From 6846da7f30aadc7b3812538d1b1c9ef2e465a922 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Fri, 25 Jul 2014 12:35:04 +0530 Subject: removing unwanted: --- .../chapter_06-checkpoint_1.ipynb | 873 --------------------- 1 file changed, 873 deletions(-) delete mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb') diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb deleted file mode 100755 index 772a8bc5..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_06-checkpoint_1.ipynb +++ /dev/null @@ -1,873 +0,0 @@ -{ - "metadata": { - "name": "" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Chapter 6: Capacitors and capacitance

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

Example 1, page no. 58

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#(a) Determine the p.d. across a 4 \u03bcF capacitor when charged with 5 mC.\n", - "# (b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "C1 = 4E-6;# in Farad\n", - "C2 = 50E-12;# in Farad\n", - "Q1 = 5E-3;# in Coulomb\n", - "V2 = 2000;# in volts\n", - "\n", - "#calculation:\n", - "V1 = Q1/C1\n", - "Q2 = C2*V2\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)P.d \",V1,\" Volts(V)\\n\"\n", - "print \"\\n (b)Charge(Q) \",(Q2/1E-6),\" micro-Coulomb\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)P.d 1250.0 Volts(V)\n", - "\n", - "\n", - " (b)Charge(Q) 0.1 micro-Coulomb" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 2, page no. 58

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Determine the pd between the plates.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "I = 4;# in amperes\n", - "C = 20E-6;# in Farad\n", - "t = 3E-3;# in sec\n", - "\n", - "#calculation:\n", - "Q = I*t\n", - "V = Q/C\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)P.d \",V,\" Volts(V)\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)P.d 600.0 Volts(V)" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 3, page no. 59

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Calculate how long the capacitor can provide an average discharge current of 2 mA.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "I = 2E-3;# in amperes\n", - "C = 5E-6;# in Farad\n", - "V = 800;# in volts\n", - "\n", - "#calculation:\n", - "Q = C*V\n", - "t = Q/I\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n capacitor can provide an average discharge current for \",t,\" Sec\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " capacitor can provide an average discharge current for 2.0 Sec" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 4, page no. 60

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Calculate the electric flux density.\n", - "#determine the electric field strength.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "Q = 0.2E-6;# in Coulomb\n", - "A = 800E-4;# in m2\n", - "d = 0.005;# in m\n", - "V = 250;# in Volts\n", - "\n", - "#calculation:\n", - "D = Q/A\n", - "E = V/d\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)Electric flux density D \",(D/1E-6),\" uC/m2\\n\"\n", - "print \"\\n (b)Electric field strength E \",(E/1000),\" kV/m\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)Electric flux density D 2.5 uC/m2\n", - "\n", - "\n", - " (b)Electric field strength E 50.0 kV/m" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 5, page no. 60

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Find the voltage gradient between the plates.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "D = 2E-6;# in micro-C/m2\n", - "e0 = 8.85E-12;# in F/m\n", - "er = 5;\n", - "\n", - "#calculation:\n", - "E = D/(e0*er)\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n Electric field strength E \",round((E/1000),2),\" kV/m\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " Electric field strength E 45.2 kV/m" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 6, page no. 60

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#What is the electric field strength?\n", - "#Find also the flux density when the dielectric between the plates is\n", - "#(a) air, and (b) polythene of relative permittivity 2.3\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "d = 0.8E-3;# in m\n", - "e0 = 8.85E-12;# in F/m\n", - "era = 1;# for air\n", - "erp = 2.3;# for polythene\n", - "V =200;# in Volts\n", - "\n", - "#calculation:\n", - "E = V/d\n", - "#for air\n", - "Da = E*e0*era\n", - "#for polythene\n", - "Dp = E*e0*erp\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)Electric flux density for air \",round((Da/1E-6),2),\" micro-C/m2\\n\"\n", - "print \"\\n (b)Electric flux density for polythene \",round((Dp/1E-6),2),\" micro-C/m2\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)Electric flux density for air 2.21 micro-C/m2\n", - "\n", - "\n", - " (b)Electric flux density for polythene 5.09 micro-C/m2" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 7, page no. 62

