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author | Prashant S | 2020-04-14 10:25:32 +0530 |
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committer | GitHub | 2020-04-14 10:25:32 +0530 |
commit | 06b09e7d29d252fb2f5a056eeb8bd1264ff6a333 (patch) | |
tree | 2b1df110e24ff0174830d7f825f43ff1c134d1af /Fundamental_Of_Physics_by_D_Haliday/31-Induction_and_Inductance.ipynb | |
parent | abb52650288b08a680335531742a7126ad0fb846 (diff) | |
parent | 476705d693c7122d34f9b049fa79b935405c9b49 (diff) | |
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diff --git a/Fundamental_Of_Physics_by_D_Haliday/31-Induction_and_Inductance.ipynb b/Fundamental_Of_Physics_by_D_Haliday/31-Induction_and_Inductance.ipynb new file mode 100644 index 0000000..d79f531 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/31-Induction_and_Inductance.ipynb @@ -0,0 +1,333 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 31: Induction and Inductance" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"i = 1.5 //in A\n", +"D = 3.2*10^-2 //in meter\n", +"N = 220/10^-2 //in turns/m\n", +"n = 130\n", +"d = 2.1*10^-2 //in meter\n", +"deltaT = 25*10^-3 //in s\n", +"uo = 4*%pi*10^-7 //in SI unit\n", +"\n", +"//Sample Problem 31-1\n", +"printf('**Sample Problem 31-1**\n')\n", +"A = %pi*(d/2)^2\n", +"deltaPhi = uo*N*i*A\n", +"E = n*deltaPhi/deltaT\n", +"printf('The emf induced is equal to %eV', E)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.2: Sample_Problem_2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"r = 0.20 //in meter\n", +"t = poly(0, 't')\n", +"B = 4.0*t^2 + 2.0*t + 3.0\n", +"E = 2.0 //in Volts\n", +"R = 2 //in Ohm\n", +"\n", +"//Sample Problem 31-2a\n", +"printf('**Sample Problem 31-2a**\n')\n", +"t = 10 //in sec\n", +"flux = B*%pi*r^2/2\n", +"Et = derivat(flux)\n", +"E1 = horner(Et, t)\n", +"printf('The Emf induced is equal to %fV\n', E1)\n", +"\n", +"//Sample Problem 31-2b\n", +"printf('\n**Sample Problem 31-2b**\n')\n", +"I = (E1-E)/R\n", +"printf('The induced current is equal to %fA', I)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"t = poly(0, 't')\n", +"//B = 4*t^2*x^2\n", +"W = 3.0 //in meter\n", +"H = 2.0 //in meter\n", +"t1 = 0.10 //in sec\n", +"\n", +"//Sample Problem 31-3\n", +"printf('**Sample Problem 31-3**\n')\n", +"flux = integrate('4*x^2*H', 'x', 0, W)\n", +"E = derivat(flux*t^2)\n", +"E1 = horner(E, t1)\n", +"printf('The induced emf is equal to %fV', E1)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 8.5*10^-2 //in meter\n", +"Rb = 0.13 //in T/s\n", +"r = 5.2*10^-2 //in meter\n", +"\n", +"//Sample Problem 31-4a\n", +"printf('**Sample Problem 31-4a**\n')\n", +"//Using Faraday's law\n", +"Rf = Rb*%pi*r^2\n", +"E = Rf/(2*%pi*r)\n", +"printf('The induced electric field is equal to %eV/m\n', E)\n", +"\n", +"//Sample Problem 31-4b\n", +"printf('\n**Sample Problem 31-4b**\n')\n", +"r = 12.5*10^-2 //in meter\n", +"Rf = Rb*%pi*R^2\n", +"E = Rf/(2*%pi*r)\n", +"printf('The induced electric field is equal to %eV/m', E)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.5: Sample_Problem_5.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"R = 9.0 //in Ohm\n", +"L = 2*10^-3 //in Henery\n", +"E = 18 //in Volts\n", +"\n", +"//Sample Problem 31-5a\n", +"printf('**Sample Problem 31-5a**\n')\n", +"//As soon as switch is closed the inductor will act like current barrier\n", +"Io = E/R\n", +"printf('The current as soon as qwitch is closed is equal to %1.2fA\n', Io)\n", +"\n", +"//Sample Problem 31-5b\n", +"printf('\n**Sample Problem 31-5b**\n')\n", +"//After long time inductor will act like short circuit\n", +"Req = R/3\n", +"If = E/(R/3)\n", +"printf('The current through the battery after long time will be %1.2fA', If)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 53*10^-3 //in H\n", +"R = 0.37 //in Ohm\n", +"\n", +"//Sample Problem 31-6\n", +"printf('**Sample Problem 31-6**\n')\n", +"//i = io(1-e^(t/T))\n", +"//ln2 = t/T\n", +"T = L/R\n", +"t = T*log(2)\n", +"printf('The time taken to rach the current to half of its stedy state value is %fs', t)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.7: Sample_Problem_7.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 53*10^-3 //in H\n", +"R = 0.35 //in Ohm\n", +"V = 12 //in Volts\n", +"\n", +"//Sample Problem 31-7a\n", +"printf('**Sample Problem 31-7a**\n')\n", +"i = V/R //in steady state\n", +"E = 1/2*L*i^2\n", +"printf('The Energy stored in the inductor in steady state is %fJ\n', E)\n", +"\n", +"//Sample Problem 31-7b\n", +"printf('\n**Sample Problem 31-7b**\n')\n", +"Et = E/2\n", +"//hence It = Io/sqrt(2)\n", +"f = log(1-1/sqrt(2)) //the number of times of time constant\n", +"printf('After t=%1.1fT, the energy stored in the inductor will be half of tis steady state value', f)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"a = 1.2*10^-3 //in meter\n", +"b = 3.5*10^-3 //in meter\n", +"i = 2.7 //in Amp\n", +"l = 1 //in meter(say)\n", +"uo = 4*%pi*10^-7\n", +"\n", +"//Sample Problem 31-8\n", +"printf('**Sample Problem 31-8**\n')\n", +"B = uo*i/(2*%pi) //divided by r\n", +"Ul = B^2/(2*uo) //divided by r^2\n", +"//Energy as a funtion of r\n", +"U = Ul*2*%pi*l //divided by r by r\n", +"Energy = integrate('U/r', 'r', a, b)\n", +"printf('Energy per unit length is equal to %1.2eJ/m', Energy)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 31.9: Sample_Problem_9.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"N1 = 1200 //turns\n", +"N2 = N1\n", +"R2 = 1.1*10^-2 //in meter\n", +"R1 = 15*10^-2 //in meter\n", +"uo = 4*%pi*10^-7\n", +"\n", +"//Sample Problem 31-9\n", +"printf('**Sample Problem 31-9**\n')\n", +"//let's assume\n", +"i = 1 //in amp\n", +"B1 = uo*N1*i/(2*R1)\n", +"phi2 = B1*%pi*R2^2*N2\n", +"M = phi2/i\n", +"printf('The mutual inductance of the two coil is equal to %1.2eH', M)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |