path: root/Electrical_Circuit_Theory_And_Technology/chapter_25-checkpoint.ipynb

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   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h1>Chapter 25: Application of complex numbers to parallel a.c. circuits</h1>"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 1, page no. 446</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the admittance, conductance and susceptance\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "Z1  =  0  -  5j;#  in  ohms\n",
      "Z2  =  25  + 40j;#  in  ohms\n",
      "Z3  =  3  -  2j;#  in  ohms\n",
      "r4  =  50;#  in  ohms\n",
      "theta4  =  40;#  in  degrees\n",
      "\n",
      "#calculation:\n",
      " #admittance  Y\n",
      "Y1  =  1/Z1\n",
      " #conductance,  G\n",
      "G1  =  Y1.real\n",
      " #Suspectance,  Bc\n",
      "Bc1  =  abs(Y1.imag)\n",
      " #admittance  Y\n",
      "Y2  =  1/Z2\n",
      " #conductance,  G\n",
      "G2  =  Y2.real\n",
      " #Suspectance,  Bc\n",
      "Bc2  =  abs(Y2.imag)\n",
      " #admittance  Y\n",
      "Y3  =  1/Z3\n",
      " #conductance,  G\n",
      "G3  =  Y3.real\n",
      " #Suspectance,  Bc\n",
      "Bc3  =  abs(Y3.imag)\n",
      "Z4  =  r4*math.cos(theta4*math.pi/180)  +  1j*r4*math.sin(theta4*math.pi/180)\n",
      " #admittance  Y\n",
      "Y4  =  1/Z4\n",
      " #conductance,  G\n",
      "G4  =  Y4.real\n",
      " #Suspectance,  Bc\n",
      "Bc4  =  abs(Y4.imag)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)admittance  Y  is  (\",round(Y1.real,2),\"  +  (\",round(Y1.imag,2),\")i)  S, \"\n",
      "print   \" conductance,  G  is  \",round(G1,2),\"  S,  susceptance,Bc  is  \",round(Bc1,2),\"  S\\n\"\n",
      "print  \"\\n  (b)admittance  Y  is  (\",round(Y2.real,2),\"  +  (\",round(Y2.imag,2),\")i)  S, \"\n",
      "print   \" conductance,  G  is  \",round(G2,2),\"  S,  susceptance,Bc  is  \",round(Bc2,2),\"  S\\n\"\n",
      "print  \"\\n  (c)admittance  Y  is  (\",round(Y3.real,2),\"  +  (\",round(Y3.imag,2),\")i)  S, \"\n",
      "print   \" conductance,  G  is  \",round(G3,2),\"  S,  susceptance,Bc  is  \",round(Bc3,2),\"  S\\n\"\n",
      "print  \"\\n  (d)admittance  Y  is  (\",round(Y4.real,2),\"  +  (\",round(Y4.imag,2),\")i)  S, \"\n",
      "print   \" conductance,  G  is  \",round(G4,2),\"  S,  susceptance,Bc  is  \",round(Bc4,2),\"  S\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)admittance  Y  is  ( -0.0   +  ( 0.2 )i)  S, \n",
        " conductance,  G  is   -0.0   S,  susceptance,Bc  is   0.2   S\n",
        "\n",
        "\n",
        "  (b)admittance  Y  is  ( 0.01   +  ( -0.02 )i)  S, \n",
        " conductance,  G  is   0.01   S,  susceptance,Bc  is   0.02   S\n",
        "\n",
        "\n",
        "  (c)admittance  Y  is  ( 0.23   +  ( 0.15 )i)  S, \n",
        " conductance,  G  is   0.23   S,  susceptance,Bc  is   0.15   S\n",
        "\n",
        "\n",
        "  (d)admittance  Y  is  ( 0.02   +  ( -0.01 )i)  S, \n",
        " conductance,  G  is   0.02   S,  susceptance,Bc  is   0.01   S\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 2, page no. 447</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine expressions for the impedance\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "Y2  =  0.001  -  0.002j;#  in  S\n",
      "Y3  =  0.05  +  0.08j;#  in  S\n",
      "r1  =  0.004;#  in  S\n",
      "theta1  =  30;#  in  degrees\n",
      "\n",
      " #calculation:\n",
      " #impedance,  Z\n",
      "Z2  =  1/Y2\n",
      "Z3  =  1/Y3\n",
      "Y1  =  r1*math.cos(theta1*math.pi/180)  +  1j*r1*math.sin(theta1*math.pi/180)\n",
      "Z1  =  1/Y1\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)Impedance,Z  is  (\",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i)  ohm\\n\"\n",
      "print  \"\\n  (b)Impedance,Z  is  (\",round(Z2.real,2),\"  +  (\",round(  Z2.imag,2),\")i)  ohm\\n\"\n",
      "print  \"\\n  (c)Impedance,Z  is  (\",round(Z3.real,2),\"  +  (\",round(  Z3.imag,2),\")i)  ohm\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)Impedance,Z  is  ( 216.51   +  ( -125.0 )i)  ohm\n",
