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  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h1>Chapter 6: Capacitors and capacitance</h1>"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 1, page no. 58</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 2, page no. 58</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 3, page no. 59</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 4, page no. 60</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 5, page no. 60</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 6, page no. 60</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 7, page no. 62</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 8, page no. 62</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 9, page no. 62</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 10, page no. 65</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 11, page no. 65</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 12, page no. 65</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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),\", \",(Q3/1E-3),\",\",(Q4/1E-3),\" mC  respectively\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "Result\n",
        "\n",
        "\n",
        "\n",
        "  (a)Equivalent  Capacitance  in  Parallel   15.0   uF\n",
        "\n",
        "\n",
        "  (b)Total  charge   1.5   mC\n",
        "\n",
        "\n",
        "  (c)Charge  on  each  capacitors  (C1,  C2,  C3,  C4)\n",
        "     0.1 ,  0.3 ,  0.5 , 0.6  mC  respectively\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 13, page no. 66</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 14, page no. 67</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 15, page no. 68</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 16, page no. 68</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": [
      "<h3>Example 17, page no. 68</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\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": {}
  }
 ]
}