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21
Alternating Current Circuits and
Electromagnetic Waves
Clicker Questions
Question O1.01
Description: Relating object color to the wave nature of light.
Question
A particular object appears to be red. The reason it appears red is:
1.
2.
3.
4.
5.
6.
It absorbs only red frequencies as they hit.
It reflects only red frequencies as they hit.
It absorbs red frequencies more than other frequencies.
It reflects red frequencies more than other frequencies.
It emits red frequencies more than other frequencies.
Not enough information.
Commentary
Purpose: To hone the concept of “color” as a property of an object.
Discussion: If an object appears red when you look at it, the light emanating from it and reaching your
eye must be predominantly red in frequency. Physically, answers (2), (4), and (5) are all possible. Case (5)
would describe a glowing object such as a neon sign or red LED. Most common objects, however, do not
emit light, but rather reflect some of the light striking them. A real red object will reflect predominantly red
frequencies of light, but will not perfectly absorb all other frequencies. Thus, (4) is a better answer than (2).
If you interpret “a particular object” as a typical non-luminous object, (4) would be the best answer.
However, (6) — not enough information — is defensible, since you were not told whether the object is
luminous, which would make (5) accurate.
Key Points:
•
The apparent color of an object is determined by the mix of frequencies in the light traveling from it to
the observer; the light from a red object will be dominated by light in the red portion of the spectrum.
•
An object may emit light if it is luminous (glowing), but most objects merely reflect some of the light
incident upon them.
For Instructors Only
Questions that may support productive discussion: “What will you see if you look at a red object through a
piece of red-tinted glass? How about through blue-tinted glass? If you shine red light on it? Or blue light?
How does an RGB monitor display millions of different colors?”
252
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Alternating Current Circuits and Electromagnetic Waves
253
Question O1.02
Description: Linking optics to real-world experience of color.
Question
The Sun appears to be yellow when overhead because:
1.
2.
3.
4.
5.
6.
The human eye is most sensitive to yellow.
Yellow frequencies are transmitted better through the atmosphere.
Air does not absorb yellow frequencies.
The temperature of the Sun is 6000 K.
More than one of the above.
None of the above.
Commentary
Purpose: To probe your understanding of the colors of glowing objects, and to connect physics processes
to your everyday experience.
Discussion: Hot matter emits photons of many frequencies, but more of some frequencies than others. The
higher the temperature of the matter, the more high-frequency photons are emitted. If the temperature high
enough, the emitted light will be visible (have enough photons in the visible part of the spectrum), the color
will be determined by the mix of photon frequencies. The higher the temperature, the closer to the blue end
of the spectrum the radiated light will appear.
At a temperature of about 6000 K, the mix of radiated photon frequencies is such that the object will
appear to glow yellow. This accounts for the sun’s color when overhead. Other stars, having different
surface temperatures, have different colors: there are red stars, blue stars, etc.
Key Points:
•
Objects glow when hot enough. The mix of photon frequencies emitted, and thus the color, depends on
the temperature of the object.
•
The sun’s surface temperature is approximately 6000 K, resulting in its yellow color.
For Instructors Only
This question is most effectively used before covering the relevant material on light and black-body
radiation, to elicit student’s preconceptions and motivate learning of the material.
Question O1.03
Description: Linking optics to real-world experience of color.
Question
The sky appears to be blue during the day because:
1.
2.
56157_21_ch21_p252-280.indd 253
Air absorbs blue light less than other frequencies (i.e., acts like a blue filter).
Air molecules emit blue light after being struck by sunlight.
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254
Chapter 21
3.
4.
5.
The sky reflects blue light from the oceans.
The temperature high in the Earth’s upper atmosphere is 1000 K.
None of the above.
Commentary
Purpose: To connect physics processes to your everyday experience.
Discussion: When light travels through the atmosphere, some of the photons collide with air molecules and
change directions (a process called “scattering”), and most travels through unscattered. Some photons are
more likely to be scattered than others, however; photons toward the blue end of the visible spectrum are
far, far more likely to scatter than those near the red end. Thus, when sunlight travels through the atmosphere,
much of the blue light scatters off in various directions. An observer looking at the sun sees photons of all
colors traveling directly without scattering; an observer looking elsewhere in the sky sees photons that have
reached her after scattering off an air molecule, and these are predominantly blue.
Thus, the proper answer to “Why is the sky blue?” is that “Blue photons are much more likely to scatter
from air molecules than photons of other colors.” This is not on the list, so “none of the above” (5) is the
best choice.
Key Points:
•
When light travels through air, some of the photons collide with air molecules and scatter off in
different directions.
•
The frequency (color) of a photon strongly affects how likely it is to scatter off of an air molecule. The
closer to the blue end of the visible spectrum, the more likely scattering is.
•
The sky’s color is due to photons that reach the observer after scattering from an air molecule.
For Instructors Only
This also is a good question to ask before students should “know” the right answer, to elicit preconceptions,
stimulate discussion, and motivate subsequent learning.
Depending on your students’ mathematical sophistication, they may appreciate knowing that the scattering
probability of a photon from an air molecule goes as the fourth power of the frequency. This is not because blue
molecules are “bigger,” but rather because of the quantum-mechanical nature of the scattering interaction.
Discussion of the sky’s color near sunrise and sunset has been omitted here, as that subject is taken up in
Question O1.04 (which makes a good follow-up to this one).
We do want students to take “none of the above” answers seriously.
Question O1.04
Description: Linking optics to real-world experience of color.
Question
The Sun appears to be red as it sets because:
1.
2.
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Air absorbs red light less than other frequencies (i.e., acts like a red filter).
The sunlight has a red-shift when you’re moving fastest away from it.
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Alternating Current Circuits and Electromagnetic Waves
3.
4.
5.
6.
255
The Sun cools down to 5000 K each evening.
Light is refracted as it enters the atmosphere.
The Sun dies in a glorious fireball each evening and is reborn each morning.
None of the above.
Commentary
Purpose: To connect physics processes to your everyday experience.
Discussion: When light travels through the atmosphere, some of the photons collide with air molecules and
change directions (a process called “scattering”), and most travels through unscattered. Some photons are
more likely to be scattered than others, however; photons toward the blue end of the visible spectrum are
far, far more likely to scatter than those near the red end. Thus, when sunlight travels through the atmosphere,
much of the blue light scatters off in various directions.
At sunset, light from the sun skims along the Earth’s surface before reaching an observer, traveling through
much more air than it does at midday. This means each photon has many more chances to scatter off in
another direction, and by the time the sunlight has reached the observer, enough of the blue light has been
scattered away that the remaining light looks reddish. This explanation does not appear among the listed
answers, so “none of the above” is the best choice.
Light does refract upon entering the atmosphere, but only slightly, about half a degree. Although refraction
will cause separation of colors because of frequency dependence of the index of refraction, refraction is not
the cause of the red light at sunset.
At midday, more blue light scatters away than for other colors, but enough remains that the sun’s color is
not significantly altered from its original yellowish-white color.
Key Points:
•
When light travels through air, some of the photons collide with air molecules and scatter off in
different directions.
•
The frequency (color) of a photon strongly affects how likely it is to scatter off of an air molecule. The
closer to the blue end of the visible spectrum, the more likely scattering is.
•
As blue photons are disproportionately scattered out of the sun’s light, the remaining light is
increasingly reddish in appearance.
For Instructors Only
This question makes a good follow-up to Question O1.03: it requires the same principles as discussed there,
but asks students to reason further with them. This checks how well they grasp those principles (as opposed
to just memorizing the answer) and helps them solidify the ideas.
QUICK QUIZZES
1.
56157_21_ch21_p252-280.indd 255
(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, for which R → ∞. (b) and (d) can never
be true because iav = 0 for AC currents.
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256
Chapter 21
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 = ∆v R + ∆v L + ∆vC.)
3.
(b). Closing switch A replaces a single resistor with a parallel combination of two resistors.
Since the equivalent resistance of a parallel combination is always less than the lowest
resistance in the combination, the total resistance of the circuit decreases, which causes the
2
impedance Z = Rtotal
+ ( X L − XC ) to decrease.
2
4.
(a). Closing switch A replaces a single resistor with a parallel combination of two resistors.
Since the equivalent resistance of a parallel combination is always less than the lowest resistance
in the combination, the total resistance of the circuit decreases, which causes the phase angle,
φ = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ , to increase.
5.
(a). Closing switch B replaces a single capacitor with a parallel combination of two capacitors. Since the equivalent capacitance of a parallel combination is greater than that of either
of the individual capacitors, the total capacitance increases, which causes the capacitive
reactance XC = 1 2π fC to decrease. Thus, the net reactance, X L − XC , increases causing the
phase angle, φ = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ , to increase.
6.
(b). Inserting an iron core in the inductor increases both the self-inductance and the inductive reactance, X L = 2π fL. This means the net reactance, X L − XC, and hence the impedance,
2
Z = Rtotal
+ ( X L − XC ) , increases, causing the current (and therefore, the bulb’s brightness) to
decrease.
2
7.
(b), (c). Since pressure is force per unit area, changing the size of the area without altering the
intensity of the radiation striking that area will not cause a change in radiation pressure. In (b),
the smaller disk absorbs less radiation, resulting in a smaller force. For the same reason, the
momentum in (c) is reduced.
8.
(b), (d). The frequency and wavelength of light waves are related by the equation λ f = v or
f = v λ , where the speed of light v 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.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
∆Vmax 170 V
=
= 120 V , so (c) is the correct choice.
2
2
1.
∆Vrms =
2.
When the frequency doubles, the rms current I rms = ∆VL ,rms X L = ∆VL ,rms 2π fL is cut in half.
Thus, the new current is I rms = 3.0 A 2 = 1.5 A and (e) is the correct answer.
3.
∆VC ,rms = I rms XC =
4.
∆VL ,rms = I rms X L = I rms ( 2π fL ) = ( 2.0 A ) 2π ( 60.0 Hz ) ⎡⎣(1.0 2π ) H ⎤⎦ = 120 V, and the correct
answer is choice (c).
5.
56157_21_ch21_p252-280.indd 256
I rms
1.00 × 10 −3 A
= 16.7 V, so choice (a) is correct.
=
2π fC 2π ( 60.0 Hz ) ⎡⎣(1.00 2π ) × 10 −6 F ⎤⎦
At the resonance frequency, X L = XC and the impedance is Z = R 2 + ( X L − XC ) = R . Thus,
the rms current is I rms = ∆Vrms Z = (120 V) ( 20 Ω ) = 6.0 A, and (b) is the correct choice.
2
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Alternating Current Circuits and Electromagnetic Waves
257
6.
The battery produces a constant current in the primary coil, which generates a constant flux
through the secondary coil. With no change in flux through the secondary coil, there is no induced
voltage across the secondary coil, and choice (e) is the correct answer.
7.
The peak values of the electric and magnetic field components of an electromagnetic wave are
related by Emax Bmax = c , where c is the speed of light in vacuum. Thus,
)
)
Emax = cBmax = ( 3.0 × 10 8 m s (1.5 × 10 −7 T = 45 N C
which is choice (d).
8.
When a power source, AC or DC, is first connected to a RL combination, the presence of the
inductor impedes the buildup of a current in the circuit. The value of the current starts at zero and
increases as the back emf induced across the inductor decreases somewhat in magnitude. Thus,
the correct choice is (c).
9.
The voltage across the capacitor is proportional to the stored charge. This charge, and hence the
voltage ∆vC , is a maximum when the current has zero value and is in the process of reversing
direction after having been flowing in one direction for a half cycle. Thus, the voltage across the
capacitor lags behind the current by 90°, and (a) is the correct choice.
10.
In an RLC circuit, the instantaneous voltages ∆v R , ∆v L , and ∆vC (across the resistance, inductance, and capacitance, respectively) are not in phase with each other. The instantaneous voltage
∆v R is in phase with the current, ∆v L leads the current by 90°, while ∆vC lags behind the current
by 90°. The instantaneous values of these three voltages do add algebraically to give the instantaneous voltages across the RLC combination, but the rms voltages across these components do
not add algebraically. The rms voltages across the three components must be added as vectors
(or phasors) to obtain the correct rms voltage across the RLC combination. Among the listed
choices, choice (e) is the false statement.
11.
In an AC circuit, both an inductor and a capacitor store energy for one-half of the cycle of the current,
and return that energy to the circuit during the other half of the cycle. On the other hand, a resistor
converts electrical potential energy into thermal energy during all parts of the cycle of the current.
Thus, only the resistor has a nonzero average power loss. The power delivered to a circuit by a
generator is given by Pav = I rms ∆Vrms cos φ , where φ is the phase difference between the generator
voltage and the current. Among the listed choices, the only true statement is choice (c).
12.
If the voltage is a maximum when the current is zero, the voltage is either leading or lagging the
current by 90° (or a quarter cycle) in phase. Thus, the element could be either an inductor or a
capacitor. It could not be a resistor since the voltage across a resistor is always in phase with the
current. If the current and voltage were out of phase by 180°, one would be a maximum in one
direction when the other was a maximum value in the opposite direction. The correct choice for
this question is (d).
13.
At resonance, X L = XC , and the phase difference between the current and the applied voltage is
φ = tan −1 ⎡⎣( X L − XC ) R ⎤⎦ = tan −1 ( 0 ) = 0°
The correct answer is choice (c).
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258
14.
Chapter 21
At a frequency of f = 5.0 × 10 2 Hz, the inductive reactance, capacitive reactance, and impedance
are X L = 2π fL, XC =
1
2
, and Z = R 2 + ( X L − XC ) . This yields
2π f C
2
Z=
( 20.0 Ω )
and I rms =
2
⎡
⎤
1
+ ⎢ 2π ( 5.0 × 10 2 Hz ( 0.120 H ) −
⎥ = 51 Ω,
2π ( 5.0 × 10 2 Hz ( 0.75 × 10 −6 F ⎥⎦
⎢⎣
)
)
)
∆Vrms 120 V
=
= 2.3 A . Choice (a) is the correct answer.
Z
51 Ω
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 Z = R.
2
4.
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.
6.
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.
8.
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 squares of positive and negative values are always positive and cannot cancel.
10.
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.
12.
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.
14.
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.
PROBLEM SOLUTIONS
21.1
56157_21_ch21_p252-280.indd 258
(a)
∆VR,rms = I rms R = ( 8.00 A ) (12.0 Ω ) = 96.0 V
(b)
∆VR,max = 2 ( ∆VR,rms ) = 2 ( 96.0 V) = 136 V
(c)
I max = 2 I rms = 2 ( 8.00 A ) = 11.3 A
(d)
2
2
Pav = I rms
R = ( 8.00 A ) (12.0 Ω ) = 768 W
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Alternating Current Circuits and Electromagnetic Waves
21.2
259
)
(a)
∆VR,max = 2 ( ∆VR,rms ) = 2 (1.20 × 10 2 V = 1.70 × 10 2 V
(b)
2
Pav = I rms
R=
(c)
Because R =
2
(1.20 × 102 V
∆Vrms
∆V 2
⇒ R = rms =
Pav
60.0 W
R
2
∆Vrms
Pav
)
2
= 2.40 × 10 2 Ω
(see above), if the bulbs are designed to operate at the same
voltage, the 1.00 × 10 2 W will have the lower resistance .
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
=
= 2 .95 A
=
= 70.7 V , and I rms =
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 = Pav ∆Vrms and the resistance is R = ∆Vrms I rms .
I1, rms = I 2, rms =
I 3, rms =
21.5
150 W
120 V
= 1.25 A and R1 = R2 =
= 96.0 Ω
120 V
1.25 A
100 W
120 V
= 0.833 A and R3 =
= 144 Ω
120 V
0.833 A
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
18.6 Ω
Req
The power to the speaker is then
21.6
The general form of the generator voltage is ∆v = ( ∆Vmax ) sin (ω t ) , so by inspection
(a)
∆VR,max = 170 V
(c)
∆VR,rms =
(d)
I rms =
(e)
I max = 2 I rms = 2 ( 6.00 A ) = 8.49 A
(f)
2
2
Pav = I rms
R = ( 6.00 A ) ( 20.0 Ω ) = 720 W
(g)
∆VR,max
2
and
=
(b)
f =
ω 60π rad s
=
= 30.0 Hz
2π
2π rad
170 V
= 120 V
2
∆VR,rms 120 V
=
= 6.00 A
20.0 Ω
R
The argument of the sine function has units of [ω t ] = ( rad s ) ( s ) = radians .
