ch31-p096 - 96. (a) From Eq. 31-4, with = 2 f , C = 2.00 nF...

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96. (a) From Eq. 31-4, with ω = 2 π f , C = 2.00 nF and L = 2.00 mH, we have 4 1 7.96 10 Hz. 2 f LC == × π (b) The maximum current in the oscillator is iI V X Cv C C C max . = = × 400 10 3 A. (c) Using Eq. 30-49, we find the maximum magnetic energy: 28 ,max max 1
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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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