Solucionario Capitulo 32 - Paul E. Tippens

33
Chapter 32. Alternating-Current Circuits Physics, 6 th Edition Chapter 32. Alternating-Current Circuits The Capacitor 32-1. A series dc circuit consists of a 4-F capacitor, a 5000- resistor, and a 12-V battery. What is the time constant for this circuit? -6 (5000 )(4 x 10 F) RC ; = 20 ms 32-2. What is the time constant for a series dc circuit containing a 6-F capacitor and a 400- resistor connected to a 20-V battery? -6 (400 )(6 x 10 F) RC ; = 2.40 ms 32-3. For the circuit described in Problem 32-1, what is the initial current and the final current? How much time is needed to be assured that the capacitor is fully charged? The initial current I o (before charge builds on capacitor) is determined by Ohm’s law: 0 12 V ; 5000 V I R I 0 = 2.40 mA 442

Transcript of Solucionario Capitulo 32 - Paul E. Tippens

Page 1: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

Chapter 32. Alternating-Current Circuits

The Capacitor

32-1. A series dc circuit consists of a 4-F capacitor, a 5000- resistor, and a 12-V battery.

What is the time constant for this circuit?

-6(5000 )(4 x 10 F)RC ; = 20 ms

32-2. What is the time constant for a series dc circuit containing a 6-F capacitor and a 400-

resistor connected to a 20-V battery?

-6(400 )(6 x 10 F)RC ; = 2.40 ms

32-3. For the circuit described in Problem 32-1, what is the initial current and the final

current? How much time is needed to be assured that the capacitor is fully charged?

The initial current Io (before charge builds on capacitor) is determined by Ohm’s law:

012 V ;

5000 VIR

I0 = 2.40 mA

It is assumed that a capacitor is fully charged after five time constants.

t = 5 = 5(20 ms); t = 100 ms

The current is zero when the capacitor is fully charged; If = 0 A

32-4. For the circuit in Problem 32-2, what is the maximum charge for the capacitor and how

much time is required to fully charge the capacitor?

Qmax = CVB = (6 x 10-6 F)(20 V); Qmax = 120 F

442

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

t = 5RC = 5(2.40 ms); t = 12.0 ms

*32-5. An 8-F capacitor is connected in series with a 600- resistor and a 24-V battery. After

one time constant, what are the charge on the capacitor and the current in the circuit?

1(1 ) 1 - (0.632)e

t RCB B BQ CV e CV CV

;

(8 )(24 V)(0.632);Q F Q = 121 C

24 V 1600

t RCBVI eR e

; I = 14.7 mA

*32-6. Assume that the fully charged capacitor of Problem 32-5 is allowed to discharge. After

one time constant, what are the current in the circuit and the charge on the capacitor?

24 V 1 ;600

t RCBVI eR e

I = -14.7 mA, decreasing

1(8 F)(24 V) ;e

t RCBQ CV e

Q = 70.6 C

*32-7. Assume that the 8-F capacitor in Problem 32-5 is fully charged and is then allowed to

decay. Find the time when the current in the circuit will have decayed to 20 percent of

its initial value? [ Procedure: Let x = t/RC, solve for x, and then find t. ]

443

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

max

max

1; ; 0.20; 5.00.2

t RC x x x xBV II e I e e e eR I

Take ln of both sides: x = ln (5.0) = 1.61 Now, 1.61; 1.61tx t RC

RC

t = 1.61(600 )(24 V); t = 7.73 ms

*32-8. A 5-F capacitor is connected in series with a 12-V source of emf and a 4000

resistor. How much time is required to place 40 C of charge on the capacitor? [ Let x

= t/RC ]

(1 ); 40 C (5 F)(12 V)(1 - ); 0.667 = (1 - )t RC x xBQ CV e e e

11 0.667; 0.333; 3.00.333

x x xe e e

x = ln(3); 1.10; 1.10 ;t t RC

RC

t = 22.2 ms

The Inductor

32-9. A series dc circuit contains a 4-mH inductor and an 80- resistor in series with a 12-V

battery. What is the time constant for the circuit? What are the initial and final currents?

