nanni (arn437) – HW #8 – Erskine – (56905)
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001(part1of2)10.0points
The output of a generator is given by
V
=
V
max
sin
ω t .
If after 0
.
0134 s
,
the output is 0
.
25 times
V
max
, what is the largest possible angular
velocity of the generator?
Correct answer: 18
.
8567 rad
/
s.
Explanation:
Let :
V
= 0
.
25
V
max
,
and
t
1
= 0
.
0134 s
.
(0
.
25)
V
max
=
V
max
sin(
ω t
1
)
ω t
1
= sin
−
1
(0
.
25)
.
So the largest possible angular velocity is
ω
=
sin
−
1
(0
.
25)
t
1
=
sin
−
1
(0
.
25)
0
.
0134 s
=
18
.
8567 rad
/
s
,
since there are an infinite number of higher
harmonics.
002(part2of2)10.0points
What is the next time (value of
t
) for which
the output is 0
.
25 times
V
max
?
Correct answer: 0
.
153203 s.
Explanation:
θ
V
max
0
t
1
t
2
t
1
T
2
From the figure, we see that the next value
of
t
for which
V
= (0
.
25)
V
max
occurs at
t
2
=
T
2

t
1
,
where
T
is the period. The frequency is given
by
f
=
ω
2
π
, and the period by
T
=
1
f
=
2
π
ω
,
so the required time is
t
2
=
T
2

t
1
=
π
ω

t
1
=
π
18
.
8567 rad
/
s

0
.
0134 s
=
0
.
153203 s
.
Note:
The third time for which the output
is 0
.
25 times
V
max
is simply
t
3
=
t
1
+
T ,
where
T
is the period.
003
10.0points
The graph below depicts an oscillating
emf
.
E
max
E
min
0
φ
0
T/
2
T
Which phasor diagram correctly represents
this oscillation?
Assume counterclockwise
rotation.
1.
E
φ
2.
E
φ
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nanni (arn437) – HW #8 – Erskine – (56905)
2
3.
E
φ
4.
E
φ
5.
E
φ
correct
Explanation:
The
oscillating
emf
is
proportional
to
sin(
ωt

φ
). So the right answer is
E
φ
004(part1of3)10.0points
An
AC
voltage of the form
V
=
V
max
sin 2
πft ,
with frequency 72 Hz and maximum voltage
872 V, is applied across a 51 W light bulb.
When the voltage is first applied, what is
the current through the circuit?
1.
the
rms
current
2.
0
correct
3.
the maximum current
Explanation:
Let :
t
= 0
.
The current in the circuit is
I
=
I
max
sin(2
π f t
) =
I
max
sin 0) =
0
.
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 Fall '11
 ERSKINE
 Correct Answer, Inductor, Electrical resistance, IMAX

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