Chapter 21
Alternating Current Circuits and
Electromagnetic Waves
Quick Quizzes
1.
(a), (c). The average power is proportional to the rms current which is non-zero even
though the average current is zero. (a) is only valid for an open circuit, for which R → ∞ .
(b) and (d) can never be true because iav = 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 ≠ ∆VR + ∆VL + ∆VC even though
∆v = ∆vR + ∆vL + ∆vC ∆.)
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.
5.
(b), (d). The frequency and wavelength of light waves are related by the equation λ f = c
or f = c λ , where c is the speed of light is a constant within a given medium. Thus, the
frequency and wavelength are inversely proportional to each other, when one increases
the other must decrease.
201
202
CHAPTER 21
Answers to Even Numbered Conceptual Questions
2.
At resonance, X L = XC . This means that the impedance Z = R 2 + ( X L − XC ) reduces to
2
Z = R.
4.
The purpose of the iron core is to increase the flux and to provide a pathway in which
nearly all the flux through one coil is led through the other.
6.
The fundamental source of an electromagnetic wave is a moving charge. For example, in a
transmitting antenna of a radio station, charges are caused to move up and down at the
frequency of the radio station. These moving charges set up electric and magnetic fields,
the electromagnetic wave, in the space around the antenna.
8.
Energy moves. No matter moves. You could say that electric and magnetic fields move,
but it is nicer to say that the fields stay at that point and oscillate. The fields vary in time,
like sports fans in the grandstand when the crowd does the wave. The fields constitute the
medium for the wave, and energy moves.
10.
The average value of an alternating current is zero because its direction is positive as often
as it is negative, and its time average is zero. The average value of the square of the
current is not zero, however, since the square of positive and negative values are always
positive and cannot cancel.
12.
The brightest portion of your face shows where you radiate the most. Your nostrils and
the openings of your ear canals are particularly bright. Brighter still are the pupils of your
eyes.
14.
No, the only element that dissipates energy in an AC circuit is a resistor. Inductors and
capacitors store energy during one half of a cycle and release that energy during the other
half of the cycle, so they dissipate no net energy.
16.
The changing magnetic field of the solenoid induces eddy currents in the conducting core.
This is accompanied by I 2 R conversion of electrically-transmitted energy into internal
energy in the conductor.
18.
The voltages are not added in a scalar form, but in a vector form, as shown in the phasor
diagrams throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the
voltages across different circuit elements are not simultaneously at their maximum values.
Do not forget that an inductor can induce an emf in itself and that the voltage across it is
90° ahead of the current in the circuit in phase.
20.
Insulation and safety limit the voltage of a transmission line. For an underground cable,
the thickness and dielectric strength of the insulation between the conductors determines
the maximum voltage that can be applied, just as with a capacitor. For an overhead line on
towers, the designer must consider electrical breakdown of the surrounding air, possible
accidents, sparking across the insulating supports, ozone production, and inducing
voltages in cars, fences, and the roof gutters of nearby houses. Nuisance effects include
noise, electrical noise, and a prankster lighting a hand-held fluorescent tube under the
line.
Alternating Current Circuits and Electromagnetic Waves
Answers to Even Numbered Problems
(b)
145 Ω
(a)
4.
I1, rms = I 2, rms = 1.25 A , R1 = R2 = 96.0 Ω , I 3, rms = 0.833 A, R3 = 144 Ω
6.
(a)
106 V
(b)
60.0 Hz
8.
(a)
141 mA
(b)
235 mA
10.
100 mA
12.
224 mA
14.
2.63 A
16.
L > 7.03 H
18.
(a)
194 V
(b)
current leads by 49.9°
20.
(a) 138 V
(c) 729 V
(b)
(d)
104 V
641 V
22.
(a) 0.11 A
(b) ∆VR , max = 130 V, ∆VC , max = 110 V
(c) ∆vR = 0, ∆vC = ∆vsource = 110 V, qC = 280 µ C
(d) ∆vR = ∆vsource = 130 V, ∆vC = 0, qC = 0
24.
(a) 0.11 A
(b) ∆VR , max = 130 V, ∆VL , max = 110 V
(c) ∆vR = ∆vsource = 130 V, ∆vL = 0
(d) ∆vR = 0, ∆vL = ∆vsource = 110 V
26.
(a)
123 nF or 124 nF
(b)
51.5 kV
28.
(a)
0.492, 48.5 W
(b)
0.404, 32.7 W
30.
(a)
100 W, 0.633
(b)
156 W, 0.790
32.
(a)
(c)
0
(e)
DVL = 104 V
(d)
|DVL – DVC| = 625 V
193 Ω
2.
3.00 A
DVR = 138 V
77.6°
DV = 641 V
DVC = 729 V
∆VR , rms + ∆VL , rms + ∆VC , rms = 21 V ≠ 10 V , but accounting for phases and adding the
voltages vectorially does yield 10 V. (b) The power loss delivered to the resistor. No
power losses occur in an ideal capacitor or inductor. (c) 3.3 W
34.
(a) Z = R = 15 Ω
(c) At resonance
36.
(a) 480 W
(b) 0.192 W
(c) 30.7 mW
(d) 0.192 W
Maximum power is delivered at resonance frequency.
(b)
(d)
41 Hz
2.5 A
(e)
30.7 mW
203
204
CHAPTER 21
38.
(a)
18 turns
(b)
3.6 W
40.
(a)
Fewer turns
(b)
25 mA
42.
(a) 29.0 kW
(b) 0.580%
(c) The maximum power that can be input to the line at 4.50 kV is far less than 5.00 MW,
and it is all lost in the transmission line.
44.
2.998 × 108 m s
46.
80%
48.
3.74 × 10 26 W
50.
11.0 m
52.
Radio listeners hear the news 8.4 ms before the studio audience because radio waves
travel much faster than sound waves.
54.
6.003 6 × 1014 Hz, the frequency increases by 3.6 × 1011 Hz
56.
1.1 × 107 m s
58.
~106 J
60.
2.5 mH, 26 µ F
62.
(a)
0.63 pF
(b)
8.5 mm
64.
(a)
6.0 Ω
(b)
12 mH
66.
32
68.
X c = 3R
(c)
(c)
20 turns
25 Ω
Alternating Current Circuits and Electromagnetic Waves
Problem Solutions
21.1
(a)
∆Vmax = 2 ( ∆Vrms ) = 2 ( 100 V ) = 141 V
(b) I rms =
(c)
I max =
∆Vrms 100 V
=
= 20.0 A
R
5.00 Ω
∆Vmax 141 V
=
= 28.3 A or I max = 2 I rms = 2 ( 20.0 A ) = 28.3 A
R
5.00 Ω
2
R = ( 20.0 A ) ( 5.00 Ω ) = 2.00 × 10 3 W = 2.00 kW
(d) Pav = I rms
2
( ∆Vmax )
( ∆Vmax )
1  ∆V 
I 
, so R =
R =  max  R =  max  R =
2 R 
2R
2 Pav
 2 
2
21.2
Pav = I
2
rms
2
2
2
( 170 V )
= 193 Ω
(a) If Pav = 75.0 W , then R =
2 ( 75.0 W )
2
( 170 V )
= 145 Ω
(b) If Pav = 100 W , then R =
2 ( 100 W )
2
21.3
The meters measure the rms values of potential difference and current. These are
∆Vrms =
21.4
∆Vrms 70.7 V
∆Vmax 100 V
=
= 70.7 V , and I rms =
=
= 2.95 A
24.0 Ω
R
2
2
All lamps are connected in parallel with the voltage source, so ∆Vrms = 120 V for each
lamp. Also, the current is I rms = √av ∆Vrms and the resistance is R = ∆Vrms I rms .
I1, rms = I 2, rms =
I 3, rms =
150 W
120 V
= 96.0 Ω
= 1.25 A and R1 = R2 =
120 V
1.25 A
100 W
120 V
= 144 Ω
= 0.833 A and R3 =
120 V
0.833 A
205
206
CHAPTER 21
21.5
The total resistance (series connection) is Req = R1 + R2 = 8.20 Ω + 10.4 Ω = 18.6 Ω , so the
current in the circuit is
I rms =
∆Vrms 15.0 V
=
= 0.806 A
Req
18.6 Ω
2
The power to the speaker is then Pav = I rm
s Rspeaker = ( 0.806 A ) ( 10.4 Ω ) = 6.76 W
2
21.6
(a)
∆Vmax = 150 V , so ∆Vrms =
(b)
f=
∆Vmax 150 V
=
= 106 V
2
2
ω 377 rad s
=
= 60.0 Hz
2π
2π
(c) At t = ( 1 120 ) s , v = ( 150 V ) sin ( 377 rad s )( 1 120 s )  = ( 150 V ) sin (π rad ) = 0
(d) I max =
21.7
∆Vmax 150 V
=
= 3.00 A
50.0 Ω
R
1
, so its units are
2π fC
XC =
1
1
Volt
Volt
=
=
=
= Ohm
(1 Sec ) Farad (1 Sec )( Coulomb Volt ) Coulomb Sec Amp
21.8
I max = 2 I rms =
(a)
2 ( ∆Vrms )
XC
= 2 ( ∆Vrms ) 2π f C
I max = 2 ( 120 V ) 2π ( 60.0 Hz ) ( 2.20 × 10 − 6 C/V ) =0.141 A = 141 mA
(b) I max = 2 ( 240 V ) 2π ( 50.0 Hz ) ( 2.20 × 10 − 6 C/V ) =0.235 A = 235 mA
21.9
I rms =
f=
∆Vrms
=2π f C ( ∆Vrms ) , so
XC
I rms
0.30 A
=
= 4.0 × 10 2 Hz
2π C ( ∆Vrms ) 2π ( 4.0 × 10 −6 F ) ( 30 V )
Alternating Current Circuits and Electromagnetic Waves
21.10
I max =
207
∆Vmax
=2π f C ( ∆Vmax )
XC
= 2π ( 90.0 Hz ) ( 3.70 × 10 − 6 C/V ) ( 48.0 V ) =0.100 A= 100 mA
21.11
I rms =
so
21.12
I rms =
or
21.13
∆Vrms
 ∆V
=2π f C  max
XC
2

C=

 = π f C ( ∆Vmax ) 2

I
0.75 A
=
= 1.7 × 10 −5 F = 17 µ F
π f ( ∆Vmax ) 2 π ( 60 Hz )( 170 V ) 2
∆Vrms  ∆Vmax
=
XC
2


ω C

 140 V 
−6
I rms = 
 ( 120 π rad s ) ( 6.00 × 10 F ) = 0.224 A = 224 mA
 2 
ε ( ∆t )
 ∆I 
. The units of self inductance are
= L   , we have L =
∆I
 ∆t 
[ε ][ ∆t ] = Volt ⋅ sec . The units of inductive reactance are given by
then [ L] =
Amp
[ ∆I ]
X L = 2π f L , and from
ε
1  Volt ⋅ sec  Volt
= Ohm
=

 sec  Amp  Amp
[ XL ] = [ f ][ L] = 
21.14
The maximum current in the purely inductive circuit is
so
I max =
∆Vmax ∆Vmax
140 V
=
=
= 3.71 A
XL
ωL
(120π rad s ) ( 0.100 H )
I rms =
I max 3.71 A
=
= 2.63 A
2
2
208
CHAPTER 21
21.15
The ratio of inductive reactance at f 2 = 50.0 Hz to that at f1 = 60.0 Hz is
( X L )2
( XL )1
=
2π f 2 L f 2
f
50.0 Hz
= , so ( X L )2 = 2 ( X L )1 =
( 54.0 Ω ) = 45.0 Ω
2π f1 L f1
f1
60.0 Hz
The maximum current at f 2 = 50.0 Hz is then
I max =
21.16
2 ( ∆Vrms )
2 ( 100 V )
∆Vmax
=
=
= 3.14 A
XL
XL
45.0 Ω
The maximum current in this inductive circuit will be
I max =
( 2 )I
rms
2 ( ∆Vrms )
=
XL
=
2 ( ∆Vrms )
2π fL
Thus, if I max < 80.0 mA , it is necessary that
L>
21.17
2 ( ∆Vrms )
2π f ( 80.0 mA )
2 ( 50.0 V )
2π ( 20.0 Hz ) ( 8.00 × 10 −2 A )
or
L > 7.03 H
N ΦB
, the total flux through the coil is Φ B , total = N Φ B = L ⋅ I where Φ B is the
I
flux through a single turn on the coil. Thus,
From L =
(Φ
B , total
)
max
 ∆V 
= L ⋅ I max = L ⋅  max 
 XL 
= L⋅
21.18
=
(a)
2 ( ∆Vrms )
2π f L
=
2 ( 120 V )
= 0.450 T ⋅ m 2
2π ( 60.0 Hz )
X L = 2π f L = 2π ( 50.0 Hz )( 400 × 10 −3 H ) = 126 Ω
1
1
XC =
=
= 719 Ω
2π f C 2π ( 50.0 Hz )( 4.43 × 10 −6 F )
Z = R2 + ( X L − XC ) =
2
2
2
( 500 Ω ) + ( 126 Ω − 719 Ω ) = 776 Ω
∆Vmax = I max Z = ( 0.250 A )( 776 Ω ) = 194 V
400 mH
50.0
Hz
4.43
mF
500 W
Alternating Current Circuits and Electromagnetic Waves
 X − XC
(b) φ = tan −1  L
R

209

−1  126 Ω − 719 Ω 
 = −49.9°
 = tan 
500 Ω



Thus, the current leads the voltage by 49.9°
21.19
XC =
1
1
=
= 66.3 Ω
2π f C 2π ( 60.0 Hz ) ( 40.0 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
2
I rms =
(a)
2
2
( 50.0 Ω ) + ( 0 − 66.3 Ω ) = 83.1 Ω
∆Vrms 30.0 V
=
= 0.361 A
83.1 Ω
Z
(b) ∆VR , rms = I rms R = ( 0.361 A )( 50.0 Ω ) = 18.1 V
∆VC , rms = I rms XC = ( 0.361 A )( 66.3 Ω ) = 23.9 V
(c)
 X − XC
(d) φ = tan −1  L
R


