chapter 21 alternating current circuits and ... current circuits and electromagnetic waves ... 1.20...

317

Chapter 21

Alternating Current Circuits and Electromagnetic Waves

Problem Solutions

21.1 (a) ,rms rms 8.00 A 12.0 96.0 VRV I R

(b) ,max ,rms2 2 96.0 V 136 VR RV V

(c) max rms2 2 8.00 A 11.3 AI I

(d) 22

av rms 8.00 A 12.0 768 WI RP

21.2 (a) 2 2

,max ,rms2 2 1.20 10 V 1.70 10 VR RV V

(b)

222 2

2 2rms rmsav rms

av

1.20 10 V 2.40 10

60.0 W

V VI R R

RP

P

(c) Because 2

rms

av

VR

P (see above), if the bulbs are designed to operate at the same

voltage, 2the 1.00 10 W will have the lower resistance.

21.3 The meters measure the rms values of potential d ifference and current. These are

maxrms

100 V70.7 V

2 2

VV , and rms

rms

70.7 V2.95 A

24.0

VI

R

318 CHAPTER 21

21.4 All lamps are connected in parallel with the voltage source, so rms 120 VV for each

lamp. Also, the current is rms av rmsI P V and the resistance is

rms rmsR V I .

1, rms 2, rms

150 W1.25 A

120 VI I and 1 2

120 V96.0

1.25 AR R

3, rms

100 W0.833 A

120 VI and 3

120 V144

0.833 AR

21.5 The total resistance (series connection) is 1 2 8.20 10.4 18.6 eqR R R , so the

current in the circuit is

rmsrms

15.0 V0.806 A

18.6 eq

VI

R

The power to the speaker is then 22

av rms 0.806 A 10.4 6.76 WspeakerI RP

21.6 The general form of the generator voltage is max sinv V t , so by inspection

(a) ,max 170 VRV and (b)

60 rad s30.0 Hz

2 2 radf

(c) ,max

,rms

170 V120 V

2 2

R

R

VV

(d) ,rms

rms

120 V6.00 A

20.0

RVI

R

(e) max rms2 2 6.00 A 8.49 AI I

(f) 2

av rms 6.00 A 20.0 720 WI RP

(g) The argument of the sine function has units of rad s s radianst .

At 0.0050 st , the instantaneous current is

170 V 170 V

sin 60 rad s 0.005 0 s sin 0.94 rad 6.9 A20.0 20.0

vi

R

Alternating Current Circuits and Electromagnetic Waves 319

21.7 1

2CX

fC, so its units are

1 1 Volt VoltOhm

1 Sec Farad 1 Sec Coulomb Volt Coulomb Sec Amp

21.8 rms

max rms rms

2= 2 = = 2 2

C

VI I V f C

X

(a) 6

max = 2 120 V 2 60.0 Hz 2.20 10 C/V =0.141 A 141 mAI

(b) 6

max = 2 240 V 2 50.0 Hz 2.20 10 C/V =0.235 A 235 mAI

21.9 rmsrms rms= =2

C

VI f C V

X, so

2rms

6rms

0.30 A4.0 10 Hz

2 2 4.0 10 F 30 V

If

C V

21.10 (a) 6

1 1221

2 2 60.0 Hz 12.0 10 FCX

fC

(b) ,rms

rms

36.0 V0.163 A

221

C

C

VI

X

(c) max rms2 2 0.163 A 0.231 AI I

(d) 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 maxim um.

Consequently, the maximum charge and the maximum current do not occur at the

same time.

21.11 rms maxrms max= =2 2

2C

V VI f C f C V

X

so 5

max

0.75 A1.7 10 F 17 F

2 60 Hz 170 V 2

IC

f V

320 CHAPTER 21

21.12 (a) By inspection, ,max 98.0 VCV , so

,max

,rms

98.0 V69.3 V

2 2

C

C

VV

(b) Also by inspection, 80 rad s , so 80 rad s

40.0 Hz2 2 rad

f

(c) maxrms

0.500 A= = 0.354 A

2 2

II

(d) ,max

max

98.0 V196

0.500 A

C

C

VX

I

(e) 1 1

2CX

fC C, so

51 12.03 10 F 20.3 F

80 rad s 196 C

CX

21.13 2LX f L , and from I

Lt

, we have t

LI

. The units of self inductance

are then Volt sec

amp

tL

I. The units of inductive reactance are given by

1 Volt sec Volt

Ohmsec Amp Amp

LX f L

21.14 (a) 32 2 80.0 Hz 25.0 10 H 12.6 LX f L

(b) ,rms

rms

78.0 V6.19 A

12.6

L

L

VI

X

(c) max rms2 2 6.19 A 8.75 AI I

21.15 The required inductive reactance is ,max ,max

max rms2

L L

L

V VX

I I

and the needed inductance is

,max

3

rms

4.00 V0.750 H

2 2 2 2 2 300.0 Hz 2.00 10 A

LLVX

Lf f I

21.16 Given: 21.20 10 V sin 30 and 0.500 HLv t L


(a) By inspection, 30 rad s , so 30 rad s

15.0 Hz2 2

f

(b) Also by inspection, 2

,max 1.20 10 VLV , so that

2

,max

,rms

1.20 10 V84.9 V

2 2

L

L

VV

(c) 2 30 rad s 0.500 H 47.1 LX fL L

(d) ,rms

rms

84.9 V1.80 A

47.1

L

L

VI

X

(e) max rms2 2 1.80 A 2.55 AI I

(f) The phase d ifference between the voltage across an inductor and the current

through the inductor is 90L, so the average power delivered to the inductor is

,av rms ,rms rms ,rmscos cos 90 0L L L LI V I VP

(g) When a sinusoidal voltage with a peak value ,maxLV 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 max ,maxL LI V X .

However, the current lags behind the voltage in phase by a quarter-cycle or

2 radians . Thus, if the voltage is given by ,max sinL Lv V t , the current as a

function of time is max sin 2i I t . In the case of the given inductor, the current

through it will be 2.55 A sin 30 2i t .

(h) When 1.00 Ai , we have: sin 30 2 1.00 A 2.55 At , or

1 130 2 sin 1.00 A 2.55 A sin 0.392 0.403 radt

and 22 rad 0.403 rad2.09 10 s 20.9 ms

30 rad st

322 CHAPTER 21

21.17 From BL N I (see Section 20.6 in the textbook), the total flux through the coil is

, totalB BN L I where B

is the flux through a single turn on the coil. Thus,

max,total maxmax

rms 22 2 120 V

0.450 T m2 2 60.0 Hz

B

L

VL I L

X

VL

fL

21.18 (a) The applied voltage is max sin 80.0 V sin 150v V t t , so we have that

max 80.0 VV and 2 150 rad sf . The impedance for the circuit is

2 2 22 2 22 1 2 1L CZ R X X R fL fC R L C

or

2

2 3

6

140.0 150 rad s 80.0 10 H 57.5

150 rad s 125 10 FZ

(b) maxmax

80.0 V1.39 A

57.5

VI

Z

21.19 6

1 166.3

2 2 60.0 Hz 40.0 10 FCX

f C

2 2 22 50.0 0 66.3 83.0 L CZ R X X

(a) rmsrms

30.0 V0.361 A

Z 83.0

VI

(b) , rms rms 0.361 A 50.0 18.1 VRV I R

(c) , rms rms 0.361 A 66.3 23.9 VC CV I X

(d) 1 1 0 66.3

tan tan 53.050.0

L CX X

R

so, the voltage lags behind the current by 53°


21.20 (a) 32 2 50.0 Hz 400 10 H 126 LX f L

6

1 1719

2 2 50.0 Hz 4.43 10 FCX

f C

2 2 22 500 126 719 776 L CZ R X X

max max 0.250 A 776 194 VV I Z

(b) 1 1 126 719

tan tan 49.9500

L CX X

R

Thus, the current leads the voltage by 49.9

21.21 (a) 32 2 240 Hz 2.50 H 3.77 10 LX f L

3

6

1 12.65 10

2 2 240 Hz 0.250 10 FCX

f C

22 22 3900 3.77 2.65 10 1.44 kL CZ R X X

(b) maxmax 3

140 V0.097 2 A

1.44 10

VI

Z

(c)

3

1 13.77 2.65 10

tan tan 51.2900

L CX X

R

(d) 0 , so the voltage leads the current

21.22 2 2 60.0 Hz 0.100 H 37.7 LX fL

6

1 1265

2 2 60.0 Hz 10.0 10 FCX

fC

2 2 22 50.0 37.7 265 233 L CZ R X X

(a) ,rms rms 2.75 A 50.0 138 VRV I R (e)

324 CHAPTER 21

(b) ,rms rms 2.75 A 37.7 104 VL LV I X

(c) rms 2.75 A 265 729 VC CV I X

(d) rms rms 2.75 A 233 641 V I Z

21.23 2 2 60.0 Hz 0.400 H 151 LX f L

6

1 1884

2 2 60.0 Hz 3.00 10 FCX

f C

2 2 22 60.0 151 884 735 RLC L CZ R X X

and rmsrms

RLC

VI

Z

(a) 2

0 733 LC L C L CZ X X X X

rms, rms rms

90.0 V733 89.8 V

735 LC LC LC

RLC

VV I Z Z

Z

(b) 2 2 22 0 60.0 884 886 RC CZ R X

rms, rms rms

90.0 V886 108 V

735 RC RC RC

RLC

VV I Z Z

Z

21.24 32 2 60 Hz 2.8 H 1.1 10 LX fL

2 222 3 3 31.2 10 1.1 10 0 1.6 10 L CZ R X X

(a) maxmax 3

170 V0.11 A

1.6 10

VI

Z

(b) 3 2

, max max 0.11 A 1.2 10 1.3 10 VRV I R

3 2

, max max 0.11 A 1.1 10 1.2 10 VL LV I X


(c) When the instantaneous current is a maximum maxi I , the instantaneous voltage

across the resistor is 2

max , max 1.3 10 VR Rv iR I R V . The instantaneous

voltage across an inductor is always 90° or a quarter cycle out of phase with the

instantaneous current. Thus, when maxi I , 0Lv .

Kirchhoff’ s loop rule always applies to the instantaneous voltages around a closed

path. Thus, for this series circuit, source R Lv v v and at this instant when

maxi I

we have 2

source max 0 1.3 10 Vv I R

(d) When the instantaneous current i is zero, the instantaneous voltage across the

resistor is 0Rv iR . Again, the instantaneous voltage across an inductor is a

quarter cycle out of phase with the current. Thus, when 0i , we must have

2

, max 1.2 10 VL Lv V . Then, applying Kirchhoff’ s loop rule to the

instantaneous voltages around the series circuit at the instant when 0i gives

2

source , max0 1.2 10 VR L Lv v v V

21.25 8

12

1 11.33 10

2 2 60.0 Hz 20.0 10 FCX

f C

2 2

2 2 3 8 850.0 10 1.33 10 1.33 10 RC CZ R X

and secondary 5rms

rms 8

5 000 V3.76 10 A

1.33 10 RC

VI

Z

Therefore, 5 3

, rms rms 3.76 10 A 50.0 10 1.88 Vb bV I R

21.26 (a) 6

1 188.4

2 2 60.0 Hz 30.0 10 FCX

f C

(b) 2 2 22 2 20 60.0 88.4 107 C CZ R X R X

(c) 2

maxmax

1.20 10 V1.12 A

107

VI

Z

326 CHAPTER 21

(d) The phase angle in this RC circuit is

1 1 0 88.4

tan tan 55.860.0

L CX X

R

Since 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) 2 2 50.0 Hz 0.250 H 78.5 LX f L

(b) 3

6

1 11.59 10 1.59 k

2 2 50.0 Hz 2.00 10 FCX

f C

(c) 2 2 22 3150 78.5 1590 1.52 10 1.52 kL CZ R X X

(d) 2

maxmax 3

2.10 10 V0.138 A 138 mA

1.52 10

VI

Z

(e) 1 1 78.5 1590

tan tan 84.3150

L CX X

R

(f) ,max max 0.138 A 150 20.7 VRV I R

,max max 0.138 A 78.5 10.8 VL LV I X

3

,max max 0.138 A 1.59 10 219 VC CV I X

21.28 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 2 .

The voltage across the capacitor lags the current by 90° or 2 rad ians, which

corresponds to an argument of 0 in the sine function and a voltage of 0Cv . Similarly,

the voltage across the inductor leads the current by 90° or 2 rad ians, corresponding

to an argument in the sine function of rad ians or 180°, giving a voltage of 0Lv .


21.29 2 2 50.0 Hz 0.185 H 58.1 LX f L

6

1 149.0

2 2 50.0 Hz 65.0 10 FCX

f C

2 2 22 40.0 58.1 49.0 41.0 ad L CZ R X X

and max

rmsrms

2 150 V2.59 A

41.0 2ad ad

VVI

Z Z

(a) 40.0 abZ R , so rms rms 2.59 A 40.0 104 VababV I Z

(b) 58.1 bc LZ X , and rms rms 2.59 A 58.1 150 VbcbcV I Z

(c) 49.0 cd CZ X , and rms rms 2.59 A 49.0 127 VcdcdV I Z

(d) 9.10 bd L CZ X X , so rms rms 2.59 A 9.10 23.6 VbdbdV I Z

21.30 3

6

1 11.1 10

2 2 60 Hz 2.5 10 FCX

fC

2 222 3 3 31.2 10 0 1.1 10 1.6 10 L CZ R X X

(a) maxmax 3

170 V0.11 A

1.6 10

VI

Z

(b) 3 2

, max max 0.11 A 1.2 10 1.3 10 VRV I R

3 2

, max max 0.11 A 1.1 10 1.2 10 VC CV I X

328 CHAPTER 21

(c) When the instantaneous current i is zero, the instantaneous voltage across the

resistor is 0Rv iR . The instantaneous voltage across a capacitor is always 90°

or a quarter cycle out of phase with the instantaneous current. Thus, when 0i ,

2

, max 1.2 10 VC Cv V

and 6 2 -42.5 10 F 1.2 10 V 3.0 10 C 300 C C Cq C v

Kirchhoff’ s loop rule always applies to the instantaneous voltages around a closed

path. Thus, for this series circuit, source 0R Cv v v , and at this instant when

0i , we have 2

source , max0 1.2 10 VC Cv v V

(d) When the instantaneous current is a maximum max( )i I , the instantaneous voltage

across the resistor is 2

max , max 1.3 10 VR Rv iR I R V . Again, the

instantaneous voltage across a capacitor is a quarter cycle out of phase with the

current. Thus, when maxi I , we must have 0Cv and 0C Cq C v . Then,

applying Kirchhoff’ s loop rule to the instantaneous voltages around the series

circuit at the instant when maxi I and 0Cv gives

2

source source , max0 1.3 10 VR C R Rv v v v v V

21.31 (a) rms

rms

104 V208

0.500 A

VZ

I

(b) 2

av rmsI RP gives av

2 2

rms

10.0 W40.0

0.500 AR

I

P

(c) 2 2

LZ R X , so 2 22 2 208 40.0 204 LX Z R

and 204

0.541 H2 2 60.0 Hz

LXL

f

21.32 Given max sin 90.0 V sin 350v V t t , observe that max 90.0 VV and

350 rad s . Also, the net reactance is 2 1 2 1L CX X fL fC L C .


(a) 6

1 1350 rad s 0.200 H 44.3

350 rad s 25.0 10 FL CX X L

C

so the impedance is 2 2 22 50.0 44.3 66.8 L CZ R X X

(b) rms maxrms

2 90.0 V0.953 A

Z 2 66.8

V VI

Z

(c) The phase d ifference between the applied voltage and the current is

1 1 44.3

tan tan 41.550.0

L CX X

R

so the average power delivered to the circuit is

av

maxrms rms rms

90.0 Vcos cos 0.953 A cos 41.5 45.4 W

2 2

VI V IP

21.33 Please see the textbook for the statement of Problem 21.21 and the answers for that

problem. There, you should find that max max140 V, 0.097 2 A, and 51.2V I . The

average power delivered to the circuit is then

max maxav rms rms

0.097 2 A 140 Vcos cos cos 51.2 4.26 W

22 2

I VI VP

21.34 The rms current in the circuit is

rmsrms

160 V2.00 A

Z 80.0

VI

and the average power delivered to the circuit is

22

av rms rms rms ,rms rms rms rmscos 2.00 A 22.0 88.0 WRI V I V I I R I RP

21.35 (a) 2

av rms rms rms rms , rmsRI R I I R I VP , so avrms

, rms

14 W0.28 A

50 VR

IV

P

Thus, , rms 2

rms

50 V1.8 10

0.28 A

RVR

I

330 CHAPTER 21

(b) 2 2

LZ R X , which yields

2 22

2 2 2 2 2rms

rms

90 V1.8 10 2.7 10

0.28 AL

VX Z R R

I

and 22.7 10

0.72 H2 2 60 Hz

LXL

f

21.36 32 2 600 Hz 6.0 10 H 23 LX fL

-6

1 111

2 2 600 Hz 25 10 FCX

fC

2 2 22 25 23 11 28 L CZ R X X

(a) rms, rms rms

10 V25 8.9 V

28 R

VV I R R

Z

rms, rms rms

10 V23 8.2 V

28 L L L

VV I X X

Z

rms, rms rms

10 V11 3.9 V

28 C C C

VV I X X

Z

, rms , rms , rms No , 9.0 V 8.2 V 3.8 V 21 V 10 VR L CV V V

(b) The power delivered to the resistor is the greatest. No power losses occur in an ideal

capacitor or inductor.

(c)

2 2

2 rmsav rms

10 V25 3.2 W

28

VI R R

ZP

21.37 (a) The frequency of the station should match the resonance frequency of the tuning

circuit. At resonance, L CX X or

0 02 1 2f L f C , which gives 2 2

0 1 4f LC or

7

06 12

1 15.81 10 Hz 58.1 MHz

2 2 3.00 10 H 2.50 10 Ff

LC


(b) 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.

21.38 (a) The resonance frequency of a RLC circuit is 0 1 2f LC . Thus, the inductance is

10

2 2 22 9 12

0

1 11.56 10 H 156 pH

4 4 9.00 10 Hz 2.00 10 FL

f C

(b) 9 10

02 2 9.00 10 Hz 1.56 10 H 8.82 LX f L

21.39 0

1

2f

LC, so

2 2

0

1

4C

f L

For 5

0 0 min500 kHz 5.00 10 Hzf f

8

max 22 5 6

15.1 10 F 51 nF

4 5.00 10 Hz 2.0 10 HC C

For 6

0 0 max1600 kHz 1.60 10 Hzf f

9

min 22 6 6

14.9 10 F 4.9 nF

4 1.60 10 Hz 2.0 10 HC C

21.40 (a) At resonance, L CX X so the impedance will be

22 2 0 15 L CZ R X X R R

(b) When L CX X , we have

12

2fL

fC which yields

6

1 141 Hz

2 2 0.20 H 75 10 Ff

LC

(c) The current is a maximum at resonance where the impedance has its minimum

value of Z R .

