path: root/Electrical_Circuit_Theory_And_Technology/chapter_33-checkpoint.ipynb

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   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h1>Chapter 33: Thevenin\u2019s and Norton\u2019s theorems</h1>"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 1, page no. 578</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine (a) the current flowing in the capacitor, and (b) the p.d. across the 150 k\u0006ohmresistor.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv  =  200;#  in  volts\n",
      "thetav  =  0;#  in  degrees\n",
      "R1  =  5000;#  in  ohm\n",
      "R2  =  20000;#  in  ohm\n",
      "R3  =  -1j*120000;#  in  ohm\n",
      "R4  =  150000;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #Initially  the  (150-i120)kohm\u0006  impedance  is  removed  from  the    circuit  as  shown  in  Figure  33.13.\n",
      " #Note  that,  to  find  the  current  in  the  capacitor,  \n",
      "    #only  the  capacitor  need  have  been  initially  removed  from  the  circuit.  \n",
      "    #However,  removing  each  of  the  components  from  the  branch  through  \n",
      "    #which  the  current  is  required  will  often  result  in  a  simpler  solution.  \n",
      " #From  Figure  33.13,\n",
      " #current,  I1  \n",
      "I1  =  V/(R1  +  R2)\n",
      " #The  open-circuit  e.m.f.  E  is  equal  to  the  p.d.  across  the  20  k\u0006ohm  resistor,  i.e.\n",
      "E  =  I1*R2\n",
      " #Removing  the  V1  source  gives  the  network  shown  in  Figure  33.14.\n",
      " #The  impedance,  z,  looking  in at  the  open-circuited  terminals  is  given  by\n",
      "z  =  R1*R2/(R1  +  R2)\n",
      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.15,  where  current  iL  is  given  by\n",
      "ZL  =  R3  +  R4\n",
      "IL  =  E/(ZL  +  z)\n",
      "ILmag  =  abs(IL)\n",
      " #current  flowing  in  the  capacitor\n",
      "Ic  =  ILmag\n",
      " #P.d.  across  the  150  kohm  resistor,\n",
      "Vr150  =  ILmag*R4\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n(a)the  current  flowing  in  the  capacitor  is  \",round(Ic*1000,2),\" mA\"\n",
      "print  \"\\n(b)  the  p.d.  across  the  150  ohm  resistance  is  \",round(Vr150,2),\"  V\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "(a)the  current  flowing  in  the  capacitor  is   0.82  mA\n",
        "\n",
        "(b)  the  p.d.  across  the  150  ohm  resistance  is   122.93   V\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 2, page no. 579</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the value of current I.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V1  =  20;#  in  volts\n",
      "V2  =  10;#  in  volts\n",
      "R1  =  2;#  in  ohm\n",
      "R2  =  1.5;#  in  ohm\n",
      "L  =  235E-6;#  in  Henry\n",
      "R4  =  3;#  in  ohm\n",
      "f  =  2000;#  in  Hz\n",
      "\n",
      " #calculation:\n",
      " #The  impedance  through  which  current  I  is  flowing  is  initially  removed  from  the  network,  as  shown  in  Figure  33.17.\n",
      " #From  Figure  33.17,\n",
      " #current,  I1  \n",
      "I1  =  (V1  -  V2)/(R1  +  R4)\n",
      " #the  open  circuit  e.m.f.  E\n",
      "E  =  V1  -  I1*R1\n",
      " #When  the  sources  of  e.m.f.  are  removed  from  the  circuit,  \n",
      "    #the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
      "z  =  R1*R4/(R1  +  R4)\n",
      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.18,  where  inductive  reactance,\n",
      "XL  =  2*math.pi*f*L\n",
      "R3  =  1j*XL\n",
      " #Hence  current\n",
      "I  =  E/(R2  +  R3  +  z)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  current  I  is  \",round(I.real,2),\"  +  (\",round(I.imag,2),\")i  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  current  I  is   2.7   +  ( -2.95 )i  A\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 3, page no. 580</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the power dissipated in the 48 ohm resistor\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv  =  50;#  in  volts\n",
