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  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h1>Chapter 31: Mesh-current and nodal analysis</h1>"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 1, page no. 546</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Use mesh-current analysis to determine the current flowing in \n",
      "#(a) the 5 ohm resistance, and (b) the 1 ohm resistance of the d.c. circuit\n",
      "from __future__ import division\n",
      "import math\n",
      "import numpy\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V1  =  4;#  in  volts\n",
      "V2  =  5;#  in  volts\n",
      "R1  =  3;#  in  ohm\n",
      "R2  =  5;#  in  ohm\n",
      "R3  =  4;#  in  ohm\n",
      "R4  =  1;#  in  ohm\n",
      "R5  =  6;#  in  ohm\n",
      "R6  =  8;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      " #The  mesh  currents  I1,  I2  and  I3  are  shown  in  Figure  31.2.  Using  Kirchhoff\u2019s  voltage  law  in  3  loops\n",
      " #three  eqns  obtained\n",
      " #(R1  +  R2)*I1  -  R2*I2  =  V1\n",
      " #-1*R2*I1  +  (R2  +  R3  +  R4  +  R5)*I2  -  R4*I3  =  0\n",
      " #  -1*R4*I2  +  (R4  +  R6)*I3  =  -1*V2\n",
      " #using  determinants\n",
      "d1  =  [[V1,  -1*R2,  0],[0,  (R2  +  R3  +  R4  +  R5),  -1*R4],[-1*V2,  -1*R4,  (R4  +  R6)]]\n",
      "D1  =  numpy.linalg.det(d1)\n",
      "d2  =  [[(R1  +  R2),  V1,  0],[-1*R2,  0,  -1*R4],[0,  -1*V2,  (R4  +  R6)]]\n",
      "D2  =  numpy.linalg.det(d2)\n",
      "d3  =  [[(R1  +  R2),  -1*R2,  V1],[-1*R2,  (R2  +  R3  +  R4  +  R5),  0],[0,  -1*R4,  -1*V2]]\n",
      "D3  =  numpy.linalg.det(d3)\n",
      "d  =  [[(R1  +  R2),  -1*R2,  0],[-1*R2,  (R2  +  R3  +  R4  +  R5),  -1*R4],[0,  -1*R4,  (R4  +  R6)]]\n",
      "D  =  numpy.linalg.det(d)\n",
      "I1  =  D1/D\n",
      "I2  =  D2/D\n",
      "I3  =  D3/D  \n",
      "IR2  =  I1  -  I2\n",
      "IR4  =  I2  -  I3\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)current  in  the  5  ohm  resistance  is  \",round(IR2,2),\"  A\"\n",
      "print  \"\\n  (b)current  in  the  1  ohm  resistance  is  \",round(IR4,2),\"  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)current  in  the  5  ohm  resistance  is   0.44   A\n",
        "\n",
        "  (b)current  in  the  1  ohm  resistance  is   0.69   A"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 2, page no. 547</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine, using mesh-current analysis, \n",
      "#(a) the mesh currents I1 and I2 \n",
      "#(b) the current flowing in the capacitor,\n",
      "#(c) the active power delivered by the voltage source.\n",
      "from __future__ import division\n",
      "import math\n",
      "import numpy\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv  =  100;#  in  volts\n",
      "thetav  =  0;#  in  degrees\n",
      "R1  =  5;#  in  ohm\n",
      "R2  =  -1j*4;#  in  ohm\n",
      "R3  =  4;#  in  ohm\n",
      "R4  =  3j;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltages\n",
      "V  =  rv*math.cos(thetav*math.pi/180)  +  1j*rv*math.sin(thetav*math.pi/180)\n",
      " #Currents  I1,  I2  with  their  directions  are  shown  in  Figure  31.03.\n",
      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
      " #using  kirchoff  rule  in  2  loops\n",
      " #two  eqns  obtained\n",
      " #(R1  +  R2)*I1  -  R2*I2  =  V\n",
      " #-1*R2*I1  +  (R3  +  R2  +  R4)*I2  =  0\n",
      " #using  determinants\n",
      "d1  =  [[V,  -1*R2],[0,  (R3  +  R2  +  R4)]]\n",
      "D1  =  numpy.linalg.det(d1)\n",
      "d2  =  [[(R1  +  R2),  V],[-1*R2,  0]]\n",
      "D2  =  numpy.linalg.det(d2)\n",
