60289-Chapter_05

# 60289-Chapter_05 - CHAPTER 5 Exercises E5.1(a We are given...

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CHAPTER 5 Exercises E5.1 (a) We are given ) 30 200 cos( 150 ) ( o = t v π 200 . The angular frequency is the coefficient of so we have radian/s ω = . Then Hz 100 2 / = = f ms 10 / 1 = = T V 1 . 106 2 / 150 2 / = = = m rms V Furthermore, ( ) attains a positive peak when the argument of the cosine function is zero. Thus keeping in mind that has units of radians, the positive peak occurs when ms 8333 . 0 180 30 max max = × = (b) W 225 / 2 = = R P avg (c) A plot of ( ) is shown in Figure 5.4 in the book. E5.2 We use the trigonometric identity Thus ). 90 cos( ) sin( o = z 100 ) 30 300 cos( 100 ) 60 300 sin( o o = + E5.3 radian/s 377 2 = ms 67 . 16 / 1 = V 6 . 155 2 = The period corresponds to 360 therefore 5 ms corresponds to a phase angle of ( . Thus the voltage is o o 108 = o 360 ) 67 . 16 / 5 × ) 108 377 cos( 6 . 155 ) ( o = E5.4 (a) o o o 45 14 . 14 10 10 90 10 0 10 1 = + = j V cos( 10 ) 45 cos( 14 . 14 ) sin( 10 ) o = + (b) 10 330 . 4 5 . 2 5 660 . 8 60 5 30 1 + + + o o = I o 44 . 3 18 . 11 670 . 0 16 . 11 + cos( 10 ) 44 . 3 cos( 18 . 11 ) 30 sin( 5 ) 30 o o o + = + + + (c) 99 . 12 5 . 7 0 20 60 15 0 20 2 + + + = o o o 28 . 25 41 . 30 99 . 12 5 . 27 sin( 20 ) 28 . 25 cos( 41 . 30 ) 60 cos( 15 ) 90 o o o = + + 142

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E5.5 The phasors are V o o o 45 10 and 30 10 30 10 3 2 1 = + = = V V v 1 lags 2 by 60 (or we could say o o o 2 leads 1 by 60 ) o 1 leads 3 by 15 (or we could say 3 lags 1 by 15 ) o 2 leads 3 by 75 (or we could say 3 lags 2 by 75 ) o E5.6 (a) o 90 50 50 = = = j L Z ω o 0 100 = V o 90 2 50 / 100 / = = = V I (b) The phasor diagram is shown in Figure 5.11a in the book. E5.7 (a) o 90 50 50 / 1 = = = C o 0 100 = V o 90 2 ) 50 /( 100 / = = = V I (b) The phasor diagram is shown in Figure 5.11b in the book. E5.8 (a) o 0 50 50 = = = R o 0 100 = V o 0 2 ) 50 /( 100 / = = = 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 = + = = s V I 143
mA ) 135 500 cos( 28 . 28 ) ( o = t i o 135 07 . 7 = = I V R o 45 07 . 7 = = I V L j ω (b) The phasor diagram is shown in Figure 5.17b in the book. (c) ( ) lags v s ( ) by . 45 o E5.10 The transformed network is: = + + + = 31 . 56 47 . 55 ) 200 /( 1 ) 50 /( 1 100 / 1 1 o Z V 31 . 56 4 . 277 o = = I V A 69 . 33 547 . 5 ) 50 /( o = = C V I A 31 . 146 387 . 1 ) 200 /( o = = V I A 31 . 56 774 . 2 ) 100 /( o = = V I E5.11 The transformed network is: We write KVL equations for each of the meshes: 100 ) ( 100 100 2 1 1 = + I I I 0 ) ( 100 100 200 1 2 2 2 = + + I I I I Simplifying, we have 100 100 ) 100 100 ( 2 1 = + I I 0 ) 100 100 ( 100 2 1 = + I I Solving we find I Thus we have A. 0 1 and A 45 414 . 1 2 1 o o = = I 1000 cos( ) ( and A ) 45 1000 2 ). cos( 414 . 1 ) ( 1 = o = 144

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E5.12 (a) For a power factor of 100%, we have cos( , 1 ) = θ which implies that the current and voltage are in phase and . 0 = Thus, . 0 ) tan( = = P Q Also ( ) A. 500 /[ 5000 ] cos /[ 10 )] 0 cos( = = = rms V I Thus we have . 40 14 . 14 and 14 . 14 2 o = = I = m (b) For a power factor of 20% lagging, we have cos( , 2 . 0 ) = which implies that the current lags the voltage by Thus, . 46 . o = 78 ) 2 . 0 ( cos 1 = . kVAR 49 . 24 ) tan( = = Also, we have () ] = A. 0 . 50 cos /[ = Thus we have . 46 . 38 71 o . 70 and A 71 . 70 I 2 = = = (c) The current ratings would need to be five times higher for the load of part (b) than for that of part (a). Wiring costs would be lower for the load of part (a).
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## This note was uploaded on 09/25/2008 for the course ECEN 215 taught by Professor Styblinski during the Spring '08 term at Texas A&M.

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60289-Chapter_05 - CHAPTER 5 Exercises E5.1(a We are given...

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