physics chapter 21 solutions

Chapter 21

ALTERNATING CURRENT

Conceptual Questions

3. The current in an ac circuit oscillates between a maximum value +I and a minimum value –I. The amplitude of oscillation I is the peak current. For a sinusoidal current, all of the charge moving in one direction over the first half of a period passes in the opposite direction during the second half and therefore yields zero average current. The rms (root-mean-squared) current Irms is the value of the dc current that would dissipate energy in a resistor at the same average rate as an ac current of amplitude I.

4. The principal advantage of using a standard household voltage of 120 V rms is that it is “safer” in that less current would flow through a person if they were to touch a live wire. However, the use of 120 V rms means that a greater current must be supplied to an appliance to deliver the same amount of power, thereby increasing the power loss to resistive heating in the wires.

9. The 500 W reported on the appliance is the average power consumption, given by av rms rms cos .P I V The power factor cos = R/Z depends on the capacitance, inductance, and resistance of the circuit. Only if the power factor is one, or R = Z, would the average power consumption be 600 W.

11. The second component is a capacitor. As the frequency decreases the capacitive reactance increases causing the current to decrease.

12. An ac voltage with amplitude of 170 V has an rms value of about 120 V. For a lightbulb, which is essentially just a resistor, the power factor is one and the consumed power is given by Vrms

2/R. If connected instead to a 170-V dc power supply, the bulb would consume a power of V2/R, burn a lot brighter, and maybe even burn out. A dc voltage equal to the rms ac voltage, or about 120 V, would consume the same power and cause the bulb to burn with the same brightness.

13. Inserting a coil of wire with a soft iron core into the circuit in series with a light bulb would add inductance and increase the overall impedance of the circuit. The amplitude of the current in the circuit would be reduced and the bulb would be dimmed. Moving the soft-iron core in and out of the coil would vary the amount of dimming by changing the inductance of the coil.

14. With two sinusoidal waves, the difference in the argument of the sinusoidal functions is the phase difference. It tells us in a sense how far the waves are from having maxima or minima at the same time—i.e., from being in phase. For a phase difference of /2 rad, one wave is at maximum when the other is at zero. The figure below shows a sketch of i(t) and given that the current leads the voltage by /2 rad. C ( )v t

i(t)vC(t)

t

The current leads the voltage by / 2 rad.!

17. Since the power dissipated in a light bulb is given by Pav = Vrms2/R, the bulb would consume considerably less

power with an rms voltage of 120 V (in fact 1/4 as much) and would burn less brightly.

809

Chapter 21: Alternating Current College Physics

Problems

3. Strategy 1500 W is the average power dissipated by the heater and av rms rms.P I V Solution Calculate I, the peak current.

avrms

rms

1500 W2 2 2 18120 V

PI IV

A

7. (a) Strategy 4200 W is the average power drawn by the heater. Solution Compute the rms current.

avrms

rms

4200 W 35 A120 V

PIV

(b) Strategy Since 2 22 1 and , .P V R R R P V Form a proportion.

Solution Calculate 2.P

2 22 2

21 1

105, so (4200 W) 3.2 kW .120

P V PP V

9. Strategy Use the definition of rms. Solution Compute the amplitude.

m rms2 2(4.0 V) 5.% % 7 V

The instantaneous sinusoidal emf oscillates between 5.7 V and 5.7 V .

10. (a) Strategy and Use Eq. (21-4). Solution Calculate R.

av 1200 WP rms 120 V.V

2 2rms

av

(120 V) 12 1200 W

VRP

(b) Strategy Use Ohm’s law. Solution Compute the rms current.

rmsrms

120 V 10 A12

VIR

(c) Strategy The maximum power is twice the average power. Solution Compute the maximum instantaneous power.

max av2 2(1200 W) 2.4 kWP P

15. Strategy The reactance is C 1 ( )X C where 2 .f

Solution

810

College Physics Chapter 21: Alternating Current

(a) Solve for f.

C 3 6C

1 1 1, so 60.0 Hz .2 2 2 (6.63 10 )(0.400 10 F)

X ffC X C

(b) Compute the reactance.

C 61 1 13.3 k

2 (30.0 Hz)(0.400 10 F)X

C

16. (a) Strategy The reactance is C 1 ( )X C where 2 .f Solution Compute the reactance.

C 61 1 12.7 k

2 (50.0 Hz)(0.250 10 F)X

C

(b) Strategy The rms current is related to the reactance by rms rms C.V I X Solution Solve for rms.I

6rmsrms rms

C2 (50.0 Hz)(0.250 10 F)(220 V) 17 mAVI CV

X

17. Strategy where rms rms CV I X C 1 ( ).X C

Solution Solve for the capacitance, C. 3

rms rmsrms rms C

rms

2.3 10 A, so 53 nF .2 2 (60.0 Hz)(115 V)

I IV I X CC f V

C

Vrms

21. Strategy Use Eqs. (18-15), (21-6), and (21-7).

Solution

(a) Find I.

6 6 61 2 3

eq 1 1 1 1 1 1Ceq 2.0 10 F 3.0 10 F 6.0 10 F

2 2 (6300 Hz)(12.0 V) 0.475 AC C C

V f VI C VX

Find V as a function of C.