−1  0 − 66.3 Ω 
 = −53.0°
 = tan 
 50.0 Ω 

so, the voltage lags behind the current by 53.0°
X L = 2π f L = 2π ( 60.0 Hz )( 0.100 H ) = 37.7 Ω
XC =
1
1
=
= 265 Ω
2π f C 2π ( 60.0 Hz )( 10.0 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
2
(a)
2
2
( 50.0 Ω ) + ( 37.7 Ω − 265 Ω ) = 233 Ω
∆VR , rms = I rms R = ( 2.75 A )( 50.0 Ω ) = 138 V
(b) ∆VL , rms = I rms X L = ( 2.75 A )( 37.7 Ω ) = 104 V
(c)
∆VC , rms = I rms XC = ( 2.75 A )( 265 Ω ) = 729 V
(d) ∆Vrms = I rms Z = ( 2.75 A )( 233 Ω ) = 641 V
(e)
DVL = 104 V
|DVL – DVC| = 625 V
21.20
DVR = 138 V
77.6°
DV = 641 V
DVC = 729 V
210
CHAPTER 21
21.21
(a)
X L = 2π f L = 2π ( 240 Hz )( 2.5 H ) = 3.8 × 10 3 Ω
XC =
1
1
=
= 2.7 × 10 3 Ω
2π f C 2π ( 240 Hz )( 0.25 × 10 −6 F )
Z = R2 + ( X L − XC ) =
2
(b) I max =
(c)
2
( 900 Ω ) + ( 3.8 − 2.7 ) × 10 3 Ω  = 1.4 kΩ
2
∆Vmax
140 V
=
= 0.10 A
Z
1.4 × 10 3 Ω
 X L − XC
R

φ = tan −1 
 3.8 − 2.7 ) × 10 3 Ω 

−1 (
=
tan

 = 51°

900 Ω



(d) φ > 0 , so the voltage leads the current
21.22
XC =
1
1
=
= 1.1 × 10 3 Ω
−6
2π fC 2π ( 60 Hz ) ( 2.5 × 10 F )
Z = R2 + ( X L − XC ) =
2
(a)
I max =
(1.2 × 10
3
Ω ) + ( 0 − 1.1 × 10 3 Ω ) = 1.6 × 10 3 Ω
2
2
∆Vmax
170 V
=
= 0.11 A
Z
1.6 × 10 3 Ω
(b) ∆VR , max = I max R = ( 0.11 A ) ( 1.2 × 10 3 Ω ) = 1.3 × 10 2 V
∆VC , max = I max XC = ( 0.11 A ) ( 1.1 × 10 3 Ω ) = 1.1 × 10 2 V
(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 always 90°
or a quarter cycle out of phase with the instantaneous current. Thus, when i = 0 ,
∆vC = ∆VC , max = 1.1 × 10 2 V
and
qC = C ( ∆vC ) = ( 2.5 × 10 −6 F )( 1.1 × 10 2 V ) = 2.8 × 10 -4 C = 280 µ C
Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed
path. Thus, for this series circuit, ∆vsource = ∆vR + ∆vC and at this instant when i = 0 ,
we have ∆vsource = 0 + ∆VC , max = 110 V
Alternating Current Circuits and Electromagnetic Waves
211
(d) When the instantaneous current is a maximum ( i = I max ) , the instantaneous voltage
across the resistor is ∆vR = iR = I max R = ∆VR , max = 1.3 × 10 2 V . Again, the
instantaneous voltage across a capacitor is a quarter cycle out of phase with the
current. Thus, when i = I max , we must 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 when i = I max gives
∆vsource = ∆vR + ∆vC = ∆VR , max + 0 = 1.3 × 10 2 V
X L = 2π f L = 2π ( 60.0 Hz )( 0.400 H ) = 151 Ω
21.23
XC =
1
1
=
= 884 Ω
2π f C 2π ( 60.0 Hz )( 3.00 × 10 −6 F )
ZRLC = R2 + ( X L − XC ) =
2
and
(a)
I rms =
∆Vrms
ZRLC
ZLC = 0 + ( X L − XC ) = X L − XC = 733 Ω
2
 ∆V
∆VLC , rms = I rms ⋅ ZLC =  rms
 ZRLC
(b) ZRC = R 2 + ( 0 − XC ) =
2

 90.0 V 
 ZLC = 
 ( 733 Ω ) = 89.6 V
 736 Ω 

2
2
( 60.0 Ω ) + ( 884 Ω ) = 886 Ω
 ∆V
∆VRC , rms = I rms ⋅ ZRC =  rms
 ZRLC
21.24
2
2
( 60.0 Ω ) + ( 151 Ω − 884 Ω ) = 736 Ω

 90.0 V 
 ZRC = 
 ( 886 Ω ) = 108 V
 736 Ω 

X L = 2π fL = 2π ( 60 Hz )( 2.8 H ) = 1.1 × 10 3 Ω
Z = R 2 + ( X L − XC ) =
2
(a)
I max =
(1.2 × 10
3
Ω ) + ( 1.1 × 10 3 Ω − 0 ) = 1.6 × 10 3 Ω
2
∆Vmax
170 V
=
= 0.11 A
Z
1.6 × 10 3 Ω
2
212
CHAPTER 21
(b) ∆VR , max = I max R = ( 0.11 A ) ( 1.2 × 10 3 Ω ) = 1.3 × 10 2 V
∆VL , max = I max X L = ( 0.11 A ) ( 1.1 × 10 3 Ω ) = 1.1 × 10 2 V
(c) When the instantaneous current is a maximum ( i = I max ) , the instantaneous voltage
across the resistor is ∆vR = iR = I max R = ∆VR , max = 1.3 × 10 2 V . The instantaneous
voltage across an inductor is always 90° or a quarter cycle out of phase with the
instantaneous current. Thus, when i = I max ,
∆vL = 0 .
Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed
path. Thus, for this series circuit, ∆vsource = ∆vR + ∆vL and at this instant when i = I max
we have ∆vsource = I max R + 0 = 1.3 × 10 2 V
(d) When the instantaneous current i is zero, the instantaneous voltage across the
resistor is ∆vR = iR = 0 . Again, the instantaneous voltage across an inductor is a
quarter cycle out of phase with the current. Thus, when i = 0 , we must have
∆vL = ∆VL , max = 1.1 × 10 2 V . Then, applying Kirchhoff’s loop rule to the
instantaneous voltages around the series circuit at the instant when i = 0 gives
∆vsource = ∆vR + ∆vL = 0 + ∆VL , max = 1.1 × 10 2 V
XC =
21.25
1
1
=
= 1.33 × 108 Ω
2π f C 2π ( 60.0 Hz )( 20.0 × 10 −12 F )
ZRC = R2 + XC2 =
and
I rms =
Therefore,
( ∆V
sec ondary
ZRC
)
rms
( 50.0 × 10
=
3
Ω ) + ( 1.33 × 10 8 Ω ) = 1.33 × 10 8 Ω
2
2
5 000 V
= 3.76 × 10 −5 A
1.33 × 108 Ω
∆Vb , rms = I rms Rb = ( 3.76 × 10 −5 A )( 50.0 × 10 3 Ω ) = 1.88 V
213
Alternating Current Circuits and Electromagnetic Waves
21.26
(a)
X L = 2π f L = 2π ( 100 Hz )( 20.5 H ) = 1.29 × 10 4 Ω
Z=
∆Vrms 200 V
=
= 50.0 Ω
I rms
4.00 A
20.5 H
200 V
100 Hz
35.0 W
Thus,
2
2
( 50.0 Ω ) − ( 35.0 Ω ) = ± 35.7 Ω
X L − XC = ± Z 2 − R 2 = ±
and XC = X L ± 35.7 Ω or
1
= 1.29 × 10 4 Ω ± 35.7 Ω
2π f C
This yields
C=
(b)
1
= 123 nF or 124 nF
2π ( 100 Hz ) ( 1.29 × 10 4 Ω ± 35.7 Ω )
( ∆Vrms )coil = I rms Zcoil = I
R2 + X L2 = ( 4.00 A )
2
( 50.0 Ω ) + ( 1.29 × 10 4 Ω )
2
= 5.15 × 10 4 V = 51.5 kV
Notice that this is a very large voltage!
X L = 2π f L = 2π ( 50.0 Hz )( 0.185 H ) = 58.1 Ω
21.27
XC =
1
1
=
= 49.0 Ω
2π f C 2π ( 50.0 Hz )( 65.0 × 10 −6 F )
Zad = R 2 + ( X L − XC ) =
2
2
( 40.0 Ω ) + ( 58.1 Ω − 49.0 Ω ) = 41.0 Ω
2
and
(a)
I rms =
(
)
∆Vmax 2
∆Vrms
150 V
=
=
= 2.585 A
Zad
Zad
( 41.0 Ω ) 2
Zab = R = 40.0 Ω , so ( ∆Vrms ) ab = I rms Zab = ( 2.585 A )( 40.0 Ω ) = 103 V
(b) Zbc = X L = 58.1 Ω , and ( ∆Vrms )bc = I rms Zbc = ( 2.585 A )( 58.1 Ω ) = 150 V
(c)
Zcd = XC = 49.0 Ω , and ( ∆Vrms )cd = I rms Zcd = ( 2.585 A )( 49.0 Ω ) = 127 V
C
214
CHAPTER 21
(d) Zbd = X L − XC = 9.15 Ω , so ( ∆Vrms )bd = I rms Zbd = ( 2.585 A )( 9.15 Ω ) = 23.6 V
21.28
(a)
XC =
1
1
=
= 88.4 Ω
2π f C 2π ( 60.0 Hz )( 30.0 × 10 −6 F )
2
2
( 50.0 Ω ) + ( 88.4 Ω ) = 102 Ω
Z = R2 + XC2 =
I rms =
∆Vrms 100 V
=
= 0.984 A
Z
102 Ω
 0 − XC
 R
φ = tan −1 

−1  −88.4 Ω 
 = −60.5°
 = tan 
 50.0 Ω 

power factor = cos φ = cos ( −60.5° ) = 0.492
and
Pav = ( ∆Vrms ) I rms cos φ = ( 100 V )( 0.984 A )( 0.492 ) = 48.5 W
(b)
X L = 2π f L = 2π ( 60.0 Hz )( 0.300 H ) = 113 Ω
Z = R2 + X L2 =
I rms =
2
2
( 50.0 Ω ) + ( 113 Ω ) = 124 Ω
∆Vrms 100 V
=
= 0.809 A
Z
124 Ω
 XC − 0 
−1  113 Ω 
 = 66.1°
 = tan 
R
 50.0 Ω 


φ = tan −1 
and
power factor = cos φ = cos ( 66.1° ) = 0.404
Pav = ( ∆Vrms ) I rms cos φ = ( 100 V )( 0.809 A )( 0.404 ) = 32.7 W
21.29
∆Vrms
104 V
=
= 208 Ω
I rms
0.500 A
(a)
Z=
(b)
2
Pav = I rms
R gives R =
Pav
10.0 W
=
= 40.0 Ω
2
I rms ( 0.500 A )2
Alternating Current Circuits and Electromagnetic Waves
(c)
Z = R2 + X L2 , so X L = Z 2 − R 2 =
and
21.30
(a)
L=
2
2
( 208 Ω ) − ( 40.0 Ω ) = 204 Ω
XL
204 Ω
=
= 0.541 H
2π f 2π ( 60.0 Hz )
X L = 2π fL = 2π ( 60.0 Hz )( 0.100 H ) = 37.7 Ω
XC =
1
1
=
= 13.3 Ω
2π fC 2π ( 60.0 Hz ) ( 200 × 10 -6 F )
Z = R 2 + ( X L − XC ) =
2
2
( 20.0 Ω ) + ( 37.7 Ω − 13.3 Ω ) = 31.6 Ω
2
2
I 
2
R =  max  R =
Pav = I rms
 2 
1  ∆Vmax 

 R=
2 Z 
2
and power factor = cos φ =
2
1  100 V 

 ( 20.0 Ω ) = 100 W
2  31.6 Ω 
2
 I

Pav
I rms
R
R 20.0 Ω
=
=  rms  R = =
= 0.633
∆Vrms I rms ∆Vrms I rms  ∆Vrms 
Z 31.6 Ω
(b) The same calculations as shown in Part (a) above, with f = 50.0 Hz , give
X L = 31.4 Ω, XC = 15.9 Ω, Z = 25.3 Ω , Pav = 156 W and power factor = 0.790
21.31
(a)
(
)
2
√av = I rms
R = I rms ( I rms R ) = I rms ∆VR , rms , so I rms =
Thus,
R=
∆VR , rms
I rms
=
√av
14 W
=
= 0.28 A
∆VR , rms 50 V
50 V
= 1.8 × 10 2 Ω
0.28 A
(b) Z = R2 + X L2 , which yields
2
2
 ∆V 
2
 90 V 
2
2
X L = Z − R =  rms  − R2 = 
 − ( 1.8 × 10 Ω ) = 2.7 × 10 Ω
0.28
A
I


 rms 
2
and
L=
2
XL
2.7 × 10 2 Ω
=
= 0.71 H
2π f 2π ( 60 Hz )
215
216
21.32
CHAPTER 21
X L = 2π fL = 2π ( 600 Hz ) ( 6.0 × 10 −3 H ) = 23 Ω
XC =
1
1
=
= 11 Ω
2π fC 2π ( 600 Hz ) ( 25 × 10 -6 F )
Z = R2 + ( X L − XC ) =
2
2
( 25 Ω ) + ( 23 Ω − 11 Ω ) = 28 Ω
2
(a)
 ∆V
∆VR , rms = I rms R =  rms
 Z

