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  "name": "",
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 1 - Introduction to Electronics"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E1 - Pg 8"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Find the range of tolerance\n",
      "#soltion\n",
      "#given\n",
      "#color coding\n",
      "orange=3.#\n",
      "gold=5.#\n",
      "yellow=4.#\n",
      "violet=7.#\n",
      "#band pattern\n",
      "band1=yellow#\n",
      "band2=violet#\n",
      "band3=orange#\n",
      "band4=gold#\n",
      "#resistor color coding\n",
      "r=(band1*10.+band2)*10.**(band3)#\n",
      "tol=r*(band4/100.)#tolerance\n",
      "ulr=r+tol##upper limit of resistance\n",
      "llr=r-tol##lower limit of resistance\n",
      "print 'The range of resistance =',llr/1000. ,'kOhm','to',ulr/1000,'kOhm'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The range of resistance = 44.65 kOhm to 49.35 kOhm\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E2 - Pg 8"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Find the range of tolerance\n",
      "#color coding\n",
      "blue=6.#\n",
      "gold=-1.#\n",
      "gray=8.#\n",
      "silver=10.#\n",
      "#band pattern\n",
      "band1=gray#\n",
      "band2=blue#\n",
      "band3=gold#\n",
      "band4=silver#\n",
      "#resistor color coding\n",
      "r=(band1*10.+band2)*10.**(band3)#\n",
      "tol=r*(band4/100.)#tolerance\n",
      "ulr=r+tol##upper limit of resistance\n",
      "llr=r-tol##lower limit of resistance\n",
      "print 'The Range of resistance is',llr,'ohm','to',ulr,'ohm'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Range of resistance is 7.74 ohm to 9.46 ohm\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E3 - Pg 19"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Find the equivalent current source\n",
      "#given\n",
      "Vs=2.;#Volts             #dc voltage source\n",
      "Rs=1.;#ohm               #internal resistance\n",
      "Rl=1.;#ohm              #load resistance\n",
      "Ise=Vs/Rs;#ampere      #equivalent current source\n",
      "\n",
      "# In accordance to figure 1.23a\n",
      "Il1=Ise*(Rs/(Rs+Rl));#using current divider concept\n",
      "Vl1=Il1*Rl;\n",
      "print \"In accordance to figure 1.23a\\n\"\n",
      "print \"The Load current (current source Il=\",Il1,'A'\n",
      "print \"The Load voltage (current source Vl=\",Vl1,'V','\\n'\n",
      "\n",
      "# In accordance to figure 1.23b\n",
      "Vl2=Vs*(Rs/(Rs+Rl));#using voltage divider concept\n",
      "Il2=Vl2/Rl;\n",
      "print \"\\nIn accordance to figure 1.23b\"\n",
      "print \"\\nThe Load voltage (voltage source) Vl=\",Vl2,'V'\n",
      "print \"The Load current (voltage source) Il=\",Il2,'A'\n",
      "print \"\\nTherefore they both provide same voltage and current to load\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "In accordance to figure 1.23a\n",
        "\n",
        "The Load current (current source Il= 1.0 A\n",
        "The Load voltage (current source Vl= 1.0 V \n",
        "\n",
        "\n",
        "In accordance to figure 1.23b\n",
        "\n",
        "The Load voltage (voltage source) Vl= 1.0 V\n",
        "The Load current (voltage source) Il= 1.0 A\n",
        "\n",
        "Therefore they both provide same voltage and current to load\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E4 - Pg 19"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Find percentage variation in load current and load voltage\n",
      "#given\n",
      "Vs=10.;#volt#Supply voltage\n",
      "Rs=100.;#ohm#internal resistance\n",
      "\n",
      "# In accordance to figure 1.24a\n",
      "#For 1ohm - 10 ohm\n",
      "Rl11=1.;#ohm#min extreme value of Rl\n",
      "Rl12=10.;#ohm#max extreme value of Rl\n",
      "Il11=Vs/(Rs+Rl11);\n",
      "Il12=Vs/(Rs+Rl12);\n",
      "Pi1=(Il11-Il12)*100./Il11;#Percentage variation in current\n",
      "Vl11=Il11*Rl11;\n",
      "Vl12=Il12*Rl12;\n",
      "Pv1=(Vl12-Vl11)*100./Vl12;#Percentage variation in voltage\n",
      "print '%s' %(\"In accordance to figure 1.24a \\n\");\n",
      "print '%s %.2f %s' %(\"Percentage variation in Current(1-10 ohm)=\",Pi1,'percent');\n",
      "print '%s %.1f %s ' %(\"Percentage variation in Voltage(1-10 ohm)=\",Pv1,'percent\\n\\n');\n",
      "\n",
      "# In accordance to figure 1.24b\n",
      "#For 1kohm - 10kohm\n",
      "Rl21=1000.;#ohm#min extreme value of Rl\n",
      "Rl22=10000.;#ohm#max extreme value of Rl\n",
      "Il21=Vs/(Rs+Rl21);\n",
      "Il22=Vs/(Rs+Rl22);\n",
      "Pi2=(Il21-Il22)*100./Il21;#Percentage variation in current\n",
      "Vl21=Il21*Rl21;\n",
      "Vl22=Il22*Rl22;\n",
      "Pv2=(Vl22-Vl21)*100./Vl22;#Percentage variation in voltage\n",
      "print '%s' %(\"In accordance to figure 1.24b \\n\");\n",
      "print '%s %.f %s' %(\"Percentage variation in Current(1-10 ohm)=\",Pi2,'percent');\n",
      "print '%s %.f %s ' %(\"Percentage variation in Voltage(1-10 ohm)=\",Pv2,'percent \\n');\n",
      "print 'In book the percentage variation in voltage(1kohm-10kohm) is 9 percent due to' \n",
      "print 'the incorrect value of Il22 i.e. 0.000999 Amp correct value is 0.0009901 Amp'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "In accordance to figure 1.24a \n",
        "\n",
        "Percentage variation in Current(1-10 ohm)= 8.18 percent\n",
        "Percentage variation in Voltage(1-10 ohm)= 89.1 percent\n",
        "\n",
        " \n",
        "In accordance to figure 1.24b \n",
        "\n",
        "Percentage variation in Current(1-10 ohm)= 89 percent\n",
        "Percentage variation in Voltage(1-10 ohm)= 8 percent \n",
        " \n",
        "In book the percentage variation in voltage(1kohm-10kohm) is 9 percent due to\n",
        "the incorrect value of Il22 i.e. 0.000999 Amp correct value is 0.0009901 Amp\n"
       ]
      }
     ],
     "prompt_number": 4
    }
   ],
   "metadata": {}
  }
 ]
}