At t = 0.005 0 s, the instantaneous current is
i=
56157_21_ch21_p252-280.indd 259
2
2
Pav = I rms
Rspeaker = ( 0.806 A ) (10.4 Ω ) = 6.76 W
(170 V)
∆v (170 V)
sin ⎡⎣( 60π rad s ) ( 0.005 0 s ) ⎤⎦ =
=
sin ( 0.94 rad ) = 6.9 A
20.0 Ω
R
20.0 Ω
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260
Chapter 21
21.7
XC =
1
, so its units are
2π fC
1
1
Volt
Volt
=
= Ohm
=
=
(1 Sec ) Farad (1 Sec ) (Coulomb Volt ) Coulomb Sec Amp
21.8
21.9
I max = 2 I rms =
)
(a)
I max = 2 (120 V) 2π ( 60.0 Hz ) ( 2.2 0 × 10 − 6 C/V = 0.141 A = 141 mA
(b)
I max = 2 ( 240 V) 2π ( 50.0 Hz ) ( 2.2 0 × 10 − 6 C/V = 0.235 A = 235 mA
I rms =
∆Vrms
= 2π f C ( ∆Vrms ) , so
XC
)
f =
21.10
56157_21_ch21_p252-280.indd 260
)
XC =
1
1
=
= 221 Ω
2π fC 2π ( 60.0 Hz ) (12.0 × 10 −6 F
(b)
I rms =
∆VC ,rms 36.0 V
=
= 0.163 A
221 Ω
XC
(c)
I max = 2 I rms = 2 ( 0.163 A ) = 0.231 A
I rms =
so
21.12
0.30 A
I rms
=
= 4.0 × 10 2 Hz
2π C ( ∆Vrms ) 2π ( 4.0 × 10 −6 F ( 30 V)
(a)
(d)
21.11
2 ( ∆Vrms )
= 2 ( ∆Vrms ) 2π f C
XC
)
No. The current reaches its maximum value one-quarter cycle before the voltage reaches
its maximum value. From the definition of capacitance, the capacitor reaches its maximum
charge when the voltage across it is also a maximum. Consequently, the maximum charge
and the maximum current do not occur at the same time.
∆Vrms
⎛ ∆V ⎞
=2π f C ⎜ max ⎟ = π f C ( ∆Vmax ) 2
⎝ 2 ⎠
XC
C=
0.75 A
I
=
= 1.7 × 10 −5 F = 17 µ F
π f ( ∆Vmax ) 2 π ( 60 Hz ) (170 V) 2
∆VC ,max
98.0 V
= 69.3 V
2
(a)
By inspection, ∆VC,max = 98.0 V, so ∆VC ,rms =
(b)
Also by inspection, ω = 80π rad s, so f =
(c)
I rms =
I max 0.500 A
=
= 0.354 A
2
2
(d)
XC =
∆VC ,max
98.0 V
=
= 196 Ω
0.500 A
I max
(e)
XC =
1
1
1
1
=
= 2.03 × 10 −5 F = 20.3 µ F
=
, so C =
2π fC ω C
ω XC ( 80π rad s ) (196 Ω )
2
=
ω 80π rad s
=
= 40.0 Hz
2π
2π rad
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Alternating Current Circuits and Electromagnetic Waves
21.13
X L = 2π f L , and from
ε
⎛ ∆I ⎞
= L ⎜ ⎟ , we have L =
⎝ ∆t ⎠
ε ( ∆t )
∆I
261
. The units of self inductance are then
ε ∆t Volt ⋅ sec
. The units of inductive reactance are given by
[ L ] = [ ][ ] =
amp
[ ∆I ]
1 ⎛ Volt ⋅ sec ⎞ Voltt
=
= Ohm
Amp ⎟⎠ Amp
[ X L ] = [ f ][ L ] = ⎛⎜⎝ sec ⎞⎟⎠ ⎜⎝
21.14
21.15
)
(a)
X L = 2π f L = 2π ( 80.0 Hz ) ( 25.0 × 10 −3 H = 12.6 Ω
(b)
I rms =
(c)
I max = 2 I rms = 2 ( 6.19 A ) = 8.75 A
The required inductive reactance is X L =
L=
21.16
∆VL ,rms 78.0 V
=
= 6.19 A
12.6 Ω
XL
XL
=
2π f
∆VL ,max
2 ( 2π f ) I rms
(
≥
∆VL ,max ∆VL ,max
and the needed inductance is
=
I max
2 I rms
4.00 V
= 0.750 H
2 ( 2π ) ( 300.0 Hz ) ( 2.00 × 10 −3 A
)
)
Given: v L = 1.20 × 10 2 V sin ( 30π t ) and L = 0.500 H
ω
30π rad s
=
= 15.0 Hz
2π
2π
(a)
By inspection, ω = 30π rad s, so f =
(b)
Also by inspection, ∆VL,max = 1.20 × 10 2 V, so that
∆VL ,rms =
∆VL ,max
2
=
1.20 × 10 2 V
= 84.9 V
2
(c)
X L = 2π fL = ω L = ( 30π rad s ) ( 0.500 H ) = 47.1 Ω
(d)
I rms =
(e)
I max = 2 I rms = 2 (1.80 A ) = 2.55 A
(f)
∆VL ,rms 84.9 V
=
= 1.80 A
47.1 Ω
XL
The phase difference between the voltage across an inductor and the current through the
inductor is φ L = 90° , so the average power delivered to the inductor is
PL ,av = I rms ∆VL ,rms cos φL = I rms ∆VL ,rms cos ( 90°) = 0
(g)
When a sinusoidal voltage with a peak value ∆VL,max is applied to an inductor, the current
through the inductor also varies sinusoidally in time, with the same frequency as the
applied voltage, and has a maximum value of I max = ∆VL ,max X L. However, the current lags
behind the voltage in phase by a quarter-cycle or π 2 radians. Thus, if the voltage is given
by ∆v L = ∆VL ,max sin (ω t ) , the current as a function of time is i = I max sin (ω t − π 2 ). In the
case of the given inductor, the current through it will be i = ( 2.55 A ) sin ( 30π t − π 2 ) .
continued on next page
56157_21_ch21_p252-280.indd 261
3/19/08 5:49:41 PM
262
Chapter 21
(h)
When i = +1.00 A, we have:
sin ( 30π t − π 2 ) = (1.00 A 2.55 A ) , or
30π t − π 2 = sin −1 (1.00 A 2.55 A ) = sin −1 ( 0.392 ) = 0.403 rad
and
21.17
t=
π 2 rad + 0.403 rad
= 2.09 × 10 −2 s = 20.9 ms
30π rad s
From L = N Φ B I (see Section 20.6 in the textbook), the total flux through the coil is
Φ B, total = N Φ B = L ⋅ I , where Φ B is the flux through a single turn on the coil. Thus,
(Φ
)
B , total max
⎡ ∆V ⎤
= L ⋅ I max = L ⋅ ⎢ max ⎥
⎣ XL ⎦
=L
21.18
(a)
2 ( ∆Vrms )
2 (120 V)
=
= 0.450 T ⋅ m 2
2π fL
2π ( 60.0 Hz )
The applied voltage is ∆v = ∆Vmax sin (ω t ) = ( 80.0 V) sin (150t ), so we have that
∆Vmax = 80.0 V and ω = 2π f = 150 rad s. The impedance for the circuit is
Z = R 2 + ( X L − XC ) = R 2 + ( 2π fL − 1 2π fC ) = R 2 + (ω L − 1 ω C )
2
2
2
2
⎡
⎤
1
or Z = ( 40.0 Ω ) + ⎢(150 rad s ) 80.0 × 10 −3 H −
⎥ = 57.5 Ω
−6
(150 rad s ) 125 × 10 F ⎥⎦
⎢⎣
(
2
21.19
)
(
)
∆Vmax 80.0 V
=
= 1.39 A
57.5 Ω
Z
(b)
I max =
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
( 50.0 Ω )2 + ( 0 − 66.3 Ω )2
= 83.0 Ω
∆Vrms 30.0 V
=
= 0.361 A
Z
83.0 Ω
(a)
I rms =
(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
(d)
⎛ X − XC ⎞
−1 ⎛ 0 − 66.3 Ω ⎞
φ = tan −1 ⎜ L
⎟ = tan ⎜⎝
⎟ = −53.0°
⎝
R ⎠
50.0 Ω ⎠
so, the voltage lags behind the current by 53°
56157_21_ch21_p252-280.indd 262
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Alternating Current Circuits and Electromagnetic Waves
21.20
(a)
263
)
X L = 2π f L = 2π ( 50.0 Hz )( 400 × 10 −3 H = 126 Ω
XC =
1
1
=
= 719 Ω
2π f C 2π ( 50.0 Hz )( 4.43 × 10 −6 F
)
Z = R 2 + ( X L − XC ) =
2
( 500 Ω )2 + (126 Ω − 719 Ω )2
= 776 Ω
∆Vmax = I max Z = ( 0.250 A ) ( 776 Ω ) = 194 V
(b)
⎛ 126 Ω − 719 Ω ⎞
⎛ X − XC ⎞
= tan −1 ⎜
φ = tan −1 ⎜ L
⎟
⎟⎠ = − 49.9°
⎝
⎝
R ⎠
500 Ω
Thus, the current leads the voltage by 49.9° .
21.21
(a)
X L = 2π f L = 2π ( 240 Hz )( 2 .50 H ) = 3.77 × 10 3 Ω
XC =
1
1
=
= 2 .65 × 10 3 Ω
2π f C 2π ( 240 Hz )( 0.250 × 10 −6 F
)
Z = R 2 + ( X L − XC ) =
2
21.22
2
= 1.44 kΩ
∆Vmax
140 V
=
= 0.097 2 A
Z
1.44 × 10 3 Ω
(b)
I max =
(c)
⎡ ( 3.77 − 2 .65 ) × 10 3 Ω ⎤
⎛ X − XC ⎞
φ = tan −1 ⎜ L
= tan −1 ⎢
⎥ = 51.2°
⎟
⎝
R ⎠
900 Ω
⎣
⎦
(d)
φ > 0, so the voltage leads the current
X L = 2π fL = 2π ( 60.0 Hz ) ( 0.100 H ) = 37.7 Ω
XC =
1
1
=
= 265 Ω
2π fC 2π ( 60.0 Hz ) (10.0 × 10 −6 F
)
Z = R 2 + ( X L − XC ) =
2
56157_21_ch21_p252-280.indd 263
( 900 Ω )2 + ⎡⎣( 3.77 − 2 .65 ) × 10 3 Ω ⎤⎦
( 50.0 Ω )2 + ( 37.7 Ω − 265 Ω )2
(a)
∆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 = I rms XC = ( 2.75 A ) ( 265 Ω ) = 729 V
(d)
∆Vrms = I rms Z = ( 2.75 A ) ( 233 Ω ) = 641 Ω
= 233 Ω
(e)
3/19/08 5:49:43 PM
264
Chapter 21
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
)
Z RLC = R 2 + ( X L − XC ) =
2
( 60.0 Ω )2 + (151 Ω − 884 Ω )2
= 735 Ω
∆Vrms
Z RLC
and
I rms =
(a)
Z LC = 0 + ( X L − XC ) = X L − XC = 733 Ω
2
⎛ ∆V ⎞
⎛ 90.0 V ⎞
∆VLC , rms = I rms ⋅ Z LC = ⎜ rms ⎟ Z LC = ⎜
( 733 Ω ) = 89.8 V
⎝ 735 Ω ⎟⎠
⎝ Z RLC ⎠
(b)
Z RC = R 2 + ( 0 − XC ) =
2
( 60.0 Ω )2 + ( 884 Ω )2
= 886 Ω
⎛ ∆V ⎞
⎛ 90.0 V ⎞
∆VRC , rms = I rms ⋅ Z RC = ⎜ rms ⎟ Z RC = ⎜
( 886 Ω ) = 108 V
⎝ 735 Ω ⎟⎠
⎝ Z RLC ⎠
21.24
X L = 2π fL = 2π ( 60 Hz ) ( 2.8 H ) = 1.1 × 10 3 Ω
Z = R 2 + ( X L − XC ) =
2
(1.2 × 10 Ω ) + (1.1 × 10
3
2
3
Ω−0
)
2
= 1.6 × 10 3 Ω
∆Vmax
170 V
=
= 0.11 A
Z
1.6 × 10 3 Ω
(a)
I max =
(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 = 1.2 × 10 2 V
(c)
When the instantaneous current is a maximum ( i = I max ), the instantaneous voltage across
the resistor is ∆v R = iR = I max R = ∆VR, max = 1.3 × 10 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 , ∆v L = 0 .
2
Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed
path. Thus, for this series circuit, ∆vsource = ∆v R + ∆v L 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
∆v R = 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 ∆v L = ∆VL , max = 1.2 × 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 = ∆v R + ∆v L = 0 + ∆VL , max = 1.2 × 10 2 V .
56157_21_ch21_p252-280.indd 264
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Alternating Current Circuits and Electromagnetic Waves
1
1
=
= 1.33 × 10 8 Ω
2π f C 2π ( 60.0 Hz )( 20.0 × 10 −12 F
XC =
21.25
)
( 50.0 × 10 Ω ) + (1.33 × 10
Z RC = R 2 + XC2 =
and
265
)
( ∆V
secondary rms
I rms =
Z RC
3
=
2
8
Ω
)
2
= 1.33 × 10 8 Ω
5 000 V
= 3..76 × 10 −5 A
1.33 × 10 8 Ω
)
)
Therefore, ∆Vb, rms = I rms Rb = ( 3.76 × 10 −5 A ( 50.0 × 10 3 Ω = 1.888 V
21.26
1
1
=
= 88.4 Ω
2π f C 2π ( 60.0 Hz )( 30.0 × 10 −6 F
(a)
XC =
(b)
Z = R 2 + ( 0 − XC ) = R 2 + XC2 =
(c)
I max =
(d)
The phase angle in this RC circuit is
)
2
( 60.0 Ω )2 + ( 88.4 Ω )2
= 107 Ω
∆Vmax 1.20 × 10 2 V
=
= 1.12 A
107 Ω
Z
⎛ X − XC ⎞
−1 ⎛ 0 − 88.4 Ω ⎞
φ = tan −1 ⎜ L
⎟ = tan ⎜⎝
⎟ = −55.8°
⎝
R ⎠
60.0 Ω ⎠
Since φ < 0, the voltage lags behind the current by 55.8°° . Adding an inductor will change
the impedance and hence the current in the circuit.
21.27
(a)
X L = 2π f L = 2π ( 50.0 Hz )( 0.250 H ) = 78.5 Ω
(b)
XC =
(c)
Z = R 2 + ( X L − XC ) =
(d)
I max =
(e)
⎛ X − XC ⎞
⎛ 78.5 Ω − 1590 Ω ⎞
φ = tan −1 ⎜ L
= tan −1 ⎜
⎟
⎟⎠ = −84.3°
⎝
⎝
R ⎠
150 Ω
(f)
∆VR,max = I max R = ( 0.138 A ) (150 Ω ) = 20.7 V
1
1
=
= 1.59 × 10 3 Ω = 1.59 kΩ
2π f C 2π ( 50.0 Hz )( 2.00 × 10 − 6 F
)
2
(150 Ω )2 + ( 78.5 Ω − 1590 Ω )2
= 1.52 × 10 3 Ω = 1.52 kΩ
∆Vmax 2.10 × 10 2 V
=
= 0.138 A = 138 mA
1.52 × 10 3 Ω
Z
∆VL ,max = I max X L = ( 0.138 A ) ( 78.5 Ω ) = 10.8 V
)
∆VC ,max = I max XC = ( 0.138 A ) (1.59 × 10 3 Ω = 219 V
56157_21_ch21_p252-280.indd 265
3/19/08 5:49:45 PM
266
21.28
Chapter 21
The voltage across the resistor is a maximum when the current is a maximum, and the maximum
value of the current occurs when the argument of the sine function is π 2. The voltage across the
capacitor lags the current by 90° or π 2 radians, which corresponds to an argument of 0 in the
sine function and a voltage of vC = 0 . Similarly, the voltage across the inductor leads the current
by 90° or π 2 radians, corresponding to an argument in the sine function of π radians or 180°,
giving a voltage of v L = 0 .
