Chapter 05 - 1 CHAPTER 5 Solutions for Exercises E5.1 (a)...

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Unformatted text preview: 1 CHAPTER 5 Solutions for Exercises E5.1 (a) We are given ) 30 200 cos( 150 ) ( o = t t v . The angular frequency is the coefficient of t so we have radian/s 200 = . Then Hz 100 2 / = = f ms 10 / 1 = = f T V 1 . 106 2 / 150 2 / = = = m rms V V 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 . 180 30 max max = = t t (b) W 225 / 2 = = R V P rms avg (c) A plot of v ( t ) is shown in Figure 5.3 in the book. E5.2 We use the trigonometric identity ). 90 cos( ) sin( o = z z Thus ) 30 300 cos( 100 ) 60 300 sin( 100 o o = + t t E5.3 radian/s 377 2 = f ms 67 . 16 / 1 = f T V 6 . 155 2 = rms m V V The period corresponds to o 360 therefore 5 ms corresponds to a phase angle of o o 108 360 ) 67 . 16 / 5 ( = . Thus the voltage is ) 108 377 cos( 6 . 155 ) ( o = t t v E5.4 (a) o o o 45 14 . 14 10 10 90 10 10 1 = + = j V ) 45 cos( 14 . 14 ) sin( 10 ) cos( 10 o = + t t t (b) 330 . 4 5 . 2 5 660 . 8 60 5 30 10 1 j j + + + = o o I o 44 . 3 18 . 11 670 . 16 . 11 + j ) 44 . 3 cos( 18 . 11 ) 30 sin( 5 ) 30 cos( 10 o o o + = + + + t t t (c) 99 . 12 5 . 7 20 60 15 20 2 j j + + + = o o I o 28 . 25 41 . 30 99 . 12 5 . 27 j ) 28 . 25 cos( 41 . 30 ) 60 cos( 15 ) 90 sin( 20 o o o = + + t t t 2 E5.5 The phasors are o o o 45 10 and 30 10 30 10 3 2 1 = + = = V V V v 1 lags v 2 by o 60 (or we could say v 2 leads v 1 by ) 60 o v 1 leads v 3 by o 15 (or we could say v 3 lags v 1 by ) 15 o v 2 leads v 3 by o 75 (or we could say v 3 lags v 2 by ) 75 o E5.6 (a) o 90 50 50 = = = j L j Z L o 100 = L V o 90 2 50 / 100 / = = = j Z L L L V I (b) The phasor diagram is shown in Figure 5.10a in the book. E5.7 (a) o 90 50 50 / 1 = = = j C j Z C o 100 = C V o 90 2 ) 50 /( 100 / = = = j Z C C C V I (b) The phasor diagram is shown in Figure 5.10b in the book....
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Chapter 05 - 1 CHAPTER 5 Solutions for Exercises E5.1 (a)...

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