Chapter 21
Alternating Current Circuits and
Electromagnetic Waves
Problem Solutions
21.1
(a)
VR,rms
I rms R
(b)
VR,max
2
(c)
I max
(d ) Pav
21.2
(a)
VR,rms
2I rms
I
2
rms
8.00 A
2
R
(c) Because R
2
2
Vrms
Pav
136 V
11.3 A
12.0
768 W
2 1.20 102 V
VR,rms
2
Vrms
R
96.0 V
2 96.0 V
2 8.00 A
2
I rms
R
VR,max
(b) Pav
8.00 A 12.0
R
2
Vrms
Pav
1.70 102 V
1.20 102 V
60.0 W
2
2.40 102
(see above), if the bulbs are d esigned to operate at the sam e
voltage, the 1.00 102 W will have the lower resistance.
21.3
The m eters m easure the rm s values of potential d ifference and current. These are
Vrms
Vmax
2
100 V
2
70.7 V , and I rms
Vrms
R
70.7 V
24.0
2.95 A
317
318
21.4
21.5
CH APTER 21
All lam ps are connected in parallel w ith the voltage source, so Vrms 120 V for each
lam p. Also, the current is Irms Pav Vrms and the resistance is R
Vrms I rms .
150 W
120 V
I1, rms
I 2, rms
I 3, rms
100 W
120 V
1.25 A and R1
0.833 A and R3
R2
120 V
0.833 A
The total resistance (series connection) is Req
R1
120 V
1.25 A
96.0
144
R2
8.20
10.4
18.6
, so the
current in the circuit is
Vrms
Req
I rms
15.0 V
18.6
0.806 A
The pow er to the speaker is then Pav
21.6
2
I rms
Rspeaker
The general form of the generator voltage is
(a)
VR,max
(c)
VR ,rms
(d ) I rms
170 V
and
VR ,max
170 V
2
VR ,rms
R
(e)
I max
(f)
2
Pav I rms
R
(b)
2I rms
2
v
Vmax sin
60 rad s
2 rad
10.4
6.76 W
t , so by inspection
30.0 Hz
120 V
2
120 V
20.0
f
2
0.806 A
6.00 A
2 6.00 A
6.00 A 20.0
8.49 A
720 W
(g) The argum ent of the sine function has units of
t
rad s s
radians .
At t 0.005 0 s , the instantaneous current is
i
v
R
170 V
20.0
sin 60
rad s 0.005 0 s
170 V
20.0
sin 0.94 rad
6.9 A
A lternating Current Circuits and Electromagnetic W aves
21.7
1
, so its units are
2 fC
XC
1
1 Sec Farad
21.8
Vrms
= 2
XC
Volt
Amp
Ohm
Vrms 2 f C
60.0 Hz 2.20 10
6
C/V =0.141 A
141 mA
(b) I max = 2 240 V 2
50.0 Hz 2.20 10
6
C/V =0.235 A
235 mA
I rms =
Vrms
=2 f C
XC
I rms
2 C Vrms
Vrms , so
0.30 A
2
1
2 fC
XC
(a)
(b) I rms
21.11
2
Volt
Coulomb Sec
I max = 2 120 V 2
f
21.10
1
1 Sec Coulomb Volt
I max = 2 I rms =
(a)
21.9
319
4.0 10
6
4.0 10 2 Hz
F 30 V
1
2
VC ,rms
60.0 Hz 12.0 10
36.0 V
221
XC
6
221
F
0.163 A
(c)
I max
(d )
No. The current reaches its m axim um value one-quarter cycle before the voltage
reaches its m axim um value. From the d efinition of capacitance, the capacitor
reaches its m axim um charge w hen the voltage across it is also a m axim um .
Consequently, the m axim um charge and the m axim um current d o not occur at the
sam e tim e.
I rms =
so
2 I rms
2 0.163 A
Vrms
=2 f C
XC
C
Vmax
2
I
f
Vmax
fC
0.231 A
Vmax
2
0.75 A
2
60 Hz 170 V
2
1.7 10
5
F
17 F
320
21.12
CH APTER 21
(a) By inspection,
(b) Also by inspection,
80
I max
0.354 A
(c)
I rms =
2
0.500 A
=
2
VC ,max
(d ) X C
21.13
1
2 fC
XC
XL
2 f L , and from
t
I
(a)
XL
L
2 fL 2
(b) I rms
(c)
f
I max
VL ,rms
XL
2 I rms
VC ,max
VC ,rms
rad s , so f
98.0 V
2
2
80 rad s
2 rad
2
69.3 V
40.0 Hz
196
1
XC
1
, so C
C
(e)
XL
21.15
98.0 V
0.500 A
I max
are then L
21.14
98.0 V , so
VC ,max
80
I
t
L
1
rad s 196
2.03 10
t
, w e have L
I
5
F
20.3 F
. The units of self ind uctance
Volt sec
. The units of ind uctive reactance are given by
amp
1
sec
Volt sec
Amp
Volt
Amp
80.0 Hz 25.0 10 3 H
78.0 V
12.6
Ohm
12.6
6.19 A
2 6.19 A
8.75 A
The required ind uctive reactance is
XL
VL ,max
VL ,max
I max
2 I rms
and the need ed ind uctance is
L
21.16
Given: vL
XL
2 f
VL,max
2 2 f I rms
1.20 102 V sin 30 t
4.00 V
2 2
300.0 Hz 2.00 10 3 A
and L 0.500 H
0.750 H
A lternating Current Circuits and Electromagnetic W aves
(a) By inspection,
(c)
XL
(d ) I rms
1.20 102 V
VL ,max
2
2 fL
L
VL ,rms
XL
30
rad s
2
2
15.0 Hz
VL,max 1.20 102 V , so that
(b) Also by inspection,
VL ,rms
rad s , so f
30
84.9 V
2
30
rad s 0.500 H
84.9 V
47.1
47.1
1.80 A
(e)
I max
(f)
The phase d ifference betw een the voltage across an ind uctor and the current
through the ind uctor is L 90 , so the average pow er d elivered to the ind uctor is
2 I rms
321
2 1.80 A
PL,av I rms VL,rms cos
L
2.55 A
I rms VL,rms cos 90
0
(g) When a sinusoid al voltage w ith a peak value VL,max is applied to an ind uctor, the
current through the ind uctor also varies sinusoid ally in tim e, w ith the sam e
frequency as the applied voltage, and has a m axim um value of I max
VL,max X L .