444

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

0.004 H ;80

LR

= 50.0 s

When t = 0, ( / )(1 ) (1 1) 0R L tB BV Vi e

R R

; io = 0

When t = , ( / ) 12 V(1 ) 1 0 ; ;

80 R L tB BV Vi e i

R R

if = 150 mA

32-10. A 5-mH inductor, a 160- resistor and a 50-V battery are connected in series. What time

is required for the current in the inductor to reach 63 percent of its steady state value?

What is the current at that instant? (Current reaches 63% in one time constant.)

0.005 H ;160

LR

= 31.2 s

32-11. What is the current in Problem 32-9 after a time of one time constant?

( / ) 12 V 1(1 ) 1 ; (0.15 A)(0.632);80

R L tBVi e iR e

i = 94.8 mA

32-12. Assume that the inductor in Problem 32-10 has reached its steady state value. If the

circuit is broken how much time must pass before we can be sure the current in the

inductor is zero? (After 5 time constants: t = 5)

t = 5 = 5(31.2 ms); t = 156 s

445

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

32-13 The current in a 25-mH solenoid increases from 0 to 2 A in a time of 0.1 s. What is the

magnitude of the self-induced emf?

(2 A - 0)(0.025 H) ;0.1 s

iLt

E

E = - 500 mV

32-14. A series dc circuit contains a 0.05-H inductor and a 40- resistor in series with a 90-V

source of emf. What are the maximum current in the circuit and the time constant?

max90 V ;40

BViR

imax = 2.25 A

0.05 H ;40

LR

= 1.25 ms

*32-15. What is the instantaneous current in the inductor of Problem 32-14 after a decay time of

1.0 ms? (Assume the current decays from it’s maximum value of 2.25 A.)

Let x = (R/L)t: 0.8

40 1 1(0.001 s); 0.80; 0.4490.05 H

xxx x e

e e

( / ) 90 V; (0.449);40

R L t xB BV Vi e e iR R

i = 1.01 A

You should show that the current after a rise time of 1 ms is 1.24 A.

446

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

**32-16. A 6-mH inductor, a 50- resistor, and a 38-V battery are connected in series. At what

time after the connection will the current reach an instantaneous value of 600 mA?

Procedure: Let x = (R/L)t:, then solve for x and substitute to find the time t:

(0.6 A)(50 )(1 ); 1 ; 1 138 V

x x xB

B B

V iR iRi e e eR V V

; e-x = 0.211

1 4.75;0.211

xe Take ln of both sides: x = ln(4.75) = 1.56

1.56 1.56(0.006 H)1.56 ; ;50

Rt Lx tL R

x = 187 s

Alternating Currents

32-17. An ac voltmeter, when placed across a 12- resistor, reads 117 V. What are the

maximum values for the voltage and current?

max 2 1.414(117 V);effV V Vmax = 165 V

max117 V2 1.414 ;12 effI I Imax = 13.8 A

32-18. In an ac circuit, the voltage and current reach maximum values of 120 V and 6.00 A.

What are the effective ac values?

447

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

max 6 A ;1.4142eff

II Ieff = 4.24 A;

max 120 V ;1.4142eff

VV Veff = 84.9 V

32-19. A capacitor has a maximum voltage rating of 500 V. What is the highest effective ac

voltage that can be supplied to it without breakdown?

max 500 V ;1.4142eff

VV Veff = 354 V

32-20. A certain appliance is supplied with an effective voltage of 220 V under an effective

current of 20 A. What are the maximum and minimum values?

max 2 1.414(220 V);effV V Vmax = 311 V

max 2 1.414(20 A);effI I Imax = 28.3 A

Reactance

32-21. A 6-F capacitor is connected to a 40-V, 60 Hz ac line. What is the reactance? What is

the effective ac current in the circuit containing pure capacitance?

-6

1 12 2 (60 Hz)(6 x 10 F)CX

fC

; XC = 442

40 V ;442

effeff

C

VI

X

Ieff = 90.5 mA

448

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

32-22. A 2-H inductor of negligible resistance is connected to a 50-V, 50 Hz ac line. What is

the reactance? What is the effective ac current in the coil?

