{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "UNIT-5 Electricity"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.1,Page no:152"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate current flowing through a conductor\n",
      "\n",
      "#Variable declaration  \n",
      "n=10**6     #no. of electrons\n",
      "e=1.6*10**-19   #charge on an electron in C\n",
      "\n",
      "#Calculation\n",
      "q=n*e    #calculating total charge\n",
      "t=10**-3    #time in second\n",
      "I=q/t     #calculating current\n",
      "\n",
      "#Result\n",
      "print\"Current flowing  = \",I,\"Ampere\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Current flowing  =  1.6e-10 Ampere\n"
       ]
      }
     ],
     "prompt_number": 71
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.2,Page no:152"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#No of electrons passing through a lamp\n",
      "\n",
      "#Variable declaration  \n",
      "I=300*10**-3    #current n Ampere\n",
      "t=60    #time in second\n",
      "e=1.6*10**-19   #chatge on electron in C\n",
      "\n",
      "#Calculation\n",
      "q=I*t   #calculating charge\n",
      "n=q/e   #calculating no of electrons\n",
      "\n",
      "#Result\n",
      "print\"No. of electrons = \",n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "No. of electrons =  1.125e+20\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.3,Page no:154"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Frequency of electrons across a conductor\n",
      "\n",
      "#Variable declaration  \n",
      "V=200   #voltage in volt\n",
      "R=100   #resistance in Ohm\n",
      "e=1.6*10**-19   #charge on an electron in C\n",
      "\n",
      "#Calculation\n",
      "I=V/R      #Ohm's law\n",
      "t=1   #time in second\n",
      "q=I*t   #calculating charge\n",
      "n=q/e   #calculating no of electrons\n",
      "\n",
      "#Result\n",
      "print\"No. of electrons = \",n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "No. of electrons =  1.25e+19\n"
       ]
      }
     ],
     "prompt_number": 72
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.4,Page no:156"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Resistivity of small wire\n",
      "\n",
      "#Variable declaration  \n",
      "l=15    #length in m\n",
      "A=6*10**-7   #area in m square\n",
      "R=5   #resistance in Ohm\n",
      "\n",
      "#Calculation\n",
      "p=(A*R)/l   #calculating resistivity\n",
      "\n",
      "#Result\n",
      "print\"Resistivity= \",p,\"Ohm metre\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resistivity=  2e-07 Ohm metre\n"
       ]
      }
     ],
     "prompt_number": 73
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.5,Page no:157"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Resistivity of a normal wire\n",
      "\n",
      "#Variable declaration  \n",
      "l=0.1    #length in m\n",
      "A=10**-4   #area in m square\n",
      "R=0.01   #resistance in Ohm\n",
      "\n",
      "#Calculation\n",
      "p=(A*R)/l   #calculating resistivity\n",
      "\n",
      "#Result\n",
      "print\"Resistivity  = \",p,\"Ohm metre\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resistivity  =  1e-05 Ohm metre\n"
       ]
      }
     ],
     "prompt_number": 74
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.6,Page no:157"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Resistivity of a long wire\n",
      "\n",
      "import math  \n",
      "\n",
      "#Variable declaration\n",
      "L=1   #length in m\n",
      "r=0.2*10**-3   #radius in m\n",
      "R=2   #resistance in Ohm\n",
      "\n",
      "#Calculation\n",
      "A=math.pi*(r)**2   #calculating area\n",
      "P=(R*A)/L   #calculating resistivity\n",
      "\n",
      "#Result\n",