332 CHAPTER 21

(d) At 60 Hzf , 2 60 Hz 0.20 H 75 LX , 6

135

2 60 Hz 75 10 FCX ,

and 2 2

15 75 35 43 Z

Thus, max

rmsrms

2 150 V2.5 A

2 43

VVI

Z Z

21.41 0 0

3 6

1 12 1000 rad s

10.0 10 H 100 10 Ff

LC

Thus, 02 2 000 rad s

32 000 rad s 10.0 10 H 20.0 LX L

6

1 15.00

2 000 rad s 100 10 FCX

C

2 2 22 10.0 20.0 5.00 18.0 L CZ R X X

rmsrms

50.0 V2.78 A

18.0

VI

Z

The average power is 22

av rms 2.78 A 10.0 77.3 WI RP

and the energy converted in one period is

av av

2 J 277.3 0.243 J

s 2 000 rad sE TP P

21.42 The resonance frequency is 0 0

12 f

LC

Also, LX L and

1CX

C


(a) At resonance, 0 -6

1 3.00 H1000

3.00 10 FC L

LX X L L

CLC

Thus, 2 20Z R R , rmsrms

120 V4.00 A

30.0

VI

Z

and 22

av rms 4.00 A 30.0 480 WI RP

(b) At 0

1

2;

0

1500

2L LX X ,

0

2 2 000 C CX X

2 2 22 30.0 500 2 000 1500 L CZ R X X

and rms

120 V0.080 0 A

1500 I

so 22

av rms 0.080 0 A 30.0 0.192 WI RP

(c) At 0

1

4;

0

1250

4L LX X ,

0

4 4 000 C CX X

3750 Z , and rms

120 V0.032 0 A

3 750 I

so 22 2

av rms 0.032 0 A 30.0 3.07 10 W 30.7 mWI RP

(d) At 02 ;

0

2 2 000 L LX X , 0

1500

2C CX X

1500 Z , and rms

120 V0.080 0 A

1500 I

so 22

av rms 0.080 0 A 30.0 0.192 WI RP

334 CHAPTER 21

(e) At 04 ;

0

4 4 000 L LX X , 0

1250

4C CX X

3750 Z , and rms

120 V0.032 0 A

3 750 I

so 22 2

av rms 0.032 0 A 30.0 3.07 10 W 30.7 mWI RP

The power delivered to the circuit is a maximum w hen 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 max 2V is related to the maximum input voltage

max 1V by the expression 2

max max2 11

NV V

N, where

1N and 2N are the number of

turns on the primary coil and the secondary coil respectively. Thus, for the given

transformer,

3

max 2

1500170 V 1.02 10 V

250V

and the rms voltage across the secondary is 3

max 2rms 2

1.02 10 V721 V

2 2

VV .

21.44 (a) The output voltage of the transformer is

22,rms 1,rms

1

1120 V 9.23 V

13

NV V

N

(b) Assuming an ideal transformer, output inputP P , and the power delivered to the CD

player is

av av 1,rms 1,rms2 10.250 A 120 V 30.0 WI VP P


21.45 The power input to the transformer is

5

av 1, rms 1, rmsinput3 600 V 50 A 1.8 10 WV IP

For an ideal transformer, av 2, rms 2, rms avouput input

V IP P so the current in the long-

distance power line is

5

av input

2, rms

2, rms

1.8 10 W1.8 A

100 000 VI

V

P

The power d issipated as heat in the line is then

22 2

lost 2, rms line 1.8 A 100 3.2 10 WI RP

The percentage of the power delivered by the generator that is lost in the line is

2

lost

5

input

3.2 10 W% Lost 100% 100% 0.18%

1.8 10 W

P

P

21.46 (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, av avinput ouputP P or 1, rms 1, rms 2, rms 2, rmsV I V I so the

current in the primary will be

2, rms 2, rms

1, rms

1, rms

6.0 V 500 mA25 mA

120 V

V II

V

(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, 2 1 2 1V V N N . Thus, the number of

