Physics Chapter 21 Solutions

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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 P_av I_{rms rms}V 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 v tC( ) given that the current leads the voltage by S/2 rad.

i(t) vC(t)

t

The current leads the voltage by /π 2 rad.

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Problems

3. Strategy 1500 W is the average power dissipated by the heater and Pav Irms rmsV . Solution Calculate I, the peak current.

av rms rms 1500 W 2 2 2 18 120 V P I I V § · _§ _· ¨ ¸ ¨_© ¸_¹ © ¹ A

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

av rms rms 4200 W 35 A 120 V P I V

(b) Strategy Since P V2 R and R₂ R P V₁, v 2.Form a proportion.

Solution Calculate P 2. 2 2 2 2 2 1 1 105 , so (4200 W) 3.2 kW . 120 P V P P V § · § · ¨ ¸ ¨_© ¸_¹ © ¹

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 and Use Eq. (21-4).

Solution Calculate R. av 1200 W P Vrms 120 V. 2 2 rms av (120 V) 12 1200 W V R P :

(b) Strategy Use Ohm’s law.

Solution Compute the rms current.

rms

rms V 120 V₁₂ 10 A

I

R :

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

max 2 av 2(1200 W) 2.4 kW

P P

15. Strategy The reactance is X_C 1 (ZC) where Z 2Sf. Solution

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(a) Solve for f. C ₃ ₆ C 1 1 1 , so 60.0 Hz . 2 2 2 (6.63 10 )(0.400 10 F) X f fC X C S S S u : u

(b) Compute the reactance.

C 1 1 ₆ 13.3 k

2 (30.0 Hz)(0.400 10 F) X

C

Z S _u :

16. (a) Strategy The reactance is X_C 1 (ZC) where Z 2Sf. Solution Compute the reactance.

C 1 1 ₆ 12.7 k

2 (50.0 Hz)(0.250 10 F) X

C

Z S u :

(b) Strategy The rms current is related to the reactance by Vrms Irms CX . Solution Solve for Irms.

6 rms rms rms C 2 (50.0 Hz)(0.250 10 F)(220 V) 17 mA V I CV X Z S u 17. Strategy V_rms I_{rms C}X where X_C 1 (ZC).

Solution Solve for the capacitance, C.

3 rms rms rms rms C rms 2.3 10 A , so 53 nF . 2 2 (60.0 Hz)(115 V) I I V I X C C f V Z S S u C Vrms

21. Strategy Use Eqs. (18-15), (21-6), and (21-7). Solution (a) Find I. 6 6 6 1 2 3 eq ₁ ₁ ₁ ₁ ₁ ₁ Ceq _{2.0 10 F} _{3.0 10 F} _{6.0 10 F} 2 2 (6300 Hz)(12.0 V) _{0.475 A} C C C V f V I C V X S S Z u u u Find V as a function of C. C I _{2 (6300 Hz)}0.475 A V IX C C Z S C1 C2 C3

The table below gives the results for each capacitor. ( F)

C P V (V)

2.0 6.0

3.0 4.0

6.0 2.0

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27. Strategy Vrms Irms LX where XL ZL 2SfL. Solution Solve for f.

rms rms L rms rms 3 rms (2 ), so 151.0 V 7.33 kHz . 2 2 (0.820 A)(4.00 10 H) V I X I fL V f I L S S S u f ? 4.00 mH

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

(a) Form a proportion.

L 2 , V IX I LZ SfIL so eq eq 1 2 2 _. 2 i i i i V fIL L L V fIL L L L S S Thus, 1 2 25 (5.0 V) V H 0.10 H 0.50 H 3.0 i i i L L i V V L L § ¨ © ¹L · ¸ 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 mA} 2 ( ) 2 (126 Hz)(0.10 H 0.50 H) V V V I X ZL Sf L L S

33. Strategy %rms IrmsZ and Z R2XL2. Only the resistance dissipates power, so use Pav Irms2 R.

Solution Find Irms. 2 av av rms , so rms . P P I R I R Find f . 25.0 mH 110 V 25.0 Ω

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2 2 2 2 2 rms rms rms L rms 2 2 2 2 rms 2 rms 2 2 2 rms 2 2 rms 2 2 rms rms 2 ₂ ₂ 2 2 rms rms av av 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 L I L R I R L I R f R R L P R L P Z Z Z Z S S S § · ¨ ¸ © ¹ § · _: ¨ ¸ : ¨ ¸ © ¹ % % % % % % 2 34. Strategy Z R2XL2 30.0 : where XL ZL 2S fL. Solution Find L. 2 2 2 2 2 2 2 2 2 2 2 2 2 L (2 ) , so Z₂ R (30.0 )_{2 (50.0 Hz)}(20.0 ) 71.2 mH . R X R L R f L Z L f Z S S S : :

39. Strategy Use Eq. (21-14b) with X_L 0, and Eq. (21-7). Solution Find the impedance.

2 2 2 2 C _{2 2}1 (300.0 ) ₂ ₂1 ₆ ₂ 500 4 (159 Hz) (2.5 10 F) Z R X R C Z S : : u ! 2.5 µF 300.0 Ω

44. Strategy Pav Irms rms% cosI is the average power. Solution Find the power factor and the phase difference.

(a) av rms rms 240 W cos 0.71 (2.80 A)(120 V) P I I % (b) 1 av 1 rms rms 240 W cos cos 44 (2.80 A)(120 V) P I I % q

45. Strategy The average power is given by P_av I_{rms rms}% cosI where I_rms %_rms Z and cosI R Z. Solution Find the average power dissipated.