X L = 2π f L = 2π ( 50.0 Hz )( 0.185 H ) = 58.1 Ω
21.29
XC =
1
1
=
= 49.0 Ω
2π f C 2π ( 50.0 Hz )( 65.0 × 10 −6 F
)
Z ad = R 2 + ( X L − XC ) =
2
21.30
( 40.0 Ω )2 + ( 58.1 Ω − 49.0 Ω )2
= 41.0 Ω
)
(
∆Vmax 2
∆Vrms
150 V
=
= 2.59 A
=
Z ad
Z ad
( 41.0 Ω ) 2
and
I rms =
(a)
Z ab = R = 40.0 Ω , so ( ∆Vrms )ab = I rms Z ab = ( 2.59 A ) ( 40.0 Ω ) = 104 V
(b)
Z bc = X L = 58.1 Ω, and ( ∆Vrms )bc = I rms Z bc = ( 2.59 A ) ( 58.1 Ω ) = 150 V
(c)
Z cd = XC = 49.0 Ω, and ( ∆Vrms )cd = I rms Z cd = ( 2.59 A ) ( 49.0 Ω ) = 127 V
(d)
Z bd = X L − XC = 9.10 Ω, so ( ∆Vrms )bd = I rms Z bd = ( 2.59 A ) ( 9.10 Ω ) = 23.6 V
XC =
1
1
=
= 1.1 × 10 3 Ω
2π fC 2π ( 60 Hz ) ( 2.5 × 10 −6 F
Z = R 2 + ( X L − XC ) =
2
(1.2 × 10
3
)
Ω ) + ( 0 − 1.1 × 10 Ω )
2
3
2
= 1.6 × 10 3 Ω
∆Vmax
170 V
=
= 0.11 A
Z
1.6 × 10 3 Ω
(a)
I max =
(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 = 1.2 × 10 2 V
(c)
When the instantaneous current i is zero, the instantaneous voltage across the resistor is
∆v R = 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 = 1.2 × 10 2 V
and
)
)
qC = C ( ∆vC ) = ( 2.5 × 10 −6 F (1.2 × 10 2 V = 3.0 × 10 −4 C = 300 µC
Kirchhoff’s loop rule always applies to the instantaneous voltages around a closed path.
Thus, for this series circuit, ∆vsource + ∆v R + ∆vC = 0, and at this instant when i = 0, we
have ∆vsource = 0 + ∆vC = ∆VC , max = 1.2 × 10 2 V .
continued on next page
56157_21_ch21_p252-280.indd 266
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Alternating Current Circuits and Electromagnetic Waves
(d)
267
When the instantaneous current is a maximum (i = I max ), the instantaneous voltage across
the resistor is ∆v R = 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
and ∆vC = 0 gives
∆vsource + ∆v R + ∆vC = 0
21.31
∆vsource = ∆v R = ∆VR, maax = 1.3 × 10 2 V
104 V
∆Vrms
=
= 208 Ω
0.500 A
I rms
(a)
Z=
(b)
2
Pav = I rms
R gives R =
(c)
Z = R 2 + X L2 , so X L = Z 2 − R 2 =
L=
and
21.32
⇒
Pav
I
2
rms
=
10.0 W
= 40.0 Ω
( 0.500 A )2
( 208 Ω )2 − ( 40.0 Ω )2
= 204 Ω
XL
204 Ω
=
= 0.541 H
2π f 2π ( 60.0 Hz )
Given v = ∆Vmax sin (ω t ) = ( 90.0 V) sin ( 350t ) , observe that ∆Vmax = 90.0 V and ω = 350 rad s.
Also, the net reactance is X L − XC = 2π fL − 1 2π fC = ω L − 1 ω C .
(a)
X L − XC = ω L −
1
1
= ( 350 rad s ) ( 0.200 H ) −
= − 44.3 Ω
ωC
( 350 rad s ) ( 25.0 × 10 −6 F )
so the impedance is
∆Vrms ∆Vmax
=
Z
Z
Z = R 2 ( X L − XC ) =
2
2
=
( 50.0 Ω )2 + ( − 44.3 Ω )2
= 66.8 Ω
90.0 V
= 0.953 A
2 ( 66.8 Ω )
(b)
I rms =
(c)
The phase difference between the applied voltage and the current is
⎛ X − XC ⎞
−1 ⎛ − 44.3 Ω ⎞
φ = tan −1 ⎜ L
⎟ = tan ⎜⎝
⎟ = − 41.5°
⎝
R ⎠
50.0 Ω ⎠
so the average power delivered to the circuit is
∆Vmax ⎞
⎛ 90.0 V ⎞
cosφ = ( 0.953 A ) ⎜
⎟ cos ( − 41.5°) = 45.4 W
⎟
⎝
⎝ 2 ⎠
2 ⎠
Pav = I rms ∆Vrms cosφ = I rms ⎛⎜
21.33
Please see the textbook for the statement of Problem 21.21 and the answers for that problem.
There, you should find that ∆Vmax = 140 V, I max = 0.097 2 A, and φ = 51.2°. The average power
delivered to the circuit is then
( 0.097 2 A ) (140 V) cos ( 51.2°) = 4.26 W
I max ⎞ ⎛ ∆Vmax ⎞
cos φ =
⎟
⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠
2
Pav = I rms ∆Vrms cos φ = ⎛⎜
21.34
The rms current in the circuit is
I rms =
∆Vrms 160 V
=
= 2.00 A
Z
80.0 Ω
and the average power delivered to the circuit is
Pav = I rms ( ∆Vrms cos φ ) = I rms ∆VR,rms = I rms ( I rms R ) = I r2ms R = ( 2.00 A )2 ( 22.0 Ω ) = 88.0 W
56157_21_ch21_p252-280.indd 267
3/19/08 5:49:47 PM
268
21.35
Chapter 21
(a)
2
Pav = I rms
R = I rms ( I rms R ) = I rms ( ∆VR, rms ) , so I rms =
Thus,
(b)
R=
∆VR, rms
I rms
Pav
∆VR, rms
=
14 W
= 0.28 A
50 V
50 V
= 1.8 × 10 2 Ω
0.28 A
=
Z = R 2 + X L2 , which yields
2
⎛ ∆V ⎞
⎛ 90 V ⎞
− (1.8 × 10 2 Ω
X L = Z 2 − R 2 = ⎜ rms ⎟ − R 2 = ⎜
⎝ 0.28 A ⎟⎠
⎝ I rms ⎠
and
21.36
L=
2
)
2
= 2.7 × 10 2 Ω
2.7 × 10 2 Ω
XL
=
= 0.72 H
2π f
2π ( 60 Hz )
)
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 = R 2 + ( X L − XC ) =
2
(a)
( 25 Ω )2 + ( 23 Ω − 11 Ω )2
= 28 Ω
⎛ 10 V ⎞
⎛ ∆V ⎞
∆VR, rms = I rms R = ⎜ rms ⎟ R = ⎜
( 255 Ω ) = 8.9 V
⎝ 28 Ω ⎟⎠
⎝ Z ⎠
⎛ 10 V ⎞
⎛ ∆V ⎞
∆VL , rms = I rms X L = ⎜ rms ⎟ X L = ⎜
( 23 Ω ) = 8.2 V
⎝ 28 Ω ⎟⎠
⎝ Z ⎠
⎛ 10 V ⎞
⎛ ∆V ⎞
∆VC , rms = I rms XC = ⎜ rms ⎟ XC = ⎜
(11 Ω ) = 3.9 V
⎝ 28 Ω ⎟⎠
⎝ Z ⎠
No , ∆VR, rms + ∆VL , rms + ∆VC , rms = 9.0 V + 8.2 V + 3..8 V = 21 V ≠ 10 V
(b)
(c)
21.37
(a)
The power delivered to the resistor is the greatest. No power losses occur in an ideal
capacitor or inductor.
56157_21_ch21_p252-280.indd 268
2
( 25 Ω ) =
3.2 W
The frequency of the station should match the resonance frequency of the tuning circuit. At
resonance, X L = XC or 2π f0 L = 1 2π f0 C , which gives f02 = 1 4π 2 LC or
f0 =
(b)
2
∆Vrms ⎞
⎛ 10 V ⎞
R=⎜
⎝ 28 Ω ⎟⎠
⎝ Z ⎟⎠
⎛
2
R=⎜
Pav = I rms
1
2π LC
=
1
2π
( 3.00 × 10
−6
)
H ( 2.50 × 10
−12
F
)
= 5.81 × 10 7 Hz = 58.1 MHz
Yes, the resistance is not needed. The resonance frequency is found by simply equating
the inductive reactance to the capacitive reactance, which leads to Equation (21.19) as shown
above.
3/19/08 5:49:48 PM
Alternating Current Circuits and Electromagnetic Waves
21.38
(a)
The resonance frequency of an RLC circuit is f0 = 1 2π LC . Thus, the inductance is
L=
(b)
For
1
1
=
4π 2 f02 C 4π 2 ( 9.00 × 10 9 Hz
1
2π LC
(a)
F
)
= 1.56 × 10 −10 H = 156 pH
)
, so C =
1
4π 2 f02 L
f0 = ( f0 )min = 500 kHz = 5.00 × 10 5 Hz
1
4π
2
( 5.00 × 10
5
Hz
) ( 2 .0 × 10
2
−6
H
= 5.1 × 10 −8 F = 51 nF
)
f0 = ( f0 )max = 1600 kHz = 1.60 × 10 6 Hz
C = Cmin =
21.40
−12
)
C = Cmax =
For
) ( 2.00 × 10
2
X L = 2π f0 L = 2π ( 9.00 × 10 9 Hz (1.56 × 10 −10 H = 8.82 Ω
f0 =
21.39
269
1
4π (1.60 × 10 Hz
2
6
)
2
( 2 .0 × 10−6 H
)
= 4.9 × 10 −9 F = 4.9 nF
At resonance, X L = XC so the impedance will be
Z = R 2 + ( X L − XC ) = R 2 + 0 = R = 15 Ω
2
(b)
When X L = XC, we have 2π fL =
f =
(c)
(d)
1
2π LC
1
2π
( 0.20 H ) ( 75 × 10 −6
F
)
= 41 Hz
The current is a maximum at resonance where the impedance has its minimum value
of Z = R.
At f = 60 Hz, X L = 2π ( 60 Hz ) ( 0.20 H ) = 75 Ω , XC =
and Z =
(15 Ω )
Thus, I rms =
56157_21_ch21_p252-280.indd 269
=
1
, which yields
2π fC
2
1
= 35 Ω,
2π ( 60 Hz ) ( 75 × 10 −6 F
)
+ ( 75 Ω − 35 Ω ) = 43 Ω
2
(
∆Vmax
∆Vrms
=
Z
Z
2
)=
150 V
= 2.5 A
2 ( 43 Ω )
3/19/08 5:49:49 PM
270
21.41
Chapter 21
ω 0 = 2π f0 =
1
=
LC
1
(10.0 × 10
)
H (100 × 10 −6 F
−3
= 1 000 rad s
)
Thus, ω = 2ω 0 = 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
ω C ( 2 000 rad s ) (100 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
(10.0 Ω )2 + ( 20.0 Ω − 5.00 Ω )2
2
I rms =
= 18.0 Ω
∆Vrms 50.0 V
=
= 2.78 A
18.0 Ω
Z
The average power is
one period is
2
2
Pav = I rms
R = ( 2.78 A ) (10.0 Ω ) = 77.3 W and the energy converted in
⎞
J⎞ ⎛
2π
⎛ 2π ⎞ ⎛
= 0.243 J
E = Pav ⋅ T = Pav ⋅ ⎜ ⎟ = ⎜ 77.3 ⎟ ⋅ ⎜
⎝ω⎠ ⎝
s ⎠ ⎝ 2 000 rad s ⎟⎠
21.42
The resonance frequency is ω 0 = 2π f0 =
Also, X L = ω L and XC =
(a)
1
ωC
⎛ 1 ⎞
L=
At resonance, XC = X L = ω 0 L = ⎜
⎝ LC ⎟⎠
Thus, Z = R 2 + 0 2 = R, I rms =
and
(b)
At ω =
)
(
and I rms =
∆Vrms 120 V
=
= 4.00 A
Z
30.0 Ω
(
)
( 30.0 Ω )2 + ( 500 Ω − 2 000 Ω )2
= 1 500 Ω
120 V
= 0.080 0 A
1 500 Ω
2
Pav = I rms
R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W
2
At ω =
(
)
(
)
1
1
ω0 ; XL =
X L ω = 250 Ω, XC = 4 XC ω = 4 000 Ω
0
0
4
4
Z = 3 750 Ω, and I rms =
so
3.00 H
= 1 000 Ω
3.00 × 10 −6 F
1
1
ω 0; X L = X L ω = 500 Ω , XC = 2 XC ω = 2 000 Ω
0
0
2
2
2
so
L
=
C
2
2
Pav = I rms
R = ( 4.00 A ) ( 30.0 Ω ) = 480 W
Z = R 2 + ( X L − XC ) =
(c)
1
LC
120 V
= 0.032 0 A
3 750 Ω
2
Pav = I rms
R = ( 0.032 0 A ) ( 30.0 Ω ) = 3.07 × 10 −2 W = 30.7 mW
2
continued on next page
56157_21_ch21_p252-280.indd 270
3/19/08 5:49:50 PM
Alternating Current Circuits and Electromagnetic Waves
(d)
(
At ω = 2 ω 0; X L = 2 X L ω
0
) = 2 000 Ω , X
(e)
)
(
1
XC ω = 500 Ω
0
2
2
Pav = I rms
R = ( 0.080 0 A ) ( 30.0 Ω ) = 0.192 W
2
(
At ω = 4 ω 0; X L = 4 X L ω
0
) = 4 000 Ω, X
Z = 3 750 Ω, and I rms =
so
=
120 V
= 0.080 0 A
1 500 Ω
Z = 1 500 Ω, and I rms =
so
C
271
C
=
)
(
1
XC ω = 250 Ω
0
4
120 V
= 0.032 0 A
3 750 Ω
2
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.
21.43
The maximum output voltage ( ∆Vmax )2 is related to the maximum input voltage ( ∆Vmax )1 by the
N
expression ( ∆Vmax )2 = 2 ( ∆Vmax )1 , where N1 and N 2 are the number of turns on the primary
N1
coil and the secondary coil respectively. Thus, for the given transformer,
( ∆Vmax )2 =
1 500
(170 V) = 1.02 × 10 3 V
250
and the rms voltage across the secondary is ( ∆Vrms )2 =
21.44
(a)
( ∆Vmax )2
2
=
1.02 × 10 3 V
= 721 V .
2
The output voltage of the transformer is
⎛N ⎞
⎛ 1⎞
∆V2,rms = ⎜ 2 ⎟ ∆V1,rms = ⎜ ⎟ (120 V) = 9.23 V
⎝ 13 ⎠
⎝ N1 ⎠
(b)
Assuming an ideal transformer, Poutput = Pinput , and the power delivered to the CD player is
(Pav )2 = (Pav )1 = I1,rms ( ∆V1,rms ) = ( 0.250 A ) (120 V) =
21.45
30.0 W
The power input to the transformer is
(Pav )input = ( ∆V1, rms ) I1, rms = ( 3 600 V) ( 50 A ) = 1..8 × 10 5
For an ideal transformer,
power line is
I 2, rms =
(Pav )input
( ∆V )
2 , rms
=
W
(Pav )ouput = ( ∆V2, rms ) I 2, rms = (Pav )input so the current in the long-distance
1.8 × 10 5 W
= 1.8 A
100 000 V
The power dissipated as heat in the line is then
Plost = I 22, rms Rline = (1.8 A )2 (100 Ω ) = 3.2 × 10 2 W
The percentage of the power delivered by the generator that is lost in the line is
% Lost =
56157_21_ch21_p252-280.indd 271
⎛ 3.2 × 10 2 W ⎞
Plost
× 100% = ⎜
× 100% = 0.18%
Pinput
⎝ 1.8 × 10 5 W ⎟⎠
3/19/08 5:49:51 PM
272
21.46
Chapter 21
(a)
(b)
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.
I1, rms =
(c)
)
(
(
)
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
( ∆V ) I
2 , rms
2 , rms
∆V1, rms
=
( 6.0 V) ( 500 mA )
120 V
= 25 mA
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 ⎞
⎛ 6.0 V ⎞
= 20 turns
N 2 = N1 ⎜ 2 ⎟ = ( 400 ) ⎜
⎝ 120 V ⎟⎠
⎝ ∆V1 ⎠
21.47
(a)
(Pav )output = 0.90 (Pav )input
At 90% efficiency,
Thus, if
(Pav )output = 1 000 kW
the input power to the primary is
21.48
(b)
I1, rms =
(c)
I 2, rms =
(Pav )input
∆V1, rms
=
(Pav )output
(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
=
∆V2, rms
(Pav )input =
1 000 kW 1.0 × 10 6 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
so I rms =
Thus,
(b)
(Pav )transmitted
∆Vrms
=
(Pav )transmitted = ( ∆Vrms ) I rms
5.00 × 10 6 W
= 10.0 A
500 × 10 3 V
2
2
Rline = (10.0 A ) ( 290 Ω ) = 2 .90 × 10 4
(Pav )loss = I rms
W = 29.0 kW
The power input to the line is
(Pav )input = (Pav )transmitted + (Pav )loss = 5.00 × 10 6
W + 2.90 × 10 4 W = 5.03 × 10 6 W
and the fraction of input power lost during transmission is
fraction =
(Pav )loss
( Pav )input
=
2.90 × 10 4 W
= 0.005 77 or 0.577%
5.033 × 10 6 W
continued on next page
56157_21_ch21_p252-280.indd 272
3/19/08 5:49:52 PM
Alternating Current Circuits and Electromagnetic Waves
(c)
273
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.
21.49
From v = λ f , the wavelength is
λ=
v 3.00 × 10 8 m s
=
= 4.00 × 10 6 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.
21.50
d
6.44 × 1018 m ⎛
1y
⎞
2
=
⎜
⎟ = 6.80 × 10 y
c 3.00 × 10 8 m s ⎝ 3.156 × 10 7 s ⎠
(a)
t=
(b)
From Table C.4 (in Appendix C of the textbook), the average Earth-Sun distance is
d = 1.496 × 1011 m, giving the transit time as
t=
(c)
Also from Table C.4, the average Earth-Moon distance is d = 3.84 × 10 8 m, giving the time
for the round trip as
t=
21.51
56157_21_ch21_p252-280.indd 273
)
8
2 d 2 ( 3.84 × 10 m
=
= 2.56 s
c
3.00 × 10 8 m s
The amplitudes of the electric and magnetic components of an electromagnetic wave are related
by the expression Emax Bmax = c; thus the amplitude of the magnetic field is
Bmax =
21.52
d 1.496 × 1011 m ⎛ 1 min ⎞
=
⎟ = 8.31 min
⎜
c 3.00 × 10 8 m s ⎝ 60 s ⎠
330 V m
Emax
=
= 1.10 × 10 −6 T
3.00 × 10 8 m s
c
c=
1
=
µ0 ∈0
1
or
c = 2 .998 × 10 8 m s
( 4π × 10
−7
N ⋅s C
2
2
) ( 8.854 × 10
−12
C2 N ⋅ m 2
)
3/19/08 5:49:52 PM
274
21.53
Chapter 21
(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
fred =
Red:
c
3.00 × 10 8 m s
=
= 4.55 × 1014 Hz
λred
660 × 10 −9 m
and
fIR =
Infrared:
(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
21.54
c
3.00 × 10 8 m s
=
= 3.19 × 1014 Hz
λIR
940 × 10 −9 m
=
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 log10 ( 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
CHBO2 =
− log10 ( 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 =
− log10 ( 0.76 )
εL
[2]
Dividing equation [2] by equation [1] gives the ratio of deoxygenated hemoglobin to oxygenated
hemoglobin in the blood as
log10 ( 0.76 )
CHB
=
= 0.25
CHBO2 log10 ( 0.33)
or
CHB = 0.25CHBO2
Since all the hemoglobin in the blood is either oxygenated or deoxygenated, it is necessary that
CHB + CHBO2 = 1.00 , and we now have 0.25CHBO2 + CHBO2 = 1.0 . The fraction of hemoglobin that
is oxygenated in this blood is then
CHBO2 =
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.
21.55
From Intensity =
2
Emax Bmax
c Bmax
E
and max = c, we find Intensity =
2 µ0
Bmax
2 µ0
Thus,
Bmax =
and
56157_21_ch21_p252-280.indd 274
2 µ0
( Intensity ) =
c
2 ( 4π × 10 −7 T ⋅ m A
3.00 × 10 m s
8
)
) (1 340
)
W m 2 = 3.35 × 10 −6 T
)
Emax = Bmax c = ( 3.35 × 10 −6 T ( 3.00 × 10 8 m s = 1.01 × 10 3 V m
3/19/08 5:49:53 PM
Alternating Current Circuits and Electromagnetic Waves
21.56
(a)
275
To exert an upward force on the disk, the laser beam should be aimed vertically upward,
striking the lower surface of the disk. To just levitate the disk, the upward force exerted on
the disk by the beam should equal the weight of the disk.
The momentum that electromagnetic radiation of intensity I, incident normally on a perfectly
reflecting surface of area A, delivers to that surface in time ∆t is given by Equation (21.30)
as ∆p = 2U c = 2 ( IA∆t ) c . Thus, from the impulse-momentum theorem, the average force
exerted on the reflecting surface is F = ∆p ∆t = 2 IA c . Then, to just levitate the surface,
F = 2 IA c = mg and the required intensity of the incident radiation is I = mgc 2 A .
(b)
(c)
21.57
(
)(
(
)(
)
5.00 × 10 −3 kg 9.80 m s 2 3.00 × 10 8 m s
mgc mgc
=
=
= 1.46 × 10 9 W m 2
2
−2
2 A 2π r 2
2π 4.00 × 10 m
Propulsion by light pressure in a significant gravity field is impractical because of the enormous power requirements. In addition, no material is perfectly reflecting, so the absorbed
energy would melt the reflecting surface.
I=
)
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 × 10 8 m s
21.58
At Earth’s location, the wave fronts of the solar radiation are spheres whose radius is the SunEarth distance. Thus, from Intensity =
Pav = ( Intensity ) ( 4π r 2 ) = ⎛⎜ 1 340
⎝
21.59
From λ f = c , we find f =
21.60
λ=
21.61
(a)
=
Pav
4π r 2
, the total power is
2
W⎞⎡
4π (1.49 × 1011 m ⎤ = 3.74 × 10 26 W
2⎟ ⎣
⎦
m ⎠
)
c 3.00 × 10 8 m s
=
= 5.45 × 1014 Hz
λ
5.50 × 10 −7 m
For the AM band,
λmax =
c
fmax
c
fmin
=
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
=
3.00 × 10 8 m s
= 2 .78 m
108 × 10 6 Hz
=
3.00 × 10 8 m s
= 3.4 m
88 × 10 6 Hz
For the FM band,
λmin =
λmax =
56157_21_ch21_p252-280.indd 275
A
c 3.00 × 10 8 m s
=
= 11.0 m
f 27.33 × 10 6 Hz
λmin =
(b)
Pav
c
fmax
c
fmin
3/19/08 5:49:54 PM
276
21.62
Chapter 21
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 × 10 8 m s
and that for the sound wave is
ts =
ds
vsound
=
3.0 m
= 8.7 × 10 −3 s = 8.7 ms
343 m s
Thus, the radio listeners hear the news 8.4 ms beffore the studio audience because radio waves
travel so much faster than sound waves.
21.63
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
2.66 × 10 −26 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 infrareed region .
