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927_Physics ProblemsTechnical Physics

927_Physics ProblemsTechnical Physics - 268 P33.4...

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268 Alternating Current Circuits P33.4 (a) v V t R = max sin ω v V R = 0 250 . max b g , so sin . ω t = 0 250 , or ω t = sin . 1 0 250 a f . The smallest angle for which this is true is ω t = 0 253 . rad Thus, if t = 0 010 0 . s, ω = = 0 253 25 3 . . rad 0.010 0 s rad s . (b) The second time when v V R = 0 250 . max b g , ω t = sin . 1 0 250 a f again. For this occurrence, ω π t = = 0 253 2 89 . . rad rad (to understand why this is true, recall the identity sin sin π θ θ = a f from trigonometry). Thus, t = = 2 89 0 114 . . rad 25.3 rad s s . P33.5 i I t R = max sin ω becomes 0 600 0 007 00 . sin . = ω b g . Thus, 0 007 00 0 600 0 644 1 . sin . . b g a f ω = = and ω π = = 91 9 2 . rad s f so f = 14 6 . Hz . P33.6 P = I V rms rms b g and V rms V = 120 for each bulb (parallel circuit), so: I I V 1 2 1 150 1 25 = = = = P rms W 120 V A . , and R V I R 1 1 2 120 96 0 = = = = rms V 1.25 A . I V 3 3 100 0 833 = =
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