L 2 2 (50 Hz)(2 H) 628 X fL ; XL = 442

50 V ;628

effeff

L

VI

X

Ieff = 79.6 mA

32-23. A 50-mH inductor of negligible resistance is connected to a 120-V, 60-Hz ac line. What

is the inductive reactance? What is the effective ac current in the circuit?

L 2 2 (60 Hz)(0.050 H) 18.85 ;X fL ; XL = 18.9

120 V ;18.85

effeff

L

VI

X

Ieff = 6.37 mA

32-24. A 6-F capacitor is connected to a 24-V, 50-Hz ac source. What is the current in the

circuit?

-6

1 12 2 (50 Hz)(6 x 10 F)CX

fC

; XC = 531

24 V ;531

effeff

C

VI

X

Ieff = 45.2 mA

449

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

32-25. A 3-F capacitor connected to a 120-V, ac line draws an effective current of 0.5 A.

What is the frequency of the source?

120 V ; 240 0.5 A

effC C

eff

VX X

I

-6

1 1 1; ;2 2 2 (3 x 10 F)(240 )C

C

X ffC CX

f = 221 Hz

32-26. Find the reactance of a 60-F capacitor in a 600-Hz ac circuit. What is the reactance if

the frequency is reduced to 200 Hz?

-6

1 1 ;2 2 (600 Hz)(60 x 10 F)CX

fC

XC = 4.42

-6

1 1 ;2 2 (200 Hz)(60 x 10 F)CX

fC

XC = 13.3

32-27. The frequency of an alternating current is 200 Hz and the inductive reactance for a

single inductor is 100 . What is the inductance?

L100 2 ; ;

2 2 (200 Hz)LXX fL Lf

L = 79.6 mH.

450

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

32-28. A 20- resistor, a 2-F capacitor, and a 0.70-H inductor are available. Each of these, in

turn, is connected to a 120-V, 60 Hz ac source as the only circuit element. What is the

effective ac current in each case?

Resistance only:

120 V ;20

ViR

i = 6.00 A

Capacitor only: -6

1 120 V1326 ; 2 (60 Hz)(2 x 10 F) 1326 CX i

; i = 90.5 mA

Inductor only: L 2 2 (60 Hz)(0.7 H) 264 ;X fL

120 V ;264

i i = 455 mA

Series AC Circuits

*32-29. A 300- resistor, a 3-F capacitor, and a 4-H inductor are connected in series with a 90-

V, 50 Hz ac source. What is the net reactance of the circuit? What is the impedance?

-6

1 1061 ;2 (50 Hz)(3 x 10 F)CX

L 2 (50 Hz)(4 H) 1257 ;X

Net Reactance: XL – XC = 196

2 2 2 2( ) (300 ) (1257 1061 )L CZ R X X ; Z = 358

451

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

*32-30. What is the effective ac current delivered to the ac series circuit described in Problem

32-29? What is the peak value for this current?

90 V ;300

ViZ

i = 300 mA

max 2(300 mA);i i = 424 mA

*32-31. A series ac circuit consists of a 100- resistor, a 0.2-H inductor, and a 3-F capacitor

connected to a 110-V, 60 Hz ac source. What is the inductive reactance, the capacitative

reactance, and the impedance for the circuit?

-6

12 (60 Hz)(3 x 10 F)CX

884 ; L 2 (60 Hz)(0.2 H)X 75.4

2 2 2 2( ) (100 ) (75.4 884 )L CZ R X X ; Z = 815

*32-32. What are the phase angle and the power factor for the circuit described in Problem 32-31?

75.4 884 tan ;100

L CX XR

= -83.00

Power Factor = cos cos (-83.00); Power Factor = 0.123 or 12.3%

*32-33. Assume that all the elements of Problem 32-28 above are connected in series with the

given source. What is the effective current delivered to the circuit?

452

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

From Problem 32-28: R = 20 , XC = 1326, and XL = 264 , V = 120 V, f = 60 Hz

2 2 2 2( ) (20 ) (264 1326 )L CZ R X X ; Z = 1062

120 V ;1062

ViZ

i = 113 mA

*32-34. A series ac circuit contains a 12-mH inductor, an 8-F capacitor, and a 40- resistor

connected to a 110-V, 200 Hz ac line. What is the effective ac current in the circuit?

