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diff --git a/Electronic_Principles/Chapter_1.ipynb b/Electronic_Principles/Chapter_1.ipynb new file mode 100755 index 00000000..5245d442 --- /dev/null +++ b/Electronic_Principles/Chapter_1.ipynb @@ -0,0 +1,256 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "CHAPTER 1 INTRODUCTION"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1-1, Page 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "Rs=50 #source resistance(Ohm)\n",
+ "\n",
+ "RL=100*Rs/1000 #Minimum Load resistance(KOhm)\n",
+ "\n",
+ "print 'Minimum Load resistance =',RL,'KOhm'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum Load resistance = 5 KOhm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1-2, Page 12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "Is=0.002 #Current source(A)\n",
+ "Rs=10000000 #internal resistance(Ohm)\n",
+ "\n",
+ "Rl_Max=0.01*Rs/1000 #Maximum load resistance(KOhm)\n",
+ "\n",
+ "print 'With 100:1 Rule, Maximum Load resistance =',Rl_Max,'KOhm'\n",
+ "print 'Stiff range for current source is load resistance from 0 KOhm to',Rl_Max,'KOhm'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "With 100:1 Rule, Maximum Load resistance = 100.0 KOhm\n",
+ "Stiff range for current source is load resistance from 0 KOhm to 100.0 KOhm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1-3, Page 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "Il=0.002 #current source(A)\n",
+ "Rl=10000 #load resistance(Ohm)\n",
+ "\n",
+ "Vl=Il*Rl #load voltage(V)\n",
+ "\n",
+ "print 'Load Voltage =',Vl,'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load Voltage = 20.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1-4, Page 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "print 'As per figure 1-9a, Calculate Thevenin resistor by opening load resistor'\n",
+ "\n",
+ "Vs=72 #source voltage\n",
+ "R1=6 #Resistance (KOhm)\n",
+ "R2=3 #Resistance (KOhm)\n",
+ "R3=4 #Resistance (KOhm)\n",
+ "\n",
+ "Ro=R1+R2 #Resistance (KOhm)\n",
+ "I=Vs/Ro\n",
+ "Vab=R2*Vs/(R1+R2) #Thevenin voltage(V)\n",
+ "Rth=((R1*R2)/(R1+R2))+R3 #Thevenin Resistance(KOhm)\n",
+ "\n",
+ "print 'After removing Rl current flowing through 6KOhm is',I,'mA'\n",
+ "print 'Thevenin Voltage Vth =',Vab,'V'\n",
+ "print 'Calculating Thevenin Resistance with considering R1 & R2 in parallel'\n",
+ "print 'Thevenin Resistance =',Rth,'KOhm'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "As per figure 1-9a, Calculate Thevenin resistor by opening load resistor\n",
+ "After removing Rl current flowing through 6KOhm is 8 mA\n",
+ "Thevenin Voltage Vth = 24 V\n",
+ "Calculating Thevenin Resistance with considering R1 & R2 in parallel\n",
+ "Thevenin Resistance = 6 KOhm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1-5, Page 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "Vs=72 #source voltage\n",
+ "R1=2 #resistance1 (KOhm)\n",
+ "R2=2 #resistance1 (KOhm)\n",
+ "R3=1 #resistance1 (KOhm)\n",
+ "R4=2 #resistance1 (KOhm)\n",
+ "R5=1 #resistance1 (KOhm)\n",
+ "R6=2 #resistance1 (KOhm)\n",
+ "R7=0.5 #resistance1 (KOhm)\n",
+ "RL=1 #load resistance (KOhm)\n",
+ "\n",
+ "Vth=9 #Thevenin voltage(V)\n",
+ "Rth=1.5 #Thevenin resistance(KOhm)\n",
+ "\n",
+ "print 'Thevenin Voltage Vth =',Vth,'V'\n",
+ "print 'Thevenin Resistance =',Rth,'KOhm'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thevenin Voltage Vth = 9 V\n",
+ "Thevenin Resistance = 1.5 KOhm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1-6, Page 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "Vs=10 #source voltage\n",
+ "Rs=2 #series source resistance (KOhm)\n",
+ "\n",
+ "IN=Vs/Rs #Norton current(mA)\n",
+ "Rp=Rs #parallel source resistance(KOhm)\n",
+ "\n",
+ "print 'Norton current IN =',IN,'mA'\n",
+ "print 'Norton Resistance =',Rp,'KOhm'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Norton current IN = 5 mA\n",
+ "Norton Resistance = 2 KOhm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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