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C
HAPTER
21
Quick Quizzes
1.
(c). The average power is proportional to the rms current which is nonzero even though
the average current is zero. (a) is only valid for an open circuit. (b) and (d) can never be
true because
i
av
= 0 for AC currents.
2.
(b). Choices (a) and (c) are incorrect because the unaligned sine curves in Figure 21.9 mean
the voltages are out of phase, and so we cannot simply add the maximum (or rms)
voltages across the elements. (In other words,
∆
V
≠
∆
V
R
+
∆
V
L
+
∆
V
C
even though
∆
v
=
∆
v
R
+
∆
v
L
+
∆
v
C
.)
3.
(b). Note that this is a DC circuit. However, changing the amount of iron inside the
solenoid changes the magnetic field strength in that region and results in a changing
magnetic flux through the loops of the solenoid. This changing flux will generate a back
emf that opposes the current in the circuit and decreases the brightness of the bulb. The
effect will be present only while the rod is in motion. If the rod is held stationary at any
position, the back emf will disappear and the bulb will return to its original brightness
.
4.
(b), (c). The radiation pressure (a) does not change because pressure is force per unit area.
In (b), the smaller disk absorbs less radiation, resulting in a smaller force. For the same
reason, the momentum in (c) is reduced.
177
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View Full Document CHAPTER 21
Problem Solutions
21.1
(a)
( )
( )
max
rms
2
2 100 V
VV
∆=∆ =
=
141 V
(b)
rm s
100 V
5.00
V
I
R
∆
== =
20.0 A
Ω
(c)
V
I
R
∆
28.3 A
Ω
or
( )
2
2 20.0 A
II
=
==
(d)
()
2
23
av
20.0
A5
.
00
2.00
10 W
IR
℘=
=
Ω=
×
=
2.00 kW
21.2
( )
2
2
2
2
1
22
2
V
IV
R
R
RR
∆
∆
⎛⎞ ⎡⎤
=
=
=
⎜⎟
⎢⎥
⎝⎠
⎣⎦
, so
( )
2
2
V
R
∆
=
℘
(a) If
, then
75.0 W
2
170 V
2 75.0 W
R
=
=
193
Ω
(b) If
, then
100 W
2
2100 W
R
145
Ω
21.3
The meters measure the rms values of potential difference and current. These are
V
V
∆
∆=
=
=
70.7 V
, and
24.0
V
I
R
∆
=
Ω
2.95 A
21.4
All lamps are connected in parallel with the voltage source, so
120 V
V
∆
=
for each
lamp. Also, the current is
=
℘∆
and the resistance is
RVI
= ∆
.
1,rm s
2,rm s
150 W
=
1.25 A
and
12
=
96.0
Ω
3,rm s
I
0.833 A
and
3
R
=
=
144
Ω
178
CHAPTER 21
21.5
The total resistance (series connection) is
12
8.20
10.4
18.6
eq
RRR
=
+=
Ω
+
Ω
=
Ω
, so the
current in the circuit is
rm s
15.0 V
0.806 A
V
I
R
∆
== =
Ω
The power to the speaker is then
( ) ( )
2
2
av
ker
0.806
A1
0
.
4
spea
IR
℘
==
Ω
=
6.76 W
21.6
(a)
, so
max
150 V
V
∆=
22
V
V
∆
=
=
106 V
(b)
377 rad s
f
ω
ππ
=
60.0 Hz
(c) At
()
1120 s
t
=
,
( )
( )
( ) ( ) ( )
150
Vs
in
377 rad s 1 120
s1
5
0
i
n
rad
v
π
⎡⎤
⎣⎦
=
0
(d)
50.0
V
I
R
∆
3.00 A
Ω
21.7
1
2
C
X
fC
=
, so its units are
11V
o
l
tV
o
l
t
O hm
1SecFarad
1Sec Coulom b Vol
tC
o
u
l
ombSec Amp
=
=
21.8
( )
rms
2
=2
=
2
C
V
II
V
X
∆
∆
f
C
(a)
( ) ( )
( )
6
= 2 120
V2 6
0
.0 Hz 2.20
10
C/V =0.141 A
I
−
×
=
141
mA
(b)
( )
( )
6
= 2 240
V2 5
0
.0 Hz 2.20
C/V =0.235 A
I
−
×
=
235
21.9
2
C
V
If
C
X
∆
∆
V
, so
( )
6
0.30 A
2
24
.
01
0
F3
0
V
I
f
CV
−
=
2
4.
0 Hz
×
∆
×
179
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View Full Document CHAPTER 21
21.10
()
max
==
2
C
V
If
C
X
π
∆
∆
V
( )
( )
( )
6
29
0
.
0 Hz 3.70
10
C /V
48.0
V=0
.100 A=
−
=×
100
mA
21.11
rm s
m ax
2
2
C
VV
Cf
C
V
X
ππ
∆∆
⎛⎞
=∆
⎜⎟
⎝⎠
2
,
so
5
0.75 A
1.
71
0 F
26
0
H z 170
V2
I
C
fV
−
=
×
=
17 F
∆
µ
21.12
2
C
IC
X
ω
,
or
( )
6
140 V
=
120 rad s 6.00
F0
.
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This homework help was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.
 Spring '08
 Chilton

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