song (shs546) – Practice HW 9 Solutions – weathers – (22202)
1
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have
11
questions.
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before answering.
001
10.0points
At 45
◦
C, the resistance of a segment of gold
wire is 79 Ω.
When the wire is placed in a
liquid bath, the resistance increases to 180 Ω.
The temperature coefficient is 0
.
0034 (
◦
C)
−
1
at 20
◦
C.
What is the temperature of the bath?
Correct answer: 452
.
986
◦
C.
Explanation:
Let :
R
1
= 79 Ω
,
R
2
= 180 Ω
,
T
0
= 20
◦
C
,
T
1
= 45
◦
C
,
and
α
= 0
.
0034(
◦
C)
−
1
.
Neglecting change in the shape of the wire,
we have
R
1
=
R
0
[1 +
α
(
T
1
−
T
0
)]
R
0
=
R
1
1 +
α
(
T
1
−
T
0
)
and
R
2
=
R
0
[1 +
α
(
T
2
−
T
0
)]
,
where
T
0
= 20
◦
C
.
Thus
R
2
=
R
1
[1 +
α
(
T
2
−
T
0
)]
1 +
α
(
T
1
−
T
0
)
R
2
+
α R
2
(
T
1
−
T
0
) =
R
1
+
α R
1
(
T
2
−
T
0
)
α R
1
T
2
=
R
2
−
R
1
+
α R
2
(
T
1
−
T
0
) +
α R
1
T
0
so that
T
2
=
R
2
−
R
1
+
α R
2
(
T
1
−
T
0
) +
α R
1
T
0
α R
1
.
Since
R
2
−
R
1
+
α R
2
(
T
1
−
T
0
) +
R
1
α T
0
= 180 Ω
−
79 Ω
+ [0
.
0034(
◦
C)
−
1
]
×
(180 Ω) (45
◦
C
−
20
◦
C)
(79 Ω) [0
.
0034(
◦
C)
−
1
] (20
◦
C)
= 121
.
672 Ω
,
then
T
2
=
121
.
672 Ω
(79 Ω) (0
.
0034(
◦
C)
−
1
)
=
452
.
986
◦
C
.
002
10.0points
A variable resistor is connected across a con
stant voltage source.
Which of the following graphs represents
the power
P
dissipated by the resistor as a
function of its resistance
R
?
1.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
Power (W)
2.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
Power (W)
3.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
Power (W)
correct
4.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
Power (W)
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 Spring '00
 Littler
 Resistance, Resistor, Correct Answer, Rb

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