path: root/Electrical_Circuit_Theory_And_Technology/chapter_24-checkpoint_2.ipynb

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   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h1>Chapter 24: Application of complex numbers to series a.c. circuits</h1>"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 1, page no. 433</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances:\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "z1  =  12  +  5j;\n",
      "z2  =  -40j;\n",
      "r3  =  30;\n",
      "theta3  =  60;#  in  degrees\n",
      "r4  =  2.20E6;  \n",
      "theta4  =  -30;#  in  degrees\n",
      "f  =  50;#  in  Hz\n",
      "\n",
      "#calculation:\n",
      " #for  an  R-L  series  circuit,  impedance\n",
      " #  Z  =  R  +  iXL\n",
      "Ra  =  z1.real\n",
      "XLa  =  z1.imag\n",
      "La  =  XLa/(2*math.pi*f)\n",
      " #for  a  purely  capacitive  circuit,  impedance  Z  =  -iXc\n",
      "Xcb  =  abs(z2.imag)\n",
      "Cb  =  1/(2*math.pi*f*Xcb)\n",
      "z3  =  r3*cmath.cos(theta3*math.pi/180)  +  (r3*cmath.sin(theta3*math.pi/180))*1j\n",
      "Rc  =  z3.real\n",
      "XLc  =  z3.imag\n",
      "Lc  =  XLc/(2*math.pi*f)\n",
      "z4  =  r4*cmath.cos(theta4*math.pi/180)  +  (r4*cmath.sin(theta4*math.pi/180))*1j\n",
      "Rd  =  z4.real\n",
      "Xcd  =  abs(z4.imag)\n",
      "Cd  =  1/(2*math.pi*f*Xcd)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of  \",round(  Ra,2),\"  ohm  \"\n",
      "print   \"in  series  with  an  inductance  of  \",round(La*1000,2),\"mH\"\n",
      "print  \"\\n  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance  \",round(Cb*1E6,2),\"uF\"\n",
      "print  \"\\n  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of  \",round(Rc,2),\"  ohm  \"\n",
      "print   \"in  series  with  an  inductance  of  \",round(Lc*1000,2),\"mH\"\n",
      "print  \"\\n  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of  \",round(Rd/1000,2),\"kohm \"\n",
      "print   \" in  series  with  a  capacitor  of  capacitance  \",round(Cd*1E9,2),\"nF\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)an  impedance  (12  +  i5)ohm  represents  a  resistance  of   12.0   ohm  \n",
        "in  series  with  an  inductance  of   15.92 mH\n",
        "\n",
        "  (b)an  impedance  -40i  ohm  represents  a  pure  capacitor  of  capacitance   79.58 uF\n",
        "\n",
        "  (c)an  impedance  30/_60deg  ohm  represents  a  resistance  of   15.0   ohm  \n",
        "in  series  with  an  inductance  of   82.7 mH\n",
        "\n",
        "  (d)an  impedance  2.20  x  10^6  /_-30deg  ohm  represents  a  resistance  of   1905.26 kohm \n",
        " in  series  with  a  capacitor  of  capacitance   2.89 nF\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 2, page no. 434</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "L  =  0.1592  ;#  in  Henry\n",
      "V  =  250;#  in  Volts\n",
      "f  =  50;#  in  Hz\n",
      "R  =  0;#  in  ohms\n",
      "\n",
      "#calculation:\n",
      " #for  an  R\u00e2\u20ac\u201cL  series  circuit,  impedance\n",
      " #  Z  =  R  +  iXL\n",
      "XL  =  2*math.pi*f*L\n",
      "Z  =  R  +  1j*XL\n",
      "I  =  V/Z\n",
      "x  =  I.real\n",
      "y  =  I.imag\n",
      "r  =  (x**2  +  y**2)**0.5\n",
      "if  ((x==0)&(y<0)):\n",
      "    theta  =  -90\n",
      "elif  ((x==0)&(y>0)):\n",
      "    theta  =  +90\n",
      "else:\n",
      "    theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
      "\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  current  is  (\",round(r,2),\"/_\",theta,\"deg)  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  current  is  ( 5.0 /_ -90 deg)  A"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 3, page no. 435</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the value of the supply p.d.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "C  =  3E-6  ;#  in  farad\n",
