96. (a) From Eq. 31-4, with ω=2πf, C= 2.00 nF and L= 2.00 mH, we have 417.96 10 Hz.2fLC==×π(b) The maximum current in the oscillator is iIVXCvCCCmax.==×−400 103A. (c) Using Eq. 30-49, we find the maximum magnetic energy: 28,maxmax1
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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.