{ "metadata": { "name": "", "signature": "sha256:de2147a7b14f8ed39e81d7f13f742b47f7e864d50fa1a5733b8aeda0bb6528ab" }, "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": [ "\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": [ "\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": [ "\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": {} } ] }