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(g)
i
1
= 0, and
(h)
i
2
= 0.
56. (a) The inductor prevents a fast buildup of the current through it, so immediately
after the switch is closed, the current in the inductor is zero. It follows that
1
12
100V
3.33A.
10.0
+20.0
i
RR
ε
==
=
+Ω
Ω
(b)
21
3.33A.
ii
(c) After a suitably long time, the current reaches steady state. Then, the emf across the
inductor is zero, and we may imagine it replaced by a wire. The current in
R
3
is
i
1
–
i
2
.
Kirchhoff’s loop rule gives
()
11
2 2
1
2
3
0
0.
iR iR
iR
i i R
−
−=
−
−−
=
We solve these simultaneously for
i
1
and
i
2
, and find
( )( )
23
1
13
20.0
30.0
10.0
20.0
10.0
30.0
20.0
30.0
4.55A,
i
+
Ω
++
Ω
Ω
+
Ω
Ω
+
Ω
Ω
=
(d) and
( )( )
3
2
30.0
10.0
20.0
10.0
30.0
20.0
30.0
2.73A.
R
i
Ω
Ω
Ω
+
Ω
Ω
+
Ω
Ω
=
(e) The lefthand branch is now broken. We take the current (immediately) as zero in that
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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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