path: root/Basic_Electrical_Engineering/Chapter4.ipynb

{
 "metadata": {
  "name": ""
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 4: NETWORK THEOREMS"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.1,Page number: 105"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the current I in the circuit using superposition theorem.\"\"\"\n",
      "\n",
      "#Calculations:\n",
      "I1=-0.5*(0.3/(0.1+0.3))\n",
      "I2=80e-03/(0.1+0.3)\n",
      "I=I1+I2\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The current I in the circuit is %.3f A.\" %(I)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current I in the circuit is -0.175 A.\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.2,Page number: 106"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding current I_x in the network using superposition theorem.\"\"\"\n",
      "\n",
      "#Calculations:\n",
      "I1=10.0/(50+150)\n",
      "I2=40*(150.0/(50+150))\n",
      "I3=-120*(50.0/(150+50))\n",
      "Ix=I1+I2+I3\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The current Ix determined using superposition principle is %.2f A.\" %(Ix)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current Ix determined using superposition principle is 0.05 A.\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.3,Page number: 106"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the voltage across resistor by applying the principle of superposition.\"\"\"\n",
      "\n",
      "#Calculations:\n",
      "i=4.0*(1.0/(1.0+(2+3)))\n",
      "R=3.0\n",
      "v_4=i*R\n",
      "v_5=(-5*(1.0/(1+(2+3))))*R\n",
      "v_6=6.0*(3.0/(1+2+3))\n",
      "v=+v_4+v_5+v_6\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The total voltage(v) across the 3 ohm resistor is %.2f V.\" %(v)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The total voltage(v) across the 3 ohm resistor is 2.50 V.\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.4,Page number: 108 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the value of I_s to reduce the voltage across the 4-ohm resistor to zero. \"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "V_source=10.0         #Voltage of the source(in Volts) \n",
      "\n",
      "\n",
      "#Calculations:\n",
      "I1=V_source/(2+4+6)\n",
      "\n",
      "\"\"\" I2(from top to bottom in the 4 ohm resistor) = -Is*((2+6)/(2+6+4)) = -(2/3)*Is ;\n",
      "    \n",
      "    The voltage across the 4 ohm resistor can be zero,only if the current through this resistor is zero.  \n",
      "    \n",
      "     I1+I2=0; \"\"\"\n",
      "\n",
      "Is=I1*(3.0/2)\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The current Is to reduce the voltage across the 4 ohm resistor to zero is %.2f A.\" %(Is) "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current Is to reduce the voltage across the 4 ohm resistor to zero is 1.25 A.\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.5,Page number: 110"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question: \n",
      "\"\"\"Finding the voltage across the load resistor using Thevenin's theorem.\"\"\"\n",
      "\n",
      "#Calculations:\n",
      "I1=(50.0-10.0)/(10+10+20)\n",
      "I2=1.5*(10.0/(10.0+(10+20)))\n",
      "I=I1+I2\n",
      "V_Th=I*20\n",
      "R_Th=1.0/((1.0/20)+(1.0/(10+10)))\n",
      "R_L=5.0\n",
      "V_L=V_Th*(R_L/(R_L+R_Th))\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The voltage across the load resistor R_L is %.2f V.\" %(V_L)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage across the load resistor R_L is 9.17 V.\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.6,Page number: 111"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the voltage across the resistor by applying Thevenin's theorem.\"\"\"\n",
      "\n",
      "#Calculations:\n",
      "V_Th=5.0*1.0\n",
      "R_Th=3.0\n",
      "R_L=3.0\n",
