terry (ect328) – homework 19 – Turner – (59130)
1
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printout
should
have
9
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
After a 7
.
76 Ω resistor is connected across
a battery with a 0
.
17 Ω internal resistance,
the electric potential between the physical
battery terminals is 10 V.
What is the rated emf of the battery?
Correct answer: 10
.
2191 V.
Explanation:
Let :
R
= 7
.
76 Ω
,
r
= 0
.
17 Ω
,
and
V
= 10 V
.
The current drawn by the external resistor
is given by
I
=
V
R
=
10 V
7
.
76 Ω
= 1
.
28866 A
.
The output voltage is reduced by the inter
nal resistance of the battery by
V
=
E −
I r ,
so the electromotive force is
E
=
V
+
I r
= 10 V + (1
.
28866 A) (0
.
17 Ω)
=
10
.
2191 V
.
002
10.0 points
A certain coffeepot draws 4 A of current when
it is operated on 120 V household lines.
If
electrical
energy
costs
10
cents
per
kilowatthour, how much does it cost to oper
ate the coffeepot for 2 hours?
1.
cost = 16 cents
2.
cost = 8.0 cents
3.
cost = 9.6 cents
correct
4.
cost = 4.8 cents
5.
cost = 2.4 cents
Explanation:
Let :
I
= 4 A
,
V
= 120 V
,
r
= $0
.
1
/
kWh
,
and
t
= 2 h
.
The power that the coffeepot consumes is
P
=
I V ,
so the cost to operate the coffeepot for 2 hours
is
C
=
I V t r
= (4 A) (120 V) (2 h) ($0
.
1
/
kWh)
×
parenleftbigg
1 kW
1000 W
parenrightbigg
=
9.6 cents
.
003
10.0 points
Four identical light bulbs are connected ei
ther in series (circuit A), or in a parallelseries
combination (circuit B), to a constant voltage
battery with negligible internal resistance, as
shown.
E
Circuit A
E
Circuit B
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terry (ect328) – homework 19 – Turner – (59130)
2
Assuming the battery has no internal re
sistance and the resistance of the bulbs is
temperature independent, what is the ratio of
the total power consumed by circuit A to that
consumed by circuit B;
i.e.
,
parenleftbigg
P
A,T otal
P
B,T otal
parenrightbigg
?
1.
P
A
P
B
=
1
2
2.
P
A
P
B
=
1
√
8
3.
P
A
P
B
= 8
4.
P
A
P
B
= 2
5.
P
A
P
B
=
1
16
6.
P
A
P
B
=
1
4
correct
7.
P
A
P
B
= 16
8.
P
A
P
B
= 1
9.
P
A
P
B
= 4
Explanation:
In circuit A, the equivalent resistance is
R
A
= 4
R
, so the electric current through
each bulb is
i
A
=
V
4
R
and the power of each bulb is
P
A
=
I
2
R
=
parenleftbigg
V
4
R
parenrightbigg
2
R
=
V
2
16
R
.
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 Fall '09
 Turner
 SEPTA Regional Rail, Jaguar Racing, University City, JenkintownWyncote, equivalent resistance

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