benavides (jjb2356) – homework 16 – Turner – (59130)
1
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printout
should
have
10
questions.
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001
10.0 points
A 0
.
28
μ
F capacitor is given a charge
Q
0
.
After 1 s, the capacitor’s charge is 0
.
5
Q
0
.
What is the effective resistance across this
capacitor?
Correct answer: 5
.
15248 MΩ.
Explanation:
Let :
C
= 0
.
28
μ
F = 2
.
8
×
10

7
F
and
t
= 1 s
.
The charge on the capacitor is
Q
(
t
) =
Q
0
e

t/τ
.
e
t/τ
=
Q
0
Q
t
τ
= ln
parenleftbigg
Q
0
Q
parenrightbigg
τ
=
t
ln
parenleftbigg
Q
0
Q
parenrightbigg
.
The effective resistance is
R
=
τ
C
=
t
C
ln
parenleftbigg
Q
0
Q
parenrightbigg
=
1 s
(2
.
8
×
10

7
F) ln
parenleftbigg
Q
0
0
.
5
Q
0
parenrightbigg
·
1 MΩ
10
6
Ω
=
5
.
15248 MΩ
.
002
10.0 points
The switch S has been in the position “a” for
a long time. Then at
t
= 0, it is moved from
“a” to “b”.
C
R
1
R
2
E
S
b
a
Find the time when the charge in the ca
pacitor is reduced to
1
e
of its value at
t
= 0.
1.
τ
=
1
R
1
C
2.
τ
=
R
2
C
3.
τ
=
radicalbig
R
1
R
2
C
4.
τ
= (
R
1
+
R
2
)
C
correct
5.
τ
=
1
(
R
1
+
R
2
)
C
6.
τ
=
2
(
R
1
+
R
2
)
C
7.
τ
=
R
1
+
R
2
2
C
8.
τ
=
1
√
R
1
R
2
C
9.
τ
=
1
R
2
C
10.
τ
=
R
1
C
Explanation:
In charging an
R C
circuit, the characteris
tic time constant is given by
τ
=
R C ,
where in this problem
R
is the equivalent
resistance, or
R
=
R
1
+
R
2
.
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 Spring '08
 Turner
 Charge, Work, Electric charge, 1 m, Jaguar Racing, 5.15248 M, 1090 W, 2628 J

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