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Pi
R
R
eR
R
e
tt
LL
thermal
==
−
=
−
−−
2
2
2
2
2
2
11
εε
ττ
chch
.
We equate this to
dU
B
/
dt
, and solve for the time:
()
22
2
1
1
ln 2
37.0ms ln 2
25.6ms.
L
t
L
ee
e
t
RR
τ
−
−=
−
⇒
=
=
=
61. From Eq. 3049 and Eq. 3041, the rate at which the energy is being stored in the
inductor is
2
2
/2
1
L
L
t
t
B
L
dL
i
dU
di
Li
L
e
e
e
e
dt
dt
dt
R
R
R
ε
−
−
⎛⎞
=
−
=
−
⎜⎟
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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, Energy

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