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Version 120 – TEST02 – Tsoi – (59090)
1
This printout should have 14 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 2) 10.0 points
The circuit has been connected as shown in
the fgure For a “long” time.
7 V
S
15
μ
±
18 Ω
24 Ω
10 Ω
60 Ω
What is the magnitude oF the electric po
tential across the capacitor?
1. 2.0
2. 5.0
3. 18.0
4. 30.0
5. 16.0
6. 17.0
7. 8.0
8. 3.0
9. 1.0
10. 15.0
Correct answer: 2 V.
Explanation:
E
S
1
C
t
b
a
b
I
t
R
1
2
b
3
4
Let :
R
1
= 18 Ω
,
R
2
= 24 Ω
,
R
3
= 10 Ω
,
R
4
= 60 Ω
,
and
C
= 15
μ
±
.
AFter a “long time” implies that the ca
pacitor
C
is Fully charged and thereFore the
capacitor acts as an open circuit with no cur
rent ²owing to it. The equivalent circuit is
a
b
R
t
=
R
1
+
R
2
= 18 Ω + 24 Ω = 42 Ω
R
b
=
R
3
+
R
4
= 10 Ω + 60 Ω = 70 Ω
I
t
=
E
R
t
=
7 V
42 Ω
= 0
.
166667 A
I
b
=
E
R
b
=
7 V
70 Ω
= 0
.
1 A
Across
R
1
E
1
=
I
t
R
1
= (0
.
166667 A) (18 Ω)
= 3 V
.
Across
R
3
E
3
=
I
b
R
3
= (0
.
1 A) (10 Ω)
= 1 V
.
Since
E
1
and
E
3
are “measured” From the same
point “
a
”, the potential across C must be
E
C
=
E
3
 E
1
= 1 V

3 V
=

2 V
E
C

=
2 V
.
002
(part 2 oF 2) 10.0 points
IF the battery is disconnected, how long does it
take For the capacitor to discharge to
E
t
E
0
=
1
e
oF its initial voltage?
1. 198.0
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2
2. 308.0
3. 900.0
4. 147.0
5. 315.0
6. 91.0
7. 504.0
8. 153.0
9. 224.0
10. 1080.0
Correct answer: 315
μ
s.
Explanation:
With the battery removed, the circuit is
C
I
ℓ
R
1
r
2
3
4
ℓ
r
eq
where
R
ℓ
=
R
1
+
R
3
= 18 Ω + 10 Ω = 28 Ω
,
R
r
=
R
2
+
R
4
= 24 Ω + 60 Ω = 84 Ω
and
R
eq
=
p
1
R
ℓ
+
1
R
r
P
−
1
=
p
1
28 Ω
+
1
84 Ω
P
−
1
= 21 Ω
.
Therefore the time constant
τ
is
τ
≡
R
eq
C
= (21 Ω) (15
μ
F) = 315
μ
s
.
The equation for discharge of the capacitor is
Q
t
Q
0
=
e

t/τ
,
or
E
t
E
0
=
e

t/τ
=
1
e
.
Taking the logarithm of both sides, we have

t
τ
= ln
p
1
e
P
t
=

τ
(

ln
e
)
=

(315
μ
s) (

1)
=
315
μ
s
.
003
10.0 points
Given a spherical capacitor with radius of the
inner conducting sphere
a
and the outer shell
b
. The outer shell is grounded. The charges
are +
Q
and

Q
. A point
C
is located at
r
=
R
2
, where
R
=
a
+
b
.
a
A
B
C
+
Q

Q
b
What is the capacitance of this spherical
capacitor?
1.
C
=
1
k
e
(
a

b
)
2.
C
=
b

a
2
k
e
ln
p
b
a
P
3.
C
=
b
k
e
4.
C
=
1
k
e
p
1
a

1
b
P
correct
5.
C
=
a
k
e
6.
C
=
k
e
b
7.
C
=
k
e
a
8.
C
=
a
+
b
k
e
Version 120 – TEST02 – Tsoi – (59090)
3
9.
C
=
b
2
4
k
e
(
b

a
)
10.
C
=
1
k
e
(
a
+
b
)
Explanation:
Δ
V
=
V
a

V
b
=
k
e
Q
p
1
a

1
b
P

0
since
V
b
is grounded.
The charge on the
inside of the shell doesn’t aFect the grounded
potential.
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This note was uploaded on 06/15/2010 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner

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