      "f  =  1000;#  in  Hz\n",
      "ri  =  2.83;\n",
      "thetai  =  90;#  in  degrees\n",
      "\n",
      "#calculation:\n",
      " #Capacitive  reactance  Xc\n",
      "Xc  =  1/(2*math.pi*f*C)\n",
      " #  circuit  impedance  Z\n",
      "Z  =  -1*1j*Xc\n",
      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
      "V  =  I*Z\n",
      "x  =  V.real\n",
      "y  =  V.imag\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  supply  p.d.  is \",round(abs(V),0),\"V\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  supply  p.d.  is  150.0 V"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 4, page no. 435</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) the resistance, (b) the capacitance, \n",
      "#(c) the modulus of the impedance, and (d) the current flowing and its phase angle,\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",
      "Z  =  30  -  50j;\n",
      "\n",
      "#calculation:\n",
      " #Since  impedance  Z  =  30  -  i50,\n",
      " #resistance\n",
      "R  =  Z.real\n",
      " #capacitive  reactance\n",
      "Xc  =  abs(Z.imag)\n",
      " #capacitance\n",
      "C  =  1/(2*math.pi*f*Xc)\n",
      " #modulus  of  impedance\n",
      "modZ  =  (R**2  +  Xc**2)**0.5\n",
      "I  =  V/Z\n",
      "x  =  I.real\n",
      "y  =  I.imag\n",
      "r  =  (x**2  +  y**2)**0.5\n",
      "theta  =  cmath.phase(complex(x,y))*180/math.pi\n",
      "\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)resistance  is  \",round(  R,2),\"  ohm\"\n",
      "print  \"\\n  (b)capacitance  is  \",round(C*1E6,2),\"uFarad\"\n",
      "print  \"\\n  (c)modulus  of  impedance  is  \",round(modZ,2),\"  ohm\"\n",
      "print  \"\\n  (d)current  flowing  and  its  phase  angle  is  (\",round(  r,2),\"/_\",round(  theta,2),\"deg)  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)resistance  is   30.0   ohm\n",
        "\n",
        "  (b)capacitance  is   63.66 uFarad\n",
        "\n",
        "  (c)modulus  of  impedance  is   58.31   ohm\n",
        "\n",
        "  (d)current  flowing  and  its  phase  angle  is  ( 4.12 /_ 59.04 deg)  A"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 5, page no. 436</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) the impedance of the circuit,\n",
      "#(b) the current and circuit phase angle, \n",
      "#(c) the p.d. across the 32 ohm resistor, and (d) the p.d. across the coil\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V  =  200;#  in  Volts\n",
      "f  =  50;#  in  Hz\n",
      "R  =  32;#  in  ohms\n",
      "L  =  0.15;#  in  Henry\n",
      "\n",
      "#calculation:\n",
      " #Inductive  reactance  XL\n",
      "XL  =  2*math.pi*f*L\n",
      " #impedance,  Z\n",
      "Z  =  R  +  1j*XL\n",
      " #Current  I\n",
      "I  =  V/Z\n",
      "xi  =  I.real\n",
      "yi  =  I.imag\n",
      "ri  =  (xi**2  +  yi**2)**0.5\n",
      "if  ((xi==0)&(yi<0)):\n",
      "         thetai  =  -90\n",
      "elif  ((xi==0)&(yi>0)):\n",
      "         thetai  =  +90\n",
      "else:\n",
      "         thetai  =  cmath.phase(complex(xi,yi))*180/math.pi\n",
      "\n",
      " #P.d.  across  the  resistor\n",
      "VR  =  I*R\n",
      "xr  =  VR.real\n",
      "yr  =  VR.imag\n",
      "rr  =  (xr**2  +  yr**2)**0.5\n",
      "thetar  =  cmath.phase(complex(xr,yr))*180/math.pi\n",
      " #P.d.  across  the  coil,  VL\n",
      "VL  =  I*1j*XL\n",
      "xl  =  VL.real\n",
      "yl  =  VL.imag\n",
      "rl  =  (xl**2  +  yl**2)**0.5\n",
      "thetal  =  cmath.phase(complex(xl,yl))*180/math.pi\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)impedance  is  \",round(Z.real,2),\"  + \",round(  Z.imag,2),\")i  ohm\"\n",
      "print  \"\\n  (b)current  flowing  and  its  phase  angle  is lagging the voltage = (\",round(  ri,2),\"/_\",round(  thetai,2),\"deg)  A\"\n",
      "print  \"\\n  (c)P.d.  across  the  resistor  is  (\",round(rr,2),\"/_\",round(thetar,2),\"deg)  V\"\n",
      "print  \"\\n  (d)P.d.  across  the  coil,  VL  is  (\",round(rl,2),\"/_\",round(thetal,2),\"deg)  V\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)impedance  is   32.0   +  47.12 )i  ohm\n",
        "\n",
        "  (b)current  flowing  and  its  phase  angle  is lagging the voltage = ( 3.51 /_ -55.82 deg)  A\n",
        "\n",
        "  (c)P.d.  across  the  resistor  is  ( 112.36 /_ -55.82 deg)  V\n",
        "\n",
        "  (d)P.d.  across  the  coil,  VL  is  ( 165.46 /_ 34.18 deg)  V"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 6, page no. 436</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the value of impedance\n",
      "#determine the value of the components forming the series circuit.\n",
      "from __future__ import division\n",
      "import math\n",
      "#initializing  the  variables:\n",
      "V  =  120  +  200j;#  in  Volts\n",
      "f  =  5E6;#  in  Hz\n",
      "I  =  7  +  16j;#  in  amperes\n",
      "\n",
      "#calculation:\n",
      " #impedance,  Z\n",
      "Z  =  V/I\n",
      "R  =  Z.real\n",
      "X  =  Z.imag  \n",
      "if  ((R>0)&(X<0)):\n",
      "    C  =  -1/(2*math.pi*f*X)\n",
      "#Results\n",
      "    print  \"\\n\\n  Result  \\n\\n\"\n",
      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm  \"\n",
      "    print   \"and  a  capacitor  of  capacitive reactance\", round(X*-1,3),\"ohm and capacitance is\",round(C*1E9,2),\" nFarad\\n\"\n",
      "elif  ((R>0)&(X>0)):\n",
      "    L  =  2*math.pi*f*X\n",
      "#Results\n",
      "    print  \"\\n\\n  Result  \\n\\n\"\n",
      "    print  \"\\n  The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R,2),\"  ohm \"\n",
      "    print   \" and  a  inductor  of  insuctance  \",round(L*100,2),\" mHenry\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  The  series  circuit  thus  consists  of  a  resistor  of  resistance   13.25   ohm  \n",
        "and  a  capacitor  of  capacitive reactance 1.705 ohm and capacitance is 18.67  nFarad\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 7, page no. 437</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the value of impedance Z2.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv  =  70;#  in  volts\n",
      "thetav  =  30;#  in  degrees\n",
      "ri  =  3.5;#  in  amperes\n",
      "thetai  =  -20;#  in  degrees\n",
      " #z1  consist  of  two  resistance\n",
      "R1  =  4.36;#  in  ohms\n",
      "R2  =  -2.1j;#  in  ohms\n",
      "\n",
      " #calculation:\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
      " #impedance,  Z\n",
      "Z  =  V/I\n",
      " #Total  impedance  Z  =  z1  +  z2\n",
      "Z1  =  R1  +  R2\n",
      "Z2  =  Z  -  Z1\n",
      "x  =  Z2.real\n",
      "y  =  Z2.imag  \n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  impedance  Z2  is  \",round(x,2),\"  +  (\",round(y,2),\")i  ohm\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  impedance  Z2  is   8.5   +  ( 17.42 )i  ohm"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 8, page no. 437</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine (a) the supply voltage, (b) the voltage across the 90 \u000e resistance, \n",
      "#(c) the voltage across the inductance, and (d) the circuit phase angle.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "R  =  90;#  in  ohms\n",
      "XL  =  150;#  in  ohms\n",
      "ri  =  1.35;#  in  amperes\n",
      "thetai  =  0;#  in  degrees\n",
      "\n",
      "#calculation:\n",
      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
      " #Circuit  impedance  Z\n",
      "Z  =  R  +  1j*XL\n",
      " #Supply  voltage,  V\n",
      "V  =  I*Z\n",
      " #Voltage  across  90  ohm\u000e  resistor\n",
      "VR  =  V.real\n",
      "#Voltage  across  inductance,  VL\n",
      "VL  =  V.imag\n",
      "xv  =  V.real\n",
      "yv  =  V.imag\n",
      "rv  =  (xv**2  +  yv**2)**0.5\n",
      "thetav  = cmath.phase(complex(xv,yv))*180/math.pi\n",
      "phi  =  thetav  -  thetai\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)Supply  voltage,  V  is  \",xv,\"  +  (\",yv,\")i  V\\n\"\n",
      "print  \"\\n  (b)Voltage  across  90  ohm resistor,  VR  is  \",VR,\"  V\\n\"\n",
      "print  \"\\n  (c)Voltage  across  inductance,  VL  is  \",VL,\"  V\\n\"\n",
      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)Supply  voltage,  V  is   121.5   +  ( 202.5 )i  V\n",
        "\n",
        "\n",
        "  (b)Voltage  across  90  ohm resistor,  VR  is   121.5   V\n",
        "\n",
        "\n",
        "  (c)Voltage  across  inductance,  VL  is   202.5   V\n",
        "\n",
        "\n",
        "  (d)Circuit  phase  angle  is   59.04 deg lagging"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 9, page no. 438</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine (a) the rms value of voltage (in polar form),\n",
      "#(b) the circuit impedance, (c) the rms current flowing, and\n",
      "#(d) the circuit phase angle.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "R  =  25;#  in  ohms\n",
      "L  =  0.02;#  in  henry\n",
      "Vm  =  282.8;#  in  volts\n",
      "w  =  628.4;#  in  rad/sec\n",
      "phiv  =  math.pi/3;#  phase  angle\n",
      "\n",
      "#calculation:\n",
      " #rms  voltage\n",
      "Vrms  =  0.707*Vm*math.cos(phiv)  +  0.707*Vm*math.sin(phiv)*1j\n",
      " #frequency\n",
      "f  =  w/(2*math.pi)\n",
      " #Inductive  reactance  XL\n",
      "XL  =  2*math.pi*f*L\n",
      " #Circuit  impedance  Z\n",
      "Z  =  R  +  XL*1j\n",
      " #Rms  current\n",
      "Irms  =  Vrms/Z\n",
      "phii  =  cmath.phase(complex(Irms.real, Irms.imag))*180/math.pi\n",
      "phi  =  phiv*180/math.pi  -  phii\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)the  rms  value  of  voltage  is  \",round(Vrms.real,2),\"  +  (\",round(  Vrms.imag,2),\")i  V\\n\"\n",
      "print  \"\\n  (b)the  circuit  impedance  is  \",round(R,2),\"  +  (\",round(  XL,2),\")i  ohm\\n\"\n",
      "print  \"\\n  (c)the  rms  current  flowing  is  \",round(Irms.real,2),\"  +  (\",round(  Irms.imag,2),\")i  A\\n\"\n",
      "print  \"\\n  (d)Circuit  phase  angle  is  \",round(phi,2),\"deg lagging\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)the  rms  value  of  voltage  is   99.97   +  ( 173.15 )i  V\n",
        "\n",
        "\n",
        "  (b)the  circuit  impedance  is   25.0   +  ( 12.57 )i  ohm\n",
        "\n",
        "\n",
        "  (c)the  rms  current  flowing  is   5.97   +  ( 3.92 )i  A\n",
        "\n",
        "\n",
        "  (d)Circuit  phase  angle  is   26.69 deg lagging\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 10, page no. 438</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the current flowing in the circuit\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "R  =  12;#  in  ohms\n",
      "L  =  0.10;#  in  henry\n",
      "C  =  120E-6;#  in  Farads\n",
      "f  =  50;#  in  Hz\n",
      "V  =  240;#  in  volts\n",
      "\n",
      "#calculation:\n",
      " #Inductive  reactance,  XL\n",
      "XL  =  2*math.pi*f*L\n",
      " #Capacitive  reactance,  Xc\n",
      "Xc  =  1/(2*math.pi*f*C)\n",
      " #Circuit  impedance  Z\n",
      "Z  =  R  +  1j*(XL  -  Xc)\n",
      "I  =  V/Z\n",
      "phii  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