      "print\"Resistivity =%.2g\"%P,\"Ohm.metre\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resistivity =2.5e-07 Ohm.metre\n"
       ]
      }
     ],
     "prompt_number": 38
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.7,Page no:158"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#New resistance of a wire after drawn out\n",
      "\n",
      "#Variable declaration  \n",
      "R1=5    #resisitance in Ohm\n",
      "\n",
      "#Calculation\n",
      "#A2=A/3\n",
      "#R2/5=3l*3/A*A/l\n",
      "#R2=9*5\n",
      "\n",
      "R2=9*R1   #calculating using R2/A1=(l2/A2)*(A1/l1)\n",
      "print\"Resisitance = \",R2,\"Ohm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resisitance =  45 Ohm\n"
       ]
      }
     ],
     "prompt_number": 75
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.8,Page no:159"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Change in resistance of a wire\n",
      "\n",
      "#Variable declaration  \n",
      "R1=5    #resisitance in Ohm\n",
      "\n",
      "#Calculation\n",
      "#A2=A/2\n",
      "#R1=rho*l1/A1*R2\n",
      "#R2=rho*l2/A2\n",
      "#R2/R1=A1/l1\n",
      "R2=4*R1   #calculating using R2/A1=(l2/A2)*(A1/l1)\n",
      "\n",
      "#Result\n",
      "print\"Resisitance= \",R2,\"Ohm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resisitance=  20 Ohm\n"
       ]
      }
     ],
     "prompt_number": 70
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.9,Page no:162"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Total resistance in parallel\n",
      "\n",
      "#Variable declaration  \n",
      "R1=2    #resisitance in Ohm\n",
      "R2=4   #resistance in Ohm\n",
      "R3=5    #resistance in Ohm\n",
      "\n",
      "#Calculation\n",
      "R=(R1**-1)+(R2**-1)+(R3**-1)   #calculating parallel resistance\n",
      "Rp=(1/R) \n",
      "\n",
      "#Result\n",
      "print\"Resisitance  = \",Rp,\"Ohm\"\n",
      "print\"\\nNOTE:Incorrect answer in book\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resisitance  =  1.05263157895 Ohm\n",
        "\n",
        "NOTE:Incorrect answer in book\n"
       ]
      }
     ],
     "prompt_number": 39
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.10,Page no:163"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Series parallel combination resistance\n",
      "\n",
      "from scipy.optimize import fsolve  \n",
      "\n",
      "#Variable declaration\n",
      "Rs=40    #resisitance in Ohm\n",
      "\n",
      "#Calculation\n",
      "#R1+R2=40\n",
      "#R1*R2=256\n",
      "#R1=256/R2\n",
      "#Putting this value in eq 1:\n",
      "#(256/R2)+R2=40\n",
      "from sympy import solve, symbols, pprint\n",
      "R2= symbols('R2')\n",
      "a=1\n",
      "b=-40\n",
      "c=256\n",
      "f = a*R2**2 + b*R2 + c\n",
      "solution = solve(f, R2)\n",
      "\n",
      "#Result\n",
      "print\"When R2=\",solution[0],\"Ohm    R1=\",solution[1],\"Ohm\"\n",
      "print\"When R2=\",solution[1],\"Ohm    R1=\",solution[0],\"Ohm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "When R2= 8 Ohm    R1= 32 Ohm\n",
        "When R2= 32 Ohm    R1= 8 Ohm\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.11,Page no:164"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Find the valueof current in fig\n",
      "\n",
      "#Variable declaration  \n",
      "V=2.0      #in volts\n",
      "R1=30.0    #resisitance in Ohm\n",
      "R2=60.0   #resistance in Ohm\n",
      "\n",
      "#Calculation\n",
      "Rp=(R1*R2)/(R1+R2)   #calculating parallel resistance\n",
      "I=V/Rp      #Ohm's law\n",
      "\n",
      "#Result\n",
      "print\"Resisitance = \",Rp,\"Ohm\"\n",
      "print\"Current  = \",I,\"A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resisitance =  20.0 Ohm\n",
