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| author | hardythe1 | 2014-07-25 13:33:31 +0530 |
|---|---|---|
| committer | hardythe1 | 2014-07-25 13:33:31 +0530 |
| commit | 1c1ea29e3e213559fef5f928df109b7d17c21f24 (patch) | |
| tree | a1d38eb039e19d9e84b6e769cec04d59e7179e66 /Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb | |
| parent | efb9ead5d9758d5d0bed7a22069320b14f972e40 (diff) | |
| download | Python-Textbook-Companions-1c1ea29e3e213559fef5f928df109b7d17c21f24.tar.gz Python-Textbook-Companions-1c1ea29e3e213559fef5f928df109b7d17c21f24.tar.bz2 Python-Textbook-Companions-1c1ea29e3e213559fef5f928df109b7d17c21f24.zip | |
removing unwanted and adding book
Diffstat (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb')
| -rwxr-xr-x | Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb | 1328 |
1 files changed, 0 insertions, 1328 deletions
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb deleted file mode 100755 index 3f3b710b..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_40-checkpoint.ipynb +++ /dev/null @@ -1,1328 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 40: Field theory</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 725</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance per metre length of the system\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": [
- "<h3>Example 2, page no. 725</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of a 100 m length of the cable.\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": [
- "<h3>Example 3, page no. 726</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance per metre length of the cable\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": [
- "<h3>Example 4, page no. 727</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the internal diameter of the sheath.\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": [
- "<h3>Example 5, page no. 728</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the capacitance per kilometre length of the cable, \n",
- "#(b) the dielectric stress at a radius of 30 mm, and\n",
- "#(c) the maximum and minimum values of dielectric stress.\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": [
- "<h3>Example 6, page no. 729</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the core and inner sheath radii for the most economical cable,\n",
- "#(b) the capacitance per metre length, and (c) the charging current per kilometre run.\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": [
- "<h3>Example 7, page no. 730</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for a 1 km length of the cable (a) the capacitance, (b) the charging current and (c) the power loss.\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": [
- "<h3>Example 8, page no. 732</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the capacitance of the cable per metre length by the method of curvilinear squares\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": [
- "<h3>Example 9, page no. 736</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the capacitance of the line if the total length is 200 m.\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": [
- "<h3>Example 10, page no. 736</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine, for a 1 km length of line, (a) the capacitance of the conductors, \n",
- "#(b) the value of charge carried by each conductor, and (c) the charging current\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": [
- "<h3>Example 11, page no. 737</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the maximum value required for the capacitance per metre length,\n",
- "#and (b) the maximum diameter of each conductor if their distance between centres is 1.25 m.\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": [
- "<h3>Example 12, page no. 739</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and \n",
- "#the average power developed if this energy is dissipated in 10 \u03bcs\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": [
- "<h3>Example 13, page no. 739</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the voltage across the plates and (b) the capacitance of the capacitor.\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": [
- "<h3>Example 14, page no. 740</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the required thickness of the ceramic dielectric, \n",
- "#(b) the area of plate required if the relative permittivity of the ceramic is 10, and \n",
- "#(c) the maximum energy stored by the capacitor.\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": [
- "<h3>Example 15, page no. 740</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the energy stored per cubic metre of the dielectric.\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": [
- "<h3>Example 16, page no. 744</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the inductance of the cable per metre length.\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": [
- "<h3>Example 17, page no. 744</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the diameter of the sheath.\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": [
- "<h3>Example 18, page no. 745</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the cable (a) the inductance, assuming nonmagnetic materials, and \n",
- "#(b) the capacitance, assuming a dielectric of relative permittivity 3.\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": [
- "<h3>Example 19, page no. 748</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the inductance of the line per metre length ignoring internal linkages\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": [
- "<h3>Example 20, page no. 748</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the loop inductance, and \n",
- "#(b) the capacitance of a 1 km length of single-phase twin line having conductors of diameter 10 mm and spaced 800 mm apart in air.\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": [
- "<h3>Example 21, page no. 749</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the distance between their centres.\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": [
- "<h3>Example 22, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#What value of current would double the energy stored?\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": [
- "<h3>Example 23, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the total energy stored in the magnetic field of the airgap.\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": [
- "<h3>Example 24, page no. 752</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the strength of a uniform electric fi\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 |