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96. (a) From Eq. 314, 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. 3049, 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.
 Spring '08
 Any
 Physics, Current, Energy

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