path: root/Electrical_Circuit_Theory_And_Technology/chapter_36-checkpoint.ipynb

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   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h1>Chapter 36: Complex Waveforms</h1>"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 1, page no. 643</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#(a) Write down an expression to represent voltage v.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V = 240; # in Volts\n",
      "f = 50; # in Hz\n",
      "x = 0.2;\n",
      "phi3 = 3*math.pi/4; # in Rad\n",
      "\n",
      " #calculation:\n",
      "Vamp = V*2**0.5\n",
      "w = 2*math.pi*f\n",
      "T = 1/f\n",
      "V3 = Vamp*x\n",
      "f3 = 3*f\n",
      "w3 = 3*w\n",
      "\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  voltage, V =\",round(Vamp,1),\"sin(\",round(w,1),\"t) + \",round(V3,1),\"sin(\", round(w3,1),\"t - \", round(phi3,1),\") volts\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  voltage, V = 339.4 sin( 314.2 t) +  67.9 sin( 942.5 t -  2.4 ) volts"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 3, page no. 648</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the rms value of the current waveform\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "A1  =  0.100;#  in  amperes\n",
      "A3  =  0.020;#  in  amperes\n",
      "A5  =  0.010;#  in  amperes\n",
      "\n",
      " #calculation:\n",
      " #the  rms  value  of  current  is  given  by\n",
      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  rms  value  of  current  is  \",round(Irms*1000,2),\" mA\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  rms  value  of  current  is   72.46  mA"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 4, page no. 649</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine for the voltage, (a) the rms value, (b) the mean value and (c) the form factor.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "A1  =  10;#  in  volts\n",
      "A3  =  3;#  in  volts\n",
      "A5  =  2;#  in  volts\n",
      "\n",
      "#calculation:\n",
      " #the  rms  value  of  voltage  is  given  by\n",
      "Vrms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
      " #the  mean  value  of  voltage  is  given  by\n",
      " #x  =  wt\n",
      "Vav  =  (1/math.pi)*((10 + 1 + 2/5)-(-10 - 1 - 2/5))\n",
      " #form  factor  is  given  by\n",
      "ff  =  Vrms/Vav\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
      "print  \"\\n  (b)the  mean  value  of  voltage  is  \",round(Vav,2),\"  V\"\n",
      "print  \"\\n  (c)form  factor  is  \",round(ff,3),\"  \""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)the  rms  value  of  voltage  is   7.52   V\n",
        "\n",
        "  (b)the  mean  value  of  voltage  is   7.26   V\n",
        "\n",
        "  (c)form  factor  is   1.036   "
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 5, page no. 649</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#(a) Determine the rms value of the fundamental and each harmonic. \n",
      "#(b) Write down an expression to represent the complex voltage waveform if the frequency of the fundamental is 31.83 Hz.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V  =  240;#  in  volts\n",
      "x  =  0.3;#  for  third  harmonic\n",
      "y  =  0.1;#  for  fifth  harmonic\n",
      "f  =  31.83;#  in  Hz\n",
      "\n",
      " #calculation:\n",
      " #V3  =  x*V1\n",
      " #V5  =  y*V1\n",
      " #the  rms  value  of  the  fundamental,\n",
      "V1  =  ((V**2)/(1  +  x**2  +  y**2))**0.5\n",
      " #Rms  value  of  the  third  harmonic\n",
      "V3  =  x*V1\n",
      " #the  rms  value  of  the  fifth  harmonic,\n",