turns needed on the secondary coil of this step down transformer is

22 1

1

6.0 V400 20 turns

120 V

VN N

V

336 CHAPTER 21

21.47 (a) At 90% efficiency, av avoutput input

0.90P P

Thus, if av output1000 kWP

the input power to the primary is av output 3

av input

1000 kW1.1 10 kW

0.90 0.90

PP

(b) 3 6

av input 2

1, rms

1, rms 1, rms

1.1 10 kW 1.1 10 W3.1 10 A

3 600 VI

V V

P

(c) 6

av output 3

2, rms

2, rms 1, rms

1000 kW 1.0 10 W8.3 10 A

120 VI

V V

P

21.48 4 54.50 10 m 6.44 10 m 290 lineR

(a) The power transmitted is av rms rmstransmittedV IP

so 6

av transmittedrms 3

rms

5.00 10 W10.0 A

500 10 VI

V

P

Thus, 22 4

av rms lineloss10.0 A 290 2.90 10 W 29.0 kWI RP

(b) The power input to the line is

6 4 6

av av avinput transmitted loss5.00 10 W 2.90 10 W 5.03 10 WP P P

and the fraction of input power lost during transmission is

4

av loss

6

av input

2.90 10 W0.005 77 or 0.577%

5.03 10 Wfraction

P

P


(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 th is current is

rmsrms max

line

4 500 V15.5 A

290

VI

R

The maximum input power is then

input rms rms maxmax

3 44.50 10 V 15.5 A 6.98 10 W 6.98 kW

V IP

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 f , the wavelength is

8

63.00 10 m s4.00 10 m 4 000 km

75 Hz

v

f

The required length of the antenna is then,

4 1000 kmL , or about 621 miles. Not very practical at all.

21.50 (a) 18

2

8 7

6.44 10 m 1 y6.80 10 y

3.00 10 m s 3.156 10 s

dt

c

(b) From Table C.4 (in Appendix C of the textbook), the average Earth-Sun d istance is 111.496 10 md , giving the transit time as

11

8

1.496 10 m 1 min8.31 min

60 s3.00 10 m s

dt

c

(c) Also from Table C.4, the average Earth-Moon d istance is 83.84 10 md , giving the

time for the round trip as

8

8

2 3.84 10 m22.56 s

3.00 10 m s

dt

c

338 CHAPTER 21

21.51 The amplitudes of the electric and magnetic components of an electromagnetic wave are

related by the expression max maxE B c , thus the amplitude of the magnetic field is

6maxmax 8

330 V m1.10 10 T

3.00 10 m s

EB

c

21.52 7 2 2 12 2 2

0 0

1 1

4 10 N s C 8.854 10 C N mc

or 82.998 10 m sc

21.53 (a) The frequency of an electromagnetic wave is f c , where c is the speed of light,

and is the wavelength of the wave. The frequencies of the two light sources are

then

Red: 8

14

red -9

red

3.00 10 m s4.55 10 Hz

660 10 m

cf

and

Infrared: 8

14

IR -9

IR

3.00 10 m s3.19 10 Hz

940 10 m

cf

(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

0.33f iI I . Hence, we must have

max,

max,

0.33 0.57f f

i i

E I

E I


21.54 If 0I 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 10 0log I I CL where 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

10

HBO2

log 0.33C

L [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

10

HB

log 0.76C

L [2]

Divid ing equation [2] by equation [1] gives the ratio of deoxygenated hemoglobin to

oxygenated hemoglobin in the blood as

10HB

HB HBO2

HBO2 10

log 0.760.25 or 0.25

log 0.33

CC C

C

Since all the hemoglobin in the blood is either oxygenated or deoxygenated , it is

necessary that HB HBO2 1.00C C , and we now have

HBO2 HBO2 0.25 1.0C C . The fraction

of hemoglobin that is oxygenated in this blood is then

HBO2

1.00.80 or 80%

1.0 0.25C

Someone with only 80% oxygenated hemoglobin in the blood is probably in serious

trouble needing supplemental oxygen immediately.

21.55 From max max

02

E BIntensity and max

max

Ec

B, we find

2

max

02

c BIntensity

Thus,

7

2 60max 8

2 4 10 T m A21340 W m 3.35 10 T

3.00 10 m sB Intensity

c

and 6 8 3

max max 3.35 10 T 3.00 10 m s 1.01 10 V mE B c

340 CHAPTER 21

21.56 (a) To exert an upward force on the d isk, the laser beam should be aimed vertically

upward , striking the lower surface of the d isk. To just levitate the d isk, the upward

force exerted on the d isk by the beam should equal the weight of the d isk.

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 2 2p U c IA t c . Thus, from the impulse-momentum

theorem, the average force exerted on the reflecting surface is 2F p t IA c .

Then, to just levitate the surface, 2F IA c mg and the required intensity of the

incident rad iation is 2I mgc A .

(b)

3 2 8

9 2

2 2-2

5.00 10 kg 9.80 m s 3.00 10 m s1.46 10 W m

2 2 2 4.00 10 m

mgc mgcI

A r

(c) Propulsion by light pressure in a significant gravity field is impractica l because of

the enormous power requirements. In addition, no material is perfectly reflecting,

so the absorbed energy would melt the reflecting surface.

21.57 The d istance between adjacent antinodes in a standing wave is 2

Thus, 2 6.00 cm 12.0 cm 0.120 m , and

9 80.120 m 2.45 10 Hz 2.94 10 m sc f

21.58 At Earth’ s location, the wave fronts of the solar rad iation are spheres whose radius is the

Sun-Earth d istance. Thus, from av av

24Intensity

A r

P P, the total power is

2

2 11 26

av 2

W4 1340 4 1.49 10 m 3.74 10 W

mIntensity rP

21.59 From f c , we find 8

14

7

3.00 10 m s5.45 10 Hz

5.50 10 m

cf

21.60 8

6

3.00 10 m s11.0 m

27.33 10 Hz

c

f


21.61 (a) For the AM band,

8

min 3

max

3.00 10 m s188 m

1600 10 Hz

c

f

8

max 3

min

3.00 10 m s556 m

540 10 Hz

c

f

(b) For the FM band,

8

min 6

max

3.00 10 m s2.78 m

108 10 Hz

c

f

8

max 6

min

3.00 10 m s3.4 m

88 10 Hz

c

f

21.62 The transit time for the radio wave is

3

4

8

100 10 m3.33 10 s 0.333 ms

3.00 10 m s

RR

dt

c

and that for the sound wave is

3

sound

3.0 m8.7 10 s 8.7 ms

343 m s

ss

dt

v

Thus, the radio listeners hear the news 8.4 ms before the studio audience because radio

waves travel so much faster than sound waves.

342 CHAPTER 21

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 2f k m . Thus, the resonance frequency of the C=O

double bond will be

13

26

oxygenatom

1 1 2 800 N m5.2 10 Hz

2 2 2.66 10 kg

kf

m

and the light with this frequency has wavelength

8

6

13

3.00 10 m s5.8 10 m 5.8 m

5.2 10 Hz

c

f

The infrared region of the electromagnetic spectrum ranges from max 1 mm down to

min 700 nm 0.7 m . Thus, the required wavelength falls within the infrared region .

21.64 Since the space station and the ship are moving toward one another, the frequency after

being Doppler shifted is 1O Sf f u c , so the change in frequency is

5

14 11

8

1.8 10 m s6.0 10 Hz 3.6 10 Hz

3.0 10 m sO S S

uf f f f

c

and the frequency observed on the spaceship is

14 11 146.0 10 Hz 3.6 10 Hz 6.003 6 10 HzO Sf f f

21.65 Since you and the car ahead of you are moving away from each other (getting farther

apart) at a rate of 120 km h 80 km h 40 km hu , the Doppler shifted frequency you

will detect is 1O Sf f u c , and the change in frequency is

14 7

8

40 km h 0.278 m s4.3 10 Hz 1.6 10 Hz

1 km h3.0 10 m sO S S

uf f f f

c

The frequency you will detect will be

14 7 144.3 10 Hz 1.6 10 Hz 4.299 999 84 10 HzO Sf f f


21.66 The driver was driving toward the warning lights, so the correct form of the Doppler

shift equation is 1O Sf f u c . The frequency emitted by the yellow warning light is

8

14

9

3.00 10 m s5.17 10 Hz

580 10 mS

S

cf

and the frequency the driver claims that she observed is

8

14

9

3.00 10 m s5.36 10 Hz

560 10 mO

O

cf

The speed with which she would have to approach the light for the Doppler effect to

yield this claimed shift is

14

8 7

14

5.36 10 Hz1 3.00 10 m s 1 1.1 10 m s

5.17 10 Hz

O

S

fu c

f

21.67 (a) At 60.0 Hzf , the reactance of a 15.0- F capacitor is

6

1 1177

2 2 60.0 Hz 15.0 10 FCX

fC

and the impedance of this RC circuit is

2 22 2 50.0 177 184 CZ R X

(b) 2

rmsrms

1.20 10 V0.652 A 652 mA

Z 184

VI

344 CHAPTER 21

(c) After addition of an inductor in series with the resistor and capacitor, the

impedance of the circuit is 2 2 22 50.0 2 177 L CZ R X X fL . To

reduce the current to one-half the value found above, the impedance of the circuit

must be doubled to a value of 2 184 368 Z . Thus,

2 2 2

50.0 2 177 368 fL

or 2 2

2 177 368 50.0 177 365 fL

Since the inductance cannot be negative, the potential solution associated with the

lower sign must be d iscarded, leaving

177 365

1.44 H2 60.0 Hz

L

21.68 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 21.7 m 0.3 m cos30 0.4 m .

The intensity the radiation at Earth’ s surface is surface incoming0.6I I and only 50% of this is

absorbed. Since avE t

IA A

P, the absorbed energy is

surface incoming

2 2 5 6

0.5 0.5 0.6

0.5 0.6 1340 W m 0.4 m 3 600 s 6 10 J or ~10 J

E I A t I A t

21.69 2 22 2

22 6 2

2

200 2 60 Hz 5.0 10 F 5.7 10

CZ R X R f C

Thus,

2 2

2 3rmsav rms 2

120 V200 8.9 W 8.9 10 kW

5.7 10

VI R R

ZP

and av

38.9 10 kW 24 h 8.0 cents kWh 1.7 cents

cost E rate t rateP


21.70 LX L , so

LX L

Then, 1 1

C

L

XC C X L

which gives

12 8.0 L CL X X C C or 296 L C [1]

From 0 0

12 f

LC, we obtain

2

0

1

2LC

f

Substituting from Equation [1], this becomes 2 2

2

0

196

2C

f

or 5

2 2

0

1 12.6 10 F 26 F

2 96 2 2 000 Hz 96 C

f

Then, from Equation [1],

2 5 396 2.6 10 F 2.5 10 H 2.5 mHL

21.71 DC

DC

12.0 V19.0

0.630 A

VR

I

22 rms

rms

24.0 V2 42.1

0.570 A

VZ R f L

I

Thus,

2 22 2

242.1 19.0

9.97 10 H 99.7 mH2 2 60.0 Hz

Z RL

f

21.72 (a) The required frequency is 8

10

2

3.00 10 m s1.0 10 Hz

3.00 10 m

cf . Therefore, the

resonance frequency of the circuit is 10

0

11.0 10 Hz

2f

LC, giving

13

2 210 12

0

1 16.3 10 F 0.63 pF

2 2 10 Hz 400 10 HC

f L

346 CHAPTER 21

(b) 2

0 0AC

d d, so

13 3

3

12 2

0

6.3 10 F 1.0 10 m8.4 10 m 8.4 mm

8.85 10 C N m

C d

(c) 10 12

02 2 1.0 10 Hz 400 10 H 25 C LX X f L

21.73 (a) max

max

Ec

B, so

6

16maxmax 8

0.20 10 V m6.7 10 T

3.00 10 m s

EB

c

(b)

6 16

17 2max max

70

0.20 10 V m 6.7 10 T5.3 10 W m

2 2 4 10 T m A

E BIntensity

(c) 2

av

2

17 2 14

4

20.0 m5.3 10 W m 1.7 10 W

4

dIntensity A IntensityP

21.74 (a) rms

rms

12 V6.0

2.0 A

VZ

I

(b) DC

DC

12 V4.0

3.0 A

VR

I

From 22 2 2 2LZ R X R f L , we find

2 22 2

26.0 4.0

1.2 10 H 12 mH2 2 60 Hz

Z RL

f


21.75 (a) From Equation 21.30, the momentum imparted in time t to a perfectly reflecting

sail of area A by normally incident rad iation of intensity I is 2 2p U c IA t c .

From the impulse-momentum theorem, the average force exerted on the sail is then

2 4 2

av 8

2 1340 W m 6.00 10 m2 20.536 N

3.00 10 m s

IA t cp IAF

t t c

(b) 5 2avav

0.536 N8.93 10 m s

6 000 kg

Fa

m

(c) From 2

0

1

2x v t at , with

0 0v , the time is

8

6

5 2 4

av

2 3.84 10 m2 1 d2.93 10 s 33.9 d

8.93 10 m s 8.64 10 s

xt

a

21.76 (a) The intensity of rad iation at d istance r from a point source, which radiates total

power P , is 24I A rP P . Thus, at d istance 2.0 inr from a cell phone

radiating a total power of 32.0 W 2.0 10 mWP , the intensity is

3

2

2

2.0 10 mW6.2 mW cm

4 2.0 in 2.54 cm 1 inI

This intensity is 24% higher than the maximum allowed leakage from a microwave at this

d istance of 2.0 inches.

(b) If when using a Blue tooth headset (emitting 2.5 mW of power) in the ear at d istance

2.0 in 5.1 cmhr from center of the brain, the cell phone (emitting 2.0 W of power)

is located in the pocket at d istance 21.0 m 1.0 10 cmpr from the brain, the total

rad iation intensity at the brain is

3

2 3

total phone headset 2 2 2 22

2.0 10 mW 2.5 mW mW mW1.6 10 7.6 10

cm cm4 5.1 cm4 1.0 10 cmI I I

or 2 3 2 2

total 2 2 2

mW mW mW1.6 10 7.6 10 2.4 10 0.024 mW cm

cm cm cmI

chapter 21 alternating current circuits and ... current circuits and electromagnetic waves ... 1.20...

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