940_Physics ProblemsTechnical Physics

# 940_Physics ProblemsTechnical Physics - 281 Chapter 33...

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Chapter 33 281 P33.55 V V R R X X L C out in = + 2 2 b g (a) At 200 Hz: 1 4 8 00 8 00 400 1 400 2 2 2 = + . . a f a f π π L C . At 4 000 Hz: 8 00 8 000 1 8 000 4 8 00 2 2 2 . . a f a f + L N M O Q P = π π L C . FIG. P33.55(a) At the low frequency, X X L C < 0. This reduces to 400 1 400 13 9 π π L C = − . . [1] For the high frequency half-voltage point, 8 000 1 8000 13 9 π π L C = + . . [2] Solving Equations (1) and (2) simultaneously gives C = 54 6 . F µ and L = 580 H µ . (b) When X X L C = , V V V V out in out in = F H G I K J = max . 1 00 . (c) X X L C = requires f LC 0 4 5 1 2 1 2 5 80 10 5 46 10 894 = = × × = π π . . H F Hz e je j . (d) At 200 Hz, V V R Z out in = = 1 2 and X X C L > , so the phasor diagram is as shown: φ = − F H G I K J = − F H G I K J cos cos 1 1 1 2 R Z so V V out in leads by 60.0 ° . At f 0 , X X L C = so R Z X L - X C φ or φ V out V in R Z X L - X C φ or φ V out V in FIG. P33.55(d) V V out in and have a phase difference of 0 ° . At 4 000 Hz,
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