H ow ever, the current lags behind the voltage in phase by a quarter-cycle or
VL,max sin t , the current as a
2 radians . Thus, if the voltage is given by vL
function of tim e is i
I max sin
through it w ill be i
2.55 A sin 30 t
(h) When i
1.00 A , w e have:
30 t
and
t
t
2 sin
1
2 . In the case of the given ind uctor, the current
2 .
1.00 A 2.55 A , or
sin 30 t
2
1.00 A 2.55 A
sin
1
s
20.9 ms
2 rad 0.403 rad
30 rad s
2.09 10
2
0.392
0.403 rad
322
21.17
CH APTER 21
From L N
N
B, total
B
B
I (see Section 20.6 in the textbook), the total flux through the coil is
L I w here B is the flux through a single turn on the coil. Thus,
L I max
B ,total max
L
21.18
2
Vrms
XL
2
v
80.0 V and
R2
Z
2 120 V
2 fL
(a) The applied voltage is
Vmax
Vmax
XL
L
Vmax sin
2
80.0 V sin 150t , so w e have that
t
150 rad s . The im ped ance for the circuit is
2 f
XC
0.450 T m 2
60.0 Hz
R2
2 fL 1 2 fC
2
R2
L 1 C
2
2
or Z
40.0
Vmax
Z
(b) I max
21.19
2
XC
1
2 fC
Z
R2
(a)
I rms
150 rad s 80.0 10
80.0 V
57.5
XL
60.0 Hz 40.0 10
2
XC
Vrms
Z
(b)
VR, rms
I rms R
(c)
VC , rms
I rms X C
(d )
tan
1
50.0
30.0 V
83.0
XL
R
6
2
F
150 rad s 125 10
66.3
2
0 66.3
0.361 A
0.361 A 50.0
XC
H
1.39 A
1
2
3
1
0.361 A 66.3
tan
1
0 66.3
50.0
18.1 V
23.9 V
53.0
so, the voltage lags behind the current by 53°
83.0
6
F
57.5
A lternating Current Circuits and Electromagnetic W aves
21.20
(a)
XL
2 fL 2
XC
1
2 fC
Z
R2
Vmax
2
H
XC
2
2
500
126
XC
tan
R
719
2
719
776
194 V
126
1
F
126
0.250 A 776
XL
6
50.0 Hz 4.43 10
XL
1
3
1
I max Z
tan
(b)
50.0 Hz 400 10
719
49.9
500
Thus, the current leads the voltage by 49.9
21.21
(a)
XL
2 fL 2
XC
1
2 fC
Z
R2
21.22
1
2
240 Hz 0.250 10
XL
Vmax
Z
(b) I max
XC
2
XC
6
2
900
140 V
1.44 103
XL
1
3.77 103
240 Hz 2.50 H
0.097 2 A
1
3.77 2.65
tan
(d )
0 , so the voltage leads the current
XL
2 fL 2
XC
1
2 fC
Z
R2
(a)
VR,rms
tan
XL
60.0 Hz 0.100 H
I rms R
2
50.0
51.2
37.7
60.0 Hz 10.0 10
XC
103
900
1
2
2
103
3.77 2.65
(c)
R
F
2.65 103
2
2.75 A 50.0
6
265
F
37.7
138 V
265
2
233
(e)
1.44 k
323
324
CH APTER 21
(b)
VL,rms
(c)
VC
I rms X C
2.75 A 265
729 V
(d )
Vrms
I rms Z
2.75 A 233
641
21.23
I rms X L
XL
2 fL 2
XC
1
2 fC
(a)
60.0 Hz 0.400 H
Z LC
0
2
60.0 Hz 3.00 10
XL
XL
2
XC
XC
2
6
884
F
2
60.0
VRC , rms
XL
0 XC
2
151
884
2
735
2 fL 2
60 Hz 2.8 H
Z
R2
XC
(a)
I max
Vmax
Z
I max R
VL, max
I max X L
884
90.0 V
735
733
2
89.8 V
886
886
108 V
1.1 103
1.2 103
170 V
1.6 103
VR, max
2
Vrms
Z RC
Z RLC
XL
2
733
90.0 V
735
60.0
I rms Z RC
XL
XC
Vrms
Z LC
Z RLC
I rms Z LC
R2
(b) Z RC
(b)
151
Vrms
Z RLC
I rms
VLC , rms
21.24
104 V
1
R2
Z RLC
and
2.75 A 37.7
2
1.1 103
0
2
0.11 A
0.11 A 1.2 103
0.11 A 1.1 103
1.3 102 V
1.2 102 V
1.6 103
A lternating Current Circuits and Electromagnetic W aves
(c) When the instantaneous current is a m axim um i
across the resistor is
vR
iR
I max R
325
I max , the instantaneous voltage
1.3 102 V . The instantaneous
VR, max
voltage across an ind uctor is alw ays 90° or a quarter cycle out of phase w ith the
instantaneous current. Thus, w hen i Imax , vL 0 .