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#(a)Calculate the capacitance of the capacitor in picofarads. \n", - "#(b)what will be the pd between the plates?\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "A = 4E-4;# in m2\n", - "d = 0.1E-3;# in m\n", - "e0 = 8.85E-12;# in F/m\n", - "er = 100;\n", - "Q = 1.2E-6;# in coulomb\n", - "\n", - "#calculation:\n", - "C = e0*er*A/d\n", - "V = Q/C\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)Capacitance \",(C/1E-12),\" pF\\n\"\n", - "print \"\\n (b)P.d.= \",round(V,2),\" Volts(V)\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)Capacitance 3540.0 pF\n", - "\n", - "\n", - " (b)P.d.= 338.98 Volts(V)" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 8, page no. 62

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#determine the effective thickness of the paper if its relative permittivity is 2.5\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "A = 800E-4;# in m2\n", - "C = 4425E-12;# in Farads\n", - "e0 = 8.85E-12;# in F/m\n", - "er = 2.5;\n", - "\n", - "#calculation:\n", - "d = e0*er*A/C\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n Thickness \",(d/1E-3),\" mm\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " Thickness 0.4 mm" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 9, page no. 62

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#calculate the capacitance of the capacitor.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "n = 19;# no. of plates\n", - "L = 75E-3;# in m\n", - "B = 75E-3;# in m\n", - "d = 0.2E-3;# in m\n", - "e0 = 8.85E-12;# in F/m\n", - "er = 5;\n", - "#calculation:\n", - "A = L*B\n", - "C = e0*er*A*(n-1)/d\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n Capacitance \",round((C/1E-9),2),\" nF\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " Capacitance 22.4 nF" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 10, page no. 65

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Calculate the equivalent capacitance of two capacitors of 6 \u03bcF and 4 \u03bcF connected (a) in parallel and (b) in series\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "C1 = 6E-6;# in Farads\n", - "C2 = 4E-6;# in Farads\n", - "\n", - "#calculation:\n", - "# in Parallel\n", - "Cp = C1 + C2\n", - "# in Series\n", - "Cs = 1/((1/C1) + (1/C2))\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)Capacitance in parallel \",(Cp/1E-6),\" uF\\n\"\n", - "print \"\\n (b)Capacitance in Series \",(Cs/1E-6),\" uF\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)Capacitance in parallel 10.0 uF\n", - "\n", - "\n", - " (b)Capacitance in Series 2.4 uF" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 11, page no. 65

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#What capacitance must be connected in series\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "C1 = 30E-6;# in Farads\n", - "Cs = 12E-6;# in Farads\n", - "\n", - "#calculation:\n", - "# in Series\n", - "C2 = 1/((1/Cs) - (1/C1))\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)Capacitance in series \",(C2/1E-6),\" uF\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)Capacitance in series 20.0 uF" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 12, page no. 65

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Determine (a) the equivalent circuit capacitance, \n", - "#(b) the total charge and\n", - "#(c) the charge on each capacitor\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "C1 = 1E-6;# in Farads\n", - "C2 = 3E-6;# in Farads\n", - "C3 = 5E-6;# in Farads\n", - "C4 = 6E-6;# in Farads\n", - "Vt = 100;# in Volts\n", - "\n", - "#calculation:\n", - "# in Parallel\n", - "Cp = C1 + C2 + C3 + C4\n", - "Qt = Vt*Cp\n", - "Q1 = C1*Vt\n", - "Q2 = C2*Vt\n", - "Q3 = C3*Vt\n", - "Q4 = C4*Vt\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)Equivalent Capacitance in Parallel \",(Cp/1E-6),\" uF\\n\"\n", - "print \"\\n (b)Total charge \",(Qt/1E-3),\" mC\\n\"\n", - "print \"\\n (c)Charge on each capacitors (C1, C2, C3, C4)\\n \",(Q1/1E-3),\", \",(Q2/1E-3),\", \"\n", - "print \",(Q3/1E-3),\", \",(Q4/1E-3),\" mC respectively\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "ename": "SyntaxError", - "evalue": "invalid syntax (, line 27)", - "output_type": "pyerr", - "traceback": [ - "\u001b[1;36m File \u001b[1;32m\"\"\u001b[1;36m, line \u001b[1;32m27\u001b[0m\n\u001b[1;33m print \",(Q3/1E-3),\", \",(Q4/1E-3),\" mC respectively\\n\"\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 13, page no. 66

