Solutions to Chapter 5 Homework Problems in Electrical Engineering, Assignments of Electrical and Electronics Engineering

Download Solutions to Chapter 5 Homework Problems in Electrical Engineering and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! Chapter 5 Homework Solutions CH5.1 E5.1 in Page 194 (do not turn in) (a) We are given )30200cos(150)(  ttv  . The angular frequency is the coefficient of t so we have radian/s 200  . Then Hz 1002/  f ms 10/1  fT V 1.1062/1502/  mrms VV Furthermore, v(t) attains a positive peak when the argument of the cosine function is zero. Thus keeping in mind that t has units of radians, the positive peak occurs when ms 8333.0 180 30 maxmax  tt   (b) W 225/2  RVP rmsavg (c) A plot of v(t) is shown in Figure 5.3 in the book. P5.6* Sinusoidal voltages can be expressed in the form ).cos()( θω  tVtv m The peak voltage is 28.282022 rms  VVm V. The frequency is  Tf /1 10 kHz and the angular frequency is 41022 ππω  f radians/s. The phase corresponding to a time interval of tΔ 20 s is  72360)/(  TtΔθ . Thus we have )72102cos(28.28)( 4  ttv π V. P5.11* CH5.2 P5.17*       V 5.2 25.6 )4cos( 4 5.12 5.12 )4sin(5.125.12 )2sin(5 1 5.0 0 5.0 0 5.0 0 2 0 2                    t t T rms tt dtt dtt dttv T V                   45cos4.141 454.141100100 10090100 1000100 90cos100sin100 cos100 21s 2 1 2 1       ttv j j tttv ttv s ω ωω ω VVV V V P5.23* We are given the expression )sin(4)75cos(3)75cos(5 ttt ωωω   Converting to phasors we obtain   904753755  4)8978.27765.0(8296.42941.1 jjj 09.82763.37274.35176.0  j Thus, we have )09.82cos(763.3 )sin(4)75cos(3)75cos(5     t ttt ω ωωω E5.4 in Text Page 199 (do not turn in) (a)  4514.14101090100101  jV )45cos(14.14)sin(10)cos(10  ttt  (b) 330.45.25660.860530101 jj  I 44.318.11670.016.11  j )44.3cos(18.11)30sin(5)30cos(10   ttt  (c) 99.125.702060150202 jj  I 28.2541.3099.125.27  j )28.25cos(41.30)60cos(15)90sin(20   ttt  CH5.3 E5.6 in Text Page 205 (do not turn in) (a) 905050  jLjZ L  0100LV 90250/100/  jZ LLL VI (b) The phasor diagram is shown in Figure 5.10a in the book. E5.7 in Text Page 205 (do not turn in) (a) 905050/1  jCjZC  0100CV 902)50/(100/  jZCCC VI (b) The phasor diagram is shown in Figure 5.10b in the book.    45 by leads 45 by lags 90 by lags 2s 1s 12 VV VV VV