      "thetav  =  0;#  in  degrees\n",
      "R1  =  -1j*400;#  in  ohm\n",
      "R2  =  300;#  in  ohm\n",
      "R3  =  144j;#  in  ohm\n",
      "R4  =  48;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #The  R3  and  R4  impedance  is  initially  removed  from  the  network  as  shown  in  Figure  33.20.\n",
      " #From  Figure  33.20,\n",
      " #current,  I\n",
      "i  =  V/(R1  +  R2)\n",
      " #the  open  circuit  e.m.f.  E\n",
      "E  =  i*R2\n",
      " #When  the  V  is  removed  from  the  circuit,  the  impedance,  z,  \u2018looking  in\u2019  at  the  break  is  given  by\n",
      "z  =  R1*R2/(R1  +  R2)\n",
      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.21  connected  to  R#  and  R4,\n",
      " #Hence  current\n",
      "I  =  E/(R4  +  R3  +  z)\n",
      "Imag  =  abs(I)\n",
      " #the  power  dissipated  in  the  48  ohm  resistor\n",
      "Pr48  =  R4*Imag**2\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  power  dissipated  in  the  48  ohm  resistor  is  \",round(Pr48,2),\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  power  dissipated  in  the  48  ohm  resistor  is   0.75   W"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 4, page no. 581</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the current flowing in the 80 ohm resistor.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V  =  100;#  in  volts\n",
      "R1  =  5;#  in  ohm\n",
      "R2  =  20;#  in  ohm\n",
      "R3  =  46;#  in  ohm\n",
      "R4  =  50;#  in  ohm\n",
      "R5  =  15;#  in  ohm\n",
      "R6  =  60;#  in  ohm\n",
      "R7  =  16;#  in  ohm\n",
      "R8  =  80;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      " #One  method  of  analysing  a  multi-branch  network  as  shown  in  Figure  33.22 \n",
      "    #is  to  use  Th\u00b4evenin\u2019s  theorem  on  one  part  of  the  network  at  a  time.  \n",
      "    #For  example,  the  part  of  the  circuit  to  the  left  of  AA  may  be  reduced  to  a  Th\u00b4evenin  equivalent  circuit.\n",
      " #From  Figure  33.23,\n",
      "E1  =  (R2/(R1  +  R2))*V\n",
      "z1  =  R1*R2/(R1  +  R2)\n",
      " #Thus  the  network  of  Figure  33.22  reduces  to  that  of  Figure  33.24.  \n",
      "    #The  part  of  the  network  shown  in  Figure  33.24  to  the  left  of  BB  may  be  reduced \n",
      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
      "E2  =  (R4/(R3  +  R4  +  z1))*E1\n",
      "z2  =  R4*(z1  +  R3)/(R4  +  z1  +  R3)\n",
      " #Thus  the  original  network  reduces  to  that  shown  in  Figure  33.25.  \n",
      "    #The  part  of  the  network  shown  in  Figure  33.25  to  the  left  of  CC  may  be  reduced  \n",
      "    #to  a  Th\u00b4evenin  equivalent  circuit,  where\n",
      "E3  =  (R6/(R5  +  R6  +  z2))*E2\n",
      "z3  =  R6*(z2  +  R5)/(R5  +  z2  +  R6)\n",
      " #Thus  the  original  network  reduces  to  that  of  Figure  33.26, \n",
      "    #from  which  the  current  in  the  80  ohm  resistor  is  given  by\n",
      "I  =  E3/(z3  +  R7  +  R8)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  current  flowing  in  the  80  ohm  resistor  is  \",round(I,2),\"  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  current  flowing  in  the  80  ohm  resistor  is   0.2   A"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 5, page no. 582</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine (a) the magnitude of the current flowing in a \t(3.75 + j11) ohmimpedance connected across terminals AB, and\n",
      "#(b) the magnitude of the p.d. across the (3.75 + j11) ohmimpedance.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv  =  24;#  in  volts\n",
      "thetav  =  0;#  in  degrees\n",
      "R1  =  -1j*3;#  in  ohm\n",
      "R2  =  4;#  in  ohm\n",
      "R3  =  3j;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #Current  I1  shown  in  Figure  33.27  is  given  by\n",