        "\n",
        "\n",
        "  (b)Impedance,Z  is  ( 200.0   +  ( 400.0 )i)  ohm\n",
        "\n",
        "\n",
        "  (c)Impedance,Z  is  ( 5.62   +  ( -8.99 )i)  ohm"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 3, page no. 448</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the values of the resistance and the capacitive reactance of the circuit if they are connected \n",
      "#(a) in parallel, (b) in series.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "Y  =  0.040  -  1j*0.025;#  in  S\n",
      "\n",
      "#calculation:\n",
      " #impedance,  Z\n",
      "Z  =  1/Y\n",
      " #conductance,  G\n",
      "G  =  Y.real\n",
      " #Suspectance,  Bc\n",
      "Bc  =  abs(Y.imag)\n",
      " #parallrl  \n",
      " #resistance,  R\n",
      "Rp  =  1/G\n",
      " #capacitive  reactance\n",
      "Xcp  =  1/Bc\n",
      " #series\n",
      " #resistance,  R\n",
      "Rs  =  Z.real\n",
      " #capacitive  reactance\n",
      "Xcs  =  abs(Z.imag)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)for  parallel,  resistance,R  is  \",round(Rp,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcp,2),\"  ohm\\n\"\n",
      "print  \"\\n  (b)forseries,  resistance,R  is  \",round(Rs,2),\"  ohm  and  capacitive  reactance,  Xc  is  \",round(Xcs,2),\"  ohm\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)for  parallel,  resistance,R  is   25.0   ohm  and  capacitive  reactance,  Xc  is   40.0   ohm\n",
        "\n",
        "\n",
        "  (b)forseries,  resistance,R  is   17.98   ohm  and  capacitive  reactance,  Xc  is   11.24   ohm"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 4, page no. 449</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the values of currents I, I1 and I2.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "R1  =  8;#  in  ohm\n",
      "R  =  5;#  in  ohm\n",
      "R2  =  6;#  ohm\n",
      "rv  =  50;#  in  volts\n",
      "thetav  =  0;#  in  degrees\n",
      "\n",
      "#calculation:\n",
      " #voltage,V\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #circuit  impedance,  ZT\n",
      "ZT  =  R  +  (R1*1j*R2/(R1  +  1j*R2))\n",
      " #Current  I\n",
      "I  =  V/ZT\n",
      " #current,I1\n",
      "I1  =  I*(1j*R2/(R1  +  1j*R2))\n",
      " #current,  I2\n",
      "I2  =  I*(R1/(R1  +  1j*R2))\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A,\"\n",
      "print   \"current,I1  = \",round(abs(I1),2),\"/_\",round(cmath.phase(complex(I1.real, I1.imag))*180/math.pi,2),\"deg  A,  \"\n",
      "print   \"current,  I2  = \",round(abs(I2),2),\"/_\",round(cmath.phase(complex(I2.real, I2.imag))*180/math.pi,2),\"deg  A\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  current,  I  =  5.7 /_ -25.98 deg  A,\n",
        "current,I1  =  3.42 /_ 27.15 deg  A,  \n",
        "current,  I2  =  4.56 /_ -62.85 deg  A\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 5, page no. 450</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the value of supply current I and its phase relative to the 40 V supply.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "R1  =  5;#  in  ohm\n",
      "R2  =  3;#  in  ohm  \n",
      "R3  =  8;#  ohm\n",
      "Xc  =  4;#  in  ohms\n",
      "XL  =  12;#  in  Ohms\n",
      "V  =  40;#  in  volts\n",
      "f  =  50;#  in  Hz\n",
      "\n",
      "#calculation:\n",
      "Z1  =  R1  +  1j*XL\n",
      "Z2  =  R2  -  1j*Xc\n",
      "Z3  =  R3\n",
      " #circuit  admittance,  YT  =  1/ZT\n",
      "YT  =  (1/Z1)  +  (1/Z2)  +  (1/Z3)\n",
      " #Current  I\n",
      "I  =  V*YT\n",
      "I1  =  V/Z1\n",
      "I2  =  V/Z2\n",
      "I3  =  V/Z2\n",
      "thetav  =  0\n",
      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
      "phi  =  thetav  -  thetai  \n",
      "if  (phi>0):\n",
      "         a  =  \"lagging\"\n",
      "else:\n",
      "         a  =  \"leading\"\n",
      "\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  current,  I  is  (\",round(I.real,2),\"  +  (\",round(I.imag,2),\")i)  A,\"\n",