-6

12 (200 Hz)(8 x 10 F)CX

99.5 ; L 2 (200 Hz)(0.012 H)X 15.1

2 2 2 2( ) (40 ) (15.1 99.5 )L CZ R X X ; Z = 93.4

*32-34. (Cont.)

15.1 99.5 tan ;40

L CX XR

= -64.60

Power Factor = cos cos (-64.60);

Power Factor = 0.428 or 42.8%

*32-35. When a 6- resistor and a pure inductor are connected to a 100-V, 60 Hz ac line, the

effective current is 10 A. What is the inductance ? What is the power loss through the

resistor and what is the power loss through the inductor?

453

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Chapter 32. Alternating-Current Circuits Physics, 6th Edition

2 2 2 2110 V 11.0 ; (11.0) (6 ) ;10 A LZ X Z R

XL = 9.22

(9.22 ) ;2 2 (60 Hz)

LXLf

L = 24.5 mH

P = I2R = (10 A)2 (6.00 ); P = 600 W;

No Power lost in inductor: 0

*32-36. A capacitor is in series with a resistance of 35 and connected to a 220-V, ac line. The

reactance of the capacitor is 45 . What is the effective ac current? What is the phase

angle? What is the power factor?

2 2 2 2(0 ) (35 ) (45 ) 57.0 ;CZ R X

220 V 3.89 A;57.0

i i = 3.89 A

45 tan ;35

L CX XR

= -52.10;

Power Factor = cos 52.10 = 61.4%

*32-37. A coil having an inductance of 0.15 H and a resistance of 12 is connected to a 110-V,

25 Hz line. What is the effective current in the circuit? What is the power factor? What

power is lost in the circuit? ( XC = 0 )

454

Page 14: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

L 2 2 (25 Hz)(0.15 H) 23.56 ;X fL 2 2

LZ R X

2 2 110 V(12) (23.56) 26.44 ; ;26.44

Z i i = 4.16 A

23.56 tan ;12.0

L CX XR

= 630; Power Factor = cos 630 = 45.4%

Power is lost only through resistance: P = I2R = (4.16 A)2(12 ) = 208 W

*32-38. What is the resonant frequency for the circuit described by Problem 32-29?

-6

1 1 ;2 2 (0.2 H)(3 x 10 F)

rf LC

fr = 205 Hz

*32-39. What is the resonant frequency for the circuit described by Problem 32-34?

-6

1 1 ;2 2 (0.012 H)(8 x 10 F)

rf LC

fr = 514 Hz

Challenge Problems

32-40. A 100-V battery is connected to a dc series circuit with a 60-mH inductor and a 50-

resistor. Calculate the maximum current in the circuit and the time constant.

455

Page 15: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

When t = , ( / ) 100 V(1 ) 1 0 ; ;

50 R L tB BV Vi e i

R R

imax = 2.00 A

0.060 H ;50

LR

= 1.20 ms

32-41. If the circuit of Problem 32-40 reaches a steady state condition and is then broken, what

will be the instantaneous current after one time constant? (Decay of an inductor.)

( / ) 100 V 1 ;50

R L tBVi eR e

i = 736 mA

32-42. A resonant circuit has an inductance of 400 H and a capacitance of 100 pF. What is the

resonant frequency?

-4 -10

1 1 ;2 2 (4 x 10 H)(1 x 10 F)

rf LC

fr = 796 kHz

32-43. An LR dc circuit has a time constant of 2 ms. What is the inductance if the resistance is 2

k? Find the instantaneous current 2 ms after the circuit is connected to a 12-V battery.

; (0.002 s)(2000 );L L RR

L = 4.00 H

456

Page 16: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

( / ) 12 V 1(1 ) (1 );2000

R L tBVi eR e

i = 3.79 mA

32-44. A series dc circuit consists of a 12-V battery, a 20- resistor, and an unknown capacitor.

The time constant is 40 ms. What is the capacitance? What is the maximum charge on

the capacitor?

0.040 s; ;20

RC CR

C = 2.00 mF;

Q = CV = (2 mF)(12 V) = 24 mC

*32-45. A 2-H inductor having a resistance of 120 is connected to a 30-V battery. How much

time is required for the current to reach 63 percent of its maximum value? What is the

initial rate of current increase in amperes per second? What is the final current?