C0.475 A

2 (6300 Hz)IV IXC C

C1 C2 C3

The table below gives the results for each capacitor.

( F)C (V)V

2.0 6.0

3.0 4.0

6.0 2.0 (b) From part (a), 0.48 A .I

811


27. Strategy where rms rms LV I X L 2 .X L fL

Solution Solve for f. rms rms L rms

rms3

rms

(2 ), so151.0 V 7.33 kHz .

2 2 (0.820 A)(4.00 10 H)

V I X I fLVfI L

f ?

4.00 mH

28. Strategy Use Eqs. (21-9) and (21-10).

Solution

(a) Form a proportion.

L 2 ,V IX I L fIL so eq eq 1 2

2 .2

i i i iV fIL L LV fIL L L L

Thus,

1 2

25(5.0 V) V H0.10 H 0.50 H 3.0

i ii i

L LV VL L

L is the peak

voltage across inductor i.

0.10 H

5.0 V

0.50 H

The peak voltages are given in the table below.

L (H) V (V)

0.10 0.83

0.50 4.2 (b) Compute the peak current.

L eq 1 2

5.0 V 11 mA2 ( ) 2 (126 Hz)(0.10 H 0.50 H)

V V VIX L f L L

33. Strategy and rms rmsI Z% 2 2L .Z R X Only the resistance dissipates power, so use 2

av rms .P I R

Solution Find rms.I

2 avav rms rms, so .PP I R I

R

Find f .

25.0 mH

110 V

25.0 "

812


2 2 2 2 2rms rms rms L rms

22 2 2rms

2rms

22 2 2rms

2rms

22rms

rms2 2 2

2 2rms rms

avav

1

1 1 1 (110 V) (25.0 ) (25.0 )2 2 2 (0.0250 H) 50.0 W

470 Hz

I Z I R X I R L

R LI

L RI

RL I

Rf R RL L PP R

%%

%

%

% % 2

34. Strategy 2 2L 30.0 Z R X where L 2 .X L fL

Solution Find L.

2 22 22 2 2 2 2 2 2 2 2

L(30.0 ) (20.0 )

(2 ) , so 71.2 mH .2 2 (50.0 Hz)

Z RR X R L R f L Z Lf

39. Strategy Use Eq. (21-14b) with L 0,X and Eq. (21-7).

Solution Find the impedance.

2 2 2 2C 2 2 2 2 6 2

1 1(300.0 ) 500 4 (159 Hz) (2.5 10 F)

Z R X RC

!

2.5 µF300.0 "

44. Strategy av rms rms cosP I % is the average power.

Solution Find the power factor and the phase difference.

(a) av

rms rms

240 Wcos 0.71(2.80 A)(120 V)

PI %

(b) 1 1av

rms rms

240 Wcos cos 44(2.80 A)(120 V)

PI %

45. Strategy The average power is given by av rms rms cosP I % where rms rmsI Z% and cos .R Z

Solution Find the average power dissipated.

!

220 "

0.15 mH

8.0 µF

813


6

2 2rms rms

av rms rms rms 22 1

2

22 3 1

2 (2500 Hz)(8.0 10 F)

( 2)cos

(12 V) (220 ) 0.33 W

2 (220 ) 2 (2500 Hz)(0.15 10 H)

C

R RP I RZ Z Z R L

% % %% %

50. Strategy Use Eqs. (21-14b) and (21-16).

Solution Find the impedance.

6

22 2 2

L C

22 3 1

2 (1590 Hz)(5.00 10 F)

1( )

(12.5 ) 2 (1590 Hz)(3.60 10 H) 20.3

Z R X X R LC

Find the power factor.

!

12.5 "

3.60 mH

5.00 µF

12.5 cos 0.61720.26

RZ

Find the phase difference. 1 1 12.5 cos cos 51.9

20.26 RZ

55. (a) Strategy Use Eq. (21-18). Solution Compute the resonant frequency.

0 61 1 745 rad s

(0.300 H)(6.00 10 F)LC

(b) Strategy At resonance, L C.X X Use Eq. (21-14a). Solution Compute the resistance.

m 440 V 790 0.560 A

R ZI

%

!

R

6.00 µF

0.300 H

(c) Strategy At resonance, the voltages across the capacitor and inductor are 180 out of phase and equal in magnitude, so they cancel. Solution The peak voltage across the resistor is R m 440 V .V % Across the inductor, the peak voltage is

L L 0 60.300 H(0.560 A) 125 V .

6.00 10 FIL LV IX I L I

CLC

Since L C,X X C L 125 V .V V

57. Strategy Use Eq. (21-18).

814


Solution Find the resonant frequency.

0 31 1 22.4 rad s ,

(40.0 10 H)(0.0500 F)LC or 0

0 3.56 Hz .2

f

69. Strategy Use Eqs. (20-10) and (21-4).

Solution

(a) Calculate the turns ratio. 3

2 2

1 1

240 10 V 570420 V

NN

%%

(b) Calculate the rms current using the turns ratio. 32

1 21

570(60.0 10 A) 34 ANI IN

(c) Calculate the average power.

av rms rms (0.0600 A)(240,000 V) 14 kWP I V

70. (a) Strategy Model the coil as an RL series circuit. Use Eq. (21-14b) with C 0.X Solution Calculate the impedance.

