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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.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.ipynb')
| -rw-r--r-- | Electrical_Circuit_Theory_And_Technology/chapter_40.ipynb | 1319 |
1 files changed, 1319 insertions, 0 deletions
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_40.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_40.ipynb new file mode 100644 index 00000000..181b437b --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_40.ipynb @@ -0,0 +1,1319 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b1ccfc3b90d3e80d897264262435b8c5eeae5b8f515ec94674405489033f24fc"
+ },
+ "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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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": [
+ "\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 |