      "I1  =  V/(R1  +  R2  +  R3)\n",
      " #The  Th\u00b4evenin  equivalent  voltage,  i.e.,  the  open-circuit  voltage  across  terminals  AB,  is  given  by\n",
      "E  =  I1*(R2  +  R3)\n",
      " #When  the  voltage  source  is  removed,  the  impedance  z  \u2018looking  in\u2019  at  AB  is  given  by\n",
      "z  =  (R2  +  R3)*R1/(R1  +  R2  +  R3)\n",
      " #Thus  the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.28.\n",
      " #when  (3.75  +  i11)  ohm  impedance  connected  across  terminals  AB, \n",
      "    #the  current  I  flowing  in  the  impedance  is  given  by\n",
      "R  =  3.75  +  11j;#  in  ohms\n",
      "I  =  E/(R  +  z)\n",
      "Imag  =  abs(I)\n",
      " #the  p.d.  across  the(  3.75  +  i11)ohm  impedance.\n",
      "VR  =  I*R\n",
      "VRmag  =  abs(VR)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is  \",round(Imag,2),\"  A\"\n",
      "print  \"\\n  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is  \",round(VRmag,2),\"  V\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)  the  current  I  flowing  in  the  (3.75  +  i11)  impedance  is  given  by  is   3.0   A\n",
        "\n",
        "  (b)  the  magnitude  of  the  p.d.  across  the  impedance  is   34.86   V"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 6, page no. 583</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the current flowing in the capacitor of the network\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv  =  16.55;#  in  volts\n",
      "thetav  =  -22.62;#  in  degrees\n",
      "R1  =  4;#  in  ohm\n",
      "R2  =  1j*2;#  in  ohm\n",
      "R3  =  1j*6;#  in  ohm\n",
      "R4  =  3;#  in  ohm\n",
      "R5  =  5;#  in  ohm\n",
      "R6  =  -1*1j*8;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #The  capacitor  is  removed  from  branch  AB,  as  shown  in  Figure  33.30.\n",
      " #Impedance,  Z\n",
      "Z1  =  R3  +  R4  +  R5\n",
      "Z  =  R1  +  (Z1*R2/(R2  +  Z1))\n",
      "I1  =  V/Z\n",
      "I2  =  (R2/(R2  +Z1))*I1\n",
      " #The  open-circuit  voltage,  E\n",
      "E  =  I2*R5\n",
      " #If  the  voltage  source  is  removed  from  Figure  33.30,  the  impedance,  z,  \u2018looking  in\u2019  at  AB  is  given  by\n",
      "z  =  R5*((R1*R2/(R1  +  R2))  +  R3  +  R4)/(R5  +  ((R1*R2/(R1  +  R2))  +  R3  +  R4))\n",
      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.31,  \n",
      "    #where  the  current  flowing  in  the  capacitor,  I,  is  given  by\n",
      "I  =  E/(z  +  R6)\n",
      "Imag  =  abs(I)\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  current  flowing  in  the  capacitor  of  the  network  is  \",round(Imag,2),\"    A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  current  flowing  in  the  capacitor  of  the  network  is   0.43     A"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 7, page no. 584</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the power dissipated by a 2 ohm resistor connected across PQ.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv1  =  5;#  in  volts\n",
      "rv2  =  10;#  in  volts\n",
      "thetav1  =  45;#  in  degrees\n",
      "thetav2  =  0;#  in  degrees\n",
      "R1  =  8;#  in  ohm\n",
      "R2  =  5;#  in  ohm\n",
      "R3  =  3j;#  in  ohm\n",
      "R4  =  4;#  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",
      " #Current  I1  shown  in  Figure  33.32  is  given  by\n",
      "I1  =  V2/(R2  +  R3  +  R4)\n",
      " #Hence  the  voltage  drop  across  the  5  ohm\u0006  resistor  is  given  by  VX  is  in  the  direction  shown  in  Figure  33.32,\n",
      "Vx  =  I1*R2\n",
      " #The  open-circuit  voltage  E  across  PQ  is  the  phasor  sum  of  V1,  Vx  and  V2,  as  shown  in  Figure  33.33.\n",
      "E  =  V2  -  V1  -  Vx\n",
      " #The  impedance,  z,  \u2018looking  in\u2019  at  terminals  PQ  with  the  voltage  sources  removed  is  given  by\n",