Kirchhoff’ s loop rule alw ays applies to the instantaneous voltages around a closed
path. Thus, for this series circuit, vsource
vR
vL and at this instant w hen i Imax
w e have
vsource
1.3 102 V
I max R 0
(d ) When the instantaneous current i is zero, the instantaneous voltage across the
resistor is vR iR 0 . Again, the instantaneous voltage across an ind uctor is a
quarter cycle out of phase w ith the current. Thus, w hen i 0 , w e m ust have
vL
1.2 102 V . Then, applying Kirchhoff’ s loop rule to the
VL, max
instantaneous voltages around the series circuit at the instant w hen i 0 gives
vsource
vR
XC
1
2 fC
Z RC
R2
21.25
and
(a)
XC
(b) Z
(c)
I max
0
2
50.0 103
rms
Vb, rms
R2
12
2
5 000 V
1.33 108
I rms Rb
2
2
R2
1.20 102 V
107
X C2
1.12 A
F
3.76 10
3.76 10
60.0 Hz 30.0 10
1.33 108
6
2
1.33 108
5
5
1.33 108
A
A 50.0 103
1
0 XC
Vmax
Z
1.2 102 V
60.0 Hz 20.0 10
X C2
Z RC
1
2 fC
VL, max
1
Vsecondary
I rms
Therefore,
21.26
vL
F
60.0
1.88 V
88.4
2
88.4
2
107
326
CH APTER 21
(d ) The phase angle in this RC circuit is
tan
Since
XL
1
XC
tan
R
1
0 88.4
60.0
55.8
0 , the voltage lags behind the current by 55.8° . Ad d ing an ind uctor w ill
change the im ped ance and hence the current in the circuit.
21.27
(a)
XL
2 f L 2 50.0 Hz 0.250 H
(b)
XC
1
2 fC
(c)
Z
R2
21.28
1
1
2
XL
Vmax
Z
(d ) I max
78.5
50.0 Hz 2.00 10
XC
2
150
2.10 102 V
1.52 103
XL
XC
1
F
0.138 A
78.5
1.59 103
78.5
1590
1590
tan
2
1.52 103
1.52 k
(f)
VR,max
I max R
VL,max
I max X L
0.138 A 78.5
VC ,max
I max X C
0.138 A 1.59 103
84.3
150
0.138 A 150
1.59 k
138 mA
(e)
R
tan
2
6
20.7 V
10.8 V
219 V
The voltage across the resistor is a m axim um w hen the current is a m axim um , and the
m axim um value of the current occurs w hen the argum ent of the sine function is
2.
The voltage across the capacitor lags the current by 90° or
2 rad ians, w hich
correspond s to an argum ent of 0 in the sine function and a voltage of vC
the voltage across the ind uctor lead s the current by 90° or
to an argum ent in the sine function of
0 . Sim ilarly,
2 rad ians, correspond ing
rad ians or 180°, giving a voltage of vL
0 .
A lternating Current Circuits and Electromagnetic W aves
21.29
and
(a)
21.30
XL
2 f L 2 50.0 Hz 0.185 H
XC
1
2 fC
Z ad
R2
I rms
Vrms
Z ad
Zab
1
2
XC
Vmax
2
2
41.0
Vrms
XL
58.1
, and
Vrms
(c)
Zcd
XC
49.0
, and
Vrms
(d ) Zbd
XL
XC
Z
R2
(a)
I max
(b)
, so
9.10
Vrms
XL
60 Hz 2.5 10
XC
Vmax
Z
2
1.2 103
170 V
1.6 103
VR, max
I max R
VC , max
I max X C
6
bd
2
41.0
2.59 A
104 V
2.59 A 58.1
I rms Zcd
cd
49.0
2.59 A 40.0
I rms Zbc
bc
1
2
2
I rms Zab
ab
49.0
58.1
150 V
Z ad
, so
F
2
40.0
(b) Zbc
1
2 fC
6
50.0 Hz 65.0 10
XL
R 40.0
XC
58.1
2.59 A 49.0
I rms Zbd
2.59 A 9.10
1.1 103
F
2
0 1.1 103
2
0.11 A
0.11 A 1.2 103
0.11 A 1.1 103
1.3 102 V
1.2 102 V
1.6 103
150 V
127 V
23.6 V
327
328
CH APTER 21
(c) When the instantaneous current i is zero, the instantaneous voltage across the
resistor is vR iR 0 . The instantaneous voltage across a capacitor is alw ays 90°
or a quarter cycle out of phase w ith the instantaneous current. Thus, w hen i 0 ,
vC
and
qC
C
vC
1.2 102 V
VC , max
2.5 10
6
F 1.2 102 V
3.0 10-4 C
300 C
Kirchhoff’ s loop rule alw ays applies to the instantaneous voltages around a closed
path. Thus, for this series circuit, vsource
vR
vC 0 , and at this instant w hen
i 0 , w e have
vsource
0
vC
1.2 102 V
VC , max
(d ) When the instantaneous current is a m axim um (i
across the resistor is
vR
iR
I max R
VR, max
I max ) , the instantaneous voltage
1.3 102 V . Again, the
instantaneous voltage across a capacitor is a quarter cycle out of phase w ith the
current. Thus, w hen i Imax , w e m ust have vC
0 and qC C vC
0 . Then,
applying Kirchhoff’ s loop rule to the instantaneous voltages around the series
circuit at the instant w hen i Imax and vC 0 gives
vsource
21.31
Vrms
I rms
vC
0
104 V
0.500 A
(a)
Z
(b)
2
Pav Irms
R gives R
(c)
Z
and
21.32
vR
Given v
R2
X L2 , so X L
L
Vmax sin
XL
2 f
t
vsource
vR
VR , max
1.3 102 V
208
Pav
10.0 W
2
rms
0.500 A
I
Z2
204
2 60.0 Hz
R2
40.0
2
208
2
40.0
2
204
0.541 H
90.0 V sin 350t , observe that
350 rad s . Also, the net reactance is X L
XC
Vmax
90.0 V and
2 fL 1 2 fC
L 1 C.