21.64
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⎞
∆f = fO − fS = fS ⎜ ⎟ = ( 6.0 × 1014 Hz ⎜
= 3.6 × 1011 Hz
⎝ c⎠
⎝ 3.0 × 10 8 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
21.65
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 ⎞ ⎛ 0.278 m s ⎞
⎛ u⎞
= − 1.6 × 10 7 Hz
∆f = fO − fS = − fS ⎜ ⎟ = − ( 4.3 × 1014 Hz ⎜
⎝ c⎠
⎝ 3.0 × 10 8 m s ⎟⎠ ⎜⎝ 1 km h ⎟⎠
)
The frequency you will detect will be
fO = fS + ∆f = 4.3 × 1014 Hz − 1.6 × 10 7 Hz = 4.299 999 84 × 1014 Hz
56157_21_ch21_p252-280.indd 276
3/19/08 5:49:54 PM
Alternating Current Circuits and Electromagnetic Waves
21.66
277
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 =
3.00 × 10 8 m s
c
=
= 5.17 × 1014 Hz
λS
580 × 10 −9 m
and the frequency the driver claims that she observed is
fO =
3.00 × 10 8 m s
c
=
= 5.36 × 1014 Hz
λO
560 × 10 −9 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 ⎞
⎞
− 1 = 1.1 × 10 7 m s
u = c ⎜ O − 1⎟ = ( 3.00 × 10 8 m s ⎜
⎝ 5.17 × 1014 Hz ⎟⎠
⎠
⎝ fS
)
21.67
(a)
At f = 60.0 Hz, the reactance of a 15.0-µ F capacitor is
XC =
1
1
=
= 177 Ω
2π fC 2π ( 60.0 Hz ) (15.0 × 10 −6 F
)
and the impedance of this RC circuit is
Z = R 2 + XC2 =
( 50.0 Ω )2 + (177 Ω )2
= 184 Ω
∆Vrms 1.20 × 10 2 V
=
= 0.652 A = 652 mA
184 Ω
Z
(b)
I rms =
(c)
After addition of an inductor in series with the resistor and capacitor, the impedance of the
circuit is Z ′ = R 2 + ( X L − XC ) = ( 50.0 Ω ) + ( 2π fL − 177 Ω ) . To reduce the current
to one-half the value found above, the impedance of the circuit must be doubled to a value
of Z ′ = 2 (184 Ω ) = 368 Ω . Thus,
2
2
2
( 50.0 Ω )2 + ( 2π fL − 177 Ω )2 = ( 368 Ω )2
or
2π fL = 177 Ω ±
( 368 Ω )2 − ( 50.0 Ω )2
= 177 Ω ± 365 Ω
Since the inductance cannot be negative, the potential solution associated with the lower
sign must be discarded, leaving
L=
56157_21_ch21_p252-280.indd 277
177 Ω + 365 Ω
= 1.44 H
2π ( 60.0 Hz )
3/19/08 5:49:55 PM
278
21.68
Chapter 21
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
( ∆E ∆t ) , the absorbed energy is
P
absorbed. Since I = av =
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 or
Z = R 2 + ( XC ) = R 2 + ( 2π f C )
2
21.69
−2
( 200 Ω )2 + ⎡⎣ 2π ( 60 Hz ) ( 5.00 × 10 −6
=
2
⎛ 120 V ⎞
⎛ ∆V ⎞
2
R = ⎜ rms ⎟ R = ⎜
Thus, Pav = I rms
⎝ 5.7 × 10 2 Ω ⎟⎠
⎝ Z ⎠
and
~ 10 6 J
)
F ⎤⎦
2
−2
= 5.7 × 10 2 Ω
( 200 Ω ) = 8.9 W = 8.9 × 10 −3 kW
cost = ∆E ⋅ ( rate) = Pav ⋅ ∆t ⋅ ( rate )
)
= ( 8.9 × 10 −3 kW ( 24 h ) ( 8.0 cents kWh ) = 1.7 cents
X L = ω L , so ω = X L L
21.70
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 =
LC
( 2π f0 )2
From ω 0 = 2π f0 =
)
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
21.71
R=
( ∆V )DC
I DC
=
12 .0 V
= 19.0 Ω
0.630 A
Z = R 2 + ( 2π f L ) =
2
Thus, L =
56157_21_ch21_p252-280.indd 278
Z 2 − R2
=
2π f
∆Vrms
24.0 V
=
= 42.1 Ω
I rms
0.570 A
( 42.1 Ω )2 − (19.0 Ω )2
2π ( 60.0 Hz )
= 9.97 × 10 −2 H = 99.7 mH
3/19/08 5:49:56 PM
Alternating Current Circuits and Electromagnetic Waves
21.72
(a)
c 3.00 × 10 8 m s
=
= 1.0 × 1010 Hz. Therefore, the resoλ
3.00 × 10 −2 m
1
nance frequency of the circuit is f0 =
= 1.0 × 1010 Hz, giving
2π LC
The required frequency is f =
C=
(b)
C=
1
( 2π f0 )
2
L
=
( 2π × 10 Hz
10
)
1
2
( 400 × 10−12 H
)
= 6.3 × 10 −13 F = 0.63 pF
∈0 A ∈0 2
=
, so
d
d
=
21.73
279
C ⋅d
=
∈0
( 6.3 × 10
−13
)
F (1.0 × 10 −3 m
8.85 × 10 −12 C2 N ⋅ m
) = 8.4 × 10
)
−3
m = 8.4 mm
)
(c)
XC = X L = ( 2π f0 ) 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
3.00 × 10 8 m s
c
)
)
(b)
−6
−16
Emax Bmax ( 0.20 × 10 V m ( 6.7 × 10 T
Intensity =
=
= 5.3 × 10 −17 W m 2
−7
2 µ0
2 ( 4π × 10 T ⋅ m A
(c)
Pav = ( Intensity ) ⋅ A = ( Intensity ) ⎢
)
⎡πd2 ⎤
⎥
⎣ 4 ⎦
⎡ π ( 20.0 m )2 ⎤
−14
= ( 5.3 × 10 −17 W m 2 ⎢
⎥ = 1.7 × 10 W
4
⎢⎣
⎥⎦
)
21.74
(a)
Z=
∆Vrms 12 V
=
= 6.0 Ω
2 .0 A
I rms
(b)
R=
∆VDC 12 V
=
= 4.0 Ω
3.0 A
I DC
From
Z = R 2 + X L2 = R 2 + ( 2π f L ) , we find
2
L=
56157_21_ch21_p252-280.indd 279
Z 2 − R2
=
2π f
( 6.0 Ω )2 − ( 4.0 Ω )2
2π ( 60 Hz )
= 1.2 × 10 −2 H = 12 mH
3/19/08 5:49:56 PM
280
21.75
Chapter 21
(a)
From Equation (21.30), the momentum imparted in time ∆t to a perfectly reflecting sail of
area A by normally incident radiation of intensity I is ∆p = 2U c = 2 ( IA∆t ) c.
From the impulse-momentum theorem, the average force exerted on the sail is then
Fav =
aav =
(c)
1
From ∆x = v0 t + at 2 , with v0 = 0 , the time is
2
(a)
)
Fav 0.536 N
=
= 8.93 × 10 −5 m s 2
m 6 000 kg
(b)
t=
21.76
)
4
2
2
∆p 2 ( IA∆t ) c 2 IA 2 (1 340 W m ( 6.00 × 10 m
=
=
=
= 0.536 N
∆t
∆t
c
3.00 × 10 8 m s
2 ( ∆x )
=
aav
2 ( 3.84 × 10 8 m
) = ( 2 .93 × 10 s) ⎛ 1 d
⎜⎝
8.64 × 10
6
8.93 × 10 −5 m s 2
4
⎞
⎟ = 33.9 d
s⎠
The intensity of radiation at distance r from a point source, which radiates total power P ,
is I = P A = P 4π r 2 . Thus, at distance r = 2.0 in from a cell phone radiating a total power
of P = 2.0 W = 2.0 × 10 3 mW, the intensity is
I=
2.0 × 10 3 mW
4π ⎡⎣( 2.0 in ) ( 2.54 cm 1 in ) ⎤⎦
2
= 6.2 mW cm 2
This intensity is 24% higher than the maximuum allowed leakage from a microwave
at this distance of 2.0 inches.
(b)
If when using a Blue toothheadset (emitting 2.5 mW of power) in the ear at distance
rh = 2.0 in = 5.1 cm from center of the brain, the cell phone (emitting 2.0 W of power) is
located in the pocket at distance rp = 1.0 m = 1.0 × 10 2 cm from the brain, the total radiation intensity at the brain is
I total = I phone + I headset =
or
56157_21_ch21_p252-280.indd 280
I total = 1.6 × 10 −2
2.0 × 10 3 mW
4π (1.0 × 10 cm
2
)
2
+
2.5 mW
mW
−2 mW
+ 7.6 × 10 −3
2 = 1.6 × 10
2
cm
cm 2
4π ( 5.1 cm )
mW
mW
mW
+ 7.6 × 10 −3
= 2.4 × 10 −2
= 0.024 mW cm 2
cm 2
cm 2
cm 2
3/19/08 5:49:57 PM