      "V5  =  y*V1\n",
      " #Maximum  value  of  the  fundamental,\n",
      "V1m  =  V1*2**0.5\n",
      " #Maximum  value  of  the  third  harmonic,\n",
      "V3m  =  V3*2**0.5\n",
      " #Maximum  value  of  the  fifth  harmonic,\n",
      "V5m  =  V5*2**0.5\n",
      "w  =  2*math.pi*f\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"v  =  \",round(V1m,2),\"sin\",round(w,2),\"t  +  \",round(V3m,2),\"sin\",round((3*w),2),\"t  +  \",round(V5m,2),\"sin\",round((5*w),2),\"t  Volts\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "v  =   323.62 sin 199.99 t  +   97.08 sin 599.98 t  +   32.36 sin 999.97 t  Volts\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 6, page no. 652</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the average power in a 20 ohm resistance\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "A1  =  12;#  in  amperes\n",
      "A3  =  5;#  in  amperes\n",
      "A5  =  2;#  in  amperes\n",
      "R  =  20;#  in  ohms\n",
      "\n",
      "#calculation:\n",
      " #rms  current\n",
      "Irms  =  ((A1**2  +  A3**2  +  A5**2)/2)**0.5\n",
      " #average  power\n",
      "P  =  R*Irms**2\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  average  power  \",P,\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  average  power   1730.0   W"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 7, page no. 652</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) the total active power supplied to the circuit, and (b) the overall power factor\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "Ia1  =  2;#  in  amperes\n",
      "Ia3  =  0.3;#  in  amperes\n",
      "Ia5  =  0.1;#  in  amperes\n",
      "Va1  =  60;#  in  volts\n",
      "Va3  =  15;#  in  volts\n",
      "Va5  =  10;#  in  volts\n",
      "Phii1  =  -1*math.pi/6;#  in  radians\n",
      "Phii3  =  -1*math.pi/12;#  in  radians\n",
      "Phii5  =  -8*math.pi/9;#  in  radians\n",
      "Phiv1  =  0;#  in  radians\n",
      "Phiv3  =  math.pi/4;#  in  radians\n",
      "Phiv5  =  -1*math.pi/2;#  in  radians\n",
      "\n",
      "\n",
      " #calculation:\n",
      " #rms  values;\n",
      "I1  =  Ia1/(2**0.5);#  in  amperes\n",
      "I3  =  Ia3/(2**0.5);#  in  amperes\n",
      "I5  =  Ia5/(2**0.5);#  in  amperes\n",
      "V1  =  Va1/(2**0.5);#  in  volts\n",
      "V3  =  Va3/(2**0.5);#  in  volts\n",
      "V5  =  Va5/(2**0.5);#  in  volts\n",
      " #total  power  supplied,\n",
      "P  =  V1*I1*math.cos(Phiv1  -  Phii1)  +  V3*I3*math.cos(Phiv3  -  Phii3)  +  V5*I5*math.cos(Phiv5  -  Phii5)\n",
      " #rms  current\n",
      "Irms  =  ((I1**2  +  I3**2  +  I5**2))**0.5\n",
      " #rms  voltage\n",
      "Vrms  =  ((V1**2  +  V3**2  +  V5**2))**0.5\n",
      " #overall  power  factor\n",
      "pf  =  P/(Vrms*Irms)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n(a)the  total  active  power  supplied  to  the  circuit  \",round(P,2),\"  W\"\n",
      "print  \"\\n(b)overall  power  factor  \",round(pf,2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "(a)the  total  active  power  supplied  to  the  circuit   53.26   W\n",
        "\n",
        "(b)overall  power  factor   0.84"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 8, page no. 655</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine for each case an expression for the current flowing if the fundamental frequency is 1 kHz.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "R1  =  40;#  in  ohm\n",
      "L  =  7.96E-3;#  in  Henry\n",
      "C = 25E-6; # in Farad\n",
      "f = 1000; # in Hx\n",
      "\n",
      "#calculation:\n",
      "wL = 2*math.pi*1000*L\n",