!

220 Ω 0.15 mH

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6 2 2 rms rms av rms rms rms ₂ 2 1 2 2 2 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 R P I R Z Z Z _R _L Z S I Z S u § · ¨ ¸ © ¹ : ½ ° _:ª _u º ° ® _« _» ¾ ¬ ¼ ° ° ¯ ¿ % % % % %

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

6 2 2 2 2 L C 2 2 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 L C S Z Z S u § · _¨ _¸ © ¹ ª º : _« u _» ¬ ¼ :

Find the power factor.

! 12.5 Ω 3.60 mH 5.00 µF 12.5 cos 0.617 20.26 R Z I : : Find the phase difference.

1 1 12.5 cos cos 51.9 20.26 R Z I : : q

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

Solution Compute the resonant frequency.

0 1 1 ₆ 745 rad s

(0.300 H)(6.00 10 F) LC

Z

u

(b) Strategy At resonance, X_L X_C. Use Eq. (21-14a). Solution Compute the resistance.

m 440 V ₇₉₀ 0.560 A R Z I % : ! R 6.00 µF 0.300 H

(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 440 V .

Across the inductor, the peak voltage is

L L 0 (0.560 A) 0.300 H₆ 125 V . 6.00 10 F IL L V IX I L I C LC Z u Since X_L X_C, V_C V_L 125 V .

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Solution Find the resonant frequency. 0 1 ₃1 22.4 rad s , (40.0 10 H)(0.0500 F) LC Z u or 0 0 ₂ 3.56 Hz . f Z S 69. Strategy Use Eqs. (20-10) and (21-4).

Solution

(a) Calculate the turns ratio.

3 2 2 1 1 240 10 V 570 420 V N N u % %

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

3 2 1 2 1 570(60.0 10 A) 34 A N I I N u (c) Calculate the average power.

av rms rms (0.0600 A)(240,000 V) 14 kW

P I V

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

2 2 2 2 2 2 2 2 2

L (120 ) 4 (60.0 Hz) (12.0 H) 4.53 k

Z R X R Z L : S :

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

rms

rms ₄₅₃₀110 V 24 mA

I

Z :

%

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

6 rms rms rms C 2 (60.0 Hz)(0.025 10 F)(110 V) 1.0 mA I C X Z S u % _%

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

C 1 , X C Z so C 1 1 49 F . 2 (520 Hz)(6.20 ) C X Z S : P

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

0 2 f0 1 , so f0 1 1 11 kHz .

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

C 1 ₃ 1 ₆ 20 (1.0 10 rad s)(50.0 10 F) X C Z _u _u : 3 L (1.0 10 rad s)(0.0350 H) 35 X ZL u : ! 20.0 Ω 35.0 mH 50.0 µF

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

2 2 2 2

L C

( ) (20.0 ) (35 20 ) 25

Z R X X : : : :

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

rms

rms 100.0 V₂₅ 4.0 A

I

Z :

%

(d) Strategy Use the definition of rms.

Solution Find the current amplitude (peak current).

rms

2 2(4.0 A) 5.7 A

I I

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

1 120.0

cos , so cos cos 37 .

25

R R

Z Z

I I :

: q

(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 : V_{L rms} I_{rms L}X (4.0 A)(35 ): 140 V , and

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

V I X :

(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.

V_L

VR VL VC

37° !m

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

2 2 2 2 L C 1 ( ) Z R X X R L C Z Z § · _¨ _¸ © ¹ ! 12.0 Ω 15.2 mH 0.26 µF 2 2 3 3 3 6 1 (12.0 ) 2 (2.50 10 Hz)(15.2 10 H) 13 2 (2.50 10 Hz)(0.26 10 F) S S ª º : « u u » u u « » ¬ ¼ :

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

rms

rms _13.5240 V 18 A

I

Z :

%

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

1 112.0

cos , so cos cos 27 .

13.5

R R

Z Z

I I :

: q

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

C 2 2 3 2 6 L 1 ( ) 1 1 1 4 (2.50 10 Hz) (0.0152 H)(0.26 10 F) X C X L LC Z Z Z S u u !

Since XC!XL, 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 A I R rms rms (17.83 A)(12.0 ) 210 V V I R : 3 L rms rms L rms (17.83 A)2 (2.50 10 Hz)(0.0152 H) 4.3 kV V I X I ZL S u rms C rms rms C 17.83 A₃ ₆ 4.4 kV 2 (2.50 10 Hz)(0.26 10 F) I V I X C Z S u u

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 2 L 450 cos 0.95 (450 ) 4 (9.55 Hz) (2.47 H) R R R Z _R _X _R _L I Z S : :

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

2.47 H 450 Ω

! Electromagnet

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2 2 2 2

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

Z : S :

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

3 m 2.0 10 V _{4.2 A} 474 I Z u : %

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

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av rms rmscos 1₂ cos 1₂(4.2 A)(2.0 10 V)(0.95) 4.0 kW