From c8733e4b6b4bffcddf7eb45ff1c72ccc837aa3af Mon Sep 17 00:00:00 2001 From: Jovina Dsouza Date: Tue, 22 Jul 2014 00:00:04 +0530 Subject: adding book --- .../chapter_27-checkpoint_2.ipynb | 205 +++++++++++++++++++++ 1 file changed, 205 insertions(+) create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb') diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb new file mode 100755 index 00000000..4d37718b --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_27-checkpoint_2.ipynb @@ -0,0 +1,205 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Chapter 27: A.c. bridges

" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 1, page no. 485

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the values of Rx and Cx at balance.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "R2 = 2500;# in ohms\n", + "C2 = 0.2E-6;# IN fARADS\n", + "R3 = 1;\n", + "R4 = 1;\n", + "w = 2000*math.pi;\n", + "#calculation:\n", + "Rx = R4*(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2)\n", + "Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)Resistance Rx = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\"\n", + "print \"\\n (b)at balance Rx = \",round(Rx/1000,2),\"KOhm and Cx = \", round(Cx*1E9,2),\"nF\\n\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)Resistance Rx = R4(1 + w*w*C2*C2*R2*R2)/(R2*R3*w*w*C2*C2) and Capacitance Cx = R3*C2/(R4*(1 + w*w*C2*C2*R2*R2))\n", + "\n", + " (b)at balance Rx = 2.75 KOhm and Cx = 18.4 nF" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 2, page no. 487

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine, when the bridge is balanced, (a) the value of resistance R1, and (b) the frequency of the bridge.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "R2 = 30000;# in ohms\n", + "R3 = 30000;# in ohms\n", + "R4 = 1000;# in ohms\n", + "C2 = 1e-9;# IN fARADS\n", + "C3 = 1e-9;# IN fARADS\n", + "\n", + "#calculation:\n", + " #the bridge is balanced\n", + "R1 = R4/((R3/R2) + (C2/C3))\n", + " #frequency, f\n", + "f = 1/(2*math.pi*((C2*C3*R2*R3)**0.5))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)Resistance R1 = \",R1,\" ohm\\n\"\n", + "print \"\\n (b)frequency, f is \",round(f,2),\"Hz\\n\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)Resistance R1 = 500.0 ohm\n", + "\n", + "\n", + " (b)frequency, f is 5305.16 Hz" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 3, page no. 487

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine, when the bridge is balanced, \n", + "#(a) the value of resistance Rx, (b) the value of capacitance Cx,\n", + "#(c) the phase angle of the unknown arm, (d) the power factor of the unknown arm and (e) its loss angle.\n", + "from __future__ import division\n", + "import math\n", + "import cmath\n", + "#initializing the variables:\n", + "R3 = 600;# in ohms\n", + "R4 = 200;# in ohms\n", + "C2 = 0.2e-6;# IN fARADS\n", + "C3 = 4000e-12;# IN fARADS\n", + "f = 1500;#in Hz\n", + "\n", + "#calculation:\n", + " #the bridge is balanced\n", + " #Resistance, Rx\n", + "Rx = R4*C3/C2\n", + " #Capacitance, Cx\n", + "Cx = C2*R3/R4\n", + " #Phase angle\n", + "phi = math.atan(1/(2*math.pi*f*Cx*Rx))\n", + "phid = phi*180/math.pi# in degrees\n", + " #Power factor of capacitor\n", + "Pc = math.cos(phi)\n", + " #Loss angle,\n", + "de = 90 - phid\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)Resistance Rx = \",round(Rx,2),\" ohm\\n\"\n", + "print \"\\n (b)capacitance, Cx is \",round(Cx*1E9,2),\"pFarad\\n\"\n", + "print \"\\n (c)phasor diagram = \",round(phid,2),\"deg lead \"\n", + "print \"\\n (d)power factor is \",round(Pc,2),\" \\n\"\n", + "print \"\\n (e)Loss angle = \",round(de,2),\"deg\\n\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)Resistance Rx = 4.0 ohm\n", + "\n", + "\n", + " (b)capacitance, Cx is 600.0 pFarad\n", + "\n", + "\n", + " (c)phasor diagram = 88.7 deg lead \n", + "\n", + " (d)power factor is 0.02 \n", + "\n", + "\n", + " (e)Loss angle = 1.3 deg\n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit