Solutions to Homework Questions 7

Physics 112 Homework 7 (solutions) (2004 Fall) 

Solutions to Homework Questions 7 

Chapt21, Problem-3: An AC power supply produces a 

maximum voltage of !V max = 100 V. This power supply is connected 

to a 24.0-" resistor, and the current and resistor voltage are measured 

with an ideal AC ammeter and an ideal AC voltmeter, as shown in 

Figure P21.3. What does each meter read Recall that an ideal ammeter 

has zero resistance and an ideal voltmeter has infinite resistance. 

Solution: 

The meters measure the rms values of potential difference 

and current. These are 

!V rms = !V max 

2 

and I rms = !V rms 

R 

= 100 V 

2 = 70.7 V , 

= 70.7 V 

24.0 " = 2 .95 A 

Chapt21, Problem-5: An audio amplifier, represented by the 

AC source and the resistor R in Figure P21.5, delivers alternating voltages at 

audio frequencies to the speaker. If the source puts out an alternating voltage 

of 15.0 V (rms), resistance R is 8.20 " and the speaker is equivalent to a 

resistance of 10.4 " what is the time-averaged power delivered to the speaker 

Solution: 

The total resistance (series connection) I 

s R eq = R 1 + R 2 = 8.20 ! +10.4 ! = 18.6 ! , 

so the current in the circuit is 

I rms = !V rms 

= 15.0 V 

R eq 

18.6 " = 0.806 A 

2 

The power to the speaker is then ! av = I rms R spea ker = ( 0.806 A ) 2 ( 10.4 ")= 6.76 W 

Chapt21, Problem-11: What value of capacitor must be inserted in a 60-Hz circuit in series with a generator 

of 170 V maximum output voltage to produce an rms current output of 0.75 A 

Solution: 


X C 

so C = 

# 

=2" fC !V max 

% 

$ 2 

I 

! f "V max 

& 

( = " fC (!V max ) 2 , 

' 

( ) 2 = 0.75 A 

! ( 60 Hz) ( 170 V) 2 = 1.7 # 10$5 F = 

17 µF 

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Chapt21, Problem-16: A 2.40-µF capacitor is connected across an alternating voltage with an rms value of 

9.00 V. The rms current in the circuit is 25.0 mA. (a) What is the source frequency (b) If the capacitor is replaced by an 

ideal coil with an inductance of 0.160 H, what is the rms current in the coil 

Solution: 

(a) 


X C 

= (!V rms )( 2" fC), so 

I 

f = rms 

( )2" C = 25.0 # 10 $3 A 

( 9.00 V)2" 2.40 #10 $6 F 

!V rms 

(b) 


X L 

( ) 

= !V rms 

2" fL = 9.00 V 

2" 184 Hz 

= 184 Hz 

( )( 0.160 H ) = 4.86 # 10$2 A = 

48.6 mA 

Chapt21, Problem-20: A series AC circuit contains the following components: R = 150 !, L = 250 mH, 

C = 2.00 µF, and a generator with "V max = 210 V operating at 50.0 Hz. Calculate the (a) inductive reactance, (b) capacitive 

reactance, (c) impedance, (d) maximum current, and (e) phase angle between the current and the generator voltage. 

Solution: 

X L = 2! fL= 2! ( 50.0 Hz) ( 0.250 H)= 78.5 ! 

(a) 

1 

(b) X C = 

2! fC = 1 

2! 50.0 Hz 

( ) = 1.59 " 103 $ = 

( ) 2 .00" 10 #6 F 

(c) Z = R 2 + ( X L ! X C ) 2 = ( 150 ") 2 + ( 78.5 "!1.59 #10 3 ") 2 = 

(d) I max = !V max 

Z = 210 V 

138 mA 

1.52 "10 3 = 0.138 A = 

# 

(e) ! = tan "1 # X L 

" X C & # 

% ( = tan "1 78.5 )"1.59* 103 )& 

$ R ' 

% 

( = " 84.3° , 

$ 150 ) ' 

or 

the current leads the voltage by 84.3° 

1.59 k! 

1.52 k! 

Chapt21, Problem-25: A person is working near the 

secondary of a transformer, as shown in Figure P21.25. The primary 

voltage is 120 V (rms) at 60.0 Hz. The capacitance C s , which is the 

stray capacitance between the hand and the secondary winding, is 20.0 pF. 

Assuming the person has a body resistance to ground of R b = 50.0 k!, 

determine the rms voltage across the body. (Hint: Redraw the circuit with 

the secondary of the transformer as a simple AC source. 

Solution: 

1 

X C = 

2! fC = 1 

2! 60.0 Hz 

( ) = 1.33" 108 $ 

( ) 2 + ( 1.33! 10 8 ") 2 = 1.33 !10 8 " 

( ) 20.0 "10 #12 F 

Z RC = R 2 + X C 2 = 50.0 ! 10 3 " 

( ) rms 

20.5 H 

and I rms = !V 200 V 

secondary 5 000 V 

= 

Z RC 

1.33 "10 8 # = 3.76 "10 100 Hz 

$5 A 

35.0 ! 

Therefore, !V b,rms = I rms R b = ( 3.76 "10 #5 A) ( 50.0 "10 3 $ )= 1.88 V 

C 

2


Chapt21, Problem-33: An RLC circuit is used to tune a radio to an FM station broadcasting at 88.9 MHz. 

The resistance in the circuit is 12.0 ! and the capacitance is 1.40 pF. What inductance should be present in the circuit 

Solution: 

The resonance frequency of the circuit should match the broadcast frequency of the station. 

1 

f 0 = 

2! LC gives L = 1 

4! 2 f 2 0 

C , 

1 

or L = 

2 .29 µH 

4! 2 88.9 "10 6 Hz 

( ) 2 ( 1.40 " 10 #12 F) = 2 .29" 10#6 H = 

Chapt21, Problem-35: The AM band extends from approximately 500 kHz to 1 600 kHz. If a 

2.0-µH inductor is used in a tuning circuit for a radio, what are the extremes that a capacitor must reach in 

order to cover the complete band of frequencies 

Solution: 

1 

f 0 = 

2! LC , so C = 1 

4! 2 f 2 0 

L 

For f 0 = f 0 

For f 0 = f 0 

( ) min 

= 500 kHz = 5.00 !10 5 Hz , 

C = C max = 

1 

( ) 2 ( 2.0 " 10 #6 H) = 5.1" 10#8 F = 

4! 2 5.00 "10 5 Hz 

( ) max 

= 1600 kHz = 1.60! 10 6 Hz , 

C = C min = 

1 

( ) 2 ( 2.0 " 10 #6 H) = 4.9 " 10#9 F = 

4! 2 1.60 " 10 6 Hz 

51 nF 

4.9 nF 

Chapt21, Problem-41: A transformer on a pole near a factory steps the voltage down from 3600 V to 120V. 

The transformer is to deliver 1 000 kW to the factory at 90% efficiency. Find (a) the power delivered to the primary, (b) the 

current in the primary, and (c) the current in the secondary. 

Solution: 

(a) At 90% efficiency, (! av ) output 

= 0.90 (! av ) input 

. 

Thus, if (! av ) output 

= 1000 kW , 

( 

the input power to the primary is (! av ) input 

= ! av ) output 1000 kW 

= = 

0.90 0.90 

1.1! 103 kW 

( 

(b) I 1, rms = ! av ) input 

= 1.1 #103 kW 

= 1.1# 106 W 

"V 1, rms 

"V 1, rms 

3600 V = 3.1! 102 A 

( 

(c) I 2, rms = ! av ) output 1000 kW 

= = 1.0 #106 W 

"V 2, rms 

"V 1, rms 

120 V = 8.3 !103 A 

3


Chapt21, Problem-43: The U.S. Navy has long proposed the construction of extremely low-frequency 

(ELF) communications systems; such waves could penetrate the oceans to reach distant submarines. Calculate the length of 

a quarter-wavelength antenna for a transmitter generating ELF waves of frequency 75 Hz. How practical is this 

Solution: 

From v = !f , the wavelength is 

! = v f = 3.00" 108 ms 

75 Hz 

= 4.00" 10 6 m = 4000 km 

The required length of the antenna is then, 

L = ! 4 = 1000 km , and so rather unfeasible. 

Chapt21, Problem-47: A microwave oven is powered by an electron tube called a magnetron that generates 

electromagnetic waves of frequency 2.45 GHz. The microwaves enter the oven and are reflected by the walls. The standing 

wave pattern produced in the oven can cook food unevenly, with hot spots in the food at antinodes and cool spots at nodes, 

so a turntable is often used to rotate the food and distribute the energy. If a microwave oven is used with a cooking dish in 

a fixed position, the antinodes can appear as burn marks on foods such as carrot strips or cheese. The separation distance 

between the burns is measured to be 6.00 cm. Calculate the speed of the microwaves from these data. 

Solution: 

The distance between adjacent antinodes in a standing wave is ! 2 . 

Thus, ! = 2( 6.00 cm )= 12.0 cm = 0.120 m , and 

c = !f = ( 0.120 m )( 2 .45" 10 9 Hz)= 2.94!108 ms 

Chapt21, Problem-50: The eye is most sensitive to light of wavelength 5.50x10 –7 m, which is in the 

green-yellow region of the visible electromagnetic spectrum. What is the frequency of this light 

Solution: 

f = c 3.00 "108 ms 

= 

! 5.50 "10 #7 m = 5.45 !1014 Hz 

Chapt21, Problem-52: A diathermy machine, used in physiotherapy, generates electromagnetic radiation 

that gives the effect of “deep heat” when absorbed in tissue. One assigned frequency for diathermy is 27.33 MHz. What is 

the wavelength of this radiation 

Solution: 

! = c f = 3.00 "108 ms 

27.33 "10 6 Hz = 11.0 m 

Chapt21, Conceptual-6: If the fundamental source of a sound wave is a vibrating object, what is the 

fundamental source of the electromagnetic wave 

Solution: 

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 

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Chapt21, Conceptual-7: In radio transmission, a radio wave serves as a carrier wave and the sound wave is 

superimposed on the carrier wave. In amplitude-modulation (AM) radio, the amplitude of the carrier wave varies according to 

the sound wave. The Navy sometimes uses flashing lights to send Morse code between neighboring ships, a process that has 

similarities to radio broadcasting. Is this AM or FM What is the carrier frequency What is the ignal frequency What is the 

broadcasting antenna What is the receiving antenna 

Solution: 

The flashing of the light according to Morse code is a drastic amplitude modulation – the amplitude is 

changing from a maximum to zero. In this sense, it is similar to the on-and-off binary code used in 

computers and compact discs. The carrier frequency is that of the light, on the order of 10 14 Hz. The 

signal frequency depends on the skill of the signal operator, but it is of the order of 1 Hz, as the light 

flashes on and off. The broadcasting antenna for this modulated signal is the filament of the lightbulb in 

the signal source. The receiving antenna is the eye. 

Chapt21, Conceptual-8: When light (or other electromagnetic radiation) travels across a given region, what 

is it that moves 

Solution: 

As an electromagnetic wave travels through space, the things that move are fluctuations in electric and 

magnetic fields oriented at right angles to one another. These fluctuations in the fields transport energy 

as they travel along. 

Chapt21, Conceptual-10: How can the average valuue of an alternating current be zero, yet the square root 

of the average squared value not be zero 

Solution: 

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. 

Chapt21, Conceptual-13: Radio stations often advertise “instant news”. If whatt they mean is that you 

hear the news at the instant they speak it, is their claim true About how long would it take for a message to travel across 

this country by radio wave, assuming that these waves could travel this great distance and still be detected 

Solution: 

Radio waves move at the speed of light. They can travel around the curved surface of the Earth, bouncing 

between the ground and the ionosphere, which has an altitude that iss small compared to the radius of the 

Earth. The distance across the lower 48 states is approximately 5000 km, requiring a travel time of 

(5x10 6 m)/(3x10 8 m/s) ~10 -2 s. Likewise, radio waves take only 0.07s to travel halfway around the Earth. In 

other wordws, a speech can be heard on the other side of the world (in the form of radio waves) before it 

is heard at the back of the room (in the form of sound waves). 

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Solutions to Homework Questions 7

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