      "V_L=V_Th*(R_L/(R_L+R_Th))\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The voltage across the resistor by applying Thevenin's Theorem is %.2f V.\" %(V_L)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage across the resistor by applying Thevenin's Theorem is 2.50 V.\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.7,Page number: 113"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding Norton's equivalent circuit with respect to terminals AB.\"\"\"\n",
      "\n",
      "#Calculations:\n",
      "I1=10.0/5\n",
      "I2=5.0/10\n",
      "I_N=I1+I2\n",
      "R_N=1.0/((1.0/5)+(1.0/10))\n",
      "I_L=I_N*((10.0/3)/((10.0/3)+5))\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"When terminals AB are shorted, the current I_N is %.2f A.\" %(I_N)\n",
      "print \"The value of current that would flow through a load resistor of 5 ohm if it were connected across terminals AB is %.2f A.\" %(I_L)  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "When terminals AB are shorted, the current I_N is 2.50 A.\n",
        "The value of current that would flow through a load resistor of 5 ohm if it were connected across terminals AB is 1.00 A.\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.8,Page number: 114 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the available power from the battery.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "Voc=12.6              #Open-circuit voltage(in Volts)\n",
      "Isc=300.0             #Short-circuit voltage(in Volts)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "Ro=Voc/Isc\n",
      "P_avl=(Voc*Voc)/(4*Ro)\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The available power from the battery is %.2f W.\" %(P_avl)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The available power from the battery is 945.00 W.\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.9,Page number: 115 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the available power from the battery.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "no_of_cells=8         #Number of dry cells in the battery\n",
      "emf=1.5               #EMF of each cell(in Volts)\n",
      "int_res=0.75          #Internal resistanceof each cell(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "Voc=no_of_cells*emf\n",
      "Ro=no_of_cells*int_res\n",
      "P_avl=(Voc*Voc)/(4.0*Ro)\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The available power from the battery is %.2f W.\" %(P_avl)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The available power from the battery is 6.00 W.\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.10,Page number: 115 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Plotting a curve showing the variation of the output power Po with the load resistance R_L.\"\"\"\n",
      "\n",
      "from math import sqrt\n",
      "\n",
      "#Variable Declaration:\n",
      "P=25.0                #Power to be delivered to the speaker(in Watts)\n",
      "Ro=8.0                #Resistance of the speaker(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "V_Th=sqrt(P*4*Ro)\n",
      "Vo=V_Th/2.0\n",
      "R_Th=Ro\n",
      "R_L=0.0\n",
      "P1=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
      "R_L=2.0\n",
      "P2=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
      "R_L=4.0\n",
      "P3=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
      "R_L=6.0\n",
      "P4=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
      "R_L=8.0\n",
      "P5=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
      "R_L=16.0\n",
      "P6=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
      "R_L=32.0\n",
      "P7=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
      "P8=0.0\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"(a)The voltage provided by this amplifier to to the speaker is %.2f V.\" %(Vo)\n",
      "print \"(b)(i)If the load is a short circuit(R_L=0),the output voltage Vo would be zero and hence the output power,Po= Vo*I_L = 0*I_L =0.\"\n",
      "print \"   (ii) If the load is an open circuit(R_L=0),the load current I_L would be zero and hence the output power,Po= Vo*I_L = Vo*0 = 0.\"\n",
      "print \"(c)(i) For R_L = 0 , Po = %.2f W. \\n   (ii) For R_L = 2 Ohms, Po = %.2f W.\" %(P1,P2)  \n",