 10 V 
 ( 25 Ω ) = 9.0 V
R = 
 28 Ω 

 ∆V
∆VL , rms = I rms X L =  rms
 Z

 10 V 
 ( 23 Ω ) = 8.2 V
 XL = 
 28 Ω 

 ∆V
∆VC , rms = I rms XC =  rms
 Z

 10 V 
 ( 11 Ω ) = 3.8 V
 XC = 
 28 Ω 

No , ∆VR , rms + ∆VL , rms + ∆VC , rms = 9.0 V + 8.2 V + 3.8 V = 21 V ≠ 10 V
However, observe that if we take phases into account and add these voltages
vectorially, we find
( ∆V
) + ( ∆V
2
R , rms
L , rms
− ∆VC , rms
)
2
=
2
2
( 9.0 V ) + ( 8.2 V − 3.8 V ) = 10 V = ∆Vrms
(b) The power delivered to the resistor is the greatest. No power losses occur in an
ideal capacitor or inductor.
 ∆V 
 10 V 
2
R =  rms  R = 
P= I rms
 ( 25 Ω ) = 3.3 W
 28 Ω 
 Z 
2
(c)
21.33
2
The resonance frequency of the circuit should match the broadcast frequency of the
station.
f0 =
or
L=
1
2π LC
4π
2
gives L =
( 88.9 × 10
1
,
4π f 02 C
2
1
6
Hz ) ( 1.40 × 10
2
−12
F)
= 2.29 × 10 −6 H = 2.29 µ H
Alternating Current Circuits and Electromagnetic Waves
21.34
217
(a) At resonance, X L = XC so the impedance will be
Z = R2 + ( X L − XC ) = R 2 + 0 = R = 15 Ω
2
1
which yields
2π fC
(b) When X L = XC , we have 2π fL =
f=
1
2π LC
1
=
( 0.20 H ) ( 75 × 10 −6 F )
2π
= 41 Hz
(c) The current is a maximum at resonance where the impedance has its minimum
value of Z = R .
(d) At f = 60 Hz , X L = 2π ( 60 Hz )( 0.20 H ) = 75 Ω , XC =
and Z =
2
2
( 15 Ω ) + ( 75 Ω − 35 Ω ) = 43 Ω
Thus, I rms =
21.35
f0 =
1
= 35 Ω ,
2π ( 60 Hz ) ( 75 × 10 −6 F )
(
∆Vmax
∆Vrms
=
Z
Z
1
2π LC
, so C =
2
)=
150 V
= 2.5 A
2 ( 43 Ω )
1
4π f02 L
2
For f 0 = ( f0 )min = 500 kHz = 5.00 × 10 5 Hz
C = Cmax =
4π
2
( 5.00 × 10
1
5
Hz ) ( 2.0 × 10
2
−6
H)
= 5.1 × 10 −8 F = 51 nF
For f 0 = ( f0 )max = 1600 kHz = 1.60 × 106 Hz
C = Cmin =
21.36
4π
2
(1.60 × 10
1
6
Hz ) ( 2.0 × 10
2
The resonance frequency is ω0 = 2π f 0 =
Also, X L = ω L and XC =
1
ωC
1
LC
−6
H)
= 4.9 × 10 −9 F = 4.9 nF
218
CHAPTER 21
L
3.00 H
 1 
(a) At resonance, XC = X L = ω0 L = 
 L = C = 3.00 × 10 -6 F = 1 000 Ω
 LC 
Thus, Z = R2 + 0 2 = R , I rms =
and
∆Vrms 120 V
=
= 4.00 A
Z
30.0 Ω
2
Pav = I rms
R = ( 4.00 A ) ( 30.0 Ω ) = 480 W
2
(
(
)
)
1
1
(b) At ω = ω0 ; X L = X L ω = 500 Ω , XC = 2 XC ω = 2 000 Ω
0
0
2
2
2
( 30.0 Ω ) + ( 500 Ω − 2 000 Ω ) = 1 500 Ω
Z = R 2 + ( X L − XC ) =
2
2
and I rms =
120 V
= 0.080 0 A
1 500 Ω
2
R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W
so Pav = I rms
2
(
(
)
)
1
1
(c) At ω = ω0 ; X L = X L ω = 250 Ω , XC = 4 XC ω = 4 000 Ω
0
0
4
4
Z = 3 750 Ω , and I rms =
120 V
= 0.032 0 A
3 750 Ω
2
R = ( 0.032 0 A ) ( 30.0 Ω ) = 3.07 × 10 −2 W = 30.7 mW
so Pav = I rms
2
(
(d) At ω = 2 ω0 ; X L = 2 X L ω
0
) = 2 000 Ω , X
Z = 1 500 Ω , and I rms =
C
=
(
1
XC
2
120 V
= 0.080 0 A
1 500 Ω
2
R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W
so Pav = I rms
2
ω0
) = 500 Ω
Alternating Current Circuits and Electromagnetic Waves
(
(e) At ω = 4 ω0 ; X L = 4 X L ω
0
) = 4 000 Ω , X
Z = 3 750 Ω , and I rms =
C
=
(
)
1
XC ω = 250 Ω
0
4
120 V
= 0.032 0 A
3 750 Ω
2
so Pav = I rms
R = ( 0.032 0 A ) ( 30.0 Ω ) = 3.07 × 10 −2 W = 30.7 mW
2
The power delivered to the circuit is a maximum when the rms current is a
maximum. This occurs when the frequency of the source is equal to the resonance
frequency of the circuit.
1
=
LC
ω0 = 2π f 0 =
21.37
(10.0 × 10
1
−3
H )( 100 × 10 −6 F )
= 1 000 rad s
Thus, ω = 2ω0 = 2 000 rad s
X L = ω L = ( 2 000 rad s ) ( 10.0 × 10 −3 H ) = 20.0 Ω
XC =
1
1
=
= 5.00 Ω
ωC ( 2 000 rad s ) ( 100 × 10 −6 F )
Z = R2 + ( X L − XC ) =
2
I rms =
2
2
( 10.0 Ω ) + ( 20.0 Ω − 5.00 Ω ) = 18.0 Ω
∆Vrms 50.0 V
=
= 2.77 A
Z
18.0 Ω
2
R = ( 2.77 A ) ( 10.0 Ω ) = 76.9 W
The average power is Pav = I rms
2
and the energy converted in one period is
 2π
E = Pav ⋅ T = Pav ⋅ 
 ω
21.38
(a)
∆V2, rms =
(
N2
∆V1, rms
N1
 ∆V2, rms
so N 2 = N1 
 ∆V1, rms

J 
2π
 
 =  76.9  ⋅ 
s   2 000 rad
 

 = 0.242 J
s
)

 9.0 V 
 = ( 240 turns ) 
 = 18 turns

120
V



219
220
CHAPTER 21
(
)
(b) For an ideal transformer, ( Pav )input = ( Pav )ouput = ∆V2, rms I 2, rms
Thus,
21.39
( Pav )input = ( 9.0 V )( 0.400 A ) =
3.6 W
The power input to the transformer is
(
)
Pinput = ∆V1, rms I1, rms = ( 3 600 V ) ( 50 A ) = 1.8 × 10 5 W
(
)
For an ideal transformer, ( Pav )ouput = ∆V2, rms I 2, rms = ( Pav )input so the current in the longdistance power line is
I 2, rms =
( Pav )input
( ∆V
2, rms
)
=
1.8 × 10 5 W
= 1.8 A
100 000 V
The power dissipated as heat in the line is then
Plost = I 2,2 rms Rline = ( 1.8 A ) ( 100 Ω ) = 3.2 × 10 2 W
2
The percentage of the power delivered by the generator that is lost in the line is
% Lost =
21.40
 3.2 × 10 2 W 
Plost
× 100% = 
 × 100% = 0.18%
5
Pinput
 1.8 × 10 W 
(a) Since the transformer is to step the voltage down from 120 volts to 6.0 volts, the
secondary must have fewer turns than the primary.
(
)
(
)
(b) For an ideal transformer, ( Pav )input = ( Pav )ouput or ∆V1, rms I1, rms = ∆V2, rms I 2, rms so the
current in the primary will be
I1, rms =
( ∆V
2, rms
)I
2, rms
∆V1, rms
=
( 6.0 V )( 500 mA )
120 V
= 25 mA
(c) The ratio of the secondary to primary voltages is the same as the ratio of the number
of turns on the secondary and primary coils, ∆V2 ∆V1 = N 2 N1 . Thus, the number of
turns needed on the secondary coil of this step down transformer is
 ∆V
N 2 = N1  2
 ∆V1

 6.0 V 
 = ( 400 ) 
 = 20 turns
 120 V 

Alternating Current Circuits and Electromagnetic Waves
21.41
(a) At 90% efficiency, ( Pav )output = 0.90 ( Pav )input
Thus, if ( Pav )output = 1 000 kW
the input power to the primary is ( Pav )input =
(b) I1, rms =
(c)
21.42
I 2, rms =
( Pav )input
∆V1, rms
( Pav )output
∆V2, rms
=
( Pav )output
0.90
=
1 000 kW
= 1.1 × 10 3 kW
0.90
1.1 × 10 3 kW 1.1 × 10 6 W
=
= 3.1 × 10 2 A
∆V1, rms
3 600 V
=
1 000 kW 1.0 × 106 W
=
= 8.3 × 10 3 A
∆V1, rms
120 V
Rline = ( 4.50 × 10 −4 Ω m )( 6.44 × 10 5 m ) = 290 Ω
(a) The power transmitted is ( Pav )transmitted = ( ∆Vrms ) I rms
so I rms =
( Pav )transmitted
∆Vrms
=
5.00 × 106 W
= 10.0 A
500 × 10 3 V
2
Rline = ( 10.0 A ) ( 290 Ω ) = 2.90 × 10 4 W = 29.0 kW
Thus, ( Pav )loss = I rms
2
(b) The power input to the line is
( Pav )input = ( Pav )transmitted + ( Pav )loss = 5.00 × 106
W+2.90 × 10 4 W=5.03 × 10 6 W
and the fraction of input power lost during transmission is
( Pav )loss 2.90 × 10 4 W
fraction =
=
= 0.005 80 or 0.580%
( Pav )input 5.03 × 106 W
221
222
CHAPTER 21
(c) It is impossible to deliver the needed power with an input voltage of 4.50 kV. The
maximum line current with an input voltage of 4.50 kV occurs when the line is
shorted out at the customer’s end, and this current is
( I rms )max =
∆Vrms 4 500 V
=
= 15.5 A
290 Ω
Rline
The maximum input power is then
(P )
input
max
= ( ∆Vrms )( I rms )max
= ( 4.50 × 10 3 V ) ( 15.5 A ) = 6.98 × 10 4 W = 6.98 kW
This is far short of meeting the customer’s request, and all of this power is lost in the
transmission line.
21.43
From v = λ f , the wavelength is
λ=
v 3.00 × 108 m s
=
= 4.00 × 106 m = 4 000 km
f
75 Hz
The required length of the antenna is then,
L = λ 4 = 1 000 km , or about 621 miles. Not very practical at all.
21.44
c=
or
21.45
1
µ0 ∈0
=
( 4π × 10
1
−7
N⋅s C
2
2
)( 8.854 × 10
−12
C2 N ⋅ m 2 )
c = 2.998 × 108 m s
(a) The frequency of an electromagnetic wave is f = c λ , where c is the speed of light,
and λ is the wavelength of the wave. The frequencies of the two light sources are
then
3.00 × 108 m s
c
f red =
=
= 4.55 × 1014 Hz
Red:
660 × 10 -9 m
λred
and
3.00 × 108 m s
c
=
= 3.19 × 1014 Hz
Infrared: f IR =
-9
940 × 10 m
λIR
223
Alternating Current Circuits and Electromagnetic Waves
(b) The intensity of an electromagnetic wave is proportional to the square of its
amplitude. If 67% of the incident intensity of the red light is absorbed, then the
intensity of the emerging wave is ( 100% − 67% ) = 33% of the incident intensity, or
I f = 0.33I i . Hence, we must have
Emax, f
Emax, i
21.46
=
If
Ii
= 0.33 = 0.57
If I 0 is the incident 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 molecules, the
Beer-Lambert law states that log 10 ( I I 0 ) = −ε CL where ε is a constant.
For 660-nm light, the absorbing molecules are oxygenated hemoglobin. Thus, if 33% of
this wavelength light is transmitted through blood, the concentration of oxygenated
hemoglobin in the blood is
CHBO 2 =
− log 10 ( 0.33 )
εL
[1]
The absorbing molecules for 940-nm light are deoxygenated hemoglobin, so if 76% of
this light is transmitted through the blood, the concentration of these molecules in the
blood is
CHB =
− log 10 ( 0.76 )
εL
[2]
Dividing equation [2] by equation [1] gives the ratio of deoxygenated hemoglobin to
oxygenated hemoglobin in the blood as
log 10 ( 0.76 )
CHB
=
= 0.25
CHBO 2 log 10 ( 0.33 )
or
CHB = 0.25CHBO 2
Since all the hemoglobin in the blood is either oxygenated or deoxygenated, it is
necessary that CHB + CHBO 2 = 1.00 , and we now have 0.25CHBO 2 + CHBO 2 = 1.0 . The
fraction of hemoglobin that is oxygenated in this blood is then
CHBO 2 =
1.0
= 0.80
1.0 + 0.25
or
80%
Someone with only 80% oxygenated hemoglobin in the blood is probably in serious
trouble needing supplemental oxygen immediately.
224
CHAPTER 21
21.47
The distance between adjacent antinodes in a standing wave is λ 2
Thus, λ = 2 ( 6.00 cm ) = 12.0 cm = 0.120 m , and
c = λ f = ( 0.120 m ) ( 2.45 × 10 9 Hz ) = 2.94 × 108 m s
21.48
At Earth’s location, the wave fronts of the solar radiation are spheres whose radius is the
√
√
Sun-Earth distance. Thus, from Intensity = av = av2 , the total power is
A
4π r
2
W

Pav = ( Intensity ) ( 4π r 2 ) =  1 340 2   4π ( 1.49 × 1011 m )  = 3.74 × 10 26 W

m 

21.49
From Intensity =
c B2
Emax Bmax
E
and max = c , we find Intensity = max
2 µ0
Bmax
2 µ0
Thus,
Bmax
and
2 ( 4π × 10 −7 T ⋅ m A )
2 µ0
=
( Intensity ) =
(1 340 W m2 ) = 3.35 × 10−6 T
c
3.00 × 108 m s
Emax = Bmax c = ( 3.35 × 10 −6 T ) ( 3.00 × 108 m s ) = 1.01 × 10 3 V m
c 3.00 × 10 8 m s
=
= 11.0 m
f 27.33 × 106 Hz
21.50
λ=
21.51
(a) For the AM band,
λmin =
λmax =
c
fmax
c
f min
=
3.00 × 10 8 m s
= 188 m
1 600 × 10 3 Hz
=
3.00 × 10 8 m s
= 556 m
540 × 10 3 Hz
Alternating Current Circuits and Electromagnetic Waves
225
(b) For the FM band,
λmin =
λmax =
21.52
c
fmax
c
f min
=
3.00 × 10 8 m s
= 2.78 m
108 × 106 Hz
=
3.00 × 10 8 m s
= 3.4 m
88 × 10 6 Hz
The transit time for the radio wave is
tR =
dR
100 × 10 3 m
=
= 3.33 × 10 −4 s = 0.333 ms
c 3.00 × 108 m s
and that for the sound wave is
ts =
ds
3.0 m
=
= 8.7 × 10 −3 s = 8.7 ms
vsound 343 m s
Thus, the radio listeners hear the news 8.4 ms before the studio audience because
radio waves travel so much faster than sound waves.
21.53
If an object of mass m is attached to a spring of spring constant k, the natural frequency
of vibration of that system is f = k m 2π . Thus, the resonance frequency of the C=O
double bond will be
f=
1
2π
k
moxygen
=
1
2π
2 800 N m
= 5.2 × 1013 Hz
−26
2.66 × 10 kg
atom
and the light with this frequency has wavelength
λ=
c 3.00 × 10 8 m s
=
= 5.8 × 10 −6 m = 5.8 µ m
f
5.2 × 1013 Hz
The infrared region of the electromagnetic spectrum ranges from λmax ≈ 1 mm down to
λmin = 700 nm = 0.7 µ m . Thus, the required wavelength falls within the infrared region .
226
CHAPTER 21
21.54
Since the space station and the ship are moving toward one another, the frequency after
being Doppler shifted is fO = fS ( 1 + u c ) , so the change in frequency is
 1.8 × 10 5 m s 
u
11
∆f = fO − fS = fS   = ( 6.0 × 1014 Hz ) 
 = 3.6 × 10 Hz
8
c
3.0
10
m
s
×
 


and the frequency observed on the spaceship is
fO = fS + ∆f = 6.0 × 1014 Hz + 3.6 × 1011 Hz = 6.003 6 × 1014 Hz
21.55
Since you and the car ahead of you are moving away 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 will detect is fO = fS ( 1 − u c ) , and the change in frequency is
 40 km h
u
∆f = fO − fS = − fS   = − ( 4.3 × 1014 Hz ) 
8
c
 3.0 × 10 m
 0.278 m s 
7

 = − 1.6 × 10 Hz
s  1 km h 
The frequency you will detect will be
fO = fS + ∆f = 4.3 × 1014 Hz − 1.6 × 107 Hz = 4.299 999 84 × 1014 Hz
21.56
The driver was driving toward the warning lights, so the correct form of the Doppler
shift equation is fO = fS ( 1 + u c ) . The frequency emitted by the yellow warning light is
fS =
c
λS
=
3.00 × 10 8 m s
= 5.17 × 1014 Hz
−9
580 × 10 m
and the frequency the driver claims that she observed is
fO =
c
λO
=
3.00 × 108 m s
= 5.36 × 1014 Hz
−9
560 × 10 m
The speed with which she would have to approach the light for the Doppler effect to
yield this claimed shift is
f

 5.36 × 1014 Hz

u = c  O − 1  = ( 3.00 × 108 m s ) 
− 1  = 1.1 × 107 m s
14
 5.17 × 10 Hz

 fS

Alternating Current Circuits and Electromagnetic Waves
R=
21.57
( ∆V )DC
I DC
=
12.0 V
= 19.0 Ω
0.630 A
Z = R2 + ( 2π f L ) =
2
21.58
∆Vrms
24.0 V
=
= 42.1 Ω
I rms
0.570 A
( 42.1 Ω ) − ( 19.0 Ω )
= 9.96 × 10−2 H = 99.6 mH
2π ( 60.0 Hz )
2
Z 2 − R2
=
2π f
Thus, L =
227
2
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 ) cos 30° = 0.4 m 2 .
The intensity the radiation at Earth’s surface is I surface = 0.6 I incoming and only 50% of this is
absorbed. Since I =
Pav ( ∆E ∆t )
, the absorbed energy is
=
A
A
(
)
∆E = ( 0.5 I surface ) A ( ∆t ) =  0.5 0.6 I incoming  A ( ∆t )


= ( 0.5 )( 0.6 ) ( 1 340 W m 2 )( 0.4 m 2 ) ( 3 600 s ) = 6 × 10 5 J ~ 106 J
Z = R2 + ( XC ) = R2 + ( 2π f C )
2
21.59
=
−2
2
( 200 Ω ) +  2π ( 60 Hz ) ( 5.0 × 10 −6 F )  = 5.7 × 10 2 Ω
−2
 ∆V 
 120 V 
−3
2
R =  rms  R = 
Thus, Pav = I rms
 ( 200 Ω ) = 8.9 W = 8.9 × 10 kW
2
Z
5.7
×
10
Ω