A lternating Current Circuits and Electromagnetic W aves
(a)
XL
XC
1
C
L
so the im ped ance is
(b) I rms
Vrms
Z
1
350 rad s 0.200 H
Vmax
Z
R2 X L
Z
2
350 rad s 25.0 10
XC
90.0 V
2
2
50.0
6
329
44.3
F
2
44.3
66.8
0.953 A
2 66.8
(c) The phase d ifference betw een the applied voltage and the current is
tan
1
XL
XC
44.3
50.0
1
tan
R
41.5
so the average pow er d elivered to the circuit is
Vmax
P av I rms Vrms cos I rms
cos
0.953 A
2
21.33
2
cos
41.5
45.4 W
Please see the textbook for the statem ent of Problem 21.21 and the answ ers for that
problem . There, you should find that Vmax 140 V, Imax 0.097 2 A, and
51.2 . The
average pow er d elivered to the circuit is then
Pav I rms Vrms cos
21.34
90.0 V
I max
Vmax
2
cos
2
0.097 2 A 140 V
2
cos 51.2
4.26 W
The rm s current in the circuit is
I rms
Vrms
Z
160 V
80.0
2.00 A
and the average pow er d elivered to the circuit is
Pav I rms
21.35
(a)
Vrms cos
I rms VR,rms
2
Pav I rms
R I rms I rms R
Thus,
R
VR , rms
I rms
I rms I rms R
I rms
50 V
0.28 A
2
I rms
R
VR , rms , so I rms
1.8 102
2.00 A
Pav
VR , rms
2
22.0
14 W
50 V
88.0 W
0.28 A
330
CH APTER 21
R2
(b) Z
X L2 , w hich yield s
XL
R2
L
XL
2 f
and
21.36
Vrms
I rms
Z2
2
2
90 V
0.28 A
R2
2.7 102
2 60 Hz
1.8 102
600 Hz 6.0 10 3 H
2 fL 2
XC
1
2 fC
23
Z
R2
(a)
VR , rms
I rms R
VL , rms
I rms X L
Vrms
XL
Z
10 V
28
23
8.2 V
VC , rms
I rms X C
Vrms
XC
Z
10 V
28
11
3.9 V
No ,
VR, rms
1
XL
11
600 Hz 25 10-6 F
2
XC
25
2
Vrms
R
Z
VL, rms
2.7 102
0.72 H
XL
2
2
23
11
10 V
28
VC , rms
25
2
28
8.9 V
9.0 V 8.2 V 3.8 V 21 V 10 V
(b) The power delivered to the resistor is the greatest. N o pow er losses occur in an id eal
capacitor or ind uctor.
(c) Pav
21.37
Vrms
Z
2
I rms
R
2
R
10 V
28
2
25
3.2 W
(a) The frequency of the station should m atch the resonance frequency of the tuning
circuit. At resonance, X L X C or 2 f0 L 1 2 f0C , w hich gives f02 1 4 2 LC or
f0
1
2
1
LC
2
3.00 10
6
H 2.50 10
5.81 107 Hz
12
F
58.1 MHz
A lternating Current Circuits and Electromagnetic W aves
(b)
331
Yes, the resistance is not needed. The resonance frequency is found by sim ply
equating the ind uctive reactance to the capacitive reactance, w hich lead s to
Equation 21.19 as show n above.
21.38
(a) The resonance frequency of a RLC circuit is f0 1 2
1
4 f02C
L
(b)
XL
For f0
1
2
f0
LC
f0
max
2.00 10
9.00 109 Hz 1.56 10
10
H
12
10
H
156 pH
F
8.82
1
4 f 02 L
, so C
2
2
5
5.00 10 Hz
2
6
1.60 10 Hz
(a) At resonance, X L
R2
(b) When X L
2.0 10
6
5.1 10
8
F
51 nF
4.9 10
9
F
4.9 nF
H
XL
XC
1
2
2.0 10
2
R2
0
R
2
H
15
1
w hich yield s
2 fC
1
LC
6
X C so the im ped ance w ill be
X C , w e have 2 fL
2
2
1
4
f
9.00 10 Hz
1.56 10
2
1600 kHz 1.60 106 Hz
C Cmin
Z
9
1
4
For f0
2
500 kHz 5.00 105 Hz
min
C Cmax
21.40
4
2 f0 L 2
f0
21.39
1
2
LC . Thus, the ind uctance is
0.20 H 75 10
41 Hz
6
F
(c) The current is a m axim um at resonance w here the im ped ance has its m inim um
value of Z
R.
