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+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:c4a6fc97804f83d8aaf3c5e868daac688906bc270dd8b654ea3123010aa41bfa"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "chapter01:Introduction to Electronics"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example E1 - Pg 8"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#pg 8\n",

+      "#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",

+      "\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": {}

+  }

+ ]

+}
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