path: root/Electrical_Circuit_Theory_And_Technology/chapter_32.ipynb

{
 "metadata": {
  "name": "",
  "signature": "sha256:5db7f11d16032beee814462a087253a6b78ac4c2f805e28898833a5c147fee39"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h1>Chapter 32: The superposition theorem</h1>"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 1, page no. 564</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv1  =  100;#  in  volts\n",
      "rv2  =  50;#  in  volts\n",
      "thetav1  =  0;#  in  degrees\n",
      "thetav2  =  90;#  in  degrees\n",
      "r1  =  25;#  in  ohm\n",
      "R  =  20;#  in  ohm\n",
      "r2  =  10;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
      " #The  circuit  diagram  is  shown  in  Figure  32.7.  Following  the  above  procedure:\n",
      " #The  network  is  redrawn  with  the  50/_90\u00b0  V  source  removed  as  shown  in  Figure  32.8\n",
      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.8.\n",
      "I1  =  V1/(r1  +  r2*R/(R  +  r2))\n",
      "I2  =  (r2/(r2  +  R))*I1\n",
      "I3  =  (R/(r2  +  R))*I1\n",
      " #The  network  is  redrawn  with  the  100/_0\u00b0  V  source  removed  as  shown  in  Figure  32.9\n",
      " #Currents  I4,  I5  and  I6  are  labelled  as  shown  in  Figure  32.9.\n",
      "I4  =  V2/(r2  +  r1*R/(r1  +  R))\n",
      "I5  =  (r1/(r1  +  R))*I4\n",
      "I6  =  (R/(r1  +  R))*I4\n",
      " #Figure  32.10  shows  Figure  32.9  superimposed  on  Figure  32.8,  giving  the  currents  shown.\n",
      " #Current  in  the  20  ohm  load,\n",
      "I20  =  I2  +  I5\n",
      " #Current  in  the  100/_0\u00b0  V  source\n",
      "IV1  =  I1  -  I6\n",
      " #Current  in  the  50/_90\u00b0  V  source\n",
      "IV2  =  I4  -  I3\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)current  in  the  20  ohm  load  is  \",round(I20.real,2),\"  +  (\",round(I20.imag,2),\")i  A\"\n",
      "print  \"\\n  (b)Current  in  the  100/_0deg  V  source  is  \",round(IV1.real,2),\"  +  (\",round(IV1.imag,2),\")i  A\"\n",
      "print  \"\\n  (b)Current  in  the  50/_90deg  V  source  is  \",round(IV2.real,2),\"  +  (\",round(IV2.imag,2),\")i  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)current  in  the  20  ohm  load  is   1.05   +  ( 1.32 )i  A\n",
        "\n",
        "  (b)Current  in  the  100/_0deg  V  source  is   3.16   +  ( -1.05 )i  A\n",
        "\n",
        "  (b)Current  in  the  50/_90deg  V  source  is   -2.11   +  ( 2.37 )i  A"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 2, page no. 566</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V1  =  12;#  in  volts\n",
      "V2  =  20;#  in  volts\n",
      "R1  =  5;#  in  ohm\n",
      "R2  =  4;#  in  ohm\n",
      "R3  =  2.5;#  in  ohm\n",
      "R4  =  6;#  in  ohm\n",
      "R5  =  2;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      " #Removing  the  20  V  source  gives  the  network  shown  in  Figure  32.12.\n",
      " #Currents  I1  and  I2  are  shown  labelled  in  Figure  32.12\n",
      "Re1  =  (R4*R5/(R4  +  R5))  +  R3\n",
      "Re2  =  Re1*R2/(Re1    +  R2)  +  R1\n",
      "I1  =  V1/Re2\n",
      "I2  =  (R2/(Re1  +  R2))*I1\n",
      " #Removing  the  12  V  source  from  the  original  network  gives  the  network  shown  in  Figure  32.14.\n",
      " #Currents  I3,  I4  and  I5  are  shown  labelled  in  Figure  32.14.\n",
      "Re3  =  (R1*R2/(R1  +  R2))  +  R3\n",
      "Re4  =  Re3*R4/(Re3  +  R4)  +  R5\n",
      "I3  =  V2/Re4\n",
      "I4  =  (R4/(Re3  +  R4))*I3\n",
      "I5  =  (R1/(R1  +  R2))*I4\n",
      " #Superimposing  Figure  32.14  on  Figure  32.12  shows  that  the  current  flowing  in  the  4  ohm  resistor  is  given  by\n",
