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2
bat

40 V
8.0 10 A s .
0.050 H
L
di
dt
L
ε
==
=
×
(c) This circuit becomes equivalent to that analyzed in §309 when we replace the parallel
set of 20000
Ω
resistors with
R
= 10000
Ω
. Now, with
τ
L
=
L
/
R
= 5
×
10
–6
s, we have
t
/
L
= 3/5, and we apply Eq. 3041:
()
35
3
bat
11
.
8
1
0
A
.
ie
R
−−
=−
≈
×
(d) The rate of change of the current is figured from the loop rule (and Eq. 3035):
bat
0
.
L
iR
−
−=
Using the values from part (c), we obtain 
L

≈
22 V. Then,
2
bat
22 V
4.4 10 A s .
0.050 H
L
di
dt
L
≈
×
(e) As
t
→∞
, the circuit reaches a steady state condition, so that
di
bat
/
dt
= 0 and
L
= 0.
The loop rule then leads to
3
bat
bat
40 V
4
.
0
1
0A
.
10000
L
i
εε
−
−−=
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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

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