        "Current  =  0.1 A\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.12,Page no:165"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Combine resistance to get an equivalent resistance\n",
      "\n",
      "#Variable declaration  \n",
      "R1=2.0    #resisitance in Ohm\n",
      "R2=3.0   #resistance in Ohm\n",
      "R3=1.0    #resistance in Ohm\n",
      "\n",
      "#Calculation\n",
      "Rp=(R1*R2)/(R1+R2)   #calculating parallel resistance\n",
      "R=Rp+1.0       #1 Ohm in series\n",
      "Rs=(R1+R2+R3)   #series resistances\n",
      "Rp=(1.0/R1)+(1.0/R2)+(1.0/R3)   #calculating parallel resistance\n",
      "\n",
      "#Result\n",
      "print\"(1)Equivalent Resisitance= \",R,\"Ohm\"    \n",
      "print\"(2)All resistances in series = \",Rs,\"Ohm\"\n",
      "print\"(3)All in Parallel  = \",(1/Rp),\"Ohm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(1)Equivalent Resisitance=  2.2 Ohm\n",
        "(2)All resistances in series =  6.0 Ohm\n",
        "(3)All in Parallel  =  0.545454545455 Ohm\n"
       ]
      }
     ],
     "prompt_number": 76
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.13,Page no:166"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Total resistor of combination\n",
      "\n",
      "#Variable declaration  \n",
      "V=20     #voltage in Volts\n",
      "R1=2.0    #resisitance in Ohm\n",
      "R2=4.0   #resistance in Ohm\n",
      "R3=5.0    #resistance in Ohm\n",
      "\n",
      "#Calculation\n",
      "Rp=(1/R1)+(1/R2)+(1/R3)   #calculating parallel resistance\n",
      "R=1/Rp       #Parallel\n",
      "I1=V/R1   #calculating current through R1\n",
      "I2=V/R2   #calculating current through R2\n",
      "I3=V/R3   #calculating current through R3\n",
      "I=V/R    #calculating total current\n",
      "\n",
      "\n",
      "#Result\n",
      "print\"(a)Equivalent Resisitance  = \",R,\"Ohm\"\n",
      "print\"Current through R1  = \",I1,\"Ampere\"\n",
      "print\"Current through R2 = \",I2,\"Ampere\"    \n",
      "print\"Total current  = \",I,\"Ampere\"    "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)Equivalent Resisitance  =  1.05263157895 Ohm\n",
        "Current through R1  =  10.0 Ampere\n",
        "Current through R2 =  5.0 Ampere\n",
        "Total current  =  19.0 Ampere\n"
       ]
      }
     ],
     "prompt_number": 30
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.14,Page no:166"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate no of resistors form total resistor\n",
      "\n",
      "#Variable declaration\n",
      "R=7           #Total resistanc of combination\n",
      "\n",
      "#Calculation\n",
      "def f(n):\n",
      "    Rp = 6*(1/n)   #resistance in parallel\n",
      "    return(R-Rp-5)\n",
      "n=fsolve(f,1)\n",
      "\n",
      "#Result\n",
      "print\"n=\",n[0]"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "n= 3.0\n"
       ]
      }
     ],
     "prompt_number": 33
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.15,Page no:173"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Voltage and current in figure\n",
      "\n",
      "#Variable declaration  \n",
      "R1=2   #resistance in Ohm\n",
      "R2=6    #resistance in Ohm\n",
      "R3=3   #resistance in Ohm\n",
      "V=24    #voltage in volts\n",
      "R=8     #resistance in Ohm\n",
      "\n",
      "#Calculation\n",
      "I=V/R  #Ohm's Law\n",
      "V1=I*R1   #Ohm's Law\n",
      "V2=I*R2   #Ohm's Law\n",
      "V3=I*R3   #Ohm's Law\n",
      "\n",
      "#Result\n",
      "print\"Current  = \",I,\"Ampere\"    \n",
      "print\"Voltage drop across R1 = \",V1,\"Volts\"\n",
      "print\"Voltage drop across R2  = \",V2,\"Volts\"    \n",
      "print\"Voltage drop across R3  = \",V3,\"Volts\"\n",
      "print\"\\nNOTE:Wrong answer of R3 in book\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Current  =  3 Ampere\n",