      "d  =  [[(R1  +  R2),  -1*R2],[-1*R2,  (R3  +  R2  +  R4)]]\n",
      "D  =  numpy.linalg.det(d)\n",
      "I1  =  D1/D\n",
      "I2  =  D2/D\n",
      "I1mag  =  abs(I1)\n",
      " #Current  flowing  in  capacitor\n",
      "Ic  =  I1  -  I2\n",
      " #Source  power  P\n",
      "phi  =  cmath.phase(complex(I1.real,I1.imag))\n",
      "P  =  V*I1mag*math.cos(phi)\n",
      "Icmag  =  abs(Ic)\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"(a)current,I1 is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A,  current,  I2  is\",round(I2.real,2),\"  +  (\",round(I2.imag,2),\")i  A\"\n",
      "print  \"(b)current  in  the  capacitor  is  \",round(Icmag,2),\"  A\"\n",
      "print  \"(c)Source  power  P  is  \",round(abs(P),2),\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "(a)current,I1 is   10.17   +  ( 3.55 )i  A,  current,  I2  is 5.73   +  ( -8.74 )i  A\n",
        "(b)current  in  the  capacitor  is   13.06   A\n",
        "(c)Source  power  P  is   1017.06   W\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 3, page no. 548</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the value of the line currents IR, IY and IB using mesh-current analysis\n",
      "from __future__ import division\n",
      "import math\n",
      "import numpy\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv1  =  415;#  in  volts\n",
      "rv2  =  415;#  in  volts\n",
      "thetav1  =  120;#  in  degrees\n",
      "thetav2  =  0;#  in  degrees\n",
      "R  =  3  +  4j;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltages\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",
      " #Two  mesh  currents  I1  and  I2  are  chosen  as  shown  in  Figure  31.4.\n",
      " #Two  loops  are  chosen.  The  choice  of  loop  directions  is  arbitrary.\n",
      " #using  kirchoff  rule  in  2  loops\n",
      " #two  eqns  obtained\n",
      " #2*R*I1  -  R*I2  =  V1\n",
      " #-1*R*I1  +  2*R*I2  =  V2\n",
      " #using  determinants\n",
      "d1  =  [[V1,  -1*R],[V2, 2*R]]\n",
      "D1  =  numpy.linalg.det(d1)\n",
      "d2  =  [[2*R, V1],[-1*R, V2]]\n",
      "D2  =  numpy.linalg.det(d2)\n",
      "d  =  [[2*R, -1*R],[-1*R,  2*R]]\n",
      "D  =  numpy.linalg.det(d)\n",
      "I1  =  D1/D\n",
      "I2  =  D2/D\n",
      "I1mag  =  abs(I1)\n",
      " #line  current  IR\n",
      "IR  =  I1\n",
      " #line  current  IB\n",
      "IB  =  -1*I2\n",
      " #line  current  IY\n",
      "IY  =  I2  -  I1\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"current, IR is\",round(IR.real,2),\"  +  (\",round(  IR.imag,2),\")i  A,  current,  IB  is\",round(IB.real,2),\"  +  (\",round(  IB.imag,2),\")i  A  and  current,  IY  is  \",round(IY.real,2),\"  +  (\",round(IY.imag,2),\")i  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "current, IR is 38.34   +  ( 28.75 )i  A,  current,  IB  is -44.07   +  ( 18.82 )i  A  and  current,  IY  is   5.73   +  ( -47.58 )i  A\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 4, page no. 551</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the voltage VAB, by using nodal analysis.\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "ri  =  20;#  in  amperes\n",
      "thetai  =  0;#  in  degrees\n",
      "R1  =  10;#  in  ohm\n",
      "R2  =  3j;#  in  ohm\n",
      "R3  =  4;#  in  ohm\n",
      "R4  =  16;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      " #current\n",
      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
      " #Figure  31.8  contains  two  principal  nodes  (at  1  and  B)  and  thus  only  one  nodal  equation  is  required. \n",
      "    #B  is  taken  as  the  reference  node  and  the  equation  for  node  1  is  obtained  as  follows. \n",
