ch31-p075 - 75(a From Eq 31-4 we have L =(2C)1 =(2f)2C)1 =...

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75. (a) From Eq. 31-4, we have L = ( ω 2 C ) 1 = ((2 π f ) 2 C ) 1 = 2.41 µ H. (b) The total energy is the maximum energy on either device (see Fig. 31-4). Thus, we have U max = 1 2 LI 2 = 21.4 pJ. (c) Of several methods available to do this part, probably the one most “in the spirit” of
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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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