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howard (cah3459) – ohw24 – turner – (56705)
1
This printout should have 15 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001 (part 1 of 7) 10.0 points
Consider a parallel circuit with a 99 Ω re
sistor, a 6
.
1
μ
± capacitor, a 120 mH induc
tor, and a 718 rad
/
s
AC
supply operating at
120 V
rms
, as shown in the fgure below.
120 mH
6
.
1
μ
±
99 Ω
120 V
rms
A typical phasor diagram is shown below.
V
S
I
R
C
L
φ
ω t
The instantaneous voltages and
rms
volt
ages across the three circuit elements are the
same, and each is in phase with the current
through the resistor. The currents in
C
and
L
lead or lag behind the current in the resistor,
as shown above.
What is the resonance Frequency oF the cir
cuit?
Correct answer: 186
.
022 cycles
/
s.
Explanation:
Let :
ω
= 718 rad
/
s
,
L
= 120 mH =
l
u
,
C
=6
.
1
μ
± = 6
.
1
×
10

6
±
.
I
S
=
±
I
2
R
+[
I
C

I
L
]
2
,
=
V
S
Z
=
²
³
V
S
R
´
2
+
³
V
S
X
C

V
S
X
L
´
2
1
Z
=
²
1
R
2
+
³
1
X
C

1
X
L
´
2
Z
=
1
²
1
R
2
+
³
1
X
C

1
X
L
´
2
=
1
²
1
R
2
+
³
ω C

1
ω L
´
2
,
ThereFore,
I
S
=
V
S
²
1
R
2
+
³

1
´
2
,
and
I
S
is
minimum when
=
1
;
i.e.
, at its reso
nance Frequency.
f
=
ω
2
π
=
1
2
π
√
LC
=
1
2
π
µ
(0
.
12 H) (6
.
1
×
10

6
±)
=
186
.
022 cycles
/
s
.
002 (part 2 of 7) 10.0 points
At 718 rad
/
s, calculate the
rms
current in the
resistor.
Correct answer: 1
.
21212 A
rms
.
Explanation:
Let :
V
S
= 120 V
rms
and
R
= 99 Ω
.
I
R
=
V
S
R
=
120 V
rms
99 Ω
=
1
.
21212 A
rms
.
003 (part 3 of 7) 10.0 points
At 718 rad
/
s, calculate the
rms
current in the
inductor.
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View Full Documenthoward (cah3459) – ohw24 – turner – (56705)
2
Correct answer: 1
.
39276 A
rms
.
Explanation:
I
L
=
V
S
ω L
=
120 V
rms
(718 rad
/
s) (0
.
12 H)
=
1
.
39276 A
rms
.
004 (part 4 of 7) 10.0 points
At 718 rad
/
s, calculate the
rms
current in the
capacitor.
Correct answer: 0
.
525576 A
rms
.
Explanation:
I
C
=
V
S
ω C
= (120 V
rms
) (718 rad
/
s) (6
.
1
×
10

6
F)
=
0
.
525576 A
rms
.
005 (part 5 of 7) 10.0 points
At 718 rad
/
s, calculate the
rms
current deliv
ered by the
AC
supply.
Correct answer: 1
.
49038 A
rms
.
Explanation:
I
S
=
±
I
2
R
+[
I
C

I
L
]
2
=
²
(1
.
21212 A
rms
)
2
+
³
(0
.
525576 A
rms
)

(1
.
39276 A
rms
)
´
2
µ
1
/
2
=
1
.
49038 A
rms
.
006 (part 6 of 7) 10.0 points
In a phasor diagram, what is the magnitude
of the angle between the current with respect
to the voltage?
Correct answer: 35
.
5809
◦
.
Explanation:
φ
= arctan
¶
I
C

I
L
I
R
·
= arctan
¶
0
.
525576 A
rms
1
.
21212 A
rms

1
.
39276 A
rms
1
.
21212 A
rms
·
=

35
.
5809
◦
,
so the angle is
35
.
5809
◦
.
007 (part 7 of 7) 10.0 points
In a phasor diagram, is the current leading or
lagging behind the voltage?
1.
The current leads the voltage.
2.
The current lags behing the voltage.
cor
rect
Explanation:
φ <
0 so the current lags behing the volt
age.
008
10.0 points
Suppose you wish to use a transformer as an
impedancematching device between an au
dio ampli±er that has an output impedance
of 8
.
79 kΩ and a speaker that has an input
impedance of 6
.
5 Ω.
What should be the ratio of primary to
secondary turns on the transformer?
Correct answer: 37.
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