p33_059 - 59. We use the result of problem 54: Pavg = We...

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59. We use the result of problem 54: P avg = ( E ) 2 m R 2 Z 2 = ( E ) 2 m R 2[ R 2 +( ω d L 1 d C ) 2 ] . We use the expression Z = p R 2 +( ω d L 1 d C ) 2 for the impedance in terms of the angular frequency. (a) Considered as a function of C , P avg has its largest value when the factor R 2 +( ω d L 1 d C ) 2 has the smallest possible value. This occurs for ω d L =1 d C ,or C = 1 ω 2 d L = 1 (2 π ) 2 (60 . 0Hz) 2 (60 . 0 × 10 3 H) =1 . 17 × 10 4 F . The circuit is then at resonance. (b) In this case, we want Z 2 to be as large as possible. The impedance becomes large without bound as C becomes very small. Thus, the smallest average power occurs for C =0(wh ichisnotvery di±erent from a simple open switch). (c) When ω d L =1 d C , the expression for the average power becomes P avg = ( E ) 2 m 2 R ,
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