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  "name": "",
  "signature": "sha256:896619f98735f380fd7764ca92373e979d96248275ce12cd21f86086d05b2351"
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "CHAPTER02 : THE CIRCUIT ELEMENTS"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E1a - Pg 21"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#1a\n",
      "V = 1.; #  voltage supply \n",
      "R = 10.; #  resistance in ohms \n",
      "I = V/R # current flowing through R\n",
      "print '%s' %(\"a)\")\n",
      "print '%s %.f' %(\"voltage across the resistor (in volts)=\",V)\n",
      "print '%s %.2f' %(\"current flowing through the resistor (in amps) =\",I)\n",
      "\n",
      "#1b\n",
      "V = 1.; #  voltage supply \n",
      "R1 = 10.; #  first resistance in ohms \n",
      "R2 = 5.; # resistance of the second resistor \n",
      "Vr1 = V * (R1/(R1 + R2)); # voltage across R1\n",
      "Vr2 = V - Vr1; # voltage across R2\n",
      "Ir = Vr1/R1; # current flowing through R\n",
      "print '%s' %(\"b)\")\n",
      "print '%s %.2f' %(\"voltage across the first resistor (in volts)=\",Vr1)\n",
      "print '%s %.2f' %(\"voltage across the second resistor (in volts)=\",Vr2)\n",
      "print '%s %.2f' %(\"current flowing through the resistor (in amps) =\",Ir)\n",
      "\n",
      "#1c\n",
      "# c - a\n",
      "R1 = 10.; #  first resistance in ohms\n",
      "R2 = 10.;\n",
      "I = 1.; #  current source \n",
      "V = I*R1; #  voltage across R\n",
      "print '%s' %(\"c - a)\")\n",
      "print '%s %.f' %(\"voltage across the resistor (in volts)=\",V)\n",
      "print '%s %.f' %(\"current flowing through the resistor (in amps) =\",I)\n",
      "# c - b\n",
      "Vr1 = I*R1; #  voltage across R1\n",
      "Vr2 = I*R2; # voltage across R2\n",
      "Vr=Vr1+Vr2;\n",
      "print '%s' %(\"c - b)\")\n",
      "print '%s %.f' %(\"voltage across the resistor (in volts)=\",Vr)\n",
      "print '%s %.f' %(\"current flowing through the resistor (in amps) =\",I)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a)\n",
        "voltage across the resistor (in volts)= 1\n",
        "current flowing through the resistor (in amps) = 0.10\n",
        "b)\n",
        "voltage across the first resistor (in volts)= 0.67\n",
        "voltage across the second resistor (in volts)= 0.33\n",
        "current flowing through the resistor (in amps) = 0.07\n",
        "c - a)\n",
        "voltage across the resistor (in volts)= 10\n",
        "current flowing through the resistor (in amps) = 1\n",
        "c - b)\n",
        "voltage across the resistor (in volts)= 20\n",
        "current flowing through the resistor (in amps) = 1\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E2 - Pg 25"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "R = 100.; #  resistance in ohms\n",
      "I = 0.3; #  current in amps \n",
      "P = I**2 * R; #  power \n",
      "# power specification of the resistors available in the stock \n",
      "Pa = 5.;\n",
      "Pb = 7.5;\n",
      "Pc = 10.;\n",
      "\n",
      "if Pa > P :\n",
      "    print '%s' %(\"we should select resistor a\")\n",
      "if Pb > P :\n",
      "    print '%s' %(\"we should select resistor b\")\n",
      "if Pc > P :\n",
      "    print '%s' %(\"we should select resistor c\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "we should select resistor c\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E3 - Pg 26"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "L = 1.; # length of the copper wire in meters\n",
      "A = 1. * 10.**-4.; #  cross sectional area of the wire in meter square \n",
      "rho = 1.724 * 10.**-8.; #  resistivity of copper in ohm meter\n",
      "R = rho*L / A; #  resistance of the wire in ohm \n",
      "\n",
      "print '%s %.2e' %(\"resistance of the wire (in ohms)=\",R) "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "resistance of the wire (in ohms)= 1.72e-04\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E4 - Pg 27"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# 1 inches = 0.0254meters\n",
      "# 1 foot = 0.3048 meters\n",
      "import math \n",
      "d = 0.1*0.0254; #  diameter of the wire in meters\n",
      "L = 10.*0.3048; # length of the wire in meters \n",
      "rho = 1.724*10.**-8.; #  resistivity of the wire in ohm-meter\n",
      "A = math.pi*(d/2.)**2.; #  cross sectional area of the wire \n",
      "R = rho*L/A; # resistance of the wire in ohm \n",