2
and
2
cost = ∆E ⋅ ( rate ) = Pav ⋅∆t ⋅ ( rate )
= ( 8.9 × 10 −3 kW ) ( 24 h ) ( 8.0 cents kWh ) = 1.7 cents
228
CHAPTER 21
X L = ω L , so ω = X L L
21.60
Then, XC =
1
1
=
which gives
ω C C ( XL L )
L = ( X L ⋅ XC ) C = ( 12 Ω )( 8.0 Ω )  C or L = ( 96 Ω 2 ) C
1
1
, we obtain LC =
2
LC
( 2π f0 )
From ω0 = 2π f 0 =
Substituting from Equation (1), this becomes ( 96 Ω 2 ) C 2 =
or
C=
1
( 2π f0 )
(1)
96 Ω
2
=
1
 2π ( 2 000 π Hz )  96 Ω
2
1
( 2π f0 )
2
= 2.6 × 10 −5 F = 26 µ F
Then, from Equation (1),
L = ( 96 Ω 2 )( 2.6 × 10 −5 F ) = 2.5 × 10 −3 H = 2.5 mH
21.61
(a) The box cannot contain a capacitor since a steady DC current cannot flow in a series
circuit containing a capacitor. Since the AC and DC currents are different, even
when a 3.0 V potential difference is used in both cases, the box must contain a
reactive element. The conclusion is that the box must contain a
resistor and inductor connected in series.
(b)
R=
∆VDC
3.00 V
=
= 10 Ω
I DC
0.300 A
Z=
∆Vrms
3.00 V
=
= 15 Ω
I rms
0.200 A
Since Z = R2 + X L2 = R 2 + ( 2π f L ) , we find
2
Z 2 − R2
L=
=
2π f
( 15 Ω ) − ( 10 Ω )
= 30 mH
2π ( 60 Hz )
2
2
Alternating Current Circuits and Electromagnetic Waves
21.62
3.00 × 10 8 m s
= 1.0 × 1010 Hz . Therefore, the
λ 3.00 × 10 −2 m
1
resonance frequency of the circuit is f 0 =
= 1.0 × 1010 Hz , giving
2π LC
(a) The required frequency is f =
C=
1
( 2π f0 )
2
L
=
( 2π × 10
c
=
1
10
Hz ) ( 400 × 10
2
−12
H)
= 6.3 × 10 −13 F = 0.63 pF
∈0 A ∈0 A 2
=
, so
(b) C =
d
d
C⋅d
A=
=
∈0
21.63
( 6.3 × 10
−13
F )( 1.0 × 10 −3 m )
8.85 × 10 −12 C 2 N ⋅ m
= 8.5 × 10 −3 m = 8.5 mm
(c)
XC = X L = ( 2π f 0 ) L = 2π ( 1.0 × 1010 Hz )( 400 × 10 −12 H ) = 25 Ω
(a)
Emax
= c , so
Bmax
Bmax =
Emax 0.20 × 10 −6 V m
=
= 6.7 × 10 −16 T
8
c
3.00 × 10 m s
(b) Intensity =
Emax Bmax
2µ0
( 0.20 × 10 V m )( 6.7 × 10
2 ( 4π × 10 T ⋅ m A )
−6
=
(c)
−16
−7
T)
= 5.3 × 10 −17 W m 2
 π d2 
Pav = ( Intensity ) ⋅ A = ( Intensity ) 

 4 
 π ( 20.0 m )2 
−14
= ( 5.3 × 10 −17 W m 2 ) 
 = 1.7 × 10 W
4


21.64
(a)
Z=
∆Vrms 12 V
=
= 6.0 Ω
I rms
2.0 A
229
230
CHAPTER 21
R=
(b)
∆VDC 12 V
=
= 4.0 Ω
I DC
3.0 A
From Z = R2 + X L2 = R 2 + ( 2π f L ) , we find
2
( 6.0 Ω ) − ( 4.0 Ω )
= 1.2 × 10 −2 H = 12 mH
2π ( 60 Hz )
2
Z 2 − R2
L=
=
2π f
21.65
2
(a) The radiation pressure on the perfectly reflecting sail is
p=
2 ( Intensity )
c
=
2 ( 1 340 W m 2 )
3.00 × 10 m s
8
= 8.93 × 10 −6 N m 2
so the total force on the sail is
F = p ⋅ A = ( 8.93 × 10 −6 N m 2 )( 6.00 × 10 4 m 2 ) = 0.536 N
(b) a =
F 0.536 N
=
= 8.93 × 10 −5 m s 2
m 6 000 kg
(c) From ∆x = v0 t +
1 2
at , with v0 = 0 , the time is
2
2 ( 3.84 × 10 8 m )
2 ( ∆x )
1d


=
= ( 2.93 × 10 6 s ) 
t=
 = 33.9 d
2
4
−5
a
8.93 × 10 m s
 8.64 × 10 s 
21.66
We know that
(
(
)
)
I1, rms Z1
 I1, rms
N1 ∆V1, rms
=
=
=
 I 2, rms
N 2 ∆V2, rms
I 2, rms Z2

 Z1

 Z2

Also, for an ideal transformer,
( ∆V ) I
1, rms
1, rms
(
)
= ∆V2, rms I 2, rms which gives
 Z1
N  ∆V1, rms
, or 1 


N 2  ∆V2, rms
 Z2
I1, rms
I 2, rms
=
Therefore,
N1  ∆V2, rms
=
N 2  ∆V1, rms
 Z1
=
 Z2

This gives
8 000 Ω
N
Z1
N1  N1  Z1
, or 1 =
=
= 32

=
N2
Z2
8.0 Ω
N 2  N 2  Z2
∆V2, rms
∆V1, rms
231
Alternating Current Circuits and Electromagnetic Waves
21.67
Consider a cylindrical volume with
V = 1.00 Liter = 1.00 × 10 −3 m 3
A = 1.00 m 2
and cross-sectional area
The length of this one liter cylinder is
d=
V 1.00 × 10 −3 m 3
=
= 1.00 × 10 −3 m
2
A
1.00 m
Imagine this cylinder placed at the top of Earth’s atmosphere, with its length
perpendicular to the incident wave fronts. Then, all the energy in the one liter volume of
sunlight will strike the atmosphere in the time required for sunlight to travel the length
of the cylinder. This time is
∆t =
d 1.00 × 10 −3 m
=
= 3.33 × 10 −12 s
c 3.00 × 108 m s
The energy passing through the 1.00 m 2 area of the end of the cylinder in this time is
∆E = Pav ⋅ ∆t = ( Intensity ) ⋅ A  ⋅ ∆t
= ( 1 340 W m 2 )( 1.00 m 2 )( 3.33 × 10 −12 s ) = 4.47 × 10 −9 J
21.68
The capacitance of a parallel-plate capacitor is C = e0 A d , and its reactance in an AC
circuit is XC = 1 2π fC . Observe that reducing the plate separation to one-half of its
original value will double the capacitance and reduce the capacitive reactance to onehalf the original value.
The impedance of an RLC series circuit in which X L = R is Z = R 2 + ( R − XC ) . When
2
the applied voltage is ∆V , the current in the circuit is I = ∆V Z = ∆V
R 2 + ( R − XC ) . If
2
now the plate separation, and hence the capacitive reactance, is cut to one-half the
original value, the new impedance is Z′ = R 2 + ( R − XC 2 ) and the new current will be
2
I ′ = ∆V Z′ = ∆V
R 2 + ( R − XC 2 ) .
2
If it is observed that I ′ = 2I , then we must have
∆V
R 2 + ( R − XC 2 )
2
=
2∆V
R 2 + ( R − XC )
2
2
2
or R2 + ( R − XC ) = 4  R 2 + ( R − XC 2 ) 


232
CHAPTER 21
Expanding the last result yields
R2 + R2 − 2RXC + XC2 = 4 R 2 + 4 R2 − 4 RXC + XC2
which reduces to 0 = 6 R 2 − 2RXC and yields XC = 3R