      "print   \"and  its  phase  relative  to  the  40  V  supply  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  current,  I  is  ( 10.98   +  ( 3.56 )i)  A,\n",
        "and  its  phase  relative  to  the  40  V  supply  is   leading s  by   17.96 deg\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 6, page no. 451</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine (a) the total equivalent circuit impedance, (b) the supply current, \n",
      "#(c) the circuit phase angle, (d) the current in the coil, and (e) the current in the capacitor.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "L  =  0.07958;#  in  Henry\n",
      "R  =  18;#  in  ohm\n",
      "C  =  64.96E-6;#  in  Farad\n",
      "rv  =  250;#  in  volts\n",
      "thetav  =  0;#  in  degrees\n",
      "f  =  50;#  in  Hz\n",
      "\n",
      "#calculation:\n",
      " #Inductive  reactance\n",
      "XL  =  2*math.pi*f*L\n",
      " #capacitive  reactance\n",
      "Xc  =  1/(2*math.pi*f*C)\n",
      " #impedance  of  the  coil,\n",
      "Zcoil  =  R  +  1j*XL\n",
      " #impedance  presented  by  the  capacitor,\n",
      "Zc  =  -1j*Xc\n",
      " #Total  equivalent  circuit  impedance,\n",
      "ZT  =  Zcoil*Zc/(Zcoil  +  Zc)\n",
      " #voltage\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #current,  I\n",
      "I  =  V/ZT\n",
      "thetai  =  cmath.phase(complex(I.real,I.imag))*180/math.pi\n",
      "phi  =  thetav  -  thetai\n",
      "if  (phi>0):\n",
      "         a  =  \"lagging\"\n",
      "else:\n",
      "         a  =  \"leading\"\n",
      "\n",
      " #Current  in  the  coil,  ICOIL\n",
      "Icoil  =  V/Zcoil\n",
      " #Current  in  the  capacitor,  IC\n",
      "Ic  =  V/Zc\n",
      "\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)the  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
      "print  \"\\n  (b)supply  current,  I  = \",round(abs(I),2),\"/_\",round(cmath.phase(complex(I.real, I.imag))*180/math.pi,2),\"deg  A\\n\"\n",
      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"s  by  \",round(abs(phi),2),\"deg\\n\"\n",
      "print  \"\\n  (d)current  in  coil,  Icoil  = \",round(abs(Icoil),2),\"/_\",round(cmath.phase(complex(Icoil.real, Icoil.imag))*180/math.pi,2),\"deg  A\\n\"\n",
      "print  \"\\n  (e)current  in  capacitor,  Ic  = \",round(abs(Ic),2),\"/_\",round(cmath.phase(complex(Ic.real, Ic.imag))*180/math.pi,2),\"deg A\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)the  circuit  impedance  is   48.02   +  ( 15.03 )i  ohm\n",
        "\n",
        "\n",
        "  (b)supply  current,  I  =  4.97 /_ -17.38 deg  A\n",
        "\n",
        "\n",
        "  (c)circuit  phase  relative  is   lagging s  by   17.38 deg\n",
        "\n",
        "\n",
        "  (d)current  in  coil,  Icoil  =  8.12 /_ -54.25 deg  A\n",
        "\n",
        "\n",
        "  (e)current  in  capacitor,  Ic  =  5.1 /_ 90.0 deg A"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 7, page no. 452</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#(a)determine the value of impedance Z1 \n",
      "#(b) If the supply frequency is 5 kHz,determine the value of the components comprising impedance Z1\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "RL  =  6j;#  in  ohm\n",
      "R2  =  8;#  in  ohm\n",
      "Z3  =  10;#  in  ohm\n",
      "rv  =  50;#  in  volts\n",
      "thetav  =  30;#  in  degrees\n",
      "ri  =  31.4;#  in  amperes\n",
      "thetai  =  52.48;#  in  degrees\n",
      "f  =  5000;#  in  Hz\n",
      "\n",
      "#calculation:\n",
      " #impedance,  Z2\n",
      "Z2  =  R2  +  RL\n",
      " #voltage\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #current,  I\n",
      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
      " #Total  circuit  admittance,\n",
      "YT  =  I/V\n",
      " #admittance,  Y3\n",
      "Y3  =  1/Z3\n",
      " #admittance,  Y2\n",
      "Y2  =  1/Z2\n",
      " #admittance,  Y1\n",
      "Y1  =  YT  -  Y2  -  Y3\n",
      " #impedance,  Z1\n",