        "Voltage drop across R1 =  6 Volts\n",
        "Voltage drop across R2  =  18 Volts\n",
        "Voltage drop across R3  =  9 Volts\n",
        "\n",
        "NOTE:Wrong answer of R3 in book\n"
       ]
      }
     ],
     "prompt_number": 36
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.16,Page no:173"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Current through branches of circuit\n",
      "\n",
      "#Variable declaration  \n",
      "R=15   #resistance in Ohm\n",
      "print\"KVL: 16I1+15I2=6   (1)\"  #KVL equation\n",
      "I1=-1.66   #from(1)\n",
      "I2=2.17   #from (1)\n",
      "#Calculation\n",
      "V=(I1+I2)*R   #calculating potential difference\n",
      "\n",
      "#Result\n",
      "print\"Potential difference= \",V,\"Volt\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "KVL: 16I1+15I2=6   (1)\n",
        "Potential difference=  7.65 Volt\n"
       ]
      }
     ],
     "prompt_number": 37
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.17,Page no:174"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine current in wheatstone bridge\n",
      "\n",
      "print\"3I1-I2-1=0   (1)\"    #KVL equation\n",
      "print\"3I1-I2+2I=2  (2)\"    #KVL equation\n",
      "print\"3I1-I1+2I=2  (3)\"    #KVL equation\n",
      "\n",
      "#Variable declaration\n",
      "I1=4/17.0   #from (1)(2)(3)through AB \n",
      "I2=-2/17.0  #from (1)(2)(3)through BD\n",
      "I=3*I1+I2   #from (1)(2)(3)through main circuit\n",
      "\n",
      "#Calculation\n",
      "Ibc=I1-I2  #calculating current in BC\n",
      "Iad=I-I1   #calculating current in AD\n",
      "Idc=I-I1+I2   #calculating current in DC\n",
      "\n",
      "#Result\n",
      "print\"Current in branch BC = \",Ibc,\"Ampere\"\n",
      "print\"NOTE:Calculation mistake in book while calculating for BC\"\n",
      "print\"Current in branch AD = \",Iad,\"Ampere\"\n",
      "print\"Current in branch DC = \",Idc,\"Ampere\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "3I1-I2-1=0   (1)\n",
        "3I1-I2+2I=2  (2)\n",
        "3I1-I1+2I=2  (3)\n",
        "Current in branch BC =  0.352941176471 Ampere\n",
        "NOTE:Calculation mistake in book while calculating for BC\n",
        "Current in branch AD =  0.352941176471 Ampere\n",
        "Current in branch DC =  0.235294117647 Ampere\n"
       ]
      }
     ],
     "prompt_number": 77
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.18,Page no:176"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Current through a galvanometer in wheatstone bridge\n",
      "\n",
      "#Variable declaration  \n",
      "P=10   #Ohm\n",
      "Q=3    #Ohm\n",
      "R=12   #Ohm\n",
      "S=6   #Ohm\n",
      "G=20   #Ohm\n",
      "\n",
      "\n",
      "#Calculation\n",
      "print\"-12I+22I1+IgG=0  (1)\"   #KVL\n",
      "print\"6I-9I1+29Ig=0  (2)\"  #KVL\n",
      "print\"13I1-3Ig=2  (3)\"   #KVL\n",
      "#From above equations\n",
      "import numpy as np\n",
      "a = np.array([[-12,22,20],[6,-9,29],[0,13,-3]]) \n",
      "b = np.array([[0],[0],[2]])\n",
      "np.linalg.solve(a,b)\n",
      "\n",
      "\n",
      "#Result\n",
      "print\"Current through Galvanometer = \",round(Ig*1000,2),\"mA\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "-12I+22I1+IgG=0  (1)\n",
        "6I-9I1+29Ig=0  (2)\n",
        "13I1-3Ig=2  (3)\n",
        "Current through Galvanometer =  7.8 mA\n"
       ]
      }
     ],
     "prompt_number": 78
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.19,Page no:179"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#% drop in heat in bulb\n",
      "\n",
      "#Variable declaration  \n",
      "P=500   #power in Watts\n",
      "V=200   #voltage in Volts\n",
      "V1=160   #voltage in Volts\n",
      "\n",
      "#Calculation\n",