      "print '%s %.2f' %(\"resistance of the wire (in ohm)=\",R)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "resistance of the wire (in ohm)= 0.01\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E5 - Pg 29"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%matplotlib inline\n",
      "import math\n",
      "import numpy as np\n",
      "from matplotlib import pyplot\n",
      "L = 0.1; #  inductance of the coil in henry \n",
      "t1= np.linspace(0,0.1, num=101)\n",
      "t2= np.linspace(0.101,0.3, num=201)\n",
      "t3= np.linspace(0.301,0.6,num=301)\n",
      "t4= np.linspace(0.601,0.7,num=101)\n",
      "t5= np.linspace(0.701,0.9,num=201)\n",
      "# current variation as a function of time \n",
      "i1 = 100.*t1;\n",
      "i2 = (-50.*t2) + 15.;\n",
      "i3 = np.zeros(301)\n",
      "for i in range(0,301):\n",
      "\ti3[i] = -100.*math.sin(math.pi*(t3[i]-0.3)/0.3);\n",
      "\n",
      "i4 = (100.*t4) - 60.;\n",
      "i5 = (-50.*t5) + 45.;\n",
      "\n",
      "t = ([t1,t2,t3,t4,t5]);\n",
      "i = ([i1,i2,i3,i4,i5]);\n",
      "pyplot.plot(t1, i1);\n",
      "pyplot.plot(t2, i2);\n",
      "pyplot.plot(t3, i3);\n",
      "pyplot.plot(t4, i4);\n",
      "pyplot.plot(t5, i5);\n",
      "\n",
      "dt = 0.001;\n",
      "di1 = np.diff(i1);\n",
      "di2 = np.diff(i2);\n",
      "di3 = np.diff(i3);\n",
      "di4 = np.diff(i4);\n",
      "di5 = np.diff(i5);\n",
      "V1 =np.array((L/dt)*di1); # voltage drop appearing across the inductor terminals\n",
      "V2 =np.array((L/dt)*di2); # voltage drop appearing across the inductor terminals\n",
      "V3 =np.array((L/dt)*di3); # voltage drop appearing across the inductor terminals\n",
      "V4 = np.array((L/dt)*di4); # voltage drop appearing across the inductor terminals\n",
      "V5 = np.array((L/dt)*di5); # voltage drop appearing across the inductor terminals\n",
      "print(V2)\n",
      "Tv = np.linspace(0,0.899,num=900);\n",
      "V = []\n",
      "V.extend(V1)\n",
      "V.extend(V2)\n",
      "V.extend(V3)\n",
      "V.extend(V4)\n",
      "V.extend(V5)\n",
      "print(len(V))\n",
      "pyplot.plot(Tv, V)\n",
      "pyplot.show();"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "[-4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975\n",
        " -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975 -4.975]\n",
        "900\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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v927n2M1VStPKqlXeNNH77vMGvEVCoOmj4qNguoaam9fz7runMWnSb+K7smhO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\nf+iDjK55zTwcAAAAAElFTkSuQmCC\n",
       "text": [
        "<matplotlib.figure.Figure at 0x5cdc0b0>"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E7 - Pg 31"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# a\n",
      "Ri = 1.; \n",
      "Rf = 39.;\n",
      "A = 10.**5.; # open loop gain of the op-amp\n",
      "G = A/(1. + (A*Ri/(Ri+Rf))); # actual voltage gain of the circuit \n",
      "print '%s' %(\"a\")\n",
      "print '%s %.2f' %(\"actual voltage of the circuit =\",G)\n",
      "\n",
      "# b\n",
      "G1 = 1 + (Rf/Ri); #  voltage gain of the circuit with infinite open loop gain\n",
      "print '%s' %(\"b\")\n",
      "print '%s %.f' %(\"for ideal case the voltage gain =\",G1)\n",
      "\n",
      "# c\n",
      "er = ((G1 - G)/G)*100.; # percent error \n",
      "print '%s' %(\"c\")\n",
      "print '%s %.2f' %(\"percent error of the ideal value compared to the actual value=\",er)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a\n",
        "actual voltage of the circuit = 39.98\n",
        "b\n",
        "for ideal case the voltage gain = 40\n",
        "c\n",
        "percent error of the ideal value compared to the actual value= 0.04\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E8 - Pg 33"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "G = 4.; #  voltage gain of the circuit \n",
      "r = G -1.; #  ratio of the resistances in the non-inverting op-amp circuit\n",
      "print '%s %.2f' %(\"Rf/Ri =\",r)\n",
      "# Result:\n",
      "# A suitable choice for R1 is 10K, Hence Rf = 30K\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rf/Ri = 3.00\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E9 - Pg 34"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "G = 4.;\n",
      "r = G; #  ratio of the resistances in the inverting op-amp circuit\n",
      "print '%s %.f' %(\"Rf/Ri\",r)\n",
      "# Result;\n",
      "# A suitable choice for Rf=30K and R1=7.5K\n",
      "# therefore input resistance R1 = 7.5K\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rf/Ri 4\n"
       ]
      }
     ],
     "prompt_number": 14
    }
   ],
   "metadata": {}
  }
 ]
}