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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3cfae8c34ec0891b928806727193cd8cb859809240d182673d6b8dcfad3eeded"
+ },
+ "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",
+      "%matplotlib 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",
+        "\n",
+        "\n",
+        "The required plot is shown below: "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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DzgBQ540f/yVERX397wPeAQB6KaCvnQWTeMPbdcTeWh950XHbW2twbc/R+5Ur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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x4950910>"
+       ]
+      }
+     ],
+     "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": {}
+  }
+ ]
+}
\ No newline at end of file