diff options
Diffstat (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb')
| -rwxr-xr-x | Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb | 435 |
1 files changed, 0 insertions, 435 deletions
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb deleted file mode 100755 index f30323f6..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_45-checkpoint.ipynb +++ /dev/null @@ -1,435 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "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": [
- "#Calculate (a) the initial value of current flowing, \n",
- "#(b) the value of current 150 ms after connection, \n",
- "#(c) the value of capacitor voltage 80 ms after connection, and \n",
- "#(d) the time after connection when the resistor voltage is 35 V.\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": [
- "#Calculate the p.d. across the capacitor after 20 s\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": [
- "#Determine (a) the final value of current, (b) the value of current after 4 ms, \n",
- "#(c) the value of the voltage across the resistor after 6 ms,\n",
- "#(d) the value of the voltage across the inductance after 6 ms, and \n",
- "#(e) the time when the current reaches 15 A.\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": [
- "#Determine\n",
- "#(a) the time for the current in the 2 H inductor to fall to 200 mA,\n",
- "#and (b) the maximum voltage appearing across the resistor.\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": [
- "#(a) Determine whether the circuit is over, critical or underdamped. (b) If C D 5 nF, determine the state of damping.\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": [
- "#what value of capacitance will give critical damping ?\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": [
- "#Determine the nature of the response and obtain an expression for the current in the coil.\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 |