      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.34  with  the  2  ohm  resistance  connected  across  terminals  PQ.\n",
      " #The  current  flowing  in  the  2  ohm\u0006  resistance  is  given  by\n",
      "R  =  2;#  in  ohms\n",
      "I  =  E/(z  +  R)\n",
      "Imag  =  abs(I)\n",
      " #power  P  dissipated  in  the  2  ohm\u0006  resistor  is  given  by\n",
      "Pr2  =  R*Imag**2\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  power  P  dissipated  in  the  2  ohm  resistor  is  \",round(Pr2,2),\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  power  P  dissipated  in  the  2  ohm  resistor  is   0.07   W"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 8, page no. 585</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the current flowing in the capacitor, and its direction\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv  =  30;#  in  volts\n",
      "thetav  =  0;#  in  degrees\n",
      "R1  =  15;#  in  ohm\n",
      "R2  =  40;#  in  ohm\n",
      "R3  =  20j;#  in  ohm\n",
      "R4  =  20;#  in  ohm\n",
      "R5  =  5j;#  in  ohm\n",
      "R6  =  5;#  in  ohm\n",
      "R7  =  -1j*25;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #The  R7\u0006  is  initially  removed  from  the  network,  as  shown  in  Figure  33.36\n",
      "Z1  =  R1\n",
      "Z2  =  R2\n",
      "Z3  =  R3  +  R4\n",
      "Z4  =  R5  +  R6\n",
      " #P.d.  between  A  and  C,\n",
      "Vac  =  (Z1/(Z1  +  Z4))*V\n",
      " #P.d.  between  B  and  C,\n",
      "Vbc  =  (Z2/(Z2  +  Z3))*V\n",
      " #Assuming  that  point  A  is  at  a  higher  potential  than  point  B,  then  the  p.d.  between  A  and  B  is\n",
      "Vab  =  Vac  -  Vbc\n",
      " #the  open-circuit  voltage  across  AB  is  given  by\n",
      "E  =  Vab\n",
      " #Point  C  is  at  a  potential  of  V  .  Between  C  and  A  is  a  volt  drop  of  Vac.  Hence  the  voltage  at  point  A  is\n",
      "Va  =  V  -  Vac\n",
      " #Between  points  C  and  B  is  a  voltage  drop  of  Vbc.  Hence  the  voltage  at  point  B\n",
      "Vb  =  V  -  Vbc\n",
      " #Replacing  the  V  source  with  a  short-circuit  (i.e.,  zero  internal  impedance) \n",
      "    #gives  the  network  shown  in  Figure  33.37(a).  The  network  is  shown  redrawn  in  Figure  33.37(b)  \n",
      "    #and  simplified  in  Figure  33.37(c).  Hence  the  impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
      "z  =  Z1*Z4/(Z1  +  Z4)  +  Z2*Z3/(Z2  +  Z3)\n",
      " #The  Th\u00b4evenin  equivalent  circuit  is  shown  in  Figure  33.38,  where  current  I  is  given  by\n",
      "I  =  E/(z  +  R7)\n",
      "Imag  =  abs(I)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(Imag,2),\"  A  in  direction  from  B  to  A.\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "    the  current  flowing  in  the  capacitor  is   0.13   A  in  direction  from  B  to  A."
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 9, page no. 589</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the value of current I in the circuit\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V  =  5;#  in  volts\n",
      "R1  =  2;#  in  ohm\n",
      "R2  =  3;#  in  ohm\n",
      "R3  =  -1j*3;#  in  ohm\n",
      "R4  =  2.8;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #The  branch  containing  the  R4  is  short-circuited,  as  shown  in  Figure  33.48.\n",
      " #The  R2  in  parallel  with  a  short-circuit  is  the  same  as  R2  in  parallel  with  0  ohm \n",
      "    #giving  an  equivalent  impedance  of\n",
      "Z1  =  R2*0/(R3  +  0)\n",
      " #Hence  the  network  reduces  to  that  shown  in  Figure  33.49,  where\n",
      "Isc  =  V/R1\n",
      " #If  the  Voltage  source  is  removed  from  the  network  the  input  impedance,  z,  \u2018looking-in\u2019  \n",
      "    #at  a  break  made  in  AB  of  Figure  33.48  gives\n",
      "z  =  R1*R2/(R1  +  R2)\n",
      " #The  Norton  equivalent  network  is  shown  in  Figure  33.51,  where  current  I  is  given  by\n",