The current reaches 63% of its maximum value in a time equal to one time constant:

2 H ;120

LR

= 16.7 ms

For Loop: E = IR or B

iV L iRt

Solving for i/t with initial i = 0, we have: ;B

iV iR Lt

;Bi V

t L

457

0

L R

Page 17: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

30 V ;2 H

Bi Vt L

15 A/si

t

The final current is the maximum: max

30 V ;120

BViR

imax = 250 mA

*32-46. Consider the circuit shown in Fig. 38-13. What is the impedance? What is the effective

current? What is the power loss in the circuit?

L = 4 mH; C = 10 F; R = 20 ;

XL = 2fL = 2(600 Hz)(0.004 H) = 15.1

-6

1 26.5 ;2 (600 Hz)(10 x 10 F)CX

2 2(20 ) (15.1 26.5 ) ;Z

Impedance: Z = 23.0

50 V ;23

ViZ

Effective current: i = 2.17 A

Power loss is entirely in resistance: P = i2R = (2.17 A)2(20 ); P = 94.2 W

*32-47. A tuner circuit contains a 4-mH inductor and a variable capacitor. What must be the

capacitance if the circuit is to resonate at a frequency of 800 Hz?

22 2 2

1 1 1; ; 4 42r r

r

f f CLC LfLC

458

Source: 50 V, 600 Hz

CL R

Page 18: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

2 2

1 ;4 (0.004 H)(800 Hz)

C

C = 9.89 F

*32-48. An 8-F capacitor is in series with a 40- resistor and connected to a 117-V, 60-Hz ac

source. What is the impedance? What is the power factor? How much power is lost in

the circuit? ( Note: XL = 0, so that all reactance is due to the capacitor. )

-6

1 332 2 (60 Hz)(8 x 10 F)CX

2 2(40 ) (0 332) ;Z Z = 334

117 V ;334

ViZ

i = 350 mA

-332 tan ;40

L CX XR

=-83.10; Power Factor = cos 83.10 = 12.0%

Power is lost only through resistance: P = I2R = (0.350 A)2(40 ) = 4.90 W

*32-49. Someone wants to construct a circuit whose resonant frequency is 950 kHz. If a coil in

the circuit has an inductance of 3 mH, what capacitance should be added to the circuit?

22 2 2

1 1 1; ; 4 42r r

r

f f CLC LfLC

459

Source: 117 V, 60 Hz

CR

Page 19: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

2 2

1 ;4 (0.003 H)(950,000 Hz)

C

C = 9.36 pF

*32-50. A 50-F capacitor and a 70-resistor are connected in series across a 120-V, 60 Hz ac

line. Determine the current in the circuit, the phase angle, and the total power loss?

-6

1 53.05 2 (60 Hz)(50 x 10 F)CX

2 2(70 ) (0 53.05) ;Z Z = 87.8

120 V ;87.8

ViZ

i = 1.37 A

-53.05 tan ;70

L CX XR

=-37.20; Power Factor = cos 37.20 = 79.8%

Power is lost only through resistance: P = I2R = (1.37 A)2(70 ) = 131 W

*32-51. Refer to Problem 32-50. What is the voltage across the resistor? What is the voltage

across the capacitor? What inductance should be added to the circuit to reach resonance?

V = iR = (1.37 A)(70 ) = 95.6 V VC = iXC = (1.37 A)(53.05 ) = 72.5 V

22 2 2

1 1 1; ; L4 42r r

r

f fLC CfLC

460

Source: 120 V, 60 Hz

CR

Page 20: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

2 -6 2

1L ;4 (50 x 10 F)(60 Hz)

L = 141 mH

Critical Thinking Problems

*32-52 The resistance of an 8-H inductor is 200 . If this inductor is suddenly connected across

a potential difference of 50 V, what is the initial rate of increase of the current in A/s

(see Eq. 32-15)? What is the final steady current? At what rate is the current increasing

after one time constant. At what time after connecting the source of emf, does the

current equal one-half of its final value? (Initially, the current is zero.)