      "print \"   (iii) For R_L = 4 Ohms, Po = %.2f W. \\n   (iv) For R_L = 6 Ohms, Po = %.2f W.\" %(P3,P4)  \n",
      "print \"   (v) For R_L = 8 Ohms, Po = %.2f W. \\n   (vi) For R_L = 16 Ohms, Po = %.2f W.\" %(P5,P6) \n",
      "print \"   (vii) For R_L = 32 Ohms, Po = %.2f W. \\n   (viii) For R_L = infinity, Po = %.2f W.\" %(P7,P8)\n",
      "print \"Note: Po is the power delivered to the speaker (in Watts) and R_L is the speaker resistance(in Ohms).\"  \n",
      "\n",
      "\n",
      "from __future__ import division\n",
      "from pylab import *\n",
      "from matplotlib import *\n",
      "import numpy as np\n",
      "%pylab inline\n",
      "\n",
      "\n",
      "#Variable declaration:\n",
      "Rdata=[0.0, 2.0, 4.0, 6.0, 8.0, 16.0]     #(in ohm)\n",
      "Podata=[0, 16, 22.22, 24.49, 25, 22.22]   #(in Watt)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "R=np.array(Rdata)\n",
      "Po=np.array(Podata)\n",
      "length=len(R)\n",
      "Rmax=R[length-1]\n",
      "a=polyfit(R,Po,4)\n",
      "Rfit=[0]*100\n",
      "Pofit=[0]*100\n",
      "for n in range(1,100,1):\n",
      "    Rfit[n-1]=Rmax*(n-1)/100\n",
      "    Pofit[n-1]=a[0]*Rfit[n-1]**4+a[1]*Rfit[n-1]**3+a[2]*Rfit[n-1]**2+a[3]*Rfit[n-1]+a[4]\n",
      "\n",
      "#Plot the data and then the fit to compare (convert xfit to cm and Lfit to mH)\n",
      "plot(Rdata,Podata,'o')\n",
      "plot(np.array(Rfit),np.array(Pofit),'g.') \n",
      "xlabel('R_L (in ohm) ')\n",
      "ylabel('Po (in Watt) ')\n",
      "title('Ouput power(Po) vs. Speaker Resistance(R_L)')\n",
      "grid()\n",
      "print \"\\n\\nThe required plot is shown below: \"\n",
      "show()\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)The voltage provided by this amplifier to to the speaker is 14.14 V.\n",
        "(b)(i)If the load is a short circuit(R_L=0),the output voltage Vo would be zero and hence the output power,Po= Vo*I_L = 0*I_L =0.\n",
        "   (ii) If the load is an open circuit(R_L=0),the load current I_L would be zero and hence the output power,Po= Vo*I_L = Vo*0 = 0.\n",
        "(c)(i) For R_L = 0 , Po = 0.00 W. \n",
        "   (ii) For R_L = 2 Ohms, Po = 16.00 W.\n",
        "   (iii) For R_L = 4 Ohms, Po = 22.22 W. \n",
        "   (iv) For R_L = 6 Ohms, Po = 24.49 W.\n",
        "   (v) For R_L = 8 Ohms, Po = 25.00 W. \n",
        "   (vi) For R_L = 16 Ohms, Po = 22.22 W.\n",
        "   (vii) For R_L = 32 Ohms, Po = 16.00 W. \n",
        "   (viii) For R_L = infinity, Po = 0.00 W.\n",
        "Note: Po is the power delivered to the speaker (in Watts) and R_L is the speaker resistance(in Ohms).\n",
        "Populating the interactive namespace from numpy and matplotlib\n",
        "\n",
        "\n",
        "The required plot is shown below: "
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stderr",
       "text": [
        "WARNING: pylab import has clobbered these variables: ['streamplot', 'rc', 'tri', 'axes', 'legend', 'rc_context', 'figure', 'f', 'quiver', 'axis', 'linalg', 'draw_if_interactive', 'text', 'random', 'colors', 'stackplot', 'contour', 'colorbar', 'rcdefaults', 'table', 'power', 'info', 'fft', 'test']\n",
        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": "iVBORw0KGgoAAAANSUhEUgAAAYAAAAEaCAYAAAAYOoCaAAAABHNCSVQICAgIfAhkiAAAAAlwSFlz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\nxQ0AY4w1UdwAMMZYE8UNAGOMNVH/B1z6XURJsXa0AAAAAElFTkSuQmCC\n",
       "text": [
        "<matplotlib.figure.Figure at 0x43b9d10>"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.11,Page number: 115"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the current in branch B due to the voltage source of 36 V in branch A.\"\"\"\n",
      "\n",
      "\n",
      "#Variable Declaration:\n",
      "V_supply_a=36.0                         #Supply voltage in the network shown in figure a(in Volts)\n",
      "Req_a=2+ 1/((1.0/12)+(1.0/(3+1))) + 4   #Equivalent resistance of network shown in figure a(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "I_supply_a=V_supply_a/Req_a\n",
      "I=I_supply_a*(12.0/(12+4))\n",
      "V_supply_b=36.0\n",
      "Req_b= 3+ 1.0/((1.0/12)+(1.0/(2+4))) + 1\n",
      "I_supply_b=V_supply_b/Req_b\n",
      "I_dash= I_supply_b*(12.0/(12+6))\n",
      "R_tr=V_supply_a/I\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The current I(in figure (a)) is %.2f A and the current I'(in branch A in figure (b)) is %.2f A.\" %(I,I_dash)\n",
      "if(I==I_dash) : print \"As the two currents I and I' have the same value, the reciprocity theorem is established.\\n \"\n",
      "else : print \"As the two currents I and I' have different values, the reciprocity theorem is not established.\\n \"\n",
      "print \"The transfer resistance is %.2f Ohms.\" %(R_tr)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current I(in figure (a)) is 3.00 A and the current I'(in branch A in figure (b)) is 3.00 A.\n",
        "As the two currents I and I' have the same value, the reciprocity theorem is established.\n",
        " \n",
        "The transfer resistance is 12.00 Ohms.\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.13,Page number: 119"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the current using Thevenin's theorem.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "R_L=20.0              #Load resistance(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "\"\"\" KVL equation for the loop bacdeb, V_Th+50-20+10=0.\"\"\"\n",
      "V_Th=-50.0-10.0+20.0\n",
      "\"\"\"Shorting all the voltage sources.\"\"\"\n",
      "R_Th=0.0\n",
      "I3=V_Th/(R_L+R_Th)\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The current I3 in the circuit is %.2f A.\" %(I3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current I3 in the circuit is -2.00 A.\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.14,Page number: 120"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the current using Thevenin's theorem.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "R_L=1.0              #Load resistance(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "\"\"\" KVL equation for the loop bacdeb, V_Th+50-20+10=0.\"\"\"\n",
      "V_Th=20.0-10.0-9.0\n",
      "\"\"\"Shorting all the voltage sources.\"\"\"\n",
      "R_Th=0.0\n",
      "I2=V_Th/(R_L+R_Th)\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The current I2 in the circuit is %.2f A.\" %(I2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current I2 in the circuit is 1.00 A.\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.15,Page number: 121"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the current using superposition theorem.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "R=0.5                 #Load resistance(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "I=15.0/(1+0.6)\n",
      "I1=I*(1/(1+(R+1)))\n",
      "I2=20/((2/3.0)+2)\n",
      "I1_20=I2*(2.0/(2.0+(R+R)))\n",
      "Inet=I1-I1_20\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The current flowing through R from A to B is %.2f A.\" %(Inet) "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current flowing through R from A to B is -1.25 A.\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.16,Page number: 121"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the voltage across the resistance using Thevenin's theorem.\"\"\"\n",
      "\n",
      "#Calculations:\n",
      "\"\"\"The first step is to remove the 5 Ohms resistance.Then, find the open-circuit voltage Voc across AB. \n",
      "   Using nodal analysis,for node C we can write ((Voc-20)/2)+((Voc+10)/1)+((Voc-12)/4)=10.\"\"\"\n",
      "Voc=(10+3)/(0.5+1+0.25)\n",
      "V_Th=Voc\n",
      "R_Th=1.0/((1.0/2.0)+(1.0)+(1/4.0))\n",
      "V_AB=V_Th*(5.0/(5.0+R_Th))\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The voltage across the 5 Ohms resistance is %.2f V.\" %(V_AB)\n",
      "print \"Note:There is a calculation error in the textbook.Voc=7.43 V but not 1.74 V. Therefore V_AB=6.67 V.\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage across the 5 Ohms resistance is 6.67 V.\n",
