yang (ey942) – ohw 12 – turner – (56705)
1
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printout
should
have
17
questions.
Multiplechoice questions may continue on
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before answering.
001 (part 1 of 3) 10.0 points
A total charge of 11 mC passes through a
crosssectional area of a nichrome wire in 3
.
6 s.
a) What is the current in the wire?
Correct answer: 0
.
00305556 A.
Explanation:
Let :
Δ
Q
1
= 11 mC
and
Δ
t
1
= 3
.
6 s
.
I
=
Δ
Q
Δ
t
=
0
.
011 C
3
.
6 s
=
0
.
00305556 A
.
002 (part 2 of 3) 10.0 points
b) How many electrons pass through the
crosssectional area in 6.50 s?
Correct answer: 1
.
24132
×
10
17
.
Explanation:
Let :
q
e
= 1
.
6
×
10
−
19
C
and
Δ
t
2
= 6
.
50 s
.
Δ
Q
=
N q
e
N
=
Δ
Q
q
e
=
I
Δ
t
2
q
e
=
(0
.
00305556 A)(6
.
5 s)
1
.
6
×
10
−
19
C
=
1
.
24132
×
10
17
.
003 (part 3 of 3) 10.0 points
c) If the number of charges that pass through
the crosssectional area during the given time
interval doubles, what is the resulting cur
rent?
Correct answer: 0
.
00611111 A.
Explanation:
Let :
Δ
Q
2
= 2 Δ
Q
1
.
I
=
2 Δ
Q
1
Δ
t
1
=
2 (0
.
011 C)
3
.
6 s
=
0
.
00611111 A
.
004
10.0 points
An electrician finds that a 0
.
6 m length of a
certain type of wire has a resistance of 0
.
12 Ω.
What is the total resistance of the 115 m of
this wire he plans to use to wire a house?
Correct answer: 23 Ω.
Explanation:
Let :
ℓ
1
= 0
.
6 m
,
ℓ
2
= 115 m
,
and
R
= 0
.
12 Ω
.
A 0
.
6 m length of the wire has a resistance of
λ
=
R
ℓ
1
,
so the total resistance of the wire is
R
tot
=
λ ℓ
=
R
ℓ
1
ℓ
=
parenleftbigg
0
.
12 Ω
0
.
6 m
parenrightbigg
(115 m)
=
23 Ω
.
005 (part 1 of 2) 10.0 points
Suppose that a uniform wire of resistance
R
is stretched uniformly to
N
times its original
length.
Assume the total volume of the wire re
mains the same, and the volume is given by
the product of the cross sectional area and the
length. Denote the original area by
A
and the
area after the stretch by
A
′
.
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yang (ey942) – ohw 12 – turner – (56705)
2
The ratio
A
′
A
is
1.
A
′
A
= 1
2.
A
′
A
=
1
N
2
3.
A
′
A
=
N
2
4.
A
′
A
=
1
N
2
/
3
5.
A
′
A
= 0
6.
A
′
A
=
N
7.
A
′
A
=
√
N
8.
A
′
A
=
1
N
correct
9.
A
′
A
=
N
3
/
2
10.
A
′
A
=
1
√
N
Explanation:
Let :
ℓ
′
=
N ℓ .
The volume before the stretch is
V
=
A ℓ
where
ℓ
is the length.
The volume after the
stretch is
V
′
=
A
′
ℓ
′
=
A
′
N ℓ.
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 Spring '10
 Turner
 Potential Energy, Correct Answer, Electric power transmission

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