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Calculate (a) the equivalent circuit capacitance, \n", - "#(b) the charge on each capacitor and\n", - "#(c) the pd across each capacitor.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "C1 = 3E-6;# in Farads\n", - "C2 = 6E-6;# in Farads\n", - "C3 = 12E-6;# in Farads\n", - "Vt = 350;# in Volts\n", - "#calculation:\n", - "# in series\n", - "Cs = 1/((1/C1) + (1/C2) + (1/C3))\n", - "Qt = Vt*Cs\n", - "V1 = Qt/C1\n", - "V2 = Qt/C2\n", - "V3 = Qt/C3\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)Equivalent Capacitance in Series \",(Cs/1E-6),\" uF\\n\"\n", - "print \"\\n (b)Charge on each capacitors (C1, C2, C3) \",(Qt/1E-3),\" mC \\n\"\n", - "print \"\\n (b)P.d Across each capacitors (C1, C2, C3)\\n \",V1,\" V, \", V2,\" V, \", V3,\" V respectively\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)Equivalent Capacitance in Series 1.71428571429 uF\n", - "\n", - "\n", - " (b)Charge on each capacitors (C1, C2, C3) 0.6 mC \n", - "\n", - "\n", - " (b)P.d Across each capacitors (C1, C2, C3)\n", - " 200.0 V, 100.0 V, 50.0 V respectively" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 14, page no. 67

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Find (a) the thickness of the mica needed, and \n", - "#(b) the area of a plate assuming a two-plate construction.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "C = 0.2E-6;# in Farads\n", - "V = 1250;# in Volts\n", - "E = 50E6# in V/m\n", - "e0 = 8.85E-12;# in F/m\n", - "er = 6;\n", - "\n", - "#calculation:\n", - "d = V/E\n", - "A = C*d/e0/er\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)Thickness \",(d/1E-3),\" mm\\n\"\n", - "print \"\\n (b)Area of plate is \",round((A/1E-4),2),\" cm^2 \\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)Thickness 0.025 mm\n", - "\n", - "\n", - " (b)Area of plate is 941.62 cm^2 " - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 15, page no. 68

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#(a) Determine the energy stored in a 3 \u03bcF capacitor when charged to 400 V.\n", - "#(b) Find also the average power developed if this energy is dissipated in a time of 10 \u03bcs\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "C = 3E-6;# in Farads\n", - "V = 400;# in Volts\n", - "t = 10E-6;# in secs\n", - "\n", - "#calculation:\n", - "W = C*V*V/2\n", - "P = W/t\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)Energy stored \",W,\" J\\n\"\n", - "print \"\\n (b)Power developed \",(P/1E3),\" kW \\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)Energy stored 0.24 J\n", - "\n", - "\n", - " (b)Power developed 24.0 kW " - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 16, page no. 68

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#Find the pd to which the capacitor must be charged.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "C = 12E-6;# in Farads\n", - "W = 4;# in Joules\n", - "\n", - "#calculation:\n", - "V = (2*W/C)**0.5\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n P.d \",round(V,2),\" V\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " P.d 816.5 V" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "

Example 17, page no. 68

" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#find (a) the voltage and (b) the capacitance.\n", - "from __future__ import division\n", - "import math\n", - "#initializing the variables:\n", - "W = 1.2;# in Joules\n", - "Q = 10E-3;# in Coulomb\n", - "\n", - "#calculation:\n", - "V = 2*W/Q\n", - "C = Q/V\n", - "\n", - "#Results\n", - "print \"\\n\\nResult\\n\\n\"\n", - "print \"\\n (a)P.d \",V,\" V\\n\"\n", - "print \"\\n (b)Capacitance \",round((C/1E-6),2),\" uF\\n\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "\n", - "Result\n", - "\n", - "\n", - "\n", - " (a)P.d 240.0 V\n", - "\n", - "\n", - " (b)Capacitance 41.67 uF" - ] - } - ], - "prompt_number": 19 - } - ], - "metadata": {} - } - ] -} \ No newline at end of file -- cgit