        "Note:There is a calculation error in the textbook.Voc=7.43 V but not 1.74 V. Therefore V_AB=6.67 V.\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.17,Page number: 122"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the current using Norton's theorem.\"\"\"\n",
      "\n",
      "#Calculations:\n",
      "Isc=20.0/10.0\n",
      "I_N=Isc\n",
      "R_N=1.0/((1.0/10)+(1.0/10))\n",
      "I=I_N*(R_N/(R_N+2.0))\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The current through the 2 Ohms resistance when connected across terminals AB is %.2f A.\" %(I)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current through the 2 Ohms resistance when connected across terminals AB is 1.43 A.\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.18,Page number: 123"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finfing the power consumed by resistor using Norton's theorem.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "R_L=2.0               #Resistance of load(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "\"\"\"To determine I_N,short circuit the terminals xy and find the current Isc throgh this short circuit.\n",
      "\n",
      "   Applying node-voltage analysis to determine the voltage at node 1,\n",
      "   \n",
      "   ((V1-12)/3)+(V1/2)=4;\"\"\"\n",
      "\n",
      "V1=(4.0+(12.0/3.0))/((1.0/3.0)+0.5)\n",
      "Isc=V1/2.0\n",
      "I_N=Isc\n",
      "R_N=1.0/((1.0/(3.0+2.0))+(1.0/5.0))\n",
      "I=I_N*(R_N/(R_N+R_L))\n",
      "P=I*I*R_L\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The power consumed by the 2 Ohms load reistor is %.3f W.\" %(P)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The power consumed by the 2 Ohms load reistor is 14.222 W.\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.19,Page number: 124"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the voltage across load.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "R_L=5.0               #Load resistance(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "R_N=1.0/((1.0/20.0)+(1.0/(10.0+10.0)))\n",
      "I_N=(65.0-10.0)/(10.0+10.0)\n",
      "I_L=I_N*(R_N/(R_N+R_L))\n",
      "V_L=I_L*R_L\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The voltage across the load using Norton's theorem is %.2f V.\" %(V_L)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The voltage across the load using Norton's theorem is 9.17 V.\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.20,Page number: 125 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the reading of voltmeter.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "R=500e03              #Resistance across which voltmeter is connected(in Ohms)\n",
      "Rm=10e06              #Internal resistance of voltmeter(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "V_a=10.0*(R/(R+800e03))\n",
      "R_Th=1.0/((1/R)+(1/800e03))\n",
      "V_Th=V_a\n",
      "V_b=V_Th*(Rm/(Rm+R_Th))\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"(a)The reading of the voltmeter if it assumed to be ideal is %.2f V.\" %(V_a)\n",
      "print \"(b)The reading of the voltmeter if it has an internal resistance of 10 M Ohms is %.2f V.\" %(V_b)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)The reading of the voltmeter if it assumed to be ideal is 3.85 V.\n",
        "(b)The reading of the voltmeter if it has an internal resistance of 10 M Ohms is 3.73 V.\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.21,Page number: 125"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the load resistance for receiving maximum power.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "R_L=3e03             #Load resistance(in Ohms)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "\"\"\"First, we remove R_L and then find Voc=V_Th. We transform the 5-mA current source into voltage source.\"\"\"\n",
      "Voc=(60.0-40.0)*((6e03)/((6e03)+(12e03+18e03)))\n",
      "V_Th=Voc\n",
      "R_Th=1.0/((1.0/6e03)+(1.0/(12e03+18e03)))\n",
      "Vab_b=V_Th*(R_L/(R_L+R_Th))\n",
      "R_L_c=R_Th\n",
      "\"\"\" The current required through 6 kilo Ohms resistor is 0.1 mA.\"\"\"\n",
