Nguyen, Thanh – Homework 16 – Due: Apr 1 2008, 7:00 pm – Inst: Weathers
1
This
printout
should
have
10
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
A lightbulb is connected to a 60 Hz power
source having a maximum voltage of 173 V.
What is the resistance of the light bulb that
uses an average power of 82
.
6 W?
Correct answer: 181
.
168 Ω.
Explanation:
Let :
V
max
= 173 V
and
P
av
= 82
.
6 W
.
The
rms
voltage is
V
rms
=
V
max
√
2
=
173 V
√
2
=
122
.
329 V
.
The average power is
P
av
=
V
2
rms
R
R
=
V
2
rms
P
av
=
V
2
max
2
P
av
=
(173 V)
2
2 (82
.
6 W)
=
181
.
168 Ω
.
keywords:
002
(part 1 of 1) 10 points
An inductor has a 54
.
7 Ω reactance at 60 Hz.
What will be the maximum current if this
inductor is connected to a 50 Hz source that
produces a 176
.
9 V
rms
voltage?
Correct answer: 5
.
48829 A.
Explanation:
Let :
X
L
= 54
.
7 Ω
,
f
= 60 Hz
,
f
0
= 50 Hz
,
and
V
rms
= 176
.
9 V
.
The maximum voltage of the circuit is
V
max
=
V
rms
√
2
,
and the inductive reactance is
X
L
=
ω L ,
so
X
0
L
=
ω
0
ω
X
L
=
f
0
f
X
L
,
and the maximum current is
I
max
=
V
max
X
0
L
=
√
2
V
rms
f
f
0
X
L
=
√
2 (176
.
9 V) (60 Hz)
(50 Hz) (54
.
7 Ω)
=
5
.
48829 A
.
keywords:
003
(part 1 of 3) 10 points
A 680 Ω resistor, a 2
.
67
μ
F capacitor, and a
1
.
7 H inductor are connected in series across
a 270 Hz AC source for which the maximum
voltage is 210 V.
Calculate the impedance of the circuit.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '00
 Littler
 Work, Alternating Current, Inductor, RL circuit, LC circuit

Click to edit the document details