Chapter 05 - 1 CHAPTER 5 Exercises E5.1(a We are given 30...

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Unformatted text preview: 1 CHAPTER 5 Exercises E5.1 (a) We are given ) 30 200 cos( 150 ) ( o − = t v π . The angular frequency is the coefficient of so we have radian/s 200 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 . 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 ). 90 cos( ) sin( o − = z Thus ) 30 300 cos( 100 ) 60 300 sin( 100 o o − = + E5.3 radian/s 377 2 ≅ = ms 67 . 16 / 1 ≅ = V 6 . 155 2 ≅ = 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 − = 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 − = + (b) 330 . 4 5 . 2 5 660 . 8 60 5 30 10 1 − + + ≅ − ∠ + ∠ = o o I o 44 . 3 18 . 11 670 . 16 . 11 ∠ ≅ + ≅ ) 44 . 3 cos( 18 . 11 ) 30 sin( 5 ) 30 cos( 10 o o o + = + + + (c) 99 . 12 5 . 7 20 60 15 20 2 − + + ≅ − ∠ + ∠ = o o I o 28 . 25 41 . 30 99 . 12 5 . 27 − ∠ ≅ − ≅ ) 28 . 25 cos( 41 . 30 ) 60 cos( 15 ) 90 sin( 20 o o o − = − + + 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 2 by o 60 (or we could say 2 leads 1 by ) 60 o 1 leads 3 by o 15 (or we could say 3 lags 1 by ) 15 o 2 leads 3 by o 75 (or we could say 3 lags 2 by ) 75 o E5.6 (a) o 90 50 50 ∠ = = = j L Z ω o 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 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 50 50 ∠ = = = R o 100 ∠ = V o 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 3 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: Ω − ∠ = + + − + = 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 ) ( 100 100 200 1 2 2 2 = − + + − I I I I Simplifying, we have 100 100 ) 100 100 ( 2 1 = − + I I ) 100 100 ( 100 2 1 = − + − I I Solving we find A. 1 and A 45 414 . 1 2 1 o o ∠ = − ∠ = I I Thus we have ). 1000 cos(...
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Chapter 05 - 1 CHAPTER 5 Exercises E5.1(a We are given 30...

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