Shie, Gary – Homework 26 – Due: Nov 5 2004, 4:00 am – Inst: Turner
1
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printout
should
have
12
questions.
Multiplechoice questions may continue on
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before answering.
The due time is Central
time.
001
(part 1 of 2) 10 points
The alternating voltage of a generator is rep
resented by the equation
E
=
E
0
sin
ω t ,
where
E
is in volts,
E
0
= 332 V,
ω
= 643 rad
/
s,
and
t
is in seconds.
Find the frequency of the electric potential
E
.
Correct answer: 102
.
337 Hz.
Explanation:
We compare the given equation
E
= (332 V) sin(643 rad
/
s)
t
to the general form for such an equation,
E
=
E
max
sin
ω t.
By comparison, we see that
ω
= 2
π f
= 643 rad
/
s
.
Therefore
f
=
643 rad
/
s
2
π
= 102
.
337 Hz
.
002
(part 2 of 2) 10 points
Find the voltage output of the source.
Correct answer: 234
.
759 V.
Explanation:
The
rms
voltage is
E
rms
=
E
max
√
2
=
332 V
√
2
= 234
.
759 V
.
003
(part 1 of 1) 10 points
A lightbulb is connected to a 60 Hz power
source having a maximum voltage of 138 V.
What is the resistance of the light bulb that
uses an average power of 40
.
3 W?
Correct answer: 236
.
278 Ω.
Explanation:
The rms voltage is
V
rms
=
V
max
√
2
=
(138 V)
√
2
= 97
.
5807 V
.
and the average power is
P
av
=
V
2
rms
R
,
so the resistance is
R
=
V
2
rms
P
av
=
V
2
max
2
P
av
=
(138 V)
2
2 (40
.
3 W)
=
236
.
278 Ω
.
004
(part 1 of 2) 10 points
The output of a generator is given by
V
=
V
max
sin
ω t .
If after 0
.
0109 s
,
the output is 0
.
25 times
V
max
, what is the largest possible angular
velocity
ω
of the generator?
Correct answer: 23
.
1817 rad
/
s.
Explanation:
We are given that
V
= (0
.
25)
V
max
at
t
= 0
.
0109 s
.
Therefore
(0
.
25)
V
max
=
V
max
sin[
ω
(0
.
0109 s)]
ω
(0
.
0109 s) = sin

1
(0
.
25)
= 0
.
25268 rad
.
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 Spring '10
 ERSKINE
 Physics, Work, Alternating Current, Correct Answer, Inductor, Electrical resistance, shie

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