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Version PREVIEW – HW6 – feng – (58230)
1
This printout should have 25 questions.
Multiplechoice questions may continue on
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before answering.
Internal Resistance 07
001
10.0 points
After a 4
.
84 Ω resistor is connected across
a battery with a 0
.
1 Ω internal resistance,
the electric potential between the physical
battery terminals is 13 V.
What is the rated emf of the battery?
Correct answer: 13
.
2686 V.
Explanation:
Let :
R
= 4
.
84 Ω
,
r
= 0
.
1 Ω
,
and
V
= 13 V
.
The current drawn by the external resistor
is given by
I
=
V
R
=
13 V
4
.
84 Ω
= 2
.
68595 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
= 13 V + (2
.
68595 A) (0
.
1 Ω)
=
13
.
2686 V
.
Internal Resistance 04
002
10.0 points
The emf of a battery is
E
= 10 V
.
When the
battery delivers a current of 0
.
4 A to a load,
the potential di±erence between the terminals
of the battery is 8 V volts.
²ind the internal resistance of the battery.
Correct answer: 5 Ω.
Explanation:
Given :
E
= 10 V
,
V
load
= 8 V
,
and
I
= 0
.
4 A
.
The potential di±erence across the internal
resistance is
E −
V
load
, so the internal resis
tance is given by
r
=
E −
V
load
I
=
10 V
−
8 V
0
.
4 A
=
5 Ω
.
Battery with internal resistance
003
(part 1 of 2) 10.0 points
The internal resistance
r
of a battery with
emf
E
is connected to a load resistor with
resistance
R
.
I
10 V
A
B
22 Ω
147 Ω
internal
resistance
²ind the potential di±erence
V
BA
=
V
A
−
V
B
.
Correct answer: 8
.
69822 V.
Explanation:
Let :
E
= 10 V
,
R
= 147 Ω
,
and
r
= 22 Ω
.
The current through the circuit is
i
=
E
r
+
R
,
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View Full Document Version PREVIEW – HW6 – feng – (58230)
2
so the potential diFerence
V
A
−
V
B
is
V
A
−
V
B
=
i R
=
E
R
r
+
R
=
(10 V) (147 Ω)
22 Ω + 147 Ω
=
8
.
69822 V
.
004
(part 2 of 2) 10.0 points
What is the power
P
dissipated by the load
resistor
R
?
1.
P
=
E
2
r
+
R
2.
P
=
R
E
2
3.
P
=
r
E
2
4.
P
= (
r
+
R
)
E
2
5.
P
=
E
2
r
6.
P
=
p
E
r
+
R
P
2
R
correct
7.
P
=
p
E
R
P
2
r
8.
P
=
E
2
r R
9.
P
=
E
2
R
r
10.
P
=
p
E
r
+
R
P
2
r
Explanation:
The power dissipated by the load is
P
=
i
2
R
=
p
E
r
+
R
P
2
R .
Series Circuit 01
005
(part 1 of 3) 10.0 points
The current in the circuit below is 2 am
peres.
0
.
3 Ω
12 V
6 V
0
.
2 Ω
R
1
.
5 Ω
X
Y
What is the resistance
R
?
1.
1 Ω
correct
2.
4 Ω
3.
3 Ω
4.
5 Ω
5.
2 Ω
Explanation:
Let :
R
1
= 0
.
3 Ω
,
R
2
= 0
.
2 Ω
,
R
3
= 1
.
5 Ω
,
E
1
= 12 V
,
and
E
2
= 6 V
.
R
1
E
1
E
2
R
2
R
R
3
X
Y
±rom Ohm’s law, the total resistance of the
circuit is
R
total
=
V
I
=
12 V
−
6 V
2 A
= 3 Ω
.
Therefore, the resistance
R
is
R
=
R
total
−
0
.
3 Ω
−
1
.
5 Ω
−
0
.
2 Ω = 1 Ω
.
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This note was uploaded on 04/27/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Resistance

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