      "I  =  (z/(z  +  R4  +  R3))*Isc\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n    the  current  flowing  in  the  capacitor  is  \",round(I.real,2),\"  +  (\", round(I.imag,2),\")i  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "    the  current  flowing  in  the  capacitor  is   0.48   +  ( 0.36 )i  A"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 10, page no. 589</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V1  =  20;#  in  volts\n",
      "V2  =  10;#  in  volts\n",
      "R1  =  2;#  in  ohm\n",
      "R2  =  1.5;#  in  ohm\n",
      "R3  =  2.95j;#  in  ohm\n",
      "R4  =  3;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #The  inductive  branch  is  initially  short-circuited,  as  shown  in  Figure  33.53.\n",
      " #From  Figure  33.53,\n",
      "I1  =  V1/R1\n",
      "I2  =  V2/R4\n",
      "Isc  =  I1  +  I2\n",
      " #If  the  voltage  sources  are  removed,  the  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  in  AB  is  given  by\n",
      "z  =  R1*R4/(R1  +  R4)\n",
      " #The  Norton  equivalent  network  is  shown  in  Figure  33.54,  where  current  I  is  given  by\n",
      "I  =  (z/(z  +  R2  +  R3))*Isc\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n    the  current  flowing  in  the  inductive  branch  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "    the  current  flowing  in  the  inductive  branch  is   2.7   +  ( -2.95 )i  A"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 11, page no. 590</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the current flowing in the inductive branch by using Norton\u2019s theorem\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V  =  10;#  in  volts\n",
      "R1  =  1;#  in  ohm\n",
      "R2  =  4;#  in  ohm\n",
      "R3  =  4;#  in  ohm\n",
      "R4  =  -2j;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      "Isc = V/R3\n",
      "z = 1/(1/R2 + 1/R3 + 1/R4)\n",
      "I = (z/(R1 + z))*Isc\n",
      "pd1 = abs(I)*R1\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n the magnitude of the p.d. across the 1 ohm resistor is \",round(pd1, 2),\"V\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        " the magnitude of the p.d. across the 1 ohm resistor is  1.58 V"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 12, page no. 591</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the power dissipated in a 5 ohm resistor connected between A and B.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv  =  20;#  in  volts\n",
      "thetav  =  0;#  in  degrees\n",
      "R1  =  2;#  in  ohm\n",
      "R2  =  4;#  in  ohm\n",
      "R3  =  3j;#  in  ohm\n",
      "R4  =  -1j*3;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #Terminals  AB  are  initially  short-circuited,  as  shown  in  Figure  33.61.\n",
      " #The  circuit  impedance  Z  presented  to  the  voltage  source  is  given  by\n",
      "Z  =  R1  +  R4*(R2  +  R3)/(R2  +  R3  +  R4)\n",
      " #Thus  current  I  in  Figure  33.61  is  given  by\n",
      "I  =  V/Z\n",
      "Isc  =  ((R2  +  R3)/(R2  +  R3  +  R4))*I\n",
      " #Removing  the  voltage  source  of  Figure  33.60  gives  the  network  Figure  33.62  of  Figure  33.62. \n",
      "    #Impedance,  z,  \u2018looking  in\u2019  at  terminals  AB  is  given  by\n",
      "z  =  R4  +  R1*(R2  +  R3)/(R2  +  R3  +  R1)\n",
      " #The  Norton  equivalent  network  is  shown  in  Figure  33.63.\n",
      "R  =  5;#  in  ohms\n",
      " #Current  IL\n",
      "IL  =  (z/(z  +  R))*Isc\n",
      "ILmag  =  abs(IL)\n",
      " #the  power  dissipated  in  the  5  ohm  resistor  is\n",
      "Pr5  =  R*ILmag**2\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  power  dissipated  in  the  5  ohm  resistor  is  \",round(Pr5,2),\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  power  dissipated  in  the  5  ohm  resistor  is   22.54   W"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 13, page no. 592</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the magnitude of the current flowing in a 2 ohm resistor connected across PQ.\n",
