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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 1: Review of Basic Circuits"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1 Page No:5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Applying KCL at junction(N),we get I=1+2=3 A\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Given Data\n",
+ "#Current through Resistor R1 is 1A\n",
+ "#Current through Resistor R2 is 2A\n",
+ "#Display results\n",
+ "print\"Applying KCL at junction(N),we get I=1+2=3 A\" # KCL stands for Kirchoff's Current Law which states\n",
+ " # that sum of all currents at a junction equals to zero\n",
+ " # N is the name of the junction"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2 Page No:7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "I=0.200000 A\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Given Data\n",
+ "E1=4.;#EMF in volts\n",
+ "E2=2.;#EMF in volts\n",
+ "r1=2.;#Resistance in ohm\n",
+ "r2=3.;#Resistance in ohm\n",
+ "R=5.;#Resistance in ohm\n",
+ "#Calculations\n",
+ "#Applying KVL to the circuit,we get\n",
+ "I=(E1-E2)/(r1+r2+R);#Current in Amperes\n",
+ "#Display Results\n",
+ "print \"I=%f A\"%I;"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3 Page No:9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The current by the battery I=0.312500 Amperes\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Given Data\n",
+ "E=10.0;#EMF in volts\n",
+ "R1=20.0;#Resistance in ohm\n",
+ "R2=20.0;#Resistance in ohm\n",
+ "R3=30.0;#Resistance in ohm\n",
+ "#Calculations\n",
+ "Req=R1+(R2*R3/(R2+R3));# Equivalent resistance of the circuit\n",
+ "I=E/Req;#Current through the battery in Amperes\n",
+ "print \"The current by the battery I=%f Amperes\"%I;\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4 Page No:12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Current through R=5ohm is 0.277778 Amperes\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Naming emfs on both sides to be E1,E2 respectively\n",
+ "#and resistances as R1,R2,R3\n",
+ "#Given Data\n",
+ "E1=3.0;# volts\n",
+ "E2=1.0;# volts\n",
+ "R1=4.0;# ohms\n",
+ "R2=Rab=RL=5.0;# ohms\n",
+ "R3=2.0;# ohms\n",
+ "#Calculations\n",
+ "#We apply Thevenins Theorem to solve\n",
+ "#We have to find Vth and Rth\n",
+ "#First Apllying KVL in the outer loop,we get\n",
+ "I=2.0/6;# Amp\n",
+ "Va=3-I*4# volts\n",
+ "#Since 'B' is attached to negative potential\n",
+ "Vth=Vab=Va-0;# volts\n",
+ "#Then we find Rth\n",
+ "Rth=Rab=2*4/(2+4);# ohm\n",
+ "IL=Vth/(Rth+RL);# Load Current in Amp\n",
+ "print \"Current through R=5ohm is %f Amperes\"%IL\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5 Page No:14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The power delivered to 'R' is maximum when R=r/2\n",
+ "Maximum Power transferred is Pmax=E**2/(2*r)\n"
+ ]
+ }
+ ],
+ "source": [
+ "from sympy import Symbol\n",
+ "#Given Data\n",
+ "#Two batteries of emf E with internal resistances 'r'\n",
+ "#Calculations\n",
+ "#Case(a)\n",
+ "#Open circuiting the branch containing 'R'\n",
+ "#and short circuiting 'E'\n",
+ "E=Symbol('E');\n",
+ "r=Symbol('r');\n",
+ "Rth=r*r/(r+r);# Thevenin's Eq. Resistance in Ohm\n",
+ "print \"The power delivered to 'R' is maximum when R=%s\"%Rth; #According to the maximum power transform theorem\n",
+ "#case(b)\n",
+ "#Applying Parallel Theorem\n",
+ "Vth=E; # Vth is Thevenin's Voltage\n",
+ "#Pmax=(IL**2)*R\n",
+ "Pmax=(Vth/(2*Rth))**2*Rth;# R=Rth, Pmax is Maximum power tranferred\n",
+ "print \"Maximum Power transferred is Pmax=%s\"%Pmax;\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": false
+ },
+ "source": [
+ "## Solved Questions"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Q.1 Page No:33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Input Power=444.44 KW\n",
+ "Power Factor=-0.998612\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from numpy import degrees,arctan,cos\n",
+ "#Given\n",
+ "#For a 3ph motor\n",
+ "VL=2000; #in volts\n",
+ "f=50;# Hz\n",
+ "nFL=0.9;#Full load Efficiency\n",
+ "W1=300;#Wattmeter-1 Reading in KW\n",
+ "W2=100;#Wattmeter-2 Reading in KW\n",
+ "#Calculations\n",
+ "#(i)\n",
+ "Po=W1+W2;#Outout Power\n",
+ "Pin=Po/nFL;#Input Power\n",
+ "#(ii)\n",
+ "ph=degrees(arctan(((W2-W1)*math.sqrt(3))/(W1+W2)));\n",
+ "pf=cos(ph);\n",
+ "#Display\n",
+ "print\"Input Power=%.2f KW\"%Pin;\n",
+ "print\"Power Factor=%f\"%pf;"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Q.2 Page No:34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Line Current=23.09Amperes\n",
+ "Power factor=0.800000\n",
+ "Power=12.800000 KW\n",
+ "Total Volt Ampere=16.00 KVA\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from numpy import arctan2,degrees,cos\n",
+ "#Given\n",
+ "#A balanced star connected load\n",
+ "Zph=8+1j*6;#Ohm\n",
+ "VL=400;#Line voltage inVolts\n",
+ "#Calculations\n",
+ "#(i)\n",
+ "Vph=VL/math.sqrt(3);#Phase Voltage\n",
+ "Iph=Vph/abs(Zph);#Phase Current\n",
+ "IL=Iph;#Line current\n",
+ "#(ii)\n",
+ "pf=cos(arctan2(Zph.imag,Zph.real));#Power factor\n",
+ "#(iii)\n",
+ "P=math.sqrt(3)*VL*IL*pf*10**-3;#Power\n",
+ "#(iv)\n",
+ "VA=math.sqrt(3)*VL*IL*10**(-3);#Volt Ampere\n",
+ "#Display\n",
+ "print\"Line Current=%.2fAmperes\"%IL;\n",
+ "print\"Power factor=%f\"%pf;\n",
+ "print\"Power=%f KW\"%P;\n",
+ "print\"Total Volt Ampere=%.2f KVA\"%VA;"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Q.3 Page No:35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Line current=3.264397 amp\n",
+ "Power Absorbed=1.600000 KW\n",
+ "Total KVA=2.261641 KVA\n",
+ "Power Factor=0.707451\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from numpy import arctan2,cos,degrees\n",
+ "#Given\n",
+ "#Each phase of a star connected load has\n",
+ "R=100;# Resistance in Ohm\n",
+ "C=31.8*10**(-6);# Capacitance in Farad\n",
+ "#Connected to 3-ph\n",
+ "VL=400;#Volts\n",
+ "f=50;#Hz\n",
+ "#Calculations\n",
+ "Xc=1/(2*math.pi*f*C);#Capacitive Reactance in Ohm\n",
+ "num=R*-1j*Xc;#Ohm\n",
+ "den=(R-1j*Xc);\n",
+ "Z=(num*den.conjugate())/(den*den.conjugate()); #Impedance\n",
+ "#(i)\n",
+ "Vp=VL/math.sqrt(3);#Phase voltage\n",
+ "IL=Vp/abs(Z);#Line current\n",
+ "#(ii)\n",
+ "ph=arctan2(Z.imag,Z.real); #Phase angle\n",
+ "TKVA=math.sqrt(3)*VL*IL*10**(-3);# Total Power\n",
+ "P=TKVA*cos(ph);\n",
+ "pf=cos(ph);#Power factor\n",
+ "#Display\n",
+ "print\"Line current=%f amp\"%IL;\n",
+ "print\"Power Absorbed=%f KW\"%P;\n",
+ "print\"Total KVA=%f KVA\"%TKVA;\n",
+ "print\"Power Factor=%f\"%pf;\n",
+ "#Answer differs slightly from TEXTBOOK"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Q.4. Page No:36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Line voltage=461.88 volt\n",
+ "Phase voltage=266.67 volt\n",
+ "KVAR input=16.703293 KVAR\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from numpy import arccos,sin\n",
+ "#Given\n",
+ "#The load connectted to a 3-ph supply comprises\n",
+ "Pkva=20;#in KVA\n",
+ "Pkw=11.0;#in KW\n",
+ "IL=25;#Line Current in Amp\n",
+ "#Calculations\n",
+ "ph=arccos(Pkw/Pkva);\n",
+ "Pkvar=Pkva*sin(ph);#in KVAR\n",
+ "VL=20*10**3/(math.sqrt(3)*IL);\n",
+ "Vph=VL/math.sqrt(3);\n",
+ "#Display\n",
+ "print\"Line voltage=%.2f volt\"%VL;\n",
+ "print\"Phase voltage=%.2f volt\"%Vph;\n",
+ "print\"KVAR input=%f KVAR\"%Pkvar;"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Q.5. Page No:37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Resistance across Delta Circuit=60.000000 Ohm\n",
+ "Current flowing through them=6.67 Amp\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "#Given\n",
+ "#A star circuit and a delta circuit\n",
+ "#In star circuit\n",
+ "VL1=400.0;#volts\n",
+ "R1=20.0;#ohm\n",
+ "#In Delta Circuit\n",
+ "VL=400;#volt\n",
+ "#Calculations\n",
+ "Vph1=VL1/math.sqrt(3); #Phase Voltage-1\n",
+ "Iph1=Vph1/R1; #Phase current-1\n",
+ "IL1=Iph1; #Line current-1\n",
+ "#In Delta circuit\n",
+ "IL2=IL1; # Line current-2\n",
+ "VL2=VL1; #line voltage-2\n",
+ "Vph2=VL2; #Phase voltage-2\n",
+ "Iph2=IL2/math.sqrt(3);#Phase current-2\n",
+ "R2=Vph2/Iph2;#Resistance in Ohm\n",
+ "#Display\n",
+ "print\"Resistance across Delta Circuit=%f Ohm\"%R2;\n",
+ "print\"Current flowing through them=%.2f Amp\"%Iph2;\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Q.6. Page No:37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Phase current=27.712813 A\n",
+ "Power consumed=6.400000 KW\n",
+ "Phasor sum of all line currents= 0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "#Given\n",
+ "#A 3-ph supply connected to balanced delta load\n",
+ "VL=Vph=400; #in volts\n",
+ "Zph=15+1j*20; #Ohm\n",
+ "#Calculations\n",
+ "#(i)\n",
+ "Iph=Vph/abs(Zph); #Phase current\n",
+ "#(ii)\n",
+ "P=Iph**2*abs(Zph)*10**-3; #Power\n",
+ "#(iii)\n",
+ "IL=math.sqrt(3)*Iph; #Line current\n",
+ "#Display\n",
+ "print\"Phase current=%f A\"%IL;\n",
+ "print\"Power consumed=%f KW\"%P;\n",
+ "print\"Phasor sum of all line currents=\",0;\n",
+ "#SOLUTION IN THE TEXTBOOK IS WRONG"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Q.7 Page No:38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Line Voltage=346.41 V\n",
+ "On reversing the phase connections,Line voltage remains unaffected\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Given\n",
+ "#In a 3-ph Star connected System\n",
+ "Vp=200;#Phase voltage in volts\n",
+ "#Calculations\n",
+ "#(a)\n",
+ "VL=math.sqrt(3)*Vp; #Line Voltage in volts\n",
+ "#(b)\n",
+ "#Line voltage remains unaffected\n",
+ "#Display\n",
+ "print\"Line Voltage=%.2f V\"%VL;\n",
+ "print\"On reversing the phase connections,Line voltage remains unaffected\";"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Q.8 Page No:39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total Power drawn=600.000000 Watts\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from numpy import cos\n",
+ "import cmath\n",
+ "#Given\n",
+ "#A 3-ph supply connected to 3-ph load\n",
+ "Vp=200; #Phase Voltage in volts\n",
+ "Zp=cmath.rect(100,60*math.pi/180); #Phase Impedance\n",
+ "#Calculations\n",
+ "Ip=Vp/abs(Zp); #Phase current in Amp\n",
+ "P=Vp*Ip*cos(cmath.phase(Zp)); #Power in Watts\n",
+ "P3ph=3*P;#3-ph Power\n",
+ "#Display\n",
+ "print\"Total Power drawn=%f Watts\"%P3ph;"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Q.9 Page No:39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "W1=-4.843135 KW\n",
+ "W2=34.843135 KW\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from numpy import arccos,cos\n",
+ "#Given\n",
+ "#A 3-ph motor\n",
+ "VL=500.0; #Line voltage in watts\n",
+ "pf=0.4; #Power factor\n",
+ "P=30*10**3; #Total power in KW\n",
+ "#Calculations\n",
+ "ph=arccos(pf);\n",
+ "IL=P/(math.sqrt(3)*VL*pf);\n",
+ "W1=VL*IL*cos((30*math.pi/180)+ph)*10**(-3); #Wattmeter-1 in KW\n",
+ "W2=VL*IL*cos((30*math.pi/180)-ph)*10**(-3); #Wattmeter-2 in KW\n",
+ "#Display\n",
+ "print\"W1=%f KW\"%W1;\n",
+ "print\"W2=%f KW\"%W2;\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Q.10 Page No:39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Power=6.000000 KW\n",
+ "Power factor=0.654654\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from numpy import arctan\n",
+ "#Given\n",
+ "#In a particular measuremet\n",
+ "W1=5000; #Wattmeter-1 reading\n",
+ "W2=1000;#Wattmeter-2 reading\n",
+ "#Calculations\n",
+ "P=(W1+W2)*10**-3; #Power\n",
+ "pf=cos(arctan(math.sqrt(3)*(W2-W1)/(W1+W2))); #Power Factor\n",
+ "#display\n",
+ "print\"Power=%f KW\"%P;\n",
+ "print\"Power factor=%f\"%pf;\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Q.11 Page No:40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "W1=6.380724 watts\n",
+ "W2=50.037023 watts\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from numpy import arctan,arccos,cos\n",
+ "#Given\n",
+ "# A 3-ph motor has\n",
+ "VL=400.0; #Line voltage in volts\n",
+ "pf=0.75; #Power factor\n",
+ "P=26*10**3; #Power\n",
+ "#Calculations\n",
+ "ph=math.degrees(arccos(pf)); #Phase angle\n",
+ "IL=P/(math.sqrt(3)*VL*pf); #Line current\n",
+ "W1=VL*IL*cos(math.radians(30+ph))*10**(-3); #Wattmeter Reading-1\n",
+ "W2=VL*IL*cos(math.radians(30-ph))*10**(-3);#Wattmeter Reading-2\n",
+ "#Display\n",
+ "print\"W1=%f watts\"%W1;\n",
+ "print\"W2=%f watts\"%IL;\n",
+ "#Answers differing from Textbook"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Q.12 Page No:40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 34,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Power Factor=-0.804634\n"
+ ]
+ }
+ ],
+ "source": [
+ "#In 3-ph supply\n",
+ "#Given\n",
+ "W1=1200; #Wattmeter Reading -1\n",
+ "W2=300; #Wattmeter Reading-2\n",
+ "#Calculations\n",
+ "pf=arctan(math.sqrt(3)*(W2-W1)/(W1+W2)); #Power factor\n",
+ "#Display\n",
+ "print\"Power Factor=%f\"%pf;\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
generated by cgit (git 2.34.1) at 2025-04-03 19:39:34 +0000