Set Up:
At all times
The voltage across the resistor depends on the current through it
and the voltage across the inductor depends on the rate at which the current through it is changing. Immediately after
closing the switch the current thorough the inductor is zero. After a long time the current is no longer changing.
Solve: (a)
so
and
The ammeter reading is
(b)
After a long time,
and
and
The ammeter reading is
(c)
None of the answers in (a) and (b) depend on
L
so none of them would change.
Reflect:
The inductance
L
of the circuit affects the rate at which current reaches its final value. But after a long time
the inductor doesn’t affect the circuit and the final current does not depend on
L.
A
5
1.67A.
i
5
v
1
R
5
25.0 V
15.0
V
5
1.67 A.
v
1
5
iR
v
1
5
25.0 V.
v
2
5
0
A
5
0.
v
2
5
25.0 V.
v
1
5
0
i
5
0
v
1
1
v
2
5
25.0 V.
i
S
0.
t
S
`
,
t 5
L
R
5
300.0
3
10
2
3
H
500.0
V
5
6.00
3
10
2
4
s
5
0.600 ms
I
0
5
0.500 A
i
5
I
0
e
2
1
R
/
L
2
t
.
S
2
S
1
250.0 V
500.0
V
5
0.500 A.
S
1
i
5
0
i
5
E
R
5
6.00 V
8.00
V
5
0.750 A.
t
S
`
,
i
5
E
R
1
1
2
e
2
1
2
5
6.00 V
8.00
V
1
1
2
e
2
1
2
5
0.474 A.
t
5
0.313 s,
t 5
L
R
5
2.50 H
8.00
V
5
0.313 s.
i
5
0,
v
L
5
L
D
i
D
t
5
1
2.50 H
21
2.40 A
/
s
2
5
6.00 V.
D
i
D
t
5
2.40 A
/
s
t
5
0,
D
i
D
t
5
E
L
5
6.00 V
2.50 H
5
2.40 A
/
s.
i
5
0
t
5
0,
t
S
`
.
i
5
i
max
5
E
R
i
5
0.
t
5
0,
i
5
E
R
1
1
2
e
2
1
R
/
L
2
t
2
.
L
D
i
D
t
5
v
L
,
E
2
iR
2
L
D
i
D
t
5
0.
U
max
5
1
2
LI
2
5
1
2
1
11.0
3
10
2
3
H
21
0.0800 A
2
2
5
3.52
3
10
2
5
J
5
35.2
m
J
i
5
1
0.0800 A
21
1
2
e
2
73.3
m
s
/
73.3
m
s
2
5
0.0506 A
I
5
E
R
5
12.0 V
150.0
V
5
0.0800 A.
t
S
`
t 5
L
R
5
11.0
3
10
2
3
H
150.0
V
5
7.33
3
10
2
5
s
5
73.3
m
s
U
5
1
2
LI
2
.
i
5
E
R
1
1
2
e
2
t
/
t
2
.
t 5
L
/
R
.
P
5
I
2
R
5
1
0.300 A
2
2
1
180
V
2
5
16.2 W
5
16.2 J
/
s
U
5
1
2
LI
2
5
1
2
1
12.0 H
21
0.300 A
2
2
5
0.540 J
P
5
I
2
R
.
U
5
1
2
LI
2
.
2112
Chapter 21
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21.56.
Set Up:
Problem 21.55 shows that
initially is zero and rises to 25.0 V, that
starts at 25.0 V and
decreases to zero, and that
i
starts at zero and rises to 1.67 A.
so is proportional to
i. A
reads the current
i.
Solve:
The graphs are sketched in Figure 21.56.
Figure 21.56
21.57.
Set Up:
Current decay in an
R

L
circuit is described by Eq. (21.28).
Solve:
and
and
Reflect:
As
R
is decreased the time constant increases and it takes longer for the current to decay. For this circuit the
time constant is
After one time constant the current has decayed to about 37% of its original value.
It is reasonable for it to take a little over two time constants for the current to decay to 10% of its original value.
21.58.
Set Up:
The current as a function of time is given by
The energy stored in the inductor
is
U
reaches
its maximum value when
i
is
times its maximum value.
Solve: (a)
The maximum current is
gives
and
(b)
and
21.59.
Set Up:
The resonant angular frequency of an
L

C
circuit is
Solve:
21.60.
Set Up:
The energy stored in a capacitor is
The energy stored in an inductor is
Energy conservation requires that the total stored energy be constant. The current is a maximum when the charge on
the capacitor is zero.
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 Spring '09
 RODRIGUEZ
 Physics, Flux, Magnetic Field

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