      "from __future__ import division\n",
      "import math\n",
      "import numpy\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv1  =  5;#  in  volts\n",
      "rv2  =  10;#  in  volts\n",
      "thetav1  =  45;#  in  degrees\n",
      "thetav2  =  0;#  in  degrees\n",
      "R1  =  8;#  in  ohm\n",
      "R2  =  5;#  in  ohm\n",
      "R3  =  3j;#  in  ohm\n",
      "R4  =  4;#  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",
      " #Terminals  PQ  are  initially  short-circuited,  as  shown  in  Figure  33.65.\n",
      " #Currents  I1  and  I2  are  shown  labelled.  Kirchhoff\u2019s  laws  are  used.\n",
      " #For  loop  ABCD,  and  moving  anticlockwise,\n",
      " #I1*(R2  +  R3  +  R4)  +  I2*(R3  +  R4)  =  V2\n",
      " #For  loop  DPQC,  and  moving  clockwise,\n",
      " #R2*I1  -  R1*I2  =  V2  -  V1\n",
      " #Solving  Equations  by  using  determinants  gives\n",
      "d1  =  [[V2,  (R3  +  R4)],[(V2  -  V1),  -1*R1]]\n",
      "D1  =  numpy.linalg.det(d1)\n",
      "d2  =  [[(R2  +  R3  +  R4),  V2],[R2,  (V2  -  V1)]]\n",
      "D2  =  numpy.linalg.det(d2)\n",
      "d  =  [[(R2  +  R3  +  R4),  (R3  +  R4)],[R2,  -1*R1]]\n",
      "D  =  numpy.linalg.det(d)\n",
      "I1  =  D1/D\n",
      "I2  =  D2/D\n",
      " #the  short-circuit  current  Isc\n",
      "Isc  =  I2\n",
      " #The  impedance,  z,  \u2018looking  in\u2019  at  a  break  made  between  P  and  Q  is  given  by\n",
      "z  =  R1  +  R2*(R3  +  R4)/(R2  +  R3  +  R4)\n",
      " #The  Norton  equivalent  circuit  is  shown  in  Figure  33.66,  where  current  I  is  given  by\n",
      "R  =  2;#in  ohm\n",
      "I  =  (z/(z  +  R))*Isc\n",
      "Imag  =  abs(I)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  magnitude  of  the  current  flowing  5  ohm  resistor  is  \",round(Imag,2),\"  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  magnitude  of  the  current  flowing  5  ohm  resistor  is   0.19   A"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 15, page no. 595</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the magnitude of the current flowing in the \t(1.8 + j4) ohm impedance connected between terminals A and B\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "E1  =  12;#  in  volts\n",
      "E2  =  24;#  in  volts\n",
      "Z1  =  3;#  in  ohm\n",
      "Z2  =  2;#  in  ohm\n",
      "R1  =  4j;#  in  ohm\n",
      "R2  =  1.8;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      "Z3  =  R1  +  R2\n",
      " #For  the  branch  containing  the  E1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
      "Isc1  =  E1/Z1\n",
      " #For  the  branch  containing  the  E2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
      "Isc2  =  E2/Z2\n",
      " #Thus  Figure  33.73  shows  a  network  equivalent  to  Figure  33.72.  From  Figure  33.73,  \n",
      "    #the  total  short-circuit  current\n",
      "Isc  =  Isc1  +  Isc2\n",
      " #the  total  impedance  is  given  by\n",
      "z  =  Z1*Z2/(Z1  +  Z2)\n",
      " #Thus  Figure  33.73  simplifies  to  Figure  33.74.\n",
      " #The  open-circuit  voltage  across  AB  of  Figure  33.74,  E\n",
      "E  =  Isc*z\n",
      " #the  impedance  \u2018looking  in\u2019  at  AB,is  z\n",
      " #the  Th\u00b4evenin  equivalent  circuit  is  as  shown  in  Figure  33.75.\n",
      "R  =  1.8  +  4j;#  in  ohm\n",
      " #when  R  impedance  is  connected  to  terminals  AB  of  Figure  33.75,  the  current  I  flowing  is  given  by\n",
      "I  =  E/(z  +  R)\n",
      "Imag  =  abs(I)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is  \",round(Imag,2),\"  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  magnitude  of  the  current  flowing  (1.8  +  i4)  ohm  resistor  is   3.84   A"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 16, page no. 596</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the magnitude of the current flowing in the capacitive branch connected to terminals AB.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V1  =  5;#  in  volts\n",
