940_Physics ProblemsTechnical Physics

940_Physics ProblemsTechnical Physics - 281 Chapter 33...

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Chapter 33 281 P33.55 V V R RX X LC out in = +− 2 2 bg (a) At 200 Hz: 1 4 800 8 00 400 1 400 2 2 2 = . . a f af ππ . At 4 000 Hz: 8000 1 4800 2 2 2 .. ΩΩ L N M O Q P = π L C . FIG. P33.55(a) At the low frequency, XX −< 0. This reduces to 400 1 400 13 9 L C −= . . [1] For the high frequency half-voltage point, 1 13 9 L C + . [2] Solving Equations (1) and (2) simultaneously gives C = 54 6 . F µ and L = 580 H . (b) When = , V V V V out in out in = F H G I K J = max . 100 . (c) = requires f LC 0 45 1 2 1 2 5 80 10 5 46 10 894 == ×× = −− H F Hz ej e j . (d) At 200 Hz, V V R Z out in 1 2 and CL > , so the phasor diagram is as shown: φ =− F H G I K J F H G I K J cos cos 11 1 2 R Z so ∆∆ VV out in leads by 60.0 ° . At f 0 , = so R Z X L - X C or V out V in R Z X L - X C or V out V in FIG. P33.55(d) out in and have a phase difference of 0 ° . At 4 000 Hz, V V R Z out in 1 2 and −> 0. Thus, = F H G I K J cos . 1 1 2 60 0 or out in lags
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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