gilvin (jg47854) – 13. Circuits – meyers – (21235)
1
This printout should have 24 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
This homework is due ±riday, march 25, at
midnight Tucson time.
001 (part 1 of 2) 10.0 points
A battery with an emF oF 10
.
1 V and internal
resistance oF 1
.
1 Ω is connected across a load
resistor
R
.
IF the current in the circuit is 1
.
57 A, what
is the value oF
R
?
Correct answer: 5
.
33312 Ω.
Explanation:
Let :
E
= 10
.
1 V
,
I
= 1
.
57 A
,
and
R
i
= 1
.
1 Ω
.
The electromotive Force
E
is given by
E
=
I
(
R
+
R
i
)
R
=
E
I
−
R
i
=
10
.
1 V
1
.
57 A
−
1
.
1 Ω
=
5
.
33312 Ω
.
002 (part 2 of 2) 10.0 points
What power is dissipated in the internal re
sistance oF the battery?
Correct answer: 2
.
71139 W.
Explanation:
The power dissipation due to the internal
resistance is
P
=
I
2
R
i
= (1
.
57 A)
2
(1
.
1 Ω)
=
2
.
71139 W
.
003
10.0 points
What potential di²erence is measured across
a 16
.
1 Ω load resistor when it is connected
across a battery oF
emf
5
.
77 V and internal
resistance 0
.
388 Ω?
Correct answer: 5
.
63422 V.
Explanation:
Let :
E
= 5
.
77 V
,
R
= 16
.
1 Ω
,
and
R
i
= 0
.
388 Ω
.
The total resistance is
R
T
=
R
+
R
i
=
E
I
,
so
I
=
E
R
+
R
i
.
Thus the voltage across the resistance
R
is
V
=
I R
=
R
E
R
+
R
i
=
(16
.
1 Ω) (5
.
77 V)
16
.
1 Ω + 0
.
388 Ω
=
5
.
63422 V
.
004 (part 1 of 3) 10.0 points
A 25 Ω resistor and a 5 Ω resistor are con
nected in series to a 6 V battery.
±ind the current in each resistor.
Correct answer: 0
.
2 A.
Explanation:
Let :
R
1
= 25 Ω
,
R
2
= 5 Ω
,
and
Δ
V
2
= 6 V
.
±or the series resistors
R
eq
=
R
1
+
R
2
= 25 Ω + 5 Ω = 30 Ω
.
I
1
=
I
2
=
I
=
6 V
30 Ω
=
0
.
2 A
.
005 (part 2 of 3) 10.0 points
±ind the potential di²erence across the frst
resistor.
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View Full Documentgilvin (jg47854) – 13. Circuits – meyers – (21235)
2
Correct answer: 5 V.
Explanation:
Δ
V
1
=
I
1
R
1
= (0
.
2 A)(25 Ω)
=
5 V
.
006 (part 3 of 3) 10.0 points
Find the potential diference across the second
resistor.
Correct answer: 1 V.
Explanation:
Δ
V
2
=
I
2
R
2
= (0
.
2 A)(5 Ω)
=
1 V
.
007 (part 1 of 2) 10.0 points
Two identical light bulbs A and B are con
nected in series to a constant voltage source.
Suppose a wire is connected across bulb B as
shown.
E
A
B
Bulb A
1.
will burn hal± as brightly as be±ore.
2.
will burn twice as brightly as be±ore.
3.
will go out.
4.
will burn as brightly as be±ore.
5.
will burn nearly ±our times as brightly as
be±ore.
correct
Explanation:
The electric power is given by
P
=
I
2
R .
Be±ore the wire is connected,
I
A
=
I
B
=
V
2
R
,
so that
P
A
=
p
V
2
R
P
2
·
R
=
V
2
4
R
.
A±ter the wire is connected,
I
′
A
=
V
R
and
I
′
B
= 0
,
so
P
′
A
=
p
V
R
P
2
·
R
=
V
2
R
= 4
P
A
.
008 (part 2 of 2) 10.0 points
and bulb B
1.
will burn twice as brightly as be±ore.
2.
will burn hal± as brightly as be±ore.
3.
will burn nearly ±our times as brightly as
be±ore.
4.
will burn as brightly as be±ore.
5.
will go out.
correct
Explanation:
Since there is no potential diference be
tween the two ends o± bulb B, it goes out.
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 Spring '11
 doc
 Magnetism, Work, Resistor, Correct Answer, Ω, Series and parallel circuits, R1 R2 R1, R2 R1 R2

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