2 2 2 2 2 2 2 2 2L (120 ) 4 (60.0 Hz) (12.0 H) 4.53 kZ R X R L

(b) Strategy Use Eq. (21-14a). Solution Calculate the current.

rmsrms

110 V 24 mA4530

IZ

%

72. Strategy Use Eqs. (21-6) and (21-7). Solution Compute the rms current.

6rmsrms rms

C2 (60.0 Hz)(0.025 10 F)(110 V) 1.0 mAI C

X% %

73. Strategy Use Eq. (21-7). Solution Find the capacitance.

C1 ,XC

so C

1 1 49 F .2 (520 Hz)(6.20 )

CX

76. Strategy The maximum current flows at the resonance frequency. Use Eq. (21-18). Solution Compute the resonant frequency.

0 0 0 12

1 1 12 , so 11 kHz .2 2 (0.44 H)(520 10 F)

f fLC LC

815


79. (a) Strategy Use Eqs. (21-7) and (21-10). Solution Find the reactances.

C 3 61 1 20

(1.0 10 rad s)(50.0 10 F)X

C

3L (1.0 10 rad s)(0.0350 H) 35 X L

!

20.0 "

35.0 mH

50.0 µF

(b) Strategy Use Eq. (21-14b). Solution Find the impedance.

2 2 2 2L C( ) (20.0 ) (35 20 ) 25 Z R X X

(c) Strategy Use Eq. (21-14a) with rms values. Solution Find the rms current.

rmsrms

100.0 V 4.0 A25

IZ

%

(d) Strategy Use the definition of rms. Solution Find the current amplitude (peak current).

rms2 2(4.0 A) 5.7 AI I

(e) Strategy Use Eq. (21-16). Solution Find the phase angle.

1 1 20.0 cos , so cos cos 37 .25

R RZ Z

(f) Strategy Use Eqs. (21-6) and (21-9) and Ohm’s law (using rms values). Solution Find the rms voltage for each circuit element.

R rms rms (4.0 A)(20.0 ) 80 V ,V I R L rms rms L (4.0 A)(35 ) 140 V , andV I X

C rms rms C (4.0 A)(20 ) 80 V .V I X

(g) Strategy and Solution Since L C,X X the inductor dominates the capacitor, so the current lags the voltage since the current through an inductor lags the voltage across it.

(h) Strategy Use the values obtained for the rms voltages in part (f). Solution Draw the phasor diagram.

VL

VR

VC

VL VC37°

!m

816


81. (a) Strategy Use Eq. (21-14b). Solution Find the impedance.

22 2 2

L C1( )Z R X X R LC

!

12.0 "

15.2 mH

0.26 µF

22 3 3

3 61(12.0 ) 2 (2.50 10 Hz)(15.2 10 H) 13

2 (2.50 10 Hz)(0.26 10 F)

(b) Strategy Use Eq. (21-14a) with rms values. Solution Find the rms current.

rmsrms

240 V 18 A13.5

IZ

%

(c) Strategy Use Eq. (21-16). Solution Find the phase angle.

1 1 12.0 cos , so cos cos 27 .13.5

R RZ Z

(d) Strategy Use Eqs. (21-7) and (21-10). Form a proportion. Solution Compare the reactances.

C2 2 3 2 6

L

1 ( ) 1 1 14 (2.50 10 Hz) (0.0152 H)(0.26 10 F)

X CX L LC

Since C L,X X the capacitor dominates the inductor, so the current leads the voltage since the current through a capacitor leads the voltage across it.

(e) Strategy to four significant figures. Use Eqs. (21-6) and (21-9) and Ohm’s law (using rms values). Solution Find the rms voltage for each circuit element.

rms 17.83 AI

R rms rms (17.83 A)(12.0 ) 210 VV I R 3

L rms rms L rms (17.83 A)2 (2.50 10 Hz)(0.0152 H) 4.3 kVV I X I L

rmsC rms rms C 3 6

17.83 A 4.4 kV2 (2.50 10 Hz)(0.26 10 F)

IV I XC

82. (a) Strategy Model the coil as an RL series circuit. Use Eq. (21-16).

Solution Find the power factor.

2 2 2 2 2 2 2 2 2L

450 cos 0.95(450 ) 4 (9.55 Hz) (2.47 H)

R R RZ R X R L

(b) Strategy Use Eq. (21-14b). Solution Find the impedance.

2.47 H450 "

!

Electromagnet

817


2 2 2 2(450 ) 4 (9.55 Hz) (2.47 H) 470 Z

(c) Strategy Use Eq. (21-14a). Solution Find the peak current.

3m 2.0 10 V 4.2 A

474 I

Z%

(d) Strategy Use Eq. (21-17). Solution Find the average power.

3av rms rms

1 1cos cos (4.2 A)(2.0 10 V)(0.95) 4.0 kW2 2

P I V IV

818

physics chapter 21 solutions

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