 0.031 J" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 15, page no. 740

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "e0 = 8.85E-12; \n", + "er = 2.3;\n", + "A = 0.02;# in m2\n", + "C = 400E-12;# in Farad\n", + "V = 100;# in Volts\n", + "\n", + " #calculation:\n", + " #energy stored per unit volume of dielectric,\n", + "W = ((C*V)**2)/(2*e0*er*A**2)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n energy stored per unit volume of dielectric is \",round(W,2),\" J/m3\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " energy stored per unit volume of dielectric is 0.1 J/m3" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 16, page no. 744

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "u0 = 4*math.pi*1E-7; \n", + "ur = 1;\n", + "a = 0.001;# in m\n", + "b = 0.004;# in m\n", + "\n", + " #calculation:\n", + " #inductance L\n", + "L = (u0*ur/(2*math.pi))*(0.25 + math.log(b/a))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n inductance L is \",round(L*1E6,2),\"uH/m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " inductance L is 0.33 uH/m" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 17, page no. 744

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "u0 = 4*math.pi*1E-7; \n", + "ur = 1;\n", + "da = 0.010;# in m\n", + "L = 4E-7;# in H/m\n", + "\n", + " #calculation:\n", + " #diameter of the sheath\n", + "db = da*(math.e**(L/(u0*ur/(2*math.pi))))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n diameter of the sheath is \",round(db,2),\" m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " diameter of the sheath is 0.07 m" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 18, page no. 745

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "u0 = 4*math.pi*1E-7; \n", + "ur = 1;\n", + "e0 = 8.85E-12;\n", + "er = 3;\n", + "da = 0.010;# in m\n", + "db = 0.025;# in m\n", + "l = 7500;# in m\n", + "\n", + "#calculation:\n", + " #inductance per metre length\n", + "L = (u0*ur/(2*math.pi))*(0.25 + math.log(db/da))\n", + " #Since the cable is 7500 m long,\n", + "L7500 = L*7500\n", + " #capacitance C\n", + "C = 2*math.pi*e0*er/(math.log(db/da))\n", + " #//Since the cable is 7500 m long,\n", + "C7500 = C*7500\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\ninductance is \",round(L7500*1000,2),\" mH\"\n", + "print \"\\ncapCItance is \",round(C7500*1E6,2),\"uF\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "inductance is 1.75 mH\n", + "\n", + "capCItance is 1.37 uF" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 19, page no. 748

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "u0 = 4*math.pi*1E-7; \n", + "ur = 1;\n", + "e0 = 8.85E-12;\n", + "er = 3;\n", + "D = 1.2;# in m\n", + "a = 0.008;# in m\n", + "\n", + " #calculation:\n", + " #inductance per metre length\n", + "L = (u0*ur/(math.pi))*(math.log(D/a))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\ninductance is \",round(L*1E6,2),\"uH/m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "inductance is 2.0 uH/m" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 20, page no. 748

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "u0 = 4*math.pi*1E-7; \n", + "ur = 1;\n", + "e0 = 8.85E-12;\n", + "er = 1;\n", + "l = 1000;# in m\n", + "D = 0.8;# in m\n", + "a = 0.01/2;# in m\n", + "\n", + " #calculation:\n", + " #inductance per metre length\n", + "L = (u0*ur/(math.pi))*(0.25 + math.log(D/a))\n", + " #Since the cable is 1000 m long,\n", + "L1k = L*l\n", + " #capacitance C\n", + "C = math.pi*e0*er/(math.log(D/a))\n", + " #//Since the cable is 1000 m long,\n", + "C1k = C*l\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\ninductance is \",round(L1k*1000,2),\" mH\"\n", + "print \"\\ncapcitance is \",round(C1k*1E9,2),\"nF\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "inductance is 2.13 mH\n", + "\n", + "capcitance is 5.48 nF\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 21, page no. 749

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "L = 2.185E-6;# in H/m\n", + "u0 = 4*math.pi*1E-7; \n", + "ur = 1;\n", + "a = 0.012/2;# in m\n", + "\n", + " #calculation:\n", + " #distance D\n", + "D = a*math.e**((L*math.pi)/(u0*ur) - 0.25)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\ndistance D is \",round(D,2),\" m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "distance D is 1.1 m" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 22, page no. 752

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "L = 0.2;# in H\n", + "I = 0.05;# in Amperes\n", + "u0 = 4*math.pi*1E-7; \n", + "ur = 1;\n", + "\n", + "#calculation:\n", + " #energy stored in inductor\n", + "W = L*I*I/2\n", + " #current I\n", + "I = (2*2*W/L)**0.5\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\nenergy stored in inductor is \",round(W*1000,2),\"mJ\"\n", + "print \"\\ncurrent I is \",round(I,2),\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "energy stored in inductor is 0.25 mJ\n", + "\n", + "current I is 0.07 A" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 23, page no. 752

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "B = 0.05;# in Tesla\n", + "A = 500E-6;# in m2\n", + "l = 0.002;# in m\n", + "u0 = 4*math.pi*1E-7; \n", + "\n", + "#calculation:\n", + " #energy stored\n", + "W = (B**2)/(2*u0)\n", + " #Volume of airgap\n", + "v = A*l\n", + " #energy stored in airgap\n", + "W = W*v\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\nenergy stored in the airgap is \",round(W*1E6,2),\"uJ\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "energy stored in the airgap is 994.72 uJ" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 24, page no. 752

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "B = 0.8;# in Tesla\n", + "A = 500E-6;# in m2\n", + "l = 0.002;# in m\n", + "u0 = 4*math.pi*1E-7; \n", + "ur = 1;\n", + "e0 = 8.85E-12;\n", + "er = 1;\n", + "\n", + "#calculation:\n", + " #energy stored in mag. field\n", + "W = (B**2)/(2*u0)\n", + " #electric field\n", + "E = (2*W/(e0*er))**0.5\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\nelectric field strength is \",round(E/1E6,2),\"MV/m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "electric field strength is 239.89 MV/m" + ] + } + ], + "prompt_number": 24 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit