{
 "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"
   ]
  }
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