      "V2  =  10;#  in  volts\n",
      "i  =  0.001;#  in  Amperes\n",
      "R1  =  1000;#  in  ohm\n",
      "R2  =  4000;#  in  ohm\n",
      "R3  =  2000;#  in  ohm\n",
      "R4  =  200;#  in  ohm\n",
      "R5  =  -1j*4000;#  in  ohm\n",
      "\n",
      " #calculation:  \n",
      " #For  the  branch  containing  the  V1  source,  conversion  to  a  Norton  equivalent  network  gives\n",
      "Isc1  =  V1/R1\n",
      "z1  =  R1\n",
      " #For  the  branch  containing  the  V2  source,  conversion  to  a  Norton  equivalent  circuit  gives\n",
      "Isc2  =  V2/R2\n",
      "z2  =  R2\n",
      " #Thus  the  circuit  of  Figure  33.76  converts  to  that  of  Figure  33.77.\n",
      " #The  above  two  Norton  equivalent  networks  shown  in  Figure  33.77  may  be  combined,  \n",
      "    #since  the  total  short-circuit  current  is\n",
      "Isc  =  Isc1  +  Isc2\n",
      " #the  total  impedance  is  given  by\n",
      "Z1  =  z1*z2/(z1  +  z2)\n",
      " #Both  of  the  Norton  equivalent  networks  shown  in  Figure  33.78  may  be  converted  to  Th\u00b4evenin  equivalent  circuits. \n",
      "    #Open-circuit  voltage  across  CD  is\n",
      "Ecd  =  Isc*Z1\n",
      " #the  impedance  \u2018looking  in\u2019  at  CD  is  Z1\n",
      " #Open-circuit  voltage  across  EF\n",
      "Eef  =  i*R3\n",
      " #the  impedance  \u2018looking  in\u2019  Figure  33.79  at  EF\n",
      "Z2  =  R3\n",
      " #Thus  Figure  33.78  converts  to  Figure  33.79.\n",
      " #Combining  the  two  Th\u00b4evenin  circuits  gives  e.m.f.\n",
      "E  =  Ecd  -  Eef\n",
      " #impedance  z\n",
      "z  =  Z1  +  Z2\n",
      " #the  Th\u00b4evenin  equivalent  circuit  for  terminals  AB  of  Figure  33.76  is  as  shown  in  Figure  33.80.\n",
      "Z3  =  R4  +  R5\n",
      " #If  an  impedance  Z3  is  connected  across  terminals  AB,  then  the  current  I  flowing  is  given  by\n",
      "I  =  E/(z  +  Z3)\n",
      "Imag  =  abs(I)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  current  in  the  capacitive  branch  is  \",  Imag*1000,\"mA\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  current  in  the  capacitive  branch  is   0.8 mA"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 17, page no. 597</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate the power dissipated in a \t(600 - j800) ohm impedance connected between A and B\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V  =  5;#  in  volts\n",
      "i  =  0.004;#  in  Amperes\n",
      "R1  =  2000;#  in  ohm\n",
      "R2  =  1000j;#  in  ohm\n",
      "\n",
      "#calculation:  \n",
      " #Converting  the  Th\u00b4evenin  circuit  to  a  Norton  network  gives\n",
      "Isc1  =  V/R2\n",
      " #Thus  Figure  33.81  converts  to  that  shown  in  Figure  33.82.  \n",
      "    #The  two  Norton  equivalent  networks  may  be  combined,  giving\n",
      "Isc  =  Isc1  +  i\n",
      "z  =  R1*R2/(R1  +  R2)\n",
      " #This  results  in  the  equivalent  network  shown  in  Figure  33.83. \n",
      "    #Converting  to  an  equivalent  Th\u00b4evenin  circuit  gives  open  circuit  e.m.f.  across  AB,\n",
      "E  =  Isc*z\n",
      " #Thus  the  The\u00b4venin  equivalent  circuit  is  as  shown  in  Figure  33.84.\n",
      "R  =  600  -  800j;#  in  ohms\n",
      " #When  a  R  impedance  is  connected  across  AB,  the  current  I  flowing  is  given  by\n",
      "I  =  E/(z  +  R)\n",
      "Imag  =  abs(I)\n",
      " #the  power  dissipated  in  the  R  resistor  is\n",
      "PR  =  R.real*Imag**2\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is  \",round(PR*1000,2),\"mW\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  power  dissipated  in  the  (600  -  i800)  ohm  resistor  is   19.68 mW"
       ]
      }
     ],
     "prompt_number": 16
    }
   ],
   "metadata": {}
  }
 ]
}