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Note that we have dropped the
φ
term inside of the cosine because the problem says that the
capacitor starts oﬀ fully charged. The cosine term will have completed
N
complete cycles when
ω
0
t
= 2
πN
⇒
t
=
2
πN
ω
0
The book says the we can generally treat
ω
0
as if it were
ω
, but keeping it as
ω
0
does not
complicate this problem much so we will do so. Hence,
t
N
=
2
πN
q
ω
2

(
R
2
L
)
2
=
2
πN
q
1
LC

(
R
2
L
)
2
After a complete cycle the cosine term will return to 1 so we only need to worry about the
exponential term when computing the magnitude of the new charge.
q
N
=
q
0
exp

R
2
L
2
πN
q
1
LC

(
R
2
L
)
2
=
q
0
exp

2
πN
q
4
L
R
2
C

1
If we plug in the numbers we ﬁnd
q
5
= 5
.
85
μC
q
10
= 5
.
52
μC
q
100
= 1
.
93
μC
q
10
,
000
= 1
.
16
×
10

50
μC
Problem 31.34
An ac generator has emf
E
=
E
m
sin
ω
d
t
, with
E
m
= 25
V
and
ω
d
= 377
rad
s
. It is connected to
a 12
.
7
H
inductor. (a) What is the maximum value of the current? (b) When the current is
a maximum, what is the emf of the generator? (c) When the emf of the generator is

12
.
5
V
and increasing in magnitude, what is the current?
Although we derived it above, lets rederive the current in this circuit. The potential across the
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 Spring '08
 Any
 Physics, Charge

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