      "Z1  =  1/Y1\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)the  impedance  Z1  is  \",round(Z1.real,2),\"  +  (\",round(  Z1.imag,2),\")i  ohm\\n\"\n",
      "\n",
      " #resistance,  R1\n",
      "R1  =  Z1.real\n",
      "X1  =  Z1.imag  \n",
      "if  ((R1>0)&(X1<0)):\n",
      "    C1  =  -1/(2*math.pi*f*X1)\n",
      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm\"\n",
      "    print   \"  and  a  capacitor  of  capacitance  \",round(C1*1E6,2),\"uFarad\\n\"\n",
      "elif  ((R1>0)&(X1>0)):\n",
      "    L1  =  2*math.pi*f*X1\n",
      "    print  \"\\n  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance  \",round(R1,2),\"  ohm \"\n",
      "    print   \" and  a  inductor  of  insuctance  \",round(L1*1000,2),\"mHenry\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)the  impedance  Z1  is   1.6   +  ( -1.2 )i  ohm\n",
        "\n",
        "\n",
        "  (b)The  series  circuit  thus  consists  of  a  resistor  of  resistance   1.6   ohm\n",
        "  and  a  capacitor  of  capacitance   26.55 uFarad\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 8, page no. 453</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine (a) the equivalent series circuit impedance,\n",
      "#(b) the supply current I, (c) the circuit phase angle,\n",
      "#(d) the values of voltages V1 and V2, and (e) the values of currents IA and IB\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "RL1  =  1.02j;#  in  ohm\n",
      "R1  =  1.65;#  in  ohm\n",
      "RLa  =  7j;#  in  ohm\n",
      "Ra  =  5;#  in  ohm\n",
      "Rcb  =  -1j*15;#  in  ohm\n",
      "Rb  =  4;#  in  ohm\n",
      "rv  =  91;#  in  volts\n",
      "thetav  =  0;#  in  degree\n",
      "\n",
      "#calculation:\n",
      " #voltage\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #impedance,  Z1\n",
      "Z1  =  R1  +  RL1\n",
      " #impedance,  Za\n",
      "Za  =  Ra  +  RLa\n",
      " #impedance,  Zb\n",
      "Zb  =  Rb  +  Rcb\n",
      " #impedance,  Z,  of  the  two  branches  connected  in  parallel\n",
      "Z  =  Za*Zb/(Za  +  Zb)\n",
      " #Total  circuit  impedance\n",
      "ZT  =  Z1  +  Z\n",
      " #Supply  current,  I\n",
      "I  =  V/ZT\n",
      "thetai  =  cmath.phase(complex(I.real, I.imag))*180/math.pi\n",
      "phi  =  thetav  -  thetai  \n",
      "if  (phi>0):\n",
      "         a  =  \"lagging\"\n",
      "else:\n",
      "         a  =  \"leading\"\n",
      "\n",
      " #Voltage  V1\n",
      "V1  =  I*Z1\n",
      " #Voltage  V2\n",
      "V2  =  I*Z\n",
      " #current  Ia\n",
      "Ia  =  V2/Za\n",
      " #Current  Ib\n",
      "Ib  =  V2/Zb\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)equivalent  series  circuit  impedance  is  \",round(ZT.real,2),\"  +  (\",round(  ZT.imag,2),\")i  ohm\\n\"\n",
      "print  \"\\n  (b)supply  current,  I  is  \",round(I.real,2),\"  +  (\",round(  I.imag,2),\")i  A\\n\"\n",
      "print  \"\\n  (c)circuit  phase  relative  is  \",a,\"  by  \",round(abs(phi),2),\"deg\\n\"\n",
      "print  \"\\n  (d)voltage,  V1  is  (\",round(V1.real,2),\"  +  (\",round(V1.imag,2),\")i)  V  and  V2  is(\",round(V2.real,2),\"  +  (\",round(  V2.imag,2),\")i)  V\\n\"\n",
      "print  \"\\n  (e)current,  Ia  is  (\",round(Ia.real,2),\"  +  (\",round( Ia.imag,2),\")i)  A  and  Ib  is(\",round(Ib.real,2),\"  +  (\",round(  Ib.imag,2),\")i)  A\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)equivalent  series  circuit  impedance  is   12.0   +  ( 5.0 )i  ohm\n",
        "\n",
        "\n",
        "  (b)supply  current,  I  is   6.46   +  ( -2.69 )i  A\n",
        "\n",
        "\n",
        "  (c)circuit  phase  relative  is   lagging   by   22.61 deg\n",
        "\n",
        "\n",
        "  (d)voltage,  V1  is  ( 13.41   +  ( 2.15 )i)  V  and  V2  is( 77.59   +  ( -2.15 )i)  V\n",
        "\n",
        "\n",
        "  (e)current,  Ia  is  ( 5.04   +  ( -7.49 )i)  A  and  Ib  is( 1.42   +  ( 4.79 )i)  A"
       ]
      }
     ],
     "prompt_number": 8
    }
   ],
   "metadata": {}
  }
 ]
}