      "R=(V**2)/P   #using P=V**2*R\n",
      "P1=(V1**2)/R   #calculating power\n",
      "Dp=500-P1   #drop in heat\n",
      "D=(Dp*100)/500   #percentage drop\n",
      "\n",
      "#Result\n",
      "print\"Resistance= \",R,\"Ohm\"\n",
      "print\"% Drop in heat production = \",D,\"%\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resistance=  80 Ohm\n",
        "% Drop in heat production =  36 %\n"
       ]
      }
     ],
     "prompt_number": 79
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.20,Page no:180"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Bulbs connected in parallel\n",
      "\n",
      "#Variable declaration\n",
      "P1=100   #power in Watts\n",
      "P2=500   #power in Watts\n",
      "\n",
      "#Calculation\n",
      "P=P2/P1   #ratio\n",
      "\n",
      "#Result\n",
      "print \"P=\",P\n",
      "print\"P>0,I2=5I Therefore I2>I1\"    "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "P= 5\n",
        "P>0,I2=5I Therefore I2>I1\n"
       ]
      }
     ],
     "prompt_number": 80
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.21,Page no:181"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Heat and light energy produced by bulb on voltage drop\n",
      "\n",
      "#Variable declaration  \n",
      "t=1200   #time in second\n",
      "P=100   #power in Watts\n",
      "V=230   #voltage in Volts\n",
      "\n",
      "#Calculation\n",
      "R=(V**2)/P   #calculating resistance\n",
      "V1=115   #supply voltage in Volts\n",
      "E=((V1**2)*t)/R   #calculating energy\n",
      "\n",
      "#Result\n",
      "print\"Energy dissipated by bulb = \",E,\"J\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Energy dissipated by bulb =  30000 J\n"
       ]
      }
     ],
     "prompt_number": 81
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.22,Page no:181"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate effeciency of transmission\n",
      "\n",
      "#Variable declaration\n",
      "P=10**4   #power in Watts\n",
      "V=250   #voltage in Volts\n",
      "R=0.2   #resistance in ohm\n",
      "\n",
      "#Calculation\n",
      "Pl=((P/V)*(P/V))*R   #calculating power loss\n",
      "print P1\n",
      "E=P/(Pl+P)   #calculating efficiency\n",
      "\n",
      "#Result\n",
      "print\"Percent Efficiency = \",round(E*100),\"%\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "100\n",
        "Percent Efficiency =  97.0 %\n"
       ]
      }
     ],
     "prompt_number": 56
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.23,Page no:182"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Resistance and current capacity of a lamp\n",
      "\n",
      "#Variable declaration  \n",
      "P=100.0   #power in Watts\n",
      "V=220.0   #voltage in Volts\n",
      "\n",
      "#Calculation\n",
      "I=P/V   #Current in Ampere\n",
      "R=V/I   #resistance\n",
      "\n",
      "#Result\n",
      "print\"Current  = \",round(I,3),\"A\"  \n",
      "print\"Resistance=\",R,\"Ohm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Current  =  0.455 A\n",
        "Resistance= 484.0 Ohm\n"
       ]
      }
     ],
     "prompt_number": 59
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example no:5.24,Page no:182"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Resistance of windings of electric motor\n",
      "\n",
      "#Variable declaration  \n",
      "V=50   #voltage in Volts\n",
      "I=12   #Current in Ampere\n",
      "\n",
      "#Calculation\n",
      "P=V*I   #power\n",
      "Pd=P*0.7   #power dissipated\n",
      "R=(Pd/(I)**2) \n",
      "\n",
      "#Result\n",
      "print\"Resistance = \",round(R,2),\"Ohm\"    "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Resistance =  2.92 Ohm\n"
       ]
      }
     ],
     "prompt_number": 82
    }
   ],
   "metadata": {}
  }
 ]
}