aljabr (faa335) – Hw10 – Ross – (89251)
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printout
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have
11
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001
10.0 points
The emf of a battery is
E
= 16 V
.
When the
battery delivers a current of 1 A to a load, the
potential difference between the terminals of
the battery is 14 V volts.
Find the internal resistance of the battery.
Correct answer: 2 Ω.
Explanation:
Given :
E
= 16 V
,
V
load
= 14 V
,
and
I
= 1 A
.
The potential difference across the internal
resistance is
E 
V
load
, so the internal resis
tance is given by
r
=
E 
V
load
I
=
16 V

14 V
1 A
=
2 Ω
.
002
(part 1 of 4) 10.0 points
A battery has an emf of 12 V and an inter
nal resistance of 0
.
15 Ω.
Its terminals are
connected to a load resistance of 2 Ω.
Find the current in the circuit.
Correct answer: 5
.
5814 A.
Explanation:
Let :
E
= 12 V
,
R
= 2 Ω
,
and
r
= 0
.
15 Ω
.
The total resistance is
R
+
r
, so
I
=
E
R
+
r
=
12 V
2 Ω + 0
.
15 Ω
=
5
.
5814 A
.
003
(part 2 of 4) 10.0 points
Calculate the terminal voltage of the battery.
Correct answer: 11
.
1628 V.
Explanation:
The terminal voltage
V
of the battery is
equal to
V
r
=
E 
I r
= 12 V

(5
.
5814 A) (0
.
15 Ω)
=
11
.
1628 V
.
004
(part 3 of 4) 10.0 points
Find the power dissipated in the load resis
tor.
Correct answer: 62
.
3039 W.
Explanation:
The power dissipated in the load resistor is
P
R
=
I
2
R
= (5
.
5814 A)
2
(2 Ω)
=
62
.
3039 W
.
005
(part 4 of 4) 10.0 points
Find the power dissipated in the battery.
Correct answer: 4
.
6728 W.
Explanation:
The power dissipated in the battery is
P
r
=
I
2
r
= (5
.
5814 A)
2
(0
.
15 Ω)
=
4
.
6728 W
.
006
10.0 points
The current in a circuit is tripled by connect
ing a 747 Ω resistor parallel with the resis
tance of the circuit.
Determine the resistance of the circuit in
the absence of the 747 Ω resistor.
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 Spring '10
 Gavrine
 SEPTA Regional Rail, Jaguar Racing, Ω, Highways in Slovakia

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