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Shie, Gary – Homework 27 – Due: Nov 8 2004, 4:00 am – Inst: Turner
1
This printout should have 12 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
Consider a series
RLC
circuit.
The applied
voltage has a maximum value oF 150 V and
oscillates at a Frequency oF 82 Hz. The circuit
contains a capacitor whose capacitance can be
varied, a 1100 Ω resistor, and a 3
.
5 H inductor.
Determine the value oF the capacitor such
that the voltage across the capacitor is out oF
phase with the applied voltage by 33
◦
, with
V
L
leading
V
max
.
Correct answer: 0
.
555002
μ
±.
Explanation:
The phase relationships For the voltage
drops across the elements in the circuit are
shown in the fgure.
V
R
V
L
V
C
V
max
φ
α
±rom the fgure, we see that the phase angle
is
φ
=
α

90
◦
= (90
◦
)

(33
◦
)
= 57
◦
= 0
.
994838 rad
.
This is because the phasors representing
I
max
and
V
R
are in the same direction (they are in
phase). ±rom the equation
tan
φ
=
X
L

X
C
R
(1)
and
ω
= 2
π f
= 2
π
(82 Hz)
= 515
.
221 Hz
.
Substituting the expressions For
X
L
and
X
C
,
X
L
=
ω L
= (515
.
221 Hz) (3
.
5 H)
= 1803
.
27 Ω
.
and
X
C
=
1
ω C
,
into the above relation, we obtain
1
2
π f C
= 2
π f L
+
R
tan
φ,
(2)
we obtain
C
=
1
2
π f
•
1
2
π f L
+
R
tan
φ
‚
=
1
ω
•
X
L
+
R
tan
φ
‚

1
=
1
(515
.
221 Hz)
•
(1803
.
27 Ω)
+ (1100 Ω) tan(57
◦
)
‚

1
= 0
.
555002
μ
±
.
002
(part 1 oF 1) 10 points
The emF
E
drives the circuit shown below
at an angular velocity
ω.
R
E
C
L
The light bulb (with resistance
R
) glows
most brightly at
1.
both higher Frequecies,
1
√
LC
< ω <
∞
,
or lower Frequencies, 0
< ω <
1
√
LC
.
2.
lower Frequencies, 0
< ω <
1
√
LC
.
3.
steady DC voltage,
ω
= 0
.
correct
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View Full DocumentShie, Gary – Homework 27 – Due: Nov 8 2004, 4:00 am – Inst: Turner
2
4.
the resonant frequency,
ω
=
1
√
LC
.
5.
higher frequencies,
1
√
LC
< ω <
∞
.
Explanation:
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 Spring '10
 ERSKINE
 Physics, Work

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