332
CH APTER 21
(d ) At f
60 Hz , X L
and Z
0
Thus,
2
15
0
2
35
Vmax
75
2
150 V
Z
1
LC
3
10.0 10
60 Hz 75 10
1000 rad s
H 100 10
6
F
2 000 rad s
L
2 000 rad s 10.0 10 3 H
XC
1
C
1
I rms
2
2.5 A
2 43
XL
Z
1
, XC
43
1
2 f0
2
60 Hz 0.20 H
75
Vrms
Z
Thus, I rms
21.41
2
R2
Vrms
Z
2 000 rad s 100 10
XL
2
XC
50.0 V
18.0
The average pow er is Pav
10.0
6
20.0
5.00
F
2
20.0
5.00
2
2.78 A
2
I rms
R
2
2.78 A
10.0
77.3 W
and the energy converted in one period is
E Pav T
21.42
Pav
2
77.3
The resonance frequency is
Also, X L
L and X C
1
C
0
J
s
2 f0
2
2 000 rad s
1
LC
0.243 J
18.0
6
F
35
,
A lternating Current Circuits and Electromagnetic W aves
(a) At resonance, X C
R2 02
Thus, Z
1
2
XL
2
I rms
R
1
4
0
Z 3 750
so Pav
(d ) At
2
0
Z 1500
so Pav
2
I rms
R
XC
2
2
L
C
L
30.0
3.00 H
3.00 10-6 F
120 V
30.0
30.0
500
2
1
XL
4
2 XL
30.0
250
0
120 V
3 750
2
30.0
2 000
0
, and I rms
0.080 0 A
120 V
1 500
2
30.0
1 000
4.00 A
480 W
, XC
2
2 XC
500
2 000
0
2 000
2
1500
0.080 0 A
0.032 0 A
; XL
Vrms
Z
0
, and I rms
2
I rms
R
LC
4.00 A
0.080 0 A
; XL
1
L
R , I rms
1
XL
2
; XL
120 V
1 500
and I rms
(c) At
0
R2
Z
so Pav
0
2
Pav I rms
R
and
(b) At
XL
0.192 W
, XC
4 XC
4 000
0
0.032 0 A
3.07 10
, XC
2
W
1
XC
2
0.080 0 A
0.192 W
30.7 mW
500
0
333
334
CH APTER 21
(e) At
4
0
; XL
2
I rms
R
, XC
4 000
0
120 V
3 750
, and I rms
Z 3 750
so Pav
4 XL
2
0.032 0 A
1
XC
4
250
0
0.032 0 A
30.0
3.07 10
2
W
30.7 mW
The pow er d elivered to the circuit is a m axim um w hen the rm s current is a
m axim um . This occurs w hen the frequency of the source is equal to the resonance
frequency of the circuit.
21.43
The m axim um output voltage
Vmax
2
is related to the m axim um input voltage
N2
Vmax 1 , w here N1 and N 2 are the num ber of
N1
turns on the prim ary coil and the second ary coil respectively. Thus, for the given
transform er,
Vmax
1
by the expression
Vmax
2
1500
170 V
250
Vmax
2
1.02 103 V
Vrms
and the rm s voltage across the second ary is
21.44
Vmax
2
2
2
1.02 103 V
2
721 V .
(a) The output voltage of the transform er is
V2,rms
N2
N1
V1,rms
1
120 V
13
(b) Assum ing an id eal transform er, Poutput
9.23 V
Pinput , and the pow er d elivered to the CD
player is
Pav
2
Pav
1
I1,rms
V1,rms
0.250 A 120 V
30.0 W
A lternating Current Circuits and Electromagnetic W aves
21.45
335
The pow er input to the transform er is
Pav
V1, rms I1, rms
input
For an id eal transform er, Pav
3 600 V 50 A
1.8 105 W
Pav
V2, rms I 2, rms
ouput
input
so the current in the long-
d istance pow er line is
Pav
I 2, rms
input
V2, rms
1.8 105 W
1.8 A
100 000 V
The pow er d issipated as heat in the line is then
Plost I 2,2 rms Rline
1.8 A
2
3.2 102 W
100
The percentage of the pow er d elivered by the generator that is lost in the line is
% Lost
21.46
Plost
100%
Pinput
3.2 102 W
1.8 105 W
100%
0.18%
(a) Since the transform er is to step the voltage down from 120 volts to 6.0 volts, the
second ary m ust have fewer turns than the prim ary.
(b) For an id eal transform er, Pav
Pav
input
ouput
or
V1, rms I1, rms
V2, rms I 2, rms so the
current in the prim ary w ill be
I1, rms
V2, rms I 2, rms
6.0 V 500 mA
V1, rms
120 V
25 mA
(c) The ratio of the second ary to prim ary voltages is the sam e as the ratio of the num ber
of turns on the second ary and prim ary coils, V2 V1 N2 N1 . Thus, the num ber of
turns need ed on the second ary coil of this step d ow n transform er is
N2
N1
V2
V1
400
6.0 V
120 V
20 turns
336
21.47
CH APTER 21
(a) At 90% efficiency, Pav
Thus, if Pav
0.90 Pav
output
1000 kW
output
the input pow er to the prim ary is Pav
(b) I1, rms
(c)
21.48
Rline
I 2, rms
Pav
1.1 103 kW
V1, rms
input
V1, rms
Pav
1000 kW
V1, rms
output
V2, rms
4.50 10
4
transmitted
Vrms
Thus, Pav
loss
output
0.90
1 000 kW
0.90
1.1 103 kW
3.1 102 A
8.3 103 A
290
transmitted
5.00 106 W
500 103 V
2
I rms
Rline
input
1.0 106 W
120 V
m 6.44 105 m
Pav
Pav
1.1 106 W
3 600 V
(a) The pow er transm itted is Pav
so I rms
input
10.0 A
2
Vrms I rms
10.0 A
290
2.90 104 W
29.0 kW
(b) The pow er input to the line is
Pav
input
Pav
transmitted
Pav
loss
5.00 106 W 2.90 104 W 5.03 106 W
and the fraction of input pow er lost d uring transm ission is
fraction
Pav
Pav
loss
input
2.90 104 W
5.03 106 W
0.005 77 or 0.577%
A lternating Current Circuits and Electromagnetic W aves
337
(c) It is im possible to d eliver the need ed pow er w ith an input voltage of 4.50 kV. The
m axim um line current w ith an input voltage of 4.50 kV occurs w hen the line is
shorted out at the custom er’ s end , and th is current is
I rms
Vrms
Rline
max
4 500 V
15.5 A
290
The m axim um input pow er is then
Pinput
max
Vrms
I rms
max
4.50 103 V 15.5 A
6.98 104 W
6.98 kW
This is far short of m eeting the custom er’ s request, and all of this pow er is lost in
the transm ission line.
21.49
From v
f , the w avelength is
3.00 108 m s
75 Hz
v
f
4.00 106 m
4 000 km
The required length of the antenna is then,
L
21.50
(a) t
d
c
4
1000 km , or about 621 m iles. N ot very practical at all.
6.44 1018 m
1y
8
3.00 10 m s 3.156 107 s
6.80 102 y
(b) From Table C.4 (in Append ix C of the textbook), the average Earth-Sun d istance is
d 1.496 1011 m , giving the transit tim e as
t
d
c
1.496 1011 m 1 min
3.00 108 m s 60 s
8.31 min
(c) Also from Table C.4, the average Earth -Moon d istance is d
tim e for the round trip as
t
2d
c
2 3.84 108 m
3.00 108 m s
2.56 s
3.84 108 m , giving the
338
21.51
CH APTER 21
The am plitud es of the electric and m agnetic com ponents of an electrom agnetic w ave are
related by the expression Emax Bmax c , thus the am plitud e of the m agnetic field is
Bmax
21.52
c
330 V m
3.00 108 m s
1
or
1.10 10
6
T
1
0
21.53
Emax
c
0
c
4
10
7
N s2 C2 8.854 10
12
C2 N m 2
2.998 108 m s
(a) The frequency of an electrom agnetic w ave is f c , w here c is the speed of light,
and
is the w avelength of the w ave. The frequencies of the tw o light sources are
then
c
3.00 108 m s
f red
4.55 1014 Hz
Red :
-9
660 10 m
red
and
c
3.00 108 m s
3.19 1014 Hz
Infrared : f IR
-9
940
10
m
IR
(b) The intensity of an electrom agnetic w ave is proportional to the square of its
am plitud e. If 67% of the incid ent intensity of the red light is absorbed , then the
intensity of the em erging w ave is 100% 67% 33% of the incid ent intensity, or
If
0.33Ii . H ence, w e m ust have
Emax, f
If
Emax, i
Ii
0.33
0.57
339
A lternating Current Circuits and Electromagnetic W aves
21.54
If I0 is the incid ent intensity of a light beam , and I is the intensity of the beam after
passing through length L of a fluid having concentration C of absorbing m olecules, the
Beer-Lam bert law states that log10 I I0
is a constant.
CL w here
For 660-nm light, the absorbing m olecules are oxygenated hem oglobin. Thus, if 33% of
this w avelength light is transm itted through blood , the concentration of oxygenated
hem oglobin in the blood is
log10 0.33
CHBO2
[1]
L
The absorbing m olecules for 940-nm light are d eoxygenated hem oglobin, so if 76% of
this light is transm itted through the blood , the concentration of these m olecules in the
blood is
log10 0.76
CHB
[2]
L
Divid ing equation [2] by equation [1] gives the ratio of d eoxygenated hem o globin to
oxygenated hem oglobin in the blood as
log10 0.76
CHB
CHBO2
log10 0.33
0.25
or
CHB
0.25CHBO2
Since all the hem oglobin in the blood is either oxygenated or d eoxygenated , it is
necessary that CHB CHBO2 1.00 , and w e now have 0.25CHBO2 CHBO2 1.0 . The fraction
of hem oglobin that is oxygenated in this blood is then
1.0
1.0 0.25
CHBO2
0.80
or
80%
Som eone w ith only 80% oxygenated hem oglobin in the blood is probably in serious
trouble need ing supplem ental oxygen im m ed iately.
21.55
From Intensity
Emax Bmax
E
and max
2 0
Bmax
c , w e find Intensity
2
c Bmax
2 0
Thus,
Bmax
and
Emax
2
0
c
Bmax c
2 4
Intensity
3.35 10
10
7
T mA
8
3.00 10 m s
6
T 3.00 108 m s
1340 W m2
1.01 103 V m
3.35 10
6
T
340
21.56
CH APTER 21
(a) To exert an upw ard force on the d isk, the laser beam should be aim ed vertically
upw ard , striking the low er surface of the d isk. To just levitate the d isk, the upw ard
force exerted on the d isk by the beam should equal the w eight of the d isk.
The m om entum that electrom agnetic rad iation of intensity I, incid ent norm ally on
a perfectly reflecting surface of area A , d elivers to that surface in tim e t is given by
Equation 21.30 as p 2U c 2 IA t c . Thus, from the im pulse-m om entum
theorem , the average force exerted on the reflecting surface is F
p t 2IA c .
Then, to just levitate the surface, F 2IA c mg and the required intensity of the
incid ent rad iation is I
(b) I
mgc
2A
mgc
2 r2
mgc 2 A .
5.00 10
3
kg 9.80 m s 2 3.00 108 m s
2
-2
4.00 10 m
2
1.46 109 W m2
(c) Propulsion by light pressure in a significant gravity field is im practica l because of
the enorm ous pow er requirem ents. In ad d ition, no m aterial is perfectly reflecting,
so the absorbed energy w ould m elt the reflecting surface.
21.57
The d istance betw een ad jacent antinod es in a stand ing w ave is
Thus,
2 6.00 cm
c
21.58
f
2
12.0 cm 0.120 m , and
0.120 m 2.45 109 Hz
2.94 108 m s
At Earth’ s location, the w ave fronts of the solar rad iation are spheres w hose rad ius is the
Sun-Earth d istance. Thus, from Intensity
Pav
21.59
From
21.60
c
f
f
Intensity 4 r 2
c , w e find f
3.00 108 m s
27.33 106 Hz
c
1 340
W
m2
Pav
Pav
A
4 r2
4
3.00 108 m s
5.50 10 7 m
11.0 m
, the total pow er is
1.49 1011 m
5.45 1014 Hz
2
3.74 1026 W
A lternating Current Circuits and Electromagnetic W aves
21.61
(a) For the AM band ,
c
min
f max
c
max
f min
3.00 108 m s
1 600 103 Hz
188 m
3.00 108 m s
540 103 Hz
556 m
(b) For the FM band ,
c
min
f max
c
max
21.62
f min
3.00 108 m s
108 106 Hz
2.78 m
3.00 108 m s
88 106 Hz
3.4 m
The transit tim e for the rad io w ave is
tR
dR
c
100 103 m
3.00 108 m s
3.33 10
4
s 0.333 ms
and that for the sound w ave is
ts
ds
vsound
3.0 m
343 m s
8.7 10
3
s 8.7 ms
Thus, the radio listeners hear the news 8.4 ms before the studio audience because rad io
w aves travel so m uch faster than sound w aves.
341
342
21.63
CH APTER 21
If an object of m ass m is attached to a spring of spring constant k, the natural frequency
of vibration of that system is f
k m 2 . Thus, the resonance frequency of the C=O
d ouble bond w ill be
1
2
f
k
1
2
moxygen
2 800 N m
2.66 10 26 kg
5.2 1013 Hz
atom
and the light w ith this frequency has w avelength
3.00 108 m s
5.2 1013 Hz
c
f
5.8 10
6
m
5.8 m
The infrared region of the electrom agnetic spectrum ranges from
1 mm d ow n to
700 nm 0.7 m . Thus, the required wavelength falls within the infrared region .
min
21.64
max
Since the space station and the ship are m oving tow ard one another, the frequency after
being Doppler shifted is fO f S 1 u c , so the change in frequency is
f
fO
fS
fS
u
c
6.0 1014 Hz
1.8 105 m s
3.0 108 m s
3.6 1011 Hz
and the frequency observed on the spaceship is
fO
21.65
fS
6.0 1014 Hz 3.6 1011 Hz
f
6.003 6 1014 Hz
Since you and the car ahead of you are m oving aw ay from each other (getting farther
apart) at a rate of u 120 km h 80 km h 40 km h , the Doppler shifted frequency you
w ill d etect is fO
f
fO
fS
f S 1 u c , and the change in frequency is
fS
u
c
4.3 1014 Hz
40 km h
3.0 108 m s
0.278 m s
1 km h
The frequency you w ill d etect w ill be
fO
fS
f
4.3 1014 Hz 1.6 107 Hz
4.299 999 84 1014 Hz
1.6 107 Hz
A lternating Current Circuits and Electromagnetic W aves
21.66
The d river w as d riving tow ard the w arning lights, so the correct form of the Doppler
shift equation is fO f S 1 u c . The frequency em itted by the yellow w arning light is
c
fS
S
3.00 108 m s
580 10 9 m
5.17 1014 Hz
and the frequency the d river claim s that she observed is
c
fO
O
3.00 108 m s
560 10 9 m
5.36 1014 Hz
The speed w ith w hich she w ould have to approach the light for the Doppler effect to
yield this claim ed shift is
u
21.67
(a) At f
XC
c
fO
fS
3.00 108 m s
1
5.36 1014 Hz
1
5.17 1014 Hz
60.0 Hz , the reactance of a 15.0- F capacitor is
1
2 fC
1
2
60.0 Hz 15.0 10
6
F
177
and the im ped ance of this RC circuit is
Z
(b) I rms
R2
Vrms
Z
X C2
50.0
1.20 102 V
184
2
177
0.652 A
2
184
652 mA
1.1 107 m s
343
344
CH APTER 21
(c) After ad d ition of an ind uctor in series w ith the resistor and capacitor, the
R2
im ped ance of the circuit is Z
XL
2
XC
2
50.0
2 fL 177
2
. To
red uce the current to one-half the value found above, the im ped ance of the circuit
m ust be d oubled to a value of Z 2 184
368 . Thus,
2
50.0
or
2
2 fL 177
2 fL 177
368
2
2
368
2
50.0
177
365
Since the ind uctance cannot be negative, the potential solution associated w ith the
low er sign m ust be d iscard ed , leaving
L
21.68
177
365
2 60.0 Hz
1.44 H
Suppose you cover a 1.7 m -by-0.3 m section of beach blanket. Suppose the elevation
angle of the Sun is 60°. Then the target area you fill in the Sun’ s field of view is
1.7 m 0.3 m cos30 0.4 m2 .
The intensity the rad iation at Earth’ s surface is Isurface
absorbed . Since I
E
Pav
E
A
A
0.5 I surface A
t
t
, the absorbed energy is
0.5 0.6 I incoming
A
t
0.5 0.6 1 340 W m 2 0.4 m 2 3 600 s
21.69
Z
R2
200
Thus, Pav
and
cost
I
2
rms
2
XC
2
2
Vrms
Z
R
E rate
8.9 10
R2
3
2 fC
Pav
R
6 105 J
or
~106 J
2
60 Hz 5.0 10
2
0.6 Iincoming and only 50% of this is
120 V
5.7 102
6
2
F
5.7 102
2
200
8.9 W 8.9 10
t rate
kW 24 h 8.0 cents kWh
1.7 cents
3
kW
A lternating Current Circuits and Electromagnetic W aves
21.70
L , so
XL
1
C
Then, X C
L
From
0
XL L
1
w hich gives
C XL L
X L XC C
1
2 f0
LC
12
C or L
8.0
2
96
[1]
C
1
, w e obtain LC
2 f0
2
2
Substituting from Equation [1], this becom es 96
1
C2
2
2 f0
or
1
C
2 f0
1
2
96
2.6 10
2
2
2 000
Hz
96
F
2.5 10
3
2.5 mH
5
F
26 F
Then, from Equation [1],
L
21.71
V
R
Thus, L
2.6 10
5
H
12.0 V
19.0
0.630 A
DC
I DC
Z
21.72
2
96
R2
2 fL
2
Z 2 R2
2 f
Vrms
I rms
24.0 V
0.570 A
2
42.1
2
19.0
42.1
2
9.97 10
60.0 Hz
2
H
99.7 mH
3.00 108 m s
1.0 1010 Hz . Therefore, the
2
3.00 10 m
1
1.0 1010 Hz , giving
resonance frequency of the circuit is f 0
2 LC
(a) The required frequency is f
C
c
1
2 f0
1
2
L
2
10
10 Hz
2
6.3 10
400 10
12
H
13
F
0.63 pF
345
346
CH APTER 21
2
A
0
(b) C
0
d
, so
d
(c)
XC
(a)
Emax
Bmax
c , so
Bmax
Emax
c
XL
(c) Pav
Intensity
(a)
(b)
Vrms
I rms
Z
R
VDC
I DC
From Z
L
A
17
12 V
2.0 A
Z 2 R2
2 f
6
10
H
8.4 mm
25
T
V m 6.7 10
7
m
16
T
T mA
5.3 10
d2
4
Intensity
20.0 m
2
16
12
3
2
1.7 10
4
14
W
2
H
6.0
12 V
3.0 A
R2
6.7 10
2 4
W m
8.4 10
C N m
0.20 10
Emax Bmax
2 0
m
2
0.20 10 6 V m
3.00 108 m s
5.3 10
21.74
12
1.0 1010 Hz 400 10
2 f0 L 2
(b) Intensity
3
F 1.0 10
8.85 10
0
21.73
13
6.3 10
C d
4.0
X L2
R2
6.0
2
2 fL
2
4.0
60 Hz
2
, w e find
2
1.2 10
12 mH
17
W m2
A lternating Current Circuits and Electromagnetic W aves
21.75
347
(a) From Equation 21.30, the m om entum im parted in tim e t to a perfectly reflecting
sail of area A by norm ally incid ent rad iation of intensity I is p 2U c 2 IA t c .
From the im pulse-m om entum theorem , the average force exerted on the sail is then
Fav
Fav
m
(b) aav
(c) From
t
21.76
2 IA t c
p
t
t
0.536 N
6 000 kg
2
8.93 10
1 2
at , w ith v0
2
x v0 t
2 1340 W m2 6.00 104 m2
2IA
c
5
m s2
0 , the tim e is
2 3.84 108 m
x
aav
8.93 10
5
0.536 N
3.00 108 m s
ms
2
2.93 106 s
1d
8.64 104 s
33.9 d
(a) The intensity of rad iation at d istance r from a point source, w hich rad iates total
pow er P , is I P A P 4 r 2 . Thus, at d istance r 2.0 in from a cell phone
rad iating a total pow er of
P 2.0 W 2.0 103 mW , the intensity is
2.0 103 mW
I
4
2.0 in 2.54 cm 1 in
2
6.2 mW cm2
This intensity is 24% higher than the maximum allowed leakage from a microwave at this
d istance of 2.0 inches.
(b) If w hen using a Blue tooth head set (em itting 2.5 m W of pow er) in the ear at d istance
rh 2.0 in 5.1 cm from center of the brain, the cell phone (em itting 2.0 W of pow er)
is located in the pocket at d istance rp 1.0 m 1.0 102 cm from the brain, the total
rad iation intensity at the brain is
I total
I phone
2.0 103 mW
I headset
4
or
I total
1.6 10
2
1.0 10 cm
2
mW
cm 2
2.5 mW
2
7.6 10
4
3
5.1 cm
mW
cm 2
2
1.6 10
2.4 10
2
2
mW
cm 2
mW
7.6 10
cm2
3
mW
cm2
0.024 mW cm 2