{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 25 : Resonance" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_1 Page No. 775" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The resonant frequency = 12.58 Hertz\n", "approx 12.6 Hertz\n" ] } ], "source": [ "from math import pi,sqrt\n", "# Calculate the resonant frequency for an 8-H inductance and a 20-u\u0003F capacitance.\n", "\n", "# Given data\n", "\n", "L = 8.# # L=8 Henry\n", "C = 20.*10**-6# # C=20 uFarad\n", "\n", "fr = 1./(2.*pi*sqrt(L*C))#\n", "print 'The resonant frequency = %0.2f Hertz'%fr\n", "print 'approx 12.6 Hertz'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_2 Page No. 776" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The resonant frequency = 65007689.56 Hertz\n", "i.e 65 MHz\n" ] } ], "source": [ "# Calculate the resonant frequency for a 2-uH inductance and a 3-pF capacitance.\n", "\n", "# Given data\n", "\n", "L = 2.*10**-6# # Inductor=2 uHenry\n", "C = 3.*10**-12# # Capacitor=3 pFarad\n", "pi = 3.14#\n", "\n", "fr = 1./(2.*pi*sqrt(L*C))#\n", "print 'The resonant frequency = %0.2f Hertz'%fr\n", "print 'i.e 65 MHz'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_3 Page No. 778" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The value of Capacitor = 1.06e-10 Farads\n", "i.e 106 pF\n" ] } ], "source": [ "from math import pi\n", "# What value of C resonates with a 239-u\u0003H L at 1000 kHz?\n", "\n", "# Given data\n", "\n", "L = 239.*10**-6# # Inductor=239 uHenry\n", "fr = 1000.*10**3# # Resonant frequency=1000 kHertz\n", "\n", "A = pi*pi# # pi square\n", "B = fr*fr# # Resonant frequency square\n", "\n", "C = 1./(4.*A*B*L)#\n", "print 'The value of Capacitor = %0.2e Farads'%C\n", "print 'i.e 106 pF'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_4 Page No. 781" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The value of Inductor = 2.39e-04 Henry\n", "i.e 239 uF\n" ] } ], "source": [ "from math import pi\n", "# What value of L resonates with a 106-pF C at 1000 kHz, equal to 1 MHz?\n", "\n", "# Given data\n", "\n", "C = 106.*10**-12# # Capacitor=106 pFarad\n", "fr = 1.*10**6# # Resonant frequency=1 MHertz\n", "\n", "A = pi*pi# # pi square\n", "B = fr*fr# # Resonant frequency square\n", "\n", "C = 1./(4*A*B*C)#\n", "print 'The value of Inductor = %.2e Henry'%C\n", "print 'i.e 239 uF'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_5 Page No. 782" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Magnification factor Q =50.00\n" ] } ], "source": [ "# A series circuit resonant at 0.4 MHz develops 100 mV across a 250-u\u0003H L with a 2-mV input. Calculate Q .\n", "\n", "# Given data\n", "\n", "Vo = 100.*10**-3# # Output voltage=100 mVolts\n", "Vi = 2*10**-3# # Input voltage=2 mVolts\n", "L = 250*10**-6# # Inductor=250 uHenry\n", "f = 0.4*10**6# # Frequency=0.4 MHertz\n", "\n", "Q = Vo/Vi#\n", "print 'The Magnification factor Q =%0.2f'%Q" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_6 Page No. 784" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Ac Resistance of Coil = 12.57 Ohms\n" ] } ], "source": [ "from math import pi\n", "# What is the ac resistance of the coil in A series circuit resonant at 0.4 MHz develops 100 mV across a 250-u\u0003H L with a 2-mV input.\n", "\n", "# Given data\n", "\n", "Vo = 100.