      "    #Applying  Kirchhoff\u2019s  current  law  to  node  1  gives:\n",
      " #IX  +  IY  =  I\n",
      "V1  =  I/((1/R4)  +(1/(R2    +R3)))\n",
      "IY  =  V1/(R2  +  R3)\n",
      "VAB  =  IY*R3\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  voltage  VAB  is  \",round(VAB.real,2),\"  +  (\",round(  VAB.imag,2),\")i  V\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  voltage  VAB  is   62.59   +  ( -9.39 )i  V"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 5, page no. 552</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Determine the value of voltage VXY\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv1  =  8;#  in  volts\n",
      "rv2  =  8;#  in  volts\n",
      "thetav1  =  0;#  in  degrees\n",
      "thetav2  =  90;#  in  degrees\n",
      "R1  =  5;#  in  ohm\n",
      "R2  =  6j;#  in  ohm\n",
      "R3  =  4;#  in  ohm\n",
      "R4  =  3;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      " #voltages\n",
      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
      " #The  circuit  contains  no  principal  nodes.  \n",
      "    #However,  if  point  Y  is  chosen  as  the  reference  node  then  an  equation \n",
      "    #may  be  written  for  node  X  assuming  that  current  leaves  point  X  by  both  branches\n",
      "VX  =  ((V1/(R1  +  R3)  +  V2/(R2  +  R4))/(1/(R1  +  R3)  +  1/(R2  +  R4)))\n",
      "VXY  =  VX\n",
      "\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  voltage  VXY  = \",round(abs(VXY),2),\"/_\",round(cmath.phase(complex(VXY.real, VXY.imag))*180/math.pi,2),\"deg V\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  voltage  VXY  =  9.12 /_ 52.13 deg V"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 6, page no. 553</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Use nodal analysis to determine the current flowing in each branch of the network\n",
      "from __future__ import division\n",
      "import math\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv1  =  100;#  in  volts\n",
      "rv2  =  50;#  in  volts\n",
      "thetav1  =  0;#  in  degrees\n",
      "thetav2  =  90;#  in  degrees\n",
      "R1  =  25;#  in  ohm\n",
      "R2  =  20;#  in  ohm\n",
      "R3  =  10;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      " #voltages\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",
      " #There  are  only  two  principal  nodes  in  Figure  31.10  so  only  one  nodal  equation  is  required.  \n",
      "    #Node  2  is  taken  as  the  reference  node.\n",
      " #The  equation  at  node  1  is  I1  +  I2  +  I3  =  0\n",
      "Vn1  =  ((V1/R1  +  V2/R3)/(1/R1  +  1/R2  +  1/R3))\n",
      "I1  =  (Vn1  -  V1)/R1\n",
      "I2  =  Vn1/R2\n",
      "I3  =  (Vn1  -  V2)/R3\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  current,  I1  is  \",round(I1.real,2),\"  +  (\",round(  I1.imag,2),\")i  A, \\n current,  I2  is    \",round(I2.real,2),\"  +  (\",round( I2.imag,2),\")i  A \\n and  current,  I3  is  \",round(I3.real,2),\"  +  (\",round(I3.imag,2),\")i  A\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  current,  I1  is   -3.16   +  ( 1.05 )i  A, \n",
        " current,  I2  is     1.05   +  ( 1.32 )i  A \n",
        " and  current,  I3  is   2.11   +  ( -2.37 )i  A"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 7, page no. 554</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 ohm inductance,\n",
      "#(c) the current in the 5 ohm resistance, and (d) the magnitude of the active power dissipated in the 2.5ohm resistance.\n",
      "from __future__ import division\n",
      "import math\n",
