ch31-p045

# ch31-p045 - 45. (a) For a given amplitude m of the...

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45. (a) For a given amplitude ε m of the generator emf, the current amplitude is given by 22 . (1 / ) mm dd I Z RLC ωω == +− We find the maximum by setting the derivative with respect to ω d equal to zero: dI d ER L C L C L C d mdd d =− + L N M O Q P + L N M O Q P ()[ ( / )] . / 3 2 2 1 11 The only factor that can equal zero is LC (/ ) 1 ; it does so for d LC 1/ . For this d LC × = 100 224 ( . H)(20.0 10 F) rad / s . 6 (b) When d = , the impedance is Z = R , and the current amplitude is 30.0 V 6.00 A . 5.00 m I R = (c) We want to find the (positive) values of d for which / 2 : m I R = . 2 / ) R = This may be rearranged to yield d d L C R F H G I K J = 1 3 2 2 Taking the square root of both sides (acknowledging the two ± roots) and multiplying by d C , we obtain LC CR 2 31 0 () . ±− = di Using the quadratic formula, we find the smallest positive solution

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22 6 2 6 62 2 6 6 3 3 4 3(20.0 10
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch31-p045 - 45. (a) For a given amplitude m of the...

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