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Unformatted text preview: 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 . 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 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 . 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 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 ω ω ω 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 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 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 50 50 ∠ = = = R Z R o 100 ∠ = R V o 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: Ω − ∠ = + + − + = 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...
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This note was uploaded on 11/07/2011 for the course ENG 3373 taught by Professor Scut during the Spring '11 term at University of South Florida  Tampa.
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