      "Vab_d=(0.1e-03)*(6e03)\n",
      "V_30k=20-Vab_d\n",
      "I_30k=V_30k/(30e03)\n",
      "I_L=I_30k-(0.1e-03)\n",
      "R_L_d=Vab_d/I_L\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"(a)The Thevenin's equivalent voltage is %.2f V and the Thevenin's resistance is %.2f kilo Ohms.\" %(V_Th,(R_Th/1000.0))\n",
      "print \"(b)The Vab for R_L=3 kilo Ohms is %.3f V.\" %(Vab_b)\n",
      "print \"(c)The value of R_L which receives maximum power from the circuit is %.2f kilo Ohms.\" %(R_L_c/1000.0)\n",
      "print \"(d)The value of R_L which makes the current in the 6 kilo Ohms resistor to be 0.1 mA is %.3f kilo Ohms.\" %(R_L_d/1000.0) "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)The Thevenin's equivalent voltage is 3.33 V and the Thevenin's resistance is 5.00 kilo Ohms.\n",
        "(b)The Vab for R_L=3 kilo Ohms is 1.250 V.\n",
        "(c)The value of R_L which receives maximum power from the circuit is 5.00 kilo Ohms.\n",
        "(d)The value of R_L which makes the current in the 6 kilo Ohms resistor to be 0.1 mA is 1.098 kilo Ohms.\n"
       ]
      }
     ],
     "prompt_number": 29
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.22,Page number: 127"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the value of resistance.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "V=12.0                #Supply voltage(in Volts)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "\"\"\" To receive maximum power,the resistance R should be equal to the output resistance of the remaining circuit,which is same as \n",
      "    Thevenin's resistance.\"\"\"\n",
      "R_Th=1.0/((1.0/2.0)+(1.0/6.0))\n",
      "R=R_Th\n",
      "\"\"\" Mesh equations to find loop currents I1 and I2, \n",
      "    \n",
      "    (8*I1)-(6*I2)=4 and -(6*I1)+(7.5*I2)=8. Solving these equations, we get I1=3.25 A. \"\"\"\n",
      "I1=((4*7.5)+(8*6))/((8*7.5)-(6*6))\n",
      "P=V*I1\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"(a)The value of R to receive maximum power from the circuit is %.2f Ohms.\" %(R)\n",
      "print \"(b)The power supplied by the 12 V source is %.2f W.\" %(P) "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)The value of R to receive maximum power from the circuit is 1.50 Ohms.\n",
        "(b)The power supplied by the 12 V source is 39.00 W.\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.23,Page number: 127"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the current Is.\"\"\"\n",
      "\n",
      "#Calculations:\n",
      "I1=12.0/(80.0+120)\n",
      "\"\"\" I2=Is*(80/(80+120)); I1+I2=0;\"\"\"\n",
      "Is=-I1/(80.0/(80+120))\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The current Is such that the current through 120 Ohms resistor is zero is %.2f A.\" %(Is)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The current Is such that the current through 120 Ohms resistor is zero is -0.15 A.\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.24,Page number: 128"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Question:\n",
      "\"\"\"Finding the Thevenin equivalent circuit.\"\"\"\n",
      "\n",
      "#Variable Declaration:\n",
      "Vs=12.0               #Supply voltage(in Volts)\n",
      "\n",
      "\n",
      "#Calculations:\n",
      "\"\"\"The circuit has no independent source.So,the current I in the 12 Ohms resistor will be zero and hence the dependent voltage source\n",
      "   (8*I) will also be zero. Obviously,Thevenin's voltage V_Th=0 V. \"\"\"\n",
      "V_Th=0.0\n",
      "I=12.0/12.0\n",
      "V=8*I\n",
      "\"\"\"Applying KCL to node a,\"\"\"\n",
      "Is=(12.0/12.0)+(12.0/6.0)+((12.0-8.0)/4.0)\n",
      "R_Th=Vs/Is\n",
      "\n",
      "\n",
      "#Result:\n",
      "print \"The Thevenin's equivalent resistance is %.2f Ohms and the Thevenin's equivalent voltage is %.2f V.\" %(R_Th,V_Th)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Thevenin's equivalent resistance is 3.00 Ohms and the Thevenin's equivalent voltage is 0.00 V.\n"
       ]
      }
     ],
     "prompt_number": 3
    }
   ],
   "metadata": {}
  }
 ]
}