      "import numpy\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "rv1  =  25;#  in  volts\n",
      "rv2  =  25;#  in  volts\n",
      "thetav1  =  0;#  in  degrees\n",
      "thetav2  =  90;#  in  degrees\n",
      "R1  =  2;#  in  ohm\n",
      "R2  =  -1j*4;#  in  ohm\n",
      "R3  =  5;#  in  ohm\n",
      "R4  =  4j;#  in  ohm\n",
      "R5  =  2.5;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #voltages\n",
      "V1  =  rv1*math.cos(thetav1*math.pi/180)  +  1j*rv1*math.sin(thetav1*math.pi/180)\n",
      "V2  =  rv2*math.cos(thetav2*math.pi/180)  +  1j*rv2*math.sin(thetav2*math.pi/180)\n",
      " #The  equation  at  node  1\n",
      " #Vn1*(1/R1  +  1/R2  +  1/R3)  -  Vn2/R3  =  V1/R1\n",
      " #The  equation  at  node  2\n",
      " #Vn1*(-1/R3)  +  Vn2*(1/R4  +  1/R5  +  1/R3)  =  V2/R5\n",
      " #using  determinants\n",
      "d1  =  [[V1/R1,  -1/R3],[V2/R5,  (1/R4  +  1/R5  +  1/R3)]]\n",
      "D1  =  numpy.linalg.det(d1)\n",
      "d2  =  [[(1/R1  +  1/R2  +  1/R3),  V1/R1],[-1/R3,  V2/R5]]\n",
      "D2  =  numpy.linalg.det(d2)\n",
      "d  =  [[(1/R1  +  1/R2  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
      "D  =  numpy.linalg.det(d)\n",
      "Vn1  =  D1/D\n",
      "Vn2  =  D2/D\n",
      " #current  in  the  j4  ohm  inductance  is  given  by:\n",
      "I4  =  Vn2/R4\n",
      " #current  in  the  5  ohm  resistance  is  given  by:\n",
      "I3  =  (Vn1  -  Vn2)/R3\n",
      " #active  power  dissipated  in  the  2.5  ohm  resistor  is  given  by\n",
      "P5  =  R5*((Vn2  -  V2)/R5)**2\n",
      " #magnitude  of  the  active  power  dissipated\n",
      "P5mag  =  abs(P5)\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)  the  voltage  at  nodes  1  and  2  are \",round(abs(Vn1),1),\"/_\",round(cmath.phase(complex(Vn1.real, Vn1.imag))*180/math.pi,2),\"deg  V  and \",round(abs(Vn2),1),\"/_\",round(cmath.phase(complex(Vn2.real, Vn2.imag))*180/math.pi,1),\"deg  V\"\n",
      "print  \"\\n  (b)the  current  in  the  j4  ohm inductance = \",round(abs(I4),2),\"/_\",round(cmath.phase(complex(I4.real, I4.imag))*180/math.pi,2),\"deg  A\"\n",
      "print  \"\\n  (c)the  current  in  the  5  ohm resistance  = \",round(abs(I3),2),\"/_\",round(cmath.phase(complex(I3.real, I3.imag))*180/math.pi,2),\"deg  A\"\n",
      "print  \"\\n  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is  \",round(P5mag,2),\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)  the  voltage  at  nodes  1  and  2  are  17.1 /_ -5.3 deg  V  and  15.8 /_ 93.2 deg  V\n",
        "\n",
        "  (b)the  current  in  the  j4  ohm inductance =  3.95 /_ 3.23 deg  A\n",
        "\n",
        "  (c)the  current  in  the  5  ohm resistance  =  4.99 /_ -44.06 deg  A\n",
        "\n",
        "  (d)  magnitude  of  the  active  power  dissipated  in  the  2.5  ohm  resistance  is   34.4   W"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 8, page no. 556</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#determine the voltage VXY using nodal analysis.\n",
      "from __future__ import division\n",
      "import math\n",
      "import numpy\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "ri  =  25;#  in  amperes\n",
      "thetai  =  0;#  in  degrees\n",
      "R1  =  4;#  in  ohm\n",
      "R2  =  3j;#  in  ohm\n",
      "R3  =  5;#  in  ohm\n",
      "R4  =  10j;#  in  ohm\n",
      "R5  =  20j;#  in  ohm\n",
      "\n",
      " #calculation:\n",
      " #current\n",
      "I  =  ri*math.cos(thetai*math.pi/180)  +  1j*ri*math.sin(thetai*math.pi/180)\n",
      " #Node  3  is  taken  as  the  reference  node.\n",
      " #At  node  1,\n",
      " #V1*(1/(R1  +  R2)  +  1/R3)  -  V2/R3  =  I\n",
      " #The  equation  at  node  2\n",
      " #V1*(-1/R3)  +  V2*(1/R4  +  1/R5  +  1/R3)  =  0\n",
      " #using  determinants\n",
      "d1  =  [[I,  -1/R3],[0 , (1/R4  +  1/R5  +  1/R3)]]\n",
      "D1  =  numpy.linalg.det(d1)\n",
      "d2  =  [[(1/(R1  +  R2)  +  1/R3),  I],[-1/R3,  0]]\n",
      "D2  =  numpy.linalg.det(d2)\n",
      "d  =  [[(1/(R1  +  R2)  +  1/R3),  -1/R3],[-1/R3,  (1/R4  +  1/R5  +  1/R3)]]\n",
      "D  =  numpy.linalg.det(d)\n",
      "V1  =  D1/D\n",
      "V2  =  D2/D\n",
      " #the  voltage  between  point  X  and  node  3  is\n",
      "VX  =  V1*R2/(R1  +  R2)\n",
      " #Thus  the  voltage\n",
      "VY  =  V2\n",
      "VXY  =  VX  -  VY\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  voltage  VXY  is  \",round(VXY.real,2),\"  +  (\",round(  VXY.imag,2),\")i  V\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  voltage  VXY  is   -16.16   +  ( -15.05 )i  V"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "<h3>Example 9, page no. 557</h3>"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Use nodal analysis to determine the voltages at nodes 2 and 3 \n",
      "#determine the current flowing in the 2 ohm resistor and the power dissipated in the 3 ohm resistor.\n",
      "from __future__ import division\n",
      "import math\n",
      "import numpy\n",
      "import cmath\n",
      "#initializing  the  variables:\n",
      "V  =  8;#  in  volts\n",
      "R1  =  1;#  in  ohm\n",
      "R2  =  2;#  in  ohm\n",
      "R3  =  3;#  in  ohm\n",
      "R4  =  4;#  in  ohm\n",
      "R5  =  5;#  in  ohm\n",
      "R6  =  6;#  in  ohm\n",
      "\n",
      "#calculation:\n",
      " #In  Figure  31.13,  the  reference  node  is  shown  at  point  A.\n",
      " #At  node  1,\n",
      " #V1*(1/R1  +  1/R6  +  1/R5)  -  V2/R1  -  V3/R5  =  V/R5\n",
      " #The  equation  at  node  2\n",
      " #V1*(-1/R1)  +  V2*(1/R2  +  1/R1  +  1/R3)  -  V3/R3  =  0\n",
      " #At  node  3\n",
      " #  -  V1/R5  -  V2/R3  +  V3*(1/R4  +  1/R3  +  1/R5)  =  -1*V/R5\n",
      "#using  determinants\n",
      "d1  =  [[V/R5,  -1/R1,  -1/R5],[0, (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1*V/R5, -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
      "D1  =  numpy.linalg.det(d1)\n",
      "d2  =  [[(1/R1  +  1/R6  +  1/R5),  V/R5,  -1/R5],[-1/R1,  0,  -1/R3],[-1/R5,  -1*V/R5,  (1/R4  +  1/R3  +  1/R5)]]\n",
      "D2  =  numpy.linalg.det(d2)\n",
      "d3  =  [[(1/R1  +  1/R6  +  1/R5),  -1/R1,  V/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  0],[-1/R5,  -1/R3,  -1*V/R5]]\n",
      "D3  =  numpy.linalg.det(d3)\n",
      "d  =[[(1/R1  +  1/R6  +  1/R5),  -1/R1,  -1/R5],[-1/R1,  (1/R2  +  1/R1  +  1/R3),  -1/R3],[-1/R5,  -1/R3,  (1/R4  +  1/R3  +  1/R5)]]\n",
      "D  =  numpy.linalg.det(d)\n",
      "Vn1  =  D1/D\n",
      "Vn2  =  D2/D\n",
      "Vn3  =  D3/D  \n",
      " #the  current  in  the  2  ohm  resistor\n",
      "I2  =  Vn2/R2\n",
      " #power  dissipated  in  the  3  ohm  resistance\n",
      "P3  =  R3*((Vn2  -  Vn3)/R3)**2\n",
      "\n",
      "\n",
      "#Results\n",
      "print  \"\\n\\n  Result  \\n\\n\"\n",
      "print  \"\\n  (a)current  through  2  ohm  resistor  is  \",round(I2,2),\"  A\"\n",
      "print  \"\\n  (b)power  dissipated  in  the  3  ohm  resistor  is  \",round(P3,2),\"  W\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "\n",
        "  Result  \n",
        "\n",
        "\n",
        "\n",
        "  (a)current  through  2  ohm  resistor  is   0.19   A\n",
        "\n",
        "  (b)power  dissipated  in  the  3  ohm  resistor  is   1.27   W\n"
       ]
      }
     ],
     "prompt_number": 3
    }
   ],
   "metadata": {}
  }
 ]
}