1
Lehigh University
Physics 21, Spring 2008
March 3, 2008
HW16 Solutions
161.
(HRW 3059
) In Fig.
3070
, after switch S is closed
at time
, the emf of the source is automatically
adjusted to maintain a constant current i through S. (a)
Find the current through the inductor as a function of time.
(b) At what time is the current through the resistor equal to
the current through the inductor?
SSM
Figure 3070
Solution:
59. (a) We assume
i
is from left to right through the closed switch. We let
i
1
be the
current in the resistor and take it to be downward. Let
i
2
be the current in the inductor,
also assumed downward. The junction rule gives
i = i
1
+
i
2
and the loop rule gives
i
1
R –
L
(
di
2
/
dt
) = 0. According to the junction rule, (
di
1
/
dt
)
= –
(
di
2
/
dt
). We substitute into the
loop equation to obtain
L
di
dt
iR
1
1
0
+=
.
This equation is similar to Eq. 3046, and its solution is the function given as Eq. 3047:
ii
e
Rt L
10
=
−
,
where
i
0
is the current through the resistor at
t
= 0, just after the switch is closed. Now
just after the switch is closed, the inductor prevents the rapid buildup of current in its
branch, so at that moment
i
2
= 0 and
i
1
=
i
. Thus
i
0
=
i
, so
( )
12
1
,1
.
Rt L
Rt L
e
ii e
−−
==
−
=
−
(b) When
i
2
=
i
1
,
1
1.
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 Spring '08
 Kim
 Current, Magnetic Field, Energy density, HRW, DT DT DT

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