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 Pav I rmsVrms cos I . The power factor cos I = 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 Vrms2/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 S/2 rad, one wave is at maximum when the other is at zero. The figure below shows a sketch of i(t) and vC (t ) given that the current leads the voltage by S/2 rad.

vC(t)

i(t)

t

The current leads the voltage by π / 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 Pav Solution Calculate I, the peak current. § P · § 1500 W · 2 I rms 2 ¨ av ¸ 2¨ I ¸ © 120 V ¹ © Vrms ¹

18 A

7. (a) Strategy 4200 W is the average power drawn by the heater. Solution Compute the rms current. Pav 4200 W I rms 35 A Vrms 120 V (b) Strategy Since P V 2 R and R2

R1 , P v V 2 . Form a proportion.

Solution Calculate P2 .

P2 P1

2

§ V2 · ¨ ¸ , so P2 © V1 ¹

2

§ 105 · ¨ ¸ (4200 W) © 120 ¹

3.2 kW .

9. Strategy Use the definition of rms. Solution Compute the amplitude. %m 2 %rms 2(4.0 V) 5.7 V

The instantaneous sinusoidal emf oscillates between 5.7 V and 5.7 V . 10. (a) Strategy Pav

1200 W and Vrms

Solution Calculate R. 2 Vrms (120 V)2 R Pav 1200 W

120 V. Use Eq. (21-4).

12 :

(b) Strategy Use Ohm’s law. Solution Compute the rms current. Vrms 120 V I rms 10 A R 12 : (c) Strategy The maximum power is twice the average power. Solution Compute the maximum instantaneous power. Pmax

2 Pav

2(1200 W)

15. Strategy The reactance is X C

2.4 kW

1 (ZC ) where Z

2S f .

Solution

810

I rmsVrms .

College Physics

Chapter 21: Alternating Current

(a) Solve for f. 1 XC , so f 2S fC

1 2S X CC

1 2S (6.63 u 10 :)(0.400 u 106 F) 3

(b) Compute the reactance. 1 1 XC Z C 2S (30.0 Hz)(0.400 u 106 F) 16. (a) Strategy The reactance is X C

13.3 k:

1 (ZC ) where Z

Solution Compute the reactance. 1 1 XC Z C 2S (50.0 Hz)(0.250 u 106 F)

2S f .

12.7 k:

(b) Strategy The rms current is related to the reactance by Vrms Solution Solve for I rms . Vrms I rms Z CVrms 2S (50.0 Hz)(0.250 u 106 F)(220 V) XC 17. Strategy Vrms

60.0 Hz .

I rms X C .

17 mA

1 (ZC ).

I rms X C where X C

Solution Solve for the capacitance, C. I rms I rms Vrms I rms X C , so C 2S f Vrms ZC

C 3

2.3 u 10 A 2S (60.0 Hz)(115 V)

53 nF . Vrms

21. Strategy Use Eqs. (18-15), (21-6), and (21-7). Solution (a) Find I. V I X Ceq

ZCeqV

2S f V 1  1  1 C C C 1

2

3

2S (6300 Hz)(12.0 V) 1 1 1   6 6 6

2.0u10

F

3.0u10

Find V as a function of C. I 0.475 A V IX C ZC 2S (6300 Hz)C The table below gives the results for each capacitor.

C (PF)

V (V)

2.0

6.0

3.0

4.0

6.0

2.0

(b) From part (a), I

0.48 A .

811

F

6.0u10

C1 C2 C3

0.475 A F

Chapter 21: Alternating Current

27. Strategy Vrms

College Physics

I rms X L where X L

Z L 2S fL.

Solution Solve for f. Vrms I rms X L I rms (2S fL), so Vrms 151.0 V f 2S I rms L 2S (0.820 A)(4.00 u 103 H)

4.00 mH f?

7.33 kHz .

28. Strategy Use Eqs. (21-9) and (21-10). Solution (a) Form a proportion.

V

IX L

IZ L

V 2S fIL, so i V

2S fILi 2S fILeq

0.10 H 0.50 H

Li Leq

Li . Thus, L1  L2

5.0 V

Li Li § 25 · V (5.0 V) ¨ V H ¸ Li is the peak L1  L2 0.10 H  0.50 H 3.0 © ¹ voltage across inductor i. The peak voltages are given in the table below. Vi

L (H)

V (V)

0.10

0.83

0.50

4.2

(b) Compute the peak current. V V V I X L Z Leq 2S f ( L1  L2 ) 33. Strategy %rms

I rms Z and Z

5.0 V 2S (126 Hz)(0.10 H  0.50 H)

R 2  X L2 . Only the resistance dissipates power, so use Pav

Solution Find I rms . Pav

2 I rms R, so I rms

11 mA

25.0 mH

Pav . R

25.0 Ω

Find f .

110 V

812

2 I rms R.

College Physics

%rms

Chapter 21: Alternating Current

I rms R 2  X L2

I rms Z

%rms 2

I rms R 2  Z 2 L2

R 2  Z 2 L2

I rms 2

%rms 2

Z 2 L2

I rms

2

 R2 2

1 § %rms · 2 ¨ ¸ R L © I rms ¹

Z

2

§ % · rms ¸  R 2 ¨ 2S L ¨© Pav R ¸¹ 470 Hz 1

f

R 2  X L2

34. Strategy Z

1 2S L

%rms 2 R  R2 Pav

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

Z L 2S fL.

30.0 : where X L

Solution Find L. R 2  X L2

R 2  Z 2 L2

R 2  (2S f ) 2 L2

39. Strategy Use Eq. (21-14b) with X L Solution Find the impedance. 1 Z R 2  X C2 R2  2 2 Z C

Z 2  R2 2S f

Z 2 , so L

(30.0 :) 2  (20.0 :) 2 2S (50.0 Hz)

71.2 mH .

0, and Eq. (21-7). 300.0 Ω 2.5 µF

(300.0 :) 2 

1 4S (159 Hz) (2.5 u 106 F) 2 2

2

500 : !

44. Strategy Pav

I rms %rms cos I is the average power.

Solution Find the power factor and the phase difference. (a) cos I

(b) I

Pav I rms %rms

cos 1

Pav I rms %rms

240 W (2.80 A)(120 V) cos 1

0.71

240 W (2.80 A)(120 V)

45. Strategy The average power is given by Pav

44q

I rms %rms cos I where I rms

%rms Z and cos I

R Z.

Solution Find the average power dissipated. 220 Ω !

0.15 mH 8.0 µF

813

Chapter 21: Alternating Current

Pav

College Physics

%rms R %rms Z Z

I rms %rms cos I

2

§ %rms · ¨ Z ¸ R © ¹

(% 2)2 R



R 2  Z L  Z1C



2

(12 V)2 (220 :) °­ ª 1 2 ®(220 :)2  « 2S (2500 Hz)(0.15 u 103 H)  2S (2500 Hz)(8.0u106 ¬ ¯°

2 º °½ ¾ F) »¼ ¿°

0.33 W

50. Strategy Use Eqs. (21-14b) and (21-16). Solution Find the impedance. Z

2

R  (X L  XC)

2

1 · § R  ¨ZL  Z C ¸¹ © 2

12.5 Ω

2 !

ª 1 (12.5 :) 2  « 2S (1590 Hz)(3.60 u 103 H)  u106 S 2 (1590 Hz)(5.00 ¬ Find the power factor. R 12.5 : cos I 0.617 Z 20.26 : Find the phase difference. R 12.5 : cos 1 51.9q I cos 1 Z 20.26 :

º

2

F) »¼

20.3 :

3.60 mH 5.00 µF

55. (a) Strategy Use Eq. (21-18). R

Solution Compute the resonant frequency. 1 1 Z0 745 rad s LC (0.300 H)(6.00 u 106 F) (b) Strategy At resonance, X L

!

0.300 H 6.00 µF

X C . Use Eq. (21-14a).

Solution Compute the resistance. %m 440 V R Z 790 : I 0.560 A (c) Strategy At resonance, the voltages across the capacitor and inductor are 180q out of phase and equal in magnitude, so they cancel. Solution The peak voltage across the resistor is VR

%m

Across the inductor, the peak voltage is IL L 0.300 H VL IX L I Z0 L I (0.560 A) C LC 6.00 u 106 F Since X L

X C , VC

VL

125 V .

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

814

440 V .

125 V .

College Physics

Chapter 21: Alternating Current

Solution Find the resonant frequency. 1 1 Z0 3 LC (40.0 u 10 H)(0.0500 F)

22.4 rad s , or f0

Z0 2S

3.56 Hz .

69. Strategy Use Eqs. (20-10) and (21-4). Solution (a) Calculate the turns ratio. N 2 %2 240 u 103 V N1 %1 420 V

570

(b) Calculate the rms current using the turns ratio. N2 I1 I 2 570(60.0 u 103 A) 34 A N1 (c) Calculate the average power. Pav

I rmsVrms

(0.0600 A)(240,000 V)

14 kW

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

0.

Solution Calculate the impedance. Z

R 2  X L2

R 2  Z 2 L2

(120 :)2  4S 2 (60.0 Hz)2 (12.0 H) 2

4.53 k:

(b) Strategy Use Eq. (21-14a). Solution Calculate the current. %rms 110 V I rms 24 mA Z 4530 : 72. Strategy Use Eqs. (21-6) and (21-7). Solution Compute the rms current. %rms I rms ZC %rms 2S (60.0 Hz)(0.025 u 106 F)(110 V) XC

1.0 mA

73. Strategy Use Eq. (21-7). Solution Find the capacitance. 1 1 1 XC , so C ZC Z X C 2S (520 Hz)(6.20 :)

49 PF .

76. Strategy The maximum current flows at the resonance frequency. Use Eq. (21-18). Solution Compute the resonant frequency. 1 1 1 , so f0 Z0 2S f0 LC 2S LC 2S (0.44 H)(520 u1012 F)

815

11 kHz .

College Physics

Chapter 21: Alternating Current 79. (a) Strategy Use Eqs. (21-7) and (21-10). Solution Find the reactances. 1 1 XC Z C (1.0 u 103 rad s)(50.0 u 106 F)

Z L (1.0 u 103 rad s)(0.0350 H)

XL

50.0 µF

20.0 Ω

35.0 mH

20 :

!

35 :

(b) Strategy Use Eq. (21-14b). Solution Find the impedance. R 2  ( X L  X C )2

Z

(20.0 :) 2  (35 :  20 :) 2

25 :

(c) Strategy Use Eq. (21-14a) with rms values. Solution Find the rms current. %rms 100.0 V I rms 4.0 A Z 25 : (d) Strategy Use the definition of rms. Solution Find the current amplitude (peak current). I

2 I rms

2(4.0 A)

5.7 A

(e) Strategy Use Eq. (21-16). Solution Find the phase angle. R R 20.0 : cos I , so I cos 1 cos 1 Z Z 25 :

37q .

(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. VR rms

I rms R

VC rms

I rms X C

(4.0 A)(20.0 :) (4.0 A)(20 :)

80 V , VL rms

I rms X L

(4.0 A)(35 :)

140 V , and

80 V .

(g) Strategy and Solution Since X L ! X C , 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

!m VL VC

37°

VR

VC

816

College Physics

Chapter 21: Alternating Current

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

15.2 mH

1 · § R2  ¨ Z L  ZC ¸¹ ©

R 2  ( X L  X C )2

Z

0.26 µF

12.0 Ω

2 !

ª º 1 (12.0 :)  « 2S (2.50 u 103 Hz)(15.2 u 103 H)  » 6 3 2S (2.50 u 10 Hz)(0.26 u 10 F) »¼ «¬

2

2

13 :

(b) Strategy Use Eq. (21-14a) with rms values. Solution Find the rms current. %rms 240 V I rms 18 A Z 13.5 : (c) Strategy Use Eq. (21-16). Solution Find the phase angle. R R 12.0 : cos I , so I cos 1 cos 1 Z Z 13.5 :

27q .

(d) Strategy Use Eqs. (21-7) and (21-10). Form a proportion. Solution Compare the reactances. X C 1 (ZC ) 1 1 !1 2 2 3 2 XL ZL Z LC 4S (2.50 u10 Hz) (0.0152 H)(0.26 u 106 F)

Since X C ! X L , the capacitor dominates the inductor, so the current leads the voltage since the current through a capacitor leads the voltage across it. (e) Strategy I rms values).

17.83 A to four significant figures. Use Eqs. (21-6) and (21-9) and Ohm’s law (using rms

Solution Find the rms voltage for each circuit element. (17.83 A)(12.0 :)

VR rms

I rms R

VL rms

I rms X L

I rmsZ L

VC rms

I rms X C

I rms ZC

210 V

(17.83 A)2S (2.50 u 103 Hz)(0.0152 H)

17.83 A 2S (2.50 u 103 Hz)(0.26 u 106 F)

4.3 kV

4.4 kV

82. (a) Strategy Model the coil as an RL series circuit. Use Eq. (21-16). Solution Find the power factor. R R R cos I Z R2  X 2 R 2  Z 2 L2 L

Electromagnet

450 : 2

2

2

(450 :)  4S (9.55 Hz) (2.47 H)

(b) Strategy Use Eq. (21-14b).

2

0.95

450 Ω 2.47 H

!

Solution Find the impedance.

817

College Physics

Chapter 21: Alternating Current

Z

(450 :)2  4S 2 (9.55 Hz) 2 (2.47 H) 2

470 :

(c) Strategy Use Eq. (21-14a). Solution Find the peak current. %m 2.0 u 103 V I 4.2 A Z 474 : (d) Strategy Use Eq. (21-17). Solution Find the average power. 1 1 Pav I rmsVrms cos I IV cos I (4.2 A)(2.0 u 103 V)(0.95) 2 2

818

4.0 kW