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+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:76623e2b912c3ee572e7978144ba11bf11add72a64c001ac7cb23291cfabcb07"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 45: Transients and Laplace\n",
+ "transforms</h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1, page no. 903</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "C = 500E-9;# in Farad\n",
+ "R = 100000;# in Ohm\n",
+ "V = 50;# in VOlts\n",
+ "ti = 0.15;# in sec\n",
+ "tc = 0.08;# in sec\n",
+ "Vrt = 35;# in Volts\n",
+ "\n",
+ " #calculation:\n",
+ " #Initial current, \n",
+ "i0 = (V/R)\n",
+ " #when time t = 150ms current is\n",
+ "i150 = (V/R)*math.e**(-1*ti/(R*C))\n",
+ " #capacitor voltage, Vc\n",
+ "Vc = V*(1 - math.e**(-1*tc/(R*C)))\n",
+ " #time, t\n",
+ "tvr = -1*R*C*math.log(Vrt/V)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n initial value of current flowing is \",round(i0*1E3,2),\"mA\"\n",
+ "print \"\\n current flowing at t = 150ms is \",round(i150*1E6,2),\"uA\"\n",
+ "print \"\\n value of capacitor voltage at t = 80ms is \",round(Vc,2),\" V\"\n",
+ "print \"\\n the time after connection when the resistor voltage is 35 V is \",round(tvr*1E3,2),\"msec\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " initial value of current flowing is 0.5 mA\n",
+ "\n",
+ " current flowing at t = 150ms is 24.89 uA\n",
+ "\n",
+ " value of capacitor voltage at t = 80ms is 39.91 V\n",
+ "\n",
+ " the time after connection when the resistor voltage is 35 V is 17.83 msec\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 2, page no. 905</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "C = 5E-6;# in Farad\n",
+ "R = 2000000;# in Ohm\n",
+ "V = 200;# in VOlts\n",
+ "tc = 20;# in sec\n",
+ "\n",
+ " #calculation:\n",
+ " #capacitor voltage, Vc\n",
+ "Vc = V*(math.e**(-1*tc/(R*C)))\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n value of capacitor voltage at t = 20s is \",round(Vc,2),\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " value of capacitor voltage at t = 20s is 27.07 V"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 3, page no. 907</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "L = 0.05;# in Henry\n",
+ "R = 5;# in Ohm\n",
+ "V = 110;# in VOlts\n",
+ "ti = 0.004;# in sec\n",
+ "tvr = 0.006;# in sec\n",
+ "tvl = 0.006;# in sec\n",
+ "it = 15;# in amperes\n",
+ "\n",
+ " #calculation:\n",
+ " #steady state current i\n",
+ "i = V/R\n",
+ " #when time t = 4ms current is\n",
+ "i4 = (V/R)*(1 - math.e**(-1*ti*R/L))\n",
+ " #resistor voltage, VR\n",
+ "VR6 = V*(1 - math.e**(-1*tvr*R/L))\n",
+ " #inductor voltage, VL\n",
+ "VL6 = V*(math.e**(-1*tvl*R/L))\n",
+ " #time, t\n",
+ "ti = (-1*L/R)*math.log(1 - it*R/V)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n steady state current i is \",round(i,2),\" A\"\n",
+ "print \"\\n when time t = 4ms current is is \",round(i4,2),\" A\"\n",
+ "print \"\\n value of resistor voltage at t = 6ms is \",round(VR6,2),\" V\"\n",
+ "print \"\\n value of inductor voltage at t = 6ms is \",round(VL6,2),\" V\"\n",
+ "print \"\\n the time after connection when the current is 15 V is \",round(ti,5),\" sec\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " steady state current i is 22.0 A\n",
+ "\n",
+ " when time t = 4ms current is is 7.25 A\n",
+ "\n",
+ " value of resistor voltage at t = 6ms is 49.63 V\n",
+ "\n",
+ " value of inductor voltage at t = 6ms is 60.37 V\n",
+ "\n",
+ " the time after connection when the current is 15 V is 0.01145 sec"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4, page no. 909</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "i = 5;# in Amperes\n",
+ "L = 2# in Henry\n",
+ "i1 = 0.2;# in Amperes\n",
+ "R = 10;# in Ohm\n",
+ "\n",
+ " #calculation:\n",
+ " #time t\n",
+ "ti = (-1*L/R)*math.log(i1/i)\n",
+ " #voltage across the resistor is a maximum \n",
+ "VRm = i*R\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n time t for the current in the 2 H inductor to fall to 200 mA is \",round(ti,3),\" sec\"\n",
+ "print \"\\n max voltage across the resistor is \",VRm,\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " time t for the current in the 2 H inductor to fall to 200 mA is 0.644 sec\n",
+ "\n",
+ " max voltage across the resistor is 50 V"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5, page no. 911</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "L = 0.002# in Henry\n",
+ "R = 1000;# in Ohm\n",
+ "C1 = 5E-6;# in farad\n",
+ "C2 = 5E-9;# in farad\n",
+ "\n",
+ " #calculation:\n",
+ "a = (R/(2*L))**2\n",
+ "b = 1/(L*C1)\n",
+ "if (a>b):\n",
+ "\ts1 = \"overdamped\";\n",
+ "elif (a<b):\n",
+ "\ts1 = \"underdamped\";\n",
+ "else:\n",
+ "\ts1 = \"critically damped\";\n",
+ "c = 1/(L*C2)\n",
+ "if (a>c):\n",
+ "\ts2 = \"overdamped\";\n",
+ "elif (a<c):\n",
+ "\ts2 = \"underdamped\";\n",
+ "else:\n",
+ "\ts2 = \"critically damped\";\n",
+ "\t\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n circuit is \",s1\n",
+ "print \"\\n if C = 5 nF, circuit is \",s2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " circuit is overdamped\n",
+ "\n",
+ " if C = 5 nF, circuit is underdamped"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 6, page no. 912</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "L = 0.002# in Henry\n",
+ "R = 1000;# in Ohm\n",
+ "\n",
+ " #calculation:\n",
+ "a = (R/(2*L))**2\n",
+ " #for critically damped\n",
+ "C = 4*L/R**2\n",
+ "\t\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n capacitance C is \",C*1E9,\"nF\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " capacitance C is 8.0 nF"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 7, page no. 913</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "import cmath\n",
+ "#initializing the variables:\n",
+ "L = 1.5# in Henry\n",
+ "R = 90;# in Ohm\n",
+ "C = 5*1E-6; # in Farad\n",
+ "V = 10; # in Volts\n",
+ "\n",
+ "#calculation:\n",
+ "a = -1*R/(2*L)\n",
+ "b = (1/(L*C) - (R/(2*L))**2)**0.5\n",
+ "V0 = V\n",
+ "I0 = 0\n",
+ "A = V0\n",
+ "B = (I0 - C*a*V0)/(C*b)\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"Current, i = e^(\",a,\"t) (\",round((a*C*B - A*C*b),4),\"sin(\",round(b,1),\"t) + (\",round((-1*a*C*A + B*C*C*b),0),\"cos(\",round(b,1),\"t) Amps.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "Current, i = e^( -30.0 t) ( -0.0183 sin( 363.9 t) + ( 0.0 cos( 363.9 t) Amps.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file