      "Ir4  =  I5  -  I2\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\ncurrent  in  the  4  ohm  resistor  of  the  network  is  \",round(Ir4,2),\"  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "current  in  the  4  ohm  resistor  of  the  network  is   0.48   A"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 3, page no. 567</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv1  =  30;#  in  volts\n",
      "rv2  =  30;#  in  volts\n",
      "thetav1  =  45;#  in  degrees\n",
      "thetav2  =  -45;#  in  degrees\n",
      "R1  =  4;#  in  ohm\n",
      "R2  =  4;#  in  ohm\n",
      "R3  =  1j*3;#  in  ohm\n",
      "R4  =  -1j*10;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
      " #The  network  is  redrawn  with  V2  removed,  as  shown  in  Figure  32.17.\n",
      " #Current  I1  and  I2  are  shown  in  Figure  32.17.  From  Figure  32.17,\n",
      "Re1  =  R4*(R2  +  R3)/(R4  +  R3  +  R2)\n",
      "Re2  =  Re1  +  R1\n",
      " #current\n",
      "I1  =  V1/Re2\n",
      "I2  =  (R4/(R2  +  R3  +  R4))*I1\n",
      " #The  original  network  is  redrawn  with  V1  removed,  as  shown  in  Figure  32.18\n",
      " #Currents  I3  and  I4  are  shown  in  Figure  32.18.  From  Figure  32.18,\n",
      "Re3  =  R1*(R2  +  R3)/(R1  +  R3  +  R2)\n",
      "Re4  =  Re3  +  R4\n",
      "I3  =  V2/Re4\n",
      "I4  =  (R1/(R2  +  R3  +  R1))*I3\n",
      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
      "Ir4i3  =  I2  -  I4\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"current  in  (4  +  i3)  ohm  impedance  of  the  network  is  \",round(Ir4i3.real,2),\"  +  (\",round(  Ir4i3.imag,2),\")i  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "current  in  (4  +  i3)  ohm  impedance  of  the  network  is   2.15   +  ( 0.42 )i  A\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 4, page no. 568</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "E1  =  5  +  0j;#  in  volts\n",
      "E2  =  2  +  4j;#  in  volts\n",
      "Z1  =  3  +  4j;#  in  ohm\n",
      "Z2  =  2  -  5j;#  in  ohm\n",
      "Z3  =  6  +  8j;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      " #The  original  network  is  redrawn  with  E2  removed,  as  shown  in  Figure  32.20.\n",
      " #Currents  I1,  I2  and  I3  are  labelled  as  shown  in  Figure  32.20.\n",
      "Ze1  =  Z3*Z2/(Z3  +  Z2)\n",
      "Ze2  =  Ze1  +  Z1\n",
      " #current\n",
      "I1  =  E1/Ze2\n",
      "I2  =  (Z2/(Z3  +  Z2))*I1\n",
      "I3  =  (Z3/(Z3  +  Z2))*I1\n",
      " #The  original  network  is  redrawn  with  E1  removed,  as  shown  in  Figure  32.22\n",
      " #Currents  I4,  I5  and  I6  are  shown  labelled  in  Figure  32.22  \n",
      "    #with  I4  flowing  away  from  the  positive  terminal  of  the  E2  source.\n",
      "Ze3  =  Z3*Z1/(Z3  +  Z1)\n",
      "Ze4  =  Ze3  +  Z2\n",
      "I4  =  E2/Ze4\n",
      "I5  =  (Z1/(Z3  +  Z1))*I4\n",
      "I6  =  (Z3/(Z3  +  Z1))*I4\n",
      " #If  the  network  of  Figure  32.18  is  superimposed  on  the  network  of  Figure  32.17,  \n",
      "    #it  can  be  seen  that  the  current  in  the  (4+i3)  ohm  impedance  is  given  by\n",
      "i1  =  I1  +  I6\n",
      "i2  =  I3  +  I4\n",
      "i3  =  I2  -  I5\n",
      " #magnitude\n",
      "i1mag  =  abs(i1)\n",
      "i2mag  =  abs(i2)\n",
      "E1mag  =  abs(E1)\n",
      "E2mag  =  abs(E2)\n",
      " #phase\n",
      "phi1  =  cmath.phase(complex(i1.real,i1.imag))\n",
      "phi2  =  cmath.phase(complex(i2.real,i2.imag))\n",
      " #voltage  across  the(6  +  i8)  ohm  impedance\n",
      "V6i8  =  i3*Z3\n",
      "V6i8m  =  abs(V6i8)\n",
      " #power\n",
      "P  =  (E1mag*i1mag*math.cos(phi1))  +  (E2mag*i2mag*math.cos(phi2  -  cmath.phase(complex(E2.real,E2.imag))))\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n(a)currents  are: \\n \",round(i1.real,2),\"  +  (\",round(  i1.imag,2),\")i  A, \\n \",round(i2.real,2),\"  +  (\",round(i2.imag,2),\")i  A \\n and  \",round(i3.real,2),\"  +  (\",round(i3.imag,2),\")i  A\"\n",
      "print  \"\\n(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is  \",round(V6i8m,2),\"  V\"\n",
      "print  \"\\n(c)the  total  active  power  delivered  to  the  network  is  \",round(P,2),\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "(a)currents  are: \n",
        "  0.57   +  ( 0.62 )i  A, \n",
        "  0.56   +  ( 1.33 )i  A \n",
        " and   0.01   +  ( -0.71 )i  A\n",
        "\n",
        "(b)current  in  the  (6  +  i8)  ohm  resistor  of  the  network  is   7.09   V\n",
        "\n",
        "(c)the  total  active  power  delivered  to  the  network  is   9.29   W"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 5, page no. 571</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv1  =  50;#  in  volts\n",
      "rv2  =  30;#  in  volts\n",
      "thetav1  =  0;#  in  degrees\n",
      "thetav2  =  90;#  in  degrees\n",
      "R1  =  20;#  in  ohm\n",
      "R2  =  5;#  in  ohm\n",
      "R3  =  -1j*3;#  in  ohm\n",
      "R4  =  8;#  in  ohm\n",
      "R5  =  8;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      " #voltage\n",
      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
      " #The  network  is  redrawn  with  the  V2  source  removed,  as  shown  in  Figure  32.26.\n",
      " #Currents  I1  to  I5  are  shown  labelled  in  Figure  32.26.  \n",
      " #current\n",
      "Re1  =  R4*R5/(R5  +  R4)  +  R3\n",
      "Re2  =  Re1*R2/(R2  +  Re1)\n",
      "I1  =  V1/(Re2  +  R1)\n",
      "I2  =  (Re1/(R2  +  Re1))*I1\n",
      "I3  =  (R2/(Re1  +  R2))*I1\n",
      "I4  =  (R4/(R4  +  R5))*I3\n",
      "I5  =  I3  -  I4\n",
      " #The  original  network  is  redrawn  with  the  V1  source  removed,  as  shown  in  Figure  32.27.\n",
      " #Currents  I6  to  I10  are  shown  labelled  in  Figure  32.27\n",
      "Re3  =  R1*R2/(R1  +  R2)\n",
      "Re4  =  Re3  +  R3\n",
      "Re5  =  Re4*R4/(Re4  +  R4)\n",
      "Re6  =  Re5    +  R5\n",
      "I6  =  V2/Re6\n",
      "I7  =  (Re4/(Re4  +  R4))*I6\n",
      "I8  =  (R4/(Re4  +  R4))*I6\n",
      "I9  =  (R1/(R1  +  R2))*I8\n",
      "I10  =  (R2/(R1  +  R2))*I8\n",
      " #current  flowing  in  the  capacitor  is  given  by\n",
      "Ic  =  I3  -  I8\n",
      " #magnitude  of  the  current  in  the  capacitor\n",
      "Icmag  =  abs(Ic)\n",
      "\n",
      "i1  =  I2  +  I9\n",
      "i1mag  =  abs(i1)\n",
      " #magnitude  of  the  p.d.  across  the  5  ohm  resistance  is  given  by\n",
      "Vr5m  =  i1mag*R2\n",
      " #Active  power  dissipated  in  the  20  ohm  resistance  is  given  by\n",
      "i2  =  I1  -  I10\n",
      "i2mag  =  abs(i2)\n",
      "phii2  =  cmath.phase(complex(i2.real,i2.imag))\n",
      "Pr20  =  R1*(i2mag)**2\n",
      " #Active  power  developed  by  the  V1\n",
      "P1  =  rv1*i2mag*math.cos(phii2)\n",
      " #Active  power  developed  by  V2  source\n",
      "i3  =  I6  -  I5\n",
      "i3mag  =  abs(i3)\n",
      "phii3  =  cmath.phase(complex(i3.real,i3.imag))\n",
      "P2  =  rv2*i3mag*math.cos(phii3  -  (thetav2*math.pi/180))\n",
      " #Total  power  developed\n",
      "P  =  P1  +  P2\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
      "print  \"\\n(b)  the  p.d.  across  the  5  ohm  resistance  is  \",round(Vr5m,2),\"  V\"\n",
      "print  \"\\n(c)the  active  power  dissipated  in  the  20  ohm  resistance  is  \",round(Pr20,0),\"  W\"\n",
      "print  \"\\n(d)the  total  active  power  taken  from  the  supply  is  \",round(P,1),\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "(a)the  magnitude  of  the  current  flowing  in  the  capacitor  is   2.11   A\n",
        "\n",
        "(b)  the  p.d.  across  the  5  ohm  resistance  is   5.85   V\n",
        "\n",
        "(c)the  active  power  dissipated  in  the  20  ohm  resistance  is   111.0   W\n",
        "\n",
        "(d)the  total  active  power  taken  from  the  supply  is   191.9   W"
       ]
      }
     ],
     "prompt_number": 1
    }
   ],
   "metadata": {}
  }
 ]
}