      "wC = 2*math.pi*1000*C\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"(a)i  =  \",round(100/R1,2),\"sin(wt)  +\",round(30/R1,2),\"sin(3wt - pi/3)  +\",round(10/R1,2),\"sin(5wt - pi/6) A\"\n",
      "print  \"(b)i  =  \",round(100/wL,2),\"sin(wt - pi/2)  +\",round(30/(3*wL),2),\"sin(3wt - pi/6)  +\",round(10/(5*wL),2),\"sin(5wt - 2pi/3) A\"\n",
      "print  \"(c)i  =  \",round(100*wC,2),\"sin(wt + pi/2)  +\",round(30*3*wC,2),\"sin(3wt + 5pi/6)  +\",round(10*5*wC,2),\"sin(5wt + pi/3) A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "(a)i  =   2.5 sin(wt)  + 0.75 sin(3wt - pi/3)  + 0.25 sin(5wt - pi/6) A\n",
        "(b)i  =   2.0 sin(wt - pi/2)  + 0.2 sin(3wt - pi/6)  + 0.04 sin(5wt - 2pi/3) A\n",
        "(c)i  =   15.71 sin(wt + pi/2)  + 14.14 sin(3wt + 5pi/6)  + 7.85 sin(5wt + pi/3) A\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 9, page no. 656</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) an expression to represent the instantaneous value of the current, \n",
      "#(b) the rms voltage, (c) the rms current, (d) the power dissipated, and (e) the overall power factor.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V1m  =  240;#  in  volts\n",
      "V3m  =  40;#  in  volts\n",
      "V5m  =  30;#  in  volts\n",
      "w1  =  314;#  fundamental\n",
      "R  =  12;#  in  ohm\n",
      "L  =  0.00955;#  in  Henry\n",
      "\n",
      " #calculation:\n",
      " #fundamental  or  first  harmonic\n",
      " #inductive  reactance,\n",
      "XL1  =  w1*L\n",
      " #impedance  at  the  fundamental  frequency,\n",
      "Z1  =  R  +  1j*XL1\n",
      " #Maximum  current  at  fundamental  frequency\n",
      "I1m  =  V1m/Z1\n",
      "I1mag  =  abs(I1m)\n",
      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
      " #Third  harmonic\n",
      "XL3  =  3*XL1\n",
      " #impedance  at  the  third  harmonic  frequency,\n",
      "Z3  =  R  +  1j*XL3\n",
      " #Maximum  current  at  third  harmonic  frequency\n",
      "I3m  =  V3m/Z3\n",
      "I3mag  =  abs(I3m)\n",
      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
      " #fifth  harmonic\n",
      "XL5  =  5*XL1\n",
      " #impedance  at  the  third  harmonic  frequency,\n",
      "Z5  =  R  +  1j*XL5\n",
      " #Maximum  current  at  third  harmonic  frequency\n",
      "I5m  =  V5m/Z5\n",
      "I5mag  =  abs(I5m)\n",
      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
      " #rms  voltage\n",
      "Vrms  =  ((V1m**2  +  V3m**2  +  V5m**2)/2)**0.5\n",
      " #rms  current\n",
      "Irms  =  ((I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
      " #power  dissipated\n",
      "P  =  R*Irms**2\n",
      " #overall  power  factor\n",
      "pf  =  P/(Vrms*Irms)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n(a)i  =  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
      "print  \"\\n(c)the  rms  value  of  voltage  is  \",round(Vrms,2),\"  V\"\n",
      "print  \"\\n(d)the  total  power  dissipated  \",round(P,2),\"  W\"\n",
      "print  \"\\n(e)overall  power  factor  \",round(pf,2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "(a)i  =   19.4 sin( 314.0 t  +  ( -0.24 ))  +   2.67 sin( 942.0 t  +  ( -0.64 ))  +   1.56 sin( 1570.0 t  +  ( -0.9 ))  A\n",
        "\n",
        "(b)the  rms  value  of  current  is   13.89   A\n",
        "\n",
        "(c)the  rms  value  of  voltage  is   173.35   V\n",
        "\n",
        "(d)the  total  power  dissipated   2316.26   W\n",
        "\n",
        "(e)overall  power  factor   0.96"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 10, page no. 658</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Obtain an expression for the current flowing and hence determine the rms value of current.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "Vom  =  50;#  in  volts\n",
