p33_045

# p33_045 - 45. (a) For a given amplitude (E )m of the...

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45. (a) For a given amplitude ( E ) m of the generator emf, the current amplitude is given by I = ( E ) m Z = ( E ) m p R 2 +( ω d L 1 d C ) 2 . We ±nd the maximum by setting the derivative with respect to ω d equal to zero: dI d = ( E ) m £ R 2 +( ω d L 1 d C ) 2 ¤ 3 / 2 · ω d L 1 ω d C ¸· L + 1 ω 2 d C ¸ . The only factor that can equal zero is ω d L (1 d C ); it does so for ω d =1 / LC = ω .F o rt h i s circuit, ω d = 1 LC = 1 p (1 . 00 H)(20 . 0 × 10 6 F) = 224 rad / s . (b) When ω d = ω , the impedance is Z = R , and the current amplitude is I = ( E ) m R = 30 . 0V 5 . 00 Ω =6 . 00 A . (c) We want to ±nd the (positive) values of ω d for which I = ( E ) m 2 R : ( E ) m p R 2 +( ω d L 1 d C ) 2 = ( E ) m 2 R . This may be rearranged to yield µ ω d L 1 ω d C 2 =3 R 2 . Taking the square root of both sides (acknowledging the two ± roots) and multiplying by ω d C ,we obtain ω 2 d ( LC ) ± ω d ³ 3 CR ´ 1=0 . Using the quadratic formula, we ±nd the smallest positive solution
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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