howard (cah3459) – hw13 – turner – (56705)
1
This printout should have 15 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
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)
correct
3.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
4.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
5.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
6.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
7.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
Explanation:
The power dissipated in the resistor has
several expressions
P
=
E
I
=
E
2
R
=
I
2
R,
where the last two are simply derived From the
frst equation together with the application oF
the Ohm’s law.
Since the resistor is connected to a constant
voltage source
E
= constant
P
=
E
2
R
=
constant
R
,
tells us that the power is inversely propor
tional to the resistance
±
P
∝
1
R
²
.
0 1 2 3 4 5 6 7 8 910
0
1
2
3
4
5
Resistance (Ω)
002 (part 1 of 3) 10.0 points
A 880 Ω resistor is rated at 6
.
68 W.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documenthoward (cah3459) – hw13 – turner – (56705)
2
What is the maximum current through this
resistor?
Correct answer: 0
.
0871258 A.
Explanation:
Let :
R
= 880 Ω
and
P
=6
.
68 W
.
The power rating of the resistor is the maxi
mum allowed power dissipation of the resistor
and corresponds to a maximum current.
P
=
IV
=
I
2
R
I
=
±
P
R
=
±
6
.
68 W
880 Ω
=
0
.
0871258 A
.
003 (part 2 of 3) 10.0 points
If the maximum current has been passing
through the resistor for 19
.
4 minutes, how
many Coulombs of charge passes through the
resistor in this period?
Correct answer: 101
.
414 C.
Explanation:
Let :
t
= 19
.
4 minutes
.
Current is
I
=
Q
t
Q
=
It
= (0
.
0871258 A) (19
.
4 minutes)
²
60 s
min
³
=
101
.
414 C
.
004 (part 3 of 3) 10.0 points
Denote the amount of charge in part 2 by
Q
◦
. Consider the passage of the same maxi
mum current as in Part 2 through two 880 Ω
resistors connected in series.
How much charge passes through any cross
section in this resistor series in 19
.
4 minutes?
1.
None of these
2.
Q
=
Q
◦
2
3.
Q
=2
Q
◦
4.
Q
=
√
2
Q
◦
5.
Q
=
Q
◦
√
2
6.
Q
=
Q
◦
correct
7.
Q
=4
Q
◦
8.
Q
=
Q
◦
4
Explanation:
Since
Q
=
IT
, then when
I
is Fxed and
T
is
Fxed,
Q
is correspondingly Fxed. So
Q
=
Q
◦
.
005
10.0 points
We estimate that there are 265 million plugin
electric clocks in the United States, approxi
mately one clock for each person. The clocks
convert energy at the average rate of 8
.
1 W.
To supply this energy, how many metric
tons of coal are burned per hour in coalFred
electricgenerating plants that are, on aver
age, 27
.
5% e±cient? The heat of combustion
for coal is 33 MJ
/
kg.
Correct answer: 851
.
504 metric ton.
Explanation:
Let :
R
=8
.
1W
/
clock
,
L
= 33 MJ
/
kg = 3
.
3
×
10
7
J
/
kg
,
N
.
65
×
10
8
clocks
,
and
Δ
t
= 1 h
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Turner
 Resistance, Resistor, Correct Answer, Howard, Electrical resistance, Series and parallel circuits

Click to edit the document details