{ "metadata": { "name": "", "signature": "sha256:7e1563b9ba55a374999a76facd1396d51cd44d5e5678108b5dac64f78b8c5047" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "CHAPTER02 : THE CIRCUIT ELEMENTS" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E1 - Pg 27" ] }, { "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 35" ] }, { "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 35" ] }, { "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 36" ] }, { "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 41" ] }, { "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", " 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"text": [ "" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example E7 - Pg 60" ] }, { "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 61" ] }, { "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 61" ] }, { "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": {} } ] }