      "phiv  =  0#  in  degrees\n",
      "phi  =  phiv  -  phii\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  the  current  flowing  is  \",round(abs(I),1),\"/_\",round(cmath.phase(complex(I.real,I.imag))*180/math.pi,1),\"deg  A\\n\"\n",
      "print  \"and  Circuit  phase  angle  is  \",round(phi,1),\"deg lagging\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  the  current  flowing  is   18.5 /_ -22.2 deg  A\n",
        "\n",
        "and  Circuit  phase  angle  is   22.2 deg lagging\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 11, page no. 439</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the values of R and L, Determine also the voltage across the coil and the voltage across the capacitor.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "C  =  50E-6;#  in  Farads\n",
      "f  =  50;#  in  Hz\n",
      "V  =  225;#  in  volts\n",
      "ri  =  1.5;#  in  Amperes\n",
      "thetai  =  -30;#  in  degrees\n",
      "\n",
      "#calculation:\n",
      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
      " #Capacitive  reactance,  Xc\n",
      "Xc  =  1/(2*math.pi*f*C)\n",
      " #Circuit  impedance  Z\n",
      "Z  =  V/I\n",
      "R  =  Z.real\n",
      "XL  =  Z.imag  +  Xc\n",
      " #inductance  L\n",
      "L  =  XL/(2*math.pi*f)\n",
      " #Voltage  across  coil\n",
      "Zcoil  =  R  +  1j*XL\n",
      "Vcoil  =  I*Zcoil\n",
      " #Voltage  across  capacitor,\n",
      "Vc  =  I*(-1j*Xc)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)resistance  is  \",round(R,2),\"  ohm  and  inductance  is  \",round(  L,3),\" H\\n\"\n",
      "print  \"\\n  (b)voltage  across  the  coil  is  \",round(Vcoil.real,2),\"  +  (\",round(  Vcoil.imag,2),\")i  V\\n\"\n",
      "print  \"\\n  (c)voltage  across  the  capacitor  is  \",round(Vc.real,2),\"  +  (\",round(  Vc.imag,2),\")i  V\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)resistance  is   129.9   ohm  and  inductance  is   0.441  H\n",
        "\n",
        "\n",
        "  (b)voltage  across  the  coil  is   272.75   +  ( 82.7 )i  V\n",
        "\n",
        "\n",
        "  (c)voltage  across  the  capacitor  is   -47.75   +  ( -82.7 )i  V"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 12, page no. 440</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine also the value of the supply voltage V and the circuit phase angle.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "C  =  2.653E-6;#  in  Farads\n",
      "R1  =  8;#  in  ohms\n",
      "R2  =  5;#  in  ohms\n",
      "L  =  0.477E-3;#  in  Henry\n",
      "f  =  4000;#  in  Hz\n",
      "ri  =  6;#  in  Amperes\n",
      "thetai  =  0;#  in  degrees\n",
      "\n",
      "#calculation:\n",
      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
      " #Capacitive  reactance,  Xc\n",
      "Xc  =  1/(2*math.pi*f*C)\n",
      " #impedance  Z1\n",
      "Z1  =  R1  -  1j*Xc\n",
      " #inductive  reactance  XL\n",
      "XL  =  2*math.pi*f*L\n",
      " #impedance  Z2,\n",
      "Z2  =  R2  +  1j*XL\n",
      " #voltage  V1\n",
      "V1  =  I*Z1\n",
      " #voltage  V2\n",
      "V2  =  I*Z2\n",
      " #Supply  voltage,  V\n",
      "V  =  V1  +  V2\n",
      "phiv  =  cmath.phase(complex(V.real, V.imag))*180/math.pi\n",
      "phi  =  phiv  -  thetai\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  supply  voltage  is  \",round(V.real,2),\"  +  (\",round(  V.imag,2),\")i  V\\n\"\n",
      "print  \"and  Circuit  phase  angle  is  \",round(phi,2),\"deg leading\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  supply  voltage  is   78.0   +  ( -18.06 )i  V\n",
        "\n",
        "and  Circuit  phase  angle  is   -13.03 deg leading"
       ]
      }
     ],
     "prompt_number": 2
    }
   ],
   "metadata": {}
  }
 ]
}