Chapter 21
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 =
(d)
21.11
Irms =
so
21.15
∆Vrms 100 V
=
= 20.0 A
R
5.00 Ω
av
∆Vmax 141 V
=
= 28.3 A or I max = 2 I rms = 2 ( 20.0 A ) = 28.3 A
5.00 Ω
R
2
= I rms
R = ( 20.0 A )
∆Vrms
 ∆V
=2 π f C  max
XC
2

C=
2
( 5.00 Ω ) = 2.00 × 103 W = 2.00 kW

 = 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
The ratio of inductive reactance at f 2 = 50.0 Hz to that at f1 = 60.0 Hz is
( XL ) 2
( XL ) 1
=
f
50.0 Hz
2π f 2 L f 2
=
( 54.0 Ω ) = 45.0 Ω
, so ( X L ) 2 = 2 ( X L ) 1 =
f1
60.0 Hz
2π f 1L f 1
The maximum current at f2 = 50.0 Hz is then
Imax =
2 ( ∆ Vrms )
∆Vmax
2 ( 100 V)
=
=
= 3.14 A
XL
XL
45.0 Ω
201
202 CHAPTER 21
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 ) =
( 50.0 Ω ) 2 + ( 0 − 66.3 Ω ) 2 = 83.1 Ω
2
(a)
I rms =
∆Vrms 30.0 V
=
= 0.361 A
Z
83.1 Ω
(b) ∆VR , rms = I rms R = ( 0.361 A ) ( 50.0 Ω ) = 18.1 V
(c)
∆VC , rms = I rms XC = ( 0.361 A ) ( 66.3 Ω ) = 23.9 V
−1  X − X C
(d) φ = tan  L
R


−1  0 − 66.3 Ω 
 = tan 
 = − 53.0°

 50.0 Ω 
so, the voltage lags behind the current by 53.0°
21.33
The resonance frequency of the circuit should match the broadcast frequency of the
station.
1
f0 =
or
21.41
L=
2π LC
4π
2
gives L =
( 88.9× 10
(
Thus, if
1
6
(a) At 90% efficiency,
)
av output
(
1
,
4π f 02 C
Hz)
2
2
F)
= 0.90 (
av input
)
av output
(b) I1, rms =
(c)
I2, rms =
)
av input
∆V1, rms
(
)
av output
∆V2, rms
)
= 1 000 kW
the input power to the primary is
(
= 2 .29 × 10−6 H = 2.29 µ H
( 1.40× 10
−12
=
=
(
)
av input
=
(
)
av output
0.90
=
1 000 kW
1.1 × 103 kW 1.1× 106 W
=
= 3.1× 102 A
∆V1, rms
3 600 V
1 000 kW
∆V1, rms
=
1.0 × 106 W
= 8.3 × 10 3 A
120 V
0.90
= 1.1 × 103 kW
Alternating Current Circuits and Electromagnetic Waves
203
21.45
(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
λ red
and
3.00 × 108 m s
c
f IR =
=
= 3.19× 1014 Hz
Infrared:
λIR
940 × 10-9 m
(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
=
If
Ii
= 0.33 = 0.57
204 CHAPTER 21
21.46
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 I0 ) = −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
C HBO 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
C HB =
− 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
C HBO 2 log 10 ( 0.33)
or
C HB = 0.25C HBO 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.
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
av
A
=
av
4π r 2
, the total power is
2
W 

= ( Intensity ) ( 4π r 2 ) =  1 340 2   4π ( 1.49 × 1011 m)  = 3.74× 1026 W


m 
