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# chap5 - CHAPTER 5 Exercises E5.1(a We are given v(t = 150...

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174 CHAPTER 5 Exercises E5.1 (a) We are given ) 30 200 cos( 150 ) ( o = t t v π . The angular frequency is the coefficient of t so we have radian/s 200 π ω = . Then Hz 100 2 / = = π ω 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 ω has units of radians, the positive peak occurs when ms 8333 . 0 180 30 max max = × = t t π ω (b) W 225 / 2 = = R V P rms avg (c) A plot of v ( t ) is shown in Figure 5.4 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 π π E5.3 radian/s 377 2 = f π ω 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 0 10 1 = + = j V ) 45 cos( 14 . 14 ) sin( 10 ) cos( 10 o = + t t t ω ω ω (b) 330 . 4 5 . 2 5 660 . 8 60 5 30 10 1 j j + + + = o o I o 44 . 3 18 . 11 670 . 0 16 . 11 + j ) 44 . 3 cos( 18 . 11 ) 30 sin( 5 ) 30 cos( 10 o o o + = + + + t t t ω ω ω (c) 99 . 12 5 . 7 0 20 60 15 0 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 ω ω ω

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175 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 ω o 0 100 = L V o 90 2 50 / 100 / = = = j Z L L L V I (b) The phasor diagram is shown in Figure 5.11a in the book. E5.7 (a) o 90 50 50 / 1 = = = j C j Z C ω o 0 100 = C V o 90 2 ) 50 /( 100 / = = = j Z C C C V I (b) The phasor diagram is shown in Figure 5.11b in the book. E5.8 (a) o 0 50 50 = = = R Z R o 0 100 = R V o 0 2 ) 50 /( 100 / = = = R R R V I (b) The phasor diagram is shown in Figure 5.11c in the book. E5.9 (a) The transformed network is: mA 135 28 . 28 250 250 90 10 o o = + = = j Z s V I
176 mA ) 135 500 cos( 28 . 28 ) ( o = t t i o 135 07 . 7 = = I V R R o 45 07 . 7 = = I V L j L ω (b) The phasor diagram is shown in Figure 5.17b in the book. (c) i ( t ) lags v s ( t ) by . 45 o E5.10 The transformed network is: = + + + = 31 . 56 47 . 55 ) 200 /( 1 ) 50 /( 1 100 / 1 1 o j j Z V 31 . 56 4 . 277 o = = I V Z A 69 . 33 547 . 5 ) 50 /( o = = j C V I A 31 . 146 387 . 1 ) 200 /( o = = j L V I A 31 . 56 774 . 2 ) 100 /( o = = V I R E5.11 The transformed network is: We write KVL equations for each of the meshes: 100 ) ( 100 100 2 1 1 = + I I I j 0 ) ( 100 100 200 1 2 2 2 = + + I I I I j j Simplifying, we have 100 100 ) 100 100 ( 2 1 = + I I j 0 ) 100 100 ( 100 2 1 = + I I j Solving we find A. 0 1 and A 45 414 . 1 2 1 o o = = I I Thus we have ). 1000 cos( ) ( and A ) 45 1000 cos( 414 . 1 ) ( 2 1 t t i t t i = = o

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177 E5.12 (a) For a power factor of 100%, we have , 1 ) cos( = θ which implies that the current and voltage are in phase and . 0 = θ Thus, . 0 ) tan( = = θ P Q Also ( ) A. 10 )] 0 cos( 500 /[ 5000 ] cos /[ = = = θ rms rms V P I Thus we have . 40 14 . 14 and 14 . 14 2 o = = = I rms m I I (b) For a power factor of 20% lagging, we have , 2 . 0 ) cos( = θ which implies that the current lags the voltage by . 46 . 78 ) 2 . 0 ( cos 1 o = = θ Thus, . kVAR 49 . 24 ) tan( = = θ P Q Also, we have ( ) A. 0 . 50 ] cos /[ = = θ rms rms V P I Thus we have . 46 . 38 71 . 70 and A 71 . 70 2 o = = = I rms m I I (c)
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chap5 - CHAPTER 5 Exercises E5.1(a We are given v(t = 150...

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