8 H ;200

LR

= 40.0 ms

For Loop: E = IR or B

iV L iRt

Solving for i/t with initial i = 0, we have: ;B

iV iR Lt

;Bi V

t L

461

0

L R

Page 21: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

50 V ;8 H

Bi Vt L

6.25 A/si

t

The final current is the maximum: max

50 V ;200

BViR

imax = 250 mA

To find the rate of increase after one time constant, we need to know i at t = :

/max

1(1 ) (250 mA) 1- ;e

ti i e i = 158 mA

i i; L ; t t

BB B

i V iRV L iR V iRt L

i 50 V - (0.158 A)(200 )t 8 H

BV iRL

2.30 A/si

t

max

, : (1 ) 0.50; 0.5; 2.0x x xRt iLet x then e e eL i

0.693(8 H)0.693 ; ;200

Rtx tL

t = 27.7 ms

*32-53. In Chapter 29, we found that the magnetic flux density within a solenoid was given by:

0(1) NIBA l

462

Page 22: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

Here we have used l for the length of the solenoid in order to avoid confusion with the

symbol L used for inductance. We also know that induced emf is found in two ways:

(2) and (3) iN Lt t

E E

Prove the following relationships for a solenoid of length l having N turns of area A:

L NI

L N Al

and 2

Combining Eqs (2) and (3) above, we obtain: N L i

Now, at maximum current I: or NN LI L

I

From (1): 0 0 : NIA NIAN Nwhich gives Ll I I l

And finally:

20 N AL

l

Note: L is not dependent on current or voltage.

*32-54. An inductor consists of a coil 30 cm long with 300 turns of area 0.004 m2. Find the

inductance (see previous problem). A battery is connected and the current builds from

zero to 2.00 A in 0.001 s. What is the average induced emf in the coil?

463

Page 23: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

2 -7 2 20 (4 x 10 T m/A)(300) (0.004 m ) ;

0.300 mN ALl

L = 1.51 mH

-3 2 A - 0(1.51 x 10 H)0.001 s

iLt

E

; E = -3.02 V

*32-55. An inductor, resistor, and capacitor are connected in series with a 60-Hz ac line. A

voltmeter connected to each element in the circuit gives the following readings: VR

= 60 V, VL = 100 V, and VC = 160 V. What is the total voltage drop in the circuit? What

is the phase angle?

2 2 2 2( ) (60 V) (100 V - 160 V) ;T R L CV V V V VT = 84.6 V

(100 V - 160 V)tan ;60 V

L C

R

V VV

= -45.00

*32-56. The antenna circuit in a radio receiver consists of a variable capacitor and a 9-mH coil.

The resistance of the circuit is 40 . A 980-kHz radio waves produces a potential

difference of 0.2 mV across the circuit. Determine the capacitance required for

resonance. . What is the current at resonance?

22 2 2

1 1 1; ; 4 42r r

r

f f CLC LfLC

464

Page 24: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

2 2

1 ;4 (0.009 H)(980,000 Hz)

C

C = 2.93 F

At fr , XC = XL and: Z = R = 40 ;

0.2 mV ;40

i i = 5.00 mA

*32-57. A series RLC circuit has elements as follows: L = 0.6 H, C = 3.5 F, and R = 250 . The

circuit is driven by an generator that produces a maximum emf of 150 V at 60 Hz. What

is the effective ac current? What is the phase angle? What is the average power loss in

the circuit? ( First find Veff )

max 150 V ;2 2eff

VV Veff = 106 V

L = 0.600 H; C = 3.5 F; R = 250 ;

XL = 2fL = 2(60 Hz)(0.6 H) = 226

-6

1 758 ;2 (60 Hz)(3.5 x 10 F)CX

2 2(250 ) (226 758 ) ;Z

465

Source: 150 V, 60 Hz

CL R

Page 25: Solucionario Capitulo 32 - Paul E. Tippens

Chapter 32. Alternating-Current Circuits Physics, 6th Edition

Impedance: Z = 588

106 V ;588

ViZ

Effective current: i = 181 mA

226 768 tan250

L CX XR

; = 42.70

Power loss is entirely in resistance: P = i2R = (0.181 A)2(250 ); P = 8.19 W

466