*10**-3# # Output voltage=100 mVolts\n", "Vi = 2.0*10**-3# # Input voltage=2 mVolts\n", "L = 250.0*10**-6# # Inductor=250 uHenry\n", "f = 0.4*10**6# # Frequency=0.4 MHertz\n", "\n", "Q = Vo/Vi#\n", "Xl = 2*pi*f*L#\n", "\n", "rs = Xl/Q#\n", "print 'The Ac Resistance of Coil = %0.2f Ohms'%rs" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_7 Page No. 785" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Because they divide Vt equally\n", "The Equivalent Impedence = 225000 Ohms\n", "i.e 225 kOhms\n", "The Q =150.00\n" ] } ], "source": [ "# In Fig. 25–9, assume that with a 4-mVac input signal for VT, the voltage across R1 is 2 mV when R1 is 225-kOhms\u0003. Determine Zeq and Q.\n", "\n", "# Given data\n", "\n", "vin = 4.*10**-3# # Input AC signal=4 mVac\n", "R1 = 225.*10**3# # Resistance1=225 kOhms\n", "vR1 = 2.*10**-3# # Voltage across Resistor1=2 mVac\n", "xl = 1.5*10**3# # Inductive Reactance=1.5 kOhms\n", "\n", "print 'Because they divide Vt equally'\n", "\n", "Zeq = R1#\n", "print 'The Equivalent Impedence = %0.f Ohms'%Zeq\n", "print 'i.e 225 kOhms'\n", "\n", "Q = Zeq/xl#\n", "print 'The Q =%0.2f'%Q" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_8 Page No. 786" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Magnification factor Q = 40.02\n" ] } ], "source": [ "from math import pi\n", "# A parallel LC circuit tuned to 200 kHz with a 350-u\u0003H L has a measured ZEQ of 17,600. Calculate Q.\n", "\n", "# Given data\n", "\n", "L = 350.*10**-6# # Inductor=350 uHenry\n", "f = 200.*10**3# # Frequency=200 kHertz\n", "Zeq = 17600.# # Equivalent Impedence=17600 Ohms\n", "\n", "Xl = 2*pi*f*L#\n", "\n", "Q = Zeq/Xl#\n", "print 'The Magnification factor Q = %0.2f'%Q" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_9 Page No. 788" ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Bandwidth BW or Delta f = 20000 Hertz\n", "i.e 20 kHz\n", "The Edge Frequency f1 = 1990000 Hertz\n", "i.e 1990 kHz\n", "The Edge Frequency f2 = 2010000 Hertz\n", "i.e 2010 kHz\n" ] } ], "source": [ "# An LC circuit resonant at 2000 kHz has a Q of 100. Find the total bandwidth delta f and the edge frequencies f1 and f2.\n", "\n", "# Given data\n", "\n", "fr = 2000.*10**3# # Resonant frequency=2000 kHertz\n", "Q = 100.# # Magnification factor=100\n", "\n", "Bw = fr/Q#\n", "print 'The Bandwidth BW or Delta f = %0.f Hertz'%Bw\n", "print 'i.e 20 kHz'\n", "\n", "f1 = fr-Bw/2#\n", "print 'The Edge Frequency f1 = %0.f Hertz'%f1\n", "print 'i.e 1990 kHz'\n", "\n", "f2 = fr+Bw/2#\n", "print 'The Edge Frequency f2 = %0.f Hertz'%f2\n", "print 'i.e 2010 kHz'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 25_10 Page No. 789" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Bandwidth BW or Delta f = 60000 Hertz\n", "i.e 60 kHz\n", "The Edge Frequency f1 = 5970000 Hertz\n", "i.e 5970 kHz\n", "The Edge Frequency f2 = 6030000 Hertz\n", "i.e 6030 kHz\n" ] } ], "source": [ "# An LC circuit resonant at 6000 kHz has a Q of 100. Find the total bandwidth delta f and the edge frequencies f1 and f2.\n", "\n", "# Given data\n", "\n", "fr = 6000.*10**3# # Resonant frequency=6000 kHertz\n", "Q = 100.# # Magnification factor=100\n", "\n", "Bw = fr/Q#\n", "print 'The Bandwidth BW or Delta f = %0.f Hertz'%Bw\n", "print 'i.e 60 kHz'\n", "\n", "f1 = fr-Bw/2#\n", "print 'The Edge Frequency f1 = %0.f Hertz'%f1\n", "print 'i.e 5970 kHz'\n", "\n", "f2 = fr+Bw/2.0#\n", "print 'The Edge Frequency f2 = %0.f Hertz'%f2\n", "print 'i.e 6030 kHz'" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.9" } }, "nbformat": 4, "nbformat_minor": 0 }