      "V1m  =  200;#  in  volts\n",
      "V2m  =  40;#  in  volts\n",
      "V4m  =  5;#  in  volts\n",
      "f  =  50;#  in  Hz\n",
      "R  =  50;#  in  ohm\n",
      "C  =  100E-6;#  in  farad\n",
      "phiv1  =  0;#  in  rad\n",
      "phiv2  =  -1*math.pi/2;#  in  rad\n",
      "phiv4  =  math.pi/4;#  in  rad\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
      "V2  =  V2m*math.cos(phiv2)  +  1j*V2m*math.sin(phiv2)\n",
      "V4  =  V4m*math.cos(phiv4)  +  1j*V4m*math.sin(phiv4)\n",
      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
      "Iom  =  0\n",
      " #fundamental  or  first  harmonic\n",
      "w1  =  2*math.pi*f\n",
      " #inductive  reactance,\n",
      "Xc1  =  1/(w1*C)\n",
      " #impedance  at  the  fundamental  frequency,\n",
      "Z1  =  R  +  1j*Xc1\n",
      " #Maximum  current  at  fundamental  frequency\n",
      "I1m  =  V1/Z1\n",
      "I1mag  =  abs(I1m)\n",
      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
      " #second  harmonic\n",
      "Xc2  =  Xc1/2\n",
      " #impedance  at  the  third  harmonic  frequency,\n",
      "Z2  =  R  +  1j*Xc2\n",
      " #Maximum  current  at  third  harmonic  frequency\n",
      "I2m  =  V2/Z2\n",
      "I2mag  =  abs(I2m)\n",
      "phii2  =  cmath.phase(complex(I2m.real,I2m.imag))\n",
      " #fourth  harmonic\n",
      "Xc4  =  Xc1/4\n",
      " #impedance  at  the  third  harmonic  frequency,\n",
      "Z4  =  R  +  1j*Xc4\n",
      " #Maximum  current  at  third  harmonic  frequency\n",
      "I4m  =  V4/Z4\n",
      "I4mag  =  abs(I4m)\n",
      "phii4  =  cmath.phase(complex(I4m.real,I4m.imag))\n",
      " #rms  current\n",
      "Irms  =  (Iom**2  +  (I1mag**2  +  I2mag**2  +  I4mag**2)/2)**0.5\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"(a)i = \",round(Iom,2),\" + \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I2mag,2),\"sin(\",round((w1*2),2),\"t  +  (\",round(phii2,2),\"))  +  \",round(I4mag,2),\"sin(\",round((w1*4),2),\"t  +  (\",round(phii4,2),\"))  A\"\n",
      "print  \"(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "(a)i =  0.0  +  3.37 sin( 314.16 t  +  ( -0.57 ))  +   0.76 sin( 628.32 t  +  ( -1.88 ))  +   0.1 sin( 1256.64 t  +  ( 0.63 ))  A\n",
        "(b)the  rms  value  of  current  is   2.45   A\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 11, page no. 659</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) an expression to represent the current flowing in the circuit, \n",
      "#(b) the rms value of current, correct to two decimal places, and \n",
      "#(c) the power dissipated in the circuit, correct to three significant figures.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "Vom  =  25;#  in  volts\n",
      "V1m  =  100;#  in  volts\n",
      "V3m  =  40;#  in  volts\n",
      "V5m  =  20;#  in  volts\n",
      "w1  =  10000;#  fundamental\n",
      "R  =  5;#  in  ohm\n",
      "L  =  500E-6;#  in  Henry\n",
      "phiv1  =  0;#  in  rad\n",
      "phiv3  =  math.pi/6;#  in  rad\n",
      "phiv5  =  math.pi/12;#  in  rad\n",
      "\n",
      " #calculation:\n",
      " #voltage\n",
      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
      "V3  =  V3m*math.cos(phiv3)  +  1j*V3m*math.sin(phiv3)\n",
      "V5  =  V5m*math.cos(phiv5)  +  1j*V5m*math.sin(phiv5)\n",
      " #Inductance  has  no  effect  on  a  steady  current.  Hence  the  d.c.  component  of  the  current,  i0,  is  given  by\n",
      "Iom  =  Vom/R\n",
      " #fundamental  or  first  harmonic\n",
      " #inductive  reactance,\n",
      "XL1  =  w1*L\n",
      " #impedance  at  the  fundamental  frequency,\n",
      "Z1  =  R  +  1j*XL1\n",
      " #Maximum  current  at  fundamental  frequency\n",
      "I1m  =  V1/Z1\n",
      "I1mag  =  abs(I1m)\n",
      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
      "#Third  harmonic\n",
      "XL3  =  3*XL1\n",
      " #impedance  at  the  third  harmonic  frequency,\n",
      "Z3  =  R  +  1j*XL3\n",
      " #Maximum  current  at  third  harmonic  frequency\n",
      "I3m  =  V3/Z3\n",
      "I3mag  =  abs(I3m)\n",
      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
      " #fifth  harmonic\n",
      "XL5  =  5*XL1\n",
      " #impedance  at  the  third  harmonic  frequency,\n",
      "Z5  =  R  +  1j*XL5\n",
      " #Maximum  current  at  third  harmonic  frequency\n",
      "I5m  =  V5/Z5\n",
      "I5mag  =  abs(I5m)\n",
      "phii5  =  cmath.phase(complex(I5m.real,I5m.imag))\n",
      " #rms  current\n",
      "Irms  =  (Iom**2  +  (I1mag**2  +  I3mag**2  +  I5mag**2)/2)**0.5\n",
      " #power  dissipated\n",
      "P  =  R*Irms**2\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n(a)i  =  \",round(Iom,2),\"  +  \",round(I1mag,2),\"sin(\",round(w1,2),\"t  +  (\",round(phii1,2),\"))  +  \",round(I3mag,2),\"sin(\",round((w1*3),2),\"t  +  (\",round(phii3,2),\"))  +  \",round(I5mag,2),\"sin(\",round((w1*5),2),\"t  +  (\",round(phii5,2),\"))  A\"\n",
      "print  \"\\n(b)the  rms  value  of  current  is  \",round(Irms,2),\"  A\"\n",
      "print  \"\\n(c)the  total  power  dissipated  \",round(P,3),\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "(a)i  =   5.0   +   14.14 sin( 10000.0 t  +  ( -0.79 ))  +   2.53 sin( 30000.0 t  +  ( -0.73 ))  +   0.78 sin( 50000.0 t  +  ( -1.11 ))  A\n",
        "\n",
        "(b)the  rms  value  of  current  is   11.34   A\n",
        "\n",
        "(c)the  total  power  dissipated   642.538   W\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 12, page no. 661</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) the average power supplied, (b) the type of components present, and (c) the values of the components.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "Vom  =  30;#  in  volts\n",
      "V1m  =  40;#  in  volts\n",
      "V2m  =  25;#  in  volts\n",
      "V4m  =  15;#  in  volts\n",
      "Iom  =  0;#  in  amperes\n",
      "I1m  =  0.743;#  in  Amperes\n",
      "I2m  =  0.781;#  in  Amperes\n",
      "I4m  =  0.636;#  in  Amperes\n",
      "phii1  =  1.190;#  in  rad\n",
      "phii2  =  0.896;#  in  rad\n",
      "phii4  =  0.559;#  in  rad\n",
      "w  =  1000;#  in  rad\n",
      "\n",
      " #calculation:\n",
      " #the  average  power  P  is  given  by\n",
      "P  =  Vom*Iom+(0.707*V1m)*(0.707*I1m)*math.cos(phii1)+(0.707*V2m)*(0.707*I2m)*math.cos(phii2) + (0.707*V4m)*(0.707*I4m)*math.cos(phii4)\n",
      " #rms  current\n",
      "Irms  =  (Iom**2  +  (I1m**2  +  I2m**2  +  I4m**2)/2)**0.5\n",
      " #resistance  R\n",
      "R  =  P/(Irms**2)\n",
      " #impedance\n",
      "Z1  =  V1m/I1m\n",
      " #Xc1\n",
      "Xc1  =  (Z1**2  -  R**2)**0.5\n",
      " #capacitance\n",
      "C  =  1/(w*Xc1)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n(a)the  average  power  P  is  \",round(P,2),\"  W\"\n",
      "print  \"\\n(c)the  resistance  R  \",round(R,2),\"  ohm  and  capacitance  \",round(C*1E6,2),\"uF\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "(a)the  average  power  P  is   15.66   W\n",
        "\n",
        "(c)the  resistance  R   19.99   ohm  and  capacitance   20.01 uF\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 13, page no. 662</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) an expression for the supply current, i, (b) the percentage harmonic content of the supply current, (c) the total power dissipated, (d) an expression for the p.d. shown as \t1, and (e) an expression for current ic.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V1m  =  300;#  in  volts\n",
      "V3m  =  120;#  in  volts\n",
      "phiv1  =  0;#  in  rad\n",
      "phiv2  =  0.698;#  in  rad\n",
      "w1  =  314;#  in  rad\n",
      "C  =  2.123E-6;#  in  farads\n",
      "R1  =  560;#  in  ohms\n",
      "R2  =  2000;#  in  Ohm\n",
      "\n",
      "#calculation:\n",
      " #voltage\n",
      "V1  =  V1m*math.cos(phiv1)  +  1j*V1m*math.sin(phiv1)\n",
      "V3  =  V3m*math.cos(phiv2)  +  1j*V3m*math.sin(phiv2)\n",
      " #capacitive  reactance,\n",
      "Xc1  =  1/(w1*C)\n",
      " #impedance  at  the  fundamental  frequency,\n",
      "Z1  =  R1  -  1j*Xc1*R2/(R2  -  1j*Xc1)\n",
      " #Maximum  current  at  fundamental  frequency\n",
      "I1m  =  V1/Z1\n",
      "I1mag  =  abs(I1m)\n",
      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
      " #Third  harmonic\n",
      "Xc3  =  Xc1/3\n",
      " #impedance  at  the  third  harmonic  frequency,\n",
      "Z3  =  R1  -  1j*Xc3*R2/(R2  -  1j*Xc3)\n",
      "I1m = V1m/Z1\n",
      "I1mag  =  abs(I1m)\n",
      "phii1  =  cmath.phase(complex(I1m.real,I1m.imag))\n",
      " #Maximum  current  at  third  harmonic  frequency\n",
      "I3m  =  V3/Z3\n",
      "I3mag  =  abs(I3m)\n",
      "phii3  =  cmath.phase(complex(I3m.real,I3m.imag))\n",
      " #Percentage  harmonic  content  of  the  supply  current  is  given  by\n",
      "percent  =  I3mag*100/I1mag\n",
      " #total  active  power\n",
      "P  =  (0.707*V1m)*(0.707*I1mag)*math.cos(phiv1  -  phii1)  +  (0.707*V3m)*(0.707*I3m)*math.cos(phiv2  -  phii3)\n",
      "\n",
      "I1 = I1m*R2/(R2 - 1j*Xc1)\n",
      "I3 = I3m*R2/(R2 - 1j*Xc3)\n",
      "\n",
      "I1nmag  =  abs(I1)\n",
      "phini1  =  cmath.phase(complex(I1.real,I1.imag))\n",
      "I3nmag  =  abs(I3)\n",
      "phini3  =  cmath.phase(complex(I3.real,I3.imag))\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n(a)supply current, i=\", round(I1mag,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
      "print  \"\\n(b)Percentage  harmonic  content  of  the  supply  current  is  \",round(percent,2),\"  percent\"\n",
      "print  \"\\n(c)total  active  power  is  \",round(abs(P),2),\"  W\"\n",
      "print  \"\\n(d)Voltage, v1 =\", round(I1mag*R1,3),\"sin(\", w1,\"t +\",round(phii1,3),\") + \",round(I3mag*R1,3),\"sin(\", 3*w1,\"t +\",round(phii3,3),\") A\"\n",
      "print  \"\\n(e)current, ic =\", round(I1nmag,3),\"sin(\", w1,\"t +\",round(phini1,3),\") + \",round(I3nmag,3),\"sin(\", 3*w1,\"t +\",round(phini3,3),\") A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "(a)supply current, i= 0.187 sin( 314 t + 0.643 ) +  0.145 sin( 942 t + 1.305 ) A\n",
        "\n",
        "(b)Percentage  harmonic  content  of  the  supply  current  is   77.57   percent\n",
        "\n",
        "(c)total  active  power  is   25.34   W\n",
        "\n",
        "(d)Voltage, v1 = 104.996 sin( 314 t + 0.643 ) +  81.45 sin( 942 t + 1.305 ) A\n",
        "\n",
        "(e)current, ic = 0.15 sin( 314 t + 1.287 ) +  0.141 sin( 942 t + 1.55 ) A"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 14, page no. 664</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) the fundamental frequency for resonance with the third harmonic, and \n",
      "#(b) the maximum value of the fundamental and third harmonic components of current\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V1m  =  400;#  in  volts\n",
      "V3m  =  10;#  in  volts\n",
      "C  =  0.2E-6;#  in  farads\n",
      "R  =  2;#  in  ohms\n",
      "L  =  0.5;#  in  Henry\n",
      "\n",
      " #calculation:\n",
      " #Resonance  with  the  third  harmonic  means  that\n",
      "w  =  (1/(9*L*C))**0.5\n",
      " #fundamental  frequency,  f\n",
      "f  =  w/(2*math.pi)\n",
      " #At  the  fundamental  frequency,\n",
      " #impedance  Z1\n",
      "Z1  =  R  +  1j*(w*L  -  1/(w*C))\n",
      "Z1mag  =  abs(Z1)\n",
      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
      "I1m  =  V1m/Z1mag\n",
      " #At  the  third  harmonic  frequency,\n",
      "Z3  =  R  +  1j*(3*w*L  -  1/(3*w*C))\n",
      "Z3mag  =  abs(Z3)\n",
      "phiZ3  =  cmath.phase(complex(Z3.real,Z3.imag))\n",
      " #Maximum  value  of  current  at  the  third  harmonic  frequency,\n",
      "I3m  =  V3m/Z3\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is  \",round(f,2),\"  Hz\"\n",
      "print  \"(b)Maximum value of current at fundamental freq. is\",round(abs(I1m),3),\"A \"\n",
      "print   \"and  at  the  third  harmonic  frequency  \",  abs(I3m),\"  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "(a)fundamental  frequency  for  resonance  with  the  third  harmonic  is   167.76   Hz\n",
        "(b)Maximum value of current at fundamental freq. is 0.095 A \n",
        "and  at  the  third  harmonic  frequency   5.0   A\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 15, page no. 665</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) the value of n, (b) the maximum value of current at the nth harmonic, \n",
      "#(c) the p.d. across the capacitor at the nth harmonic and\n",
      "#(d) the maximum value of the fundamental current.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V1m  =  800;#  in  volts\n",
      "f  =  50;#  in  Hz\n",
      "x  =  0.015;\n",
      "C  =  0.122E-6;#  in  farads\n",
      "R  =  5;#  in  ohms\n",
      "L  =  0.369;#  in  Henry\n",
      "\n",
      " #calculation:\n",
      " #voltage  at  nth  harmonic\n",
      "Vnm  =  x*V1m\n",
      "w  =  2*math.pi*f\n",
      " #For  resonance  at  the  nth  harmonic  nwL  =  1/nwC\n",
      "n  =  1/(w*(L*C)**0.5)\n",
      " #At  resonance,  impedance\n",
      "Zn  =  R\n",
      " #the  maximum  value  of  current  at  the  nth  harmonic\n",
      "Inm  =  Vnm/Zn\n",
      " #capacitive  reactance,  at  nth  harmonic\n",
      "Xcn  =  1/(n*w*C)\n",
      " #the  p.d.  across  the  capacitor  at  the  nth  harmonic\n",
      "Vcn  =  Inm*Xcn\n",
      " #At  the  fundamental  frequency,  inductive  reactance,\n",
      "XL1  =  w*L\n",
      " #capacitive  reactance\n",
      "Xc1  =  1/(w*C)\n",
      " #Impedance  at  the  fundamental  frequency,\n",
      "Z1  =  R  +  1j*(XL1  -  Xc1)\n",
      "Z1mag  =  abs(Z1)\n",
      "phiZ1  =  cmath.phase(complex(Z1.real,Z1.imag))\n",
      " #Maximum  value  of  current  at  the  fundamental  frequency,\n",
      "I1m  =  V1m/Z1mag\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n(a)n  =  \",round(n,2),\"\"\n",
      "print  \"\\n(b)the  maximum  value  of  current  at  the  nth  harmonic  \",round(Inm,2),\"  A\"\n",
      "print  \"\\n(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is  \",round(Vcn,2),\"\"\n",
      "print  \"\\n(d)the  maximum  value  of  the  fundamental  current.  \",round(I1m,2),\"  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "(a)n  =   15.0 \n",
        "\n",
        "(b)the  maximum  value  of  current  at  the  nth  harmonic   2.4   A\n",
        "\n",
        "(c)the  p.d.  across  the  capacitor  at  the  nth  harmonic  is   4173.92 \n",
        "\n",
        "(d)the  maximum  value  of  the  fundamental  current.   0.03   A"
       ]
      }
     ],
     "prompt_number": 14
    }
   ],
   "metadata": {}
  }
 ]
}