Sm chapter33

33Alternating Current Circuits

CHAPTER OUTLINE

33.1 AC Sources33.2 Resistors in an AC Circuit33.3 Inductors in an AC Circuit33.4 Capacitors in an AC Circuit33.5 The RLC Series Circuit33.6 Power in an AC Circuit33.7 Resonance in a Series RLC

Circuit33.8 The Transformer and Power

Transmission33.9 Rectifi ers and Filters

ANSWERS TO QUESTIONS

*Q33.1 (i) Answer (d). ∆∆

VV

avg = max

2 (ii) Answer (c). The average of the squared voltage is

∆∆

VV[ ]( ) =

( )2

2

2avg.max Then its square root is

∆∆

VV

rms = max

2

*Q33.2 Answer (c). AC ammeters and voltmeters read rms values. With an oscilloscope you can read a maximum voltage, or test whether the average is zero.

*Q33.3 (i) Answer (f). The voltage varies between +170 V and −170 V. (ii) Answer (d). (iii) 170V/ 2 = 120 V. Answer (c).

Q33.4 The capacitive reactance is proportional to the inverse of the frequency. At higher and higher frequen-cies, the capacitive reactance approaches zero, making a capacitor behave like a wire. As the frequency goes to zero, the capacitive reactance approaches infi nity—the resistance of an open circuit.

Q33.5 The second letter in each word stands for the circuit element. For an inductor L, the emf ε leads the current I—thus ELI. For a capacitor C, the current leads the voltage across the device. In a circuit in which the capacitive reactance is larger than the inductive reactance, the current leads the source emf—thus ICE.

Q33.6 The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagrams throughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages across different circuit elements are not simultaneously at their maximum values. Do not forget that an inductor can induce an emf in itself and that the voltage across it is 90° ahead of the current in the circuit in phase.

Q33.7 In an RLC series circuit, the phase angle depends on the source frequency. At very low frequency the capacitor dominates the impedance and the phase angle is near −90°. The phase angle is zero at the resonance frequency, where the inductive and capacitive reactances are equal. At very high frequencies f approaches +90°.

*Q33.8 (i) Inductive reactance doubles when frequency doubles. Answer (f). (ii) Capacitive reactance is cut in half when frequency doubles. Answer (b). (iii) The resistance remains unchanged. Answer (d).

*Q33.9 At resonance the inductive reactance and capacitive reactance cancel out. Answer (c).

247

ISMV2_5104_33.indd 247ISMV2_5104_33.indd 247 6/7/07 6:45:53 PM6/7/07 6:45:53 PM

248 Chapter 33

*Q33.10 At resonance the inductive reactance and capacitive reactance add to zero. tan−1(XL−X

C)/R = 0.

Answer (c).

*Q33.11 (a) The circuit is in resonance. (b) 10 Ω/20 Ω = 0.5 (c) The resistance of the load could be increased to make a greater fraction of the emf’s power go to the load. Then the emf would put out a lot less power and less power would reach the load.

Q33.12 The person is doing work at a rate of P = Fv cosθ. One can consider the emf as the “force” that moves the charges through the circuit, and the current as the “speed” of the moving charges. The cosθ factor measures the effectiveness of the cause in producing the effect. Theta is an angle in real space for the vacuum cleaner and phi is the analogous angle of phase difference between the emf and the current in the circuit.

*Q33.13 The resonance is high-Q, so at 1 000 Hz both XL and X

C are equal and much larger than R. Now X

C

at 500 Hz is twice as large as at 1 kHz. And XL at 1.5 kHz is 1.5 times larger than at 1 kHz. Again,

XC at 1 500 Hz is two-thirds as large as at 1 kHz. And X

L at 500 Hz is half as large as at 1 kHz. The

resistance does not change with frequency. The ranking is then a > f > b = e > c > d > g = h = i.

Q33.14 In 1881, an assassin shot President James Garfi eld. The bullet was lost in his body. Alexander Graham Bell invented the metal detector in an effort to save the President’s life. The coil is preserved in the Smithsonian Institution. The detector was thrown off by metal springs in Garfi eld’s mattress, a new invention itself. Surgeons went hunting for the bullet in the wrong place. Garfi eld died.

Q33.15 No. A voltage is only induced in the secondary coil if the fl ux through the core changes in time.

Q33.16 The Q factor determines the selectivity of the radio receiver. For example, a receiver with a very low Q factor will respond to a wide range of frequencies and might pick up several adjacent radio stations at the same time. To discriminate between 102.5 MHz and 102.7 MHz requires a high-Q circuit. Typically, lowering the resistance in the circuit is the way to get a higher quality resonance.

*Q33.17 In its intended use, the transformer takes in energy by electric transmission at 12 kV and puts out nearly the same energy by electric transmission at 120 V. With the small generator putting energy into the secondary side of the transformer at 120 V, the primary side has 12 kV induced across it. It is deadly dangerous for the repairman.

Section 33.1 AC Sources

Section 33.2 Resistors in an AC Circuit

P33.1 ∆ ∆ ∆v t V t V t( ) = ( ) = ( ) =max sin sin sinω ω π2 200 2 2 1rms 000 283 628t t( )⎡⎣ ⎤⎦ = ( ) ( ) V sin

P33.2 ∆Vrms

V

2V= =170

120

(a) P =( )

→ = ( )=

∆Ω

V

RRrms V

W

2 2120

75 0193

.

(b) R = ( )=120

100144

2V

WΩ

SOLUTIONS TO PROBLEMS


Alternating Current Circuits 249

P33.3 Each meter reads the rms value.

∆

∆Ω

V

IV

R

rms

rmsrms

V

2V

V

24.0

= =

= =

10070 7

70 7

.

. == 2 95. A

P33.4 (a) ∆ ∆vR V t= max sinω

∆ ∆vR V= ( )0 250. max , so, sin . ,ω t = 0 250 or ω t = ( )−sin . .1 0 250

The smallest angle for which this is true is ω t = 0 253. rad. Thus, if t = 0 010 0. s,

ω = =0 25325 3

..

rad

0.010 0 srad s .

(b) The second time when ∆ ∆vR V= ( )0 250. ,max ω t = ( )−sin .1 0 250 again. For this

occurrence,ω πt = − =0 253 2 89. .rad rad (to understand why this is true, recall the

identity sin sinπ θ θ−( ) = from trigonometry). Thus,

t = =2 890 114

..

rad

25.3 rad ss

P33.5 i I tR = max sinω becomes 0 600 0 007 00. sin .= ( )ω

Thus, 0 007 00 0 600 0 6441. sin . .( ) = ( ) =−ω

andω π= =91 9 2. rad s f so f = 14 6. Hz

P33.6 ∆Vmax .= 15 0 V and Rtotal = + =8 20 10 4 18 6. . .Ω Ω Ω

IV

RImax

max ..= = = =

∆Ωtotal

rms

V

18.6A

15 00 806 2

PPspeaker rms2

speaker

A

2= = ⎛

⎝⎜⎞⎠⎟I R

0 80610 4

2.

. WΩ( ) = 3 38.

Section 33.3 Inductors in an AC Circuit

P33.7 (a) XV

IL = = =∆

Ωmax

max ..

100

7 5013 3

LXL= =

( ) = =ω π

13 3

2 50 00 042 4 42 4

.

.. .H mH

(b) XV

IL = = =∆

Ωmax

max ..

100

2 5040 0

ω = =×

=−

X

LL 40 0

42 4 109423

.

.rad s

FIG. P33.3


250 Chapter 33

P33.8 At 50.0 Hz, X LX

LL= ( ) = ( )2 50 0 2 50 0

2 60 060 0π π

π. .

..Hz Hz Hz

HHz( )⎛

⎝⎜⎞

⎠⎟= ( ) =50 0

60 054 0 45 0

.

.. .Ω Ω

IV

X

V

XL Lmax

max

..= =

( )= ( )

=∆ ∆

Ω2 2 100

45 03 14rms V

A

P33.9 i tV

LtL ( ) = −⎛

⎝⎞⎠ =

( ) (∆ max sin. sin .

ωω π π

2

80 0 65 0V ))( ) −⎡⎣ ⎤⎦( ) ×( )−

0 015 5 2

65 0 70 0 10 3

.

. .

ππ rad s H

i tL ( ) = ( ) ( ) =5 60 1 59 5 60. sin . .A rad A

P33.10 ω π π= = ( ) =2 2 60 0 377f . s rad s

X L

IV

X

L

L

= = ( ) ⋅( ) =

=

ω 377 0 020 0 7 54s V s A

rmsrms

. . Ω∆ == =

= = ( ) =

12015 9

2 2 15 9 22

V

7.54 A

Arms

Ω.

.maxI I ..

sin . sin.

max

5

22 52 60 0

A

As

1 i t I t( ) = = ( ) ( ) ⋅ω

π ss

180 A A

⎛

⎝⎜

⎞

⎠⎟ = ( ) ° =

= =

22 5 120 19 5

1

2

12

. sin .

U Li22

0 020 0 19 5 3 802

. . .V s A A J⋅( )( ) =

P33.11 LN

IB= Φ

where ΦB is the fl ux through each turn. N LIX V

XB

L L

L

Φ∆

, max,max

max = = ( )ω

NV

fB

LΦ∆

, .max

,rms V s

2

T C m

N=

( )= ⋅

( )⋅ ⋅2

2

120

60 0π π ⋅⋅⎛⎝

⎞⎠

⋅⎛⎝

⎞⎠ ⋅

⎛⎝

⎞⎠ = ⋅

s

N m

J

J

V CT m20 450.

Section 33.4 Capacitors in an AC Circuit

P33.12 (a) Xf C

C = 1

2π:

1

2 22 0 10175

6π f . ×( ) <−

Ω

1

2 22 0 10 1756π . ×( )( )<− f f > 41 3. Hz

(b) XC

C ∝ 1, so X X44

1

222( ) = ( ): XC < 87 5. Ω

P33.13 I IV

XV f C

C

max = = ( ) = ( )22

2 2rmsrms

rms

∆∆ π

(a) Imax . .= ( ) ( ) ×( ) =−2 120 2 60 0 2 20 10 1416 V s C V mAπ

(b) Imax . .= ( ) ( ) ×( ) =−2 240 2 50 0 2 20 10 2356 V s F mAπ

P33.14 Q C V C V C Vmax max= ( ) = ( )⎡⎣

⎤⎦ = ( )∆ ∆ ∆2 2rms rms

P33.15 I V Cmax max . . .= ( ) = ( )( )( ) ×−∆ ω π48 0 2 90 0 3 70 11 V s 00 1006−( ) =F mA



P33.16 XC

C = =( ) ×( ) =

−

1 1

2 60 0 1 00 102 65

3ω π . ..

s C V Ω

∆ ∆vC t V t( ) = max sin ,ω to be zero at t = 0

iV

XtC

C

= +( ) = ( )∆Ω

max sin.

sinω φ π2 120

2 652

60V s−−

− +⎡⎣⎢

⎤⎦⎥

= ( ) +1

118090 0 64 0 120 90 0

s° A °. . sin . °°

A

( )

= −32 0.

Section 33.5 The RLC Series Circuit

P33.17 (a) X LL = = ( ) ×( ) =−ω π2 50 0 400 10 1263. Ω

XC

Z R X X

C

L C

= =( ) ×( ) =

= + −

−

1 1

2 50 0 4 43 107196

2

ω π . .Ω

(( ) = + −( ) =

= = ×

2 2 2500 126 719 776

250 1

Ω

∆V I Zmax max 00 776 1943−( )( ) = V

(b) φ = −⎛⎝

⎞⎠ = −⎛

⎝⎞⎠ = −− −tan tan .1 1 126 719

50049 9

X X

RL C °° .

Thus, the

current leads the voltage .

P33.18 ωω

ωLC LC

= → = =×( ) ×( )

= ×− −

1 1 1

57 0 10 57 0 101 75 1

6 6. .. 004 rad s

f = =ωπ2

2 79. kHz

P33.19 (a) X LL = = ( ) ×( ) =− −ω π2 50 0 250 10 78 51 3. .s H Ω

(b) XCC = = ( ) ×( )⎡⎣ ⎤⎦ =− − −1

2 50 0 2 00 10 1 591 6 1

ωπ . . .s F kΩ

(c) Z R X XL C= + −( ) =2 21 52. kΩ

(d) IV

Zmaxmax= =

×=

∆Ω

210138

V

1.52 10mA3

(e) φ = −⎡⎣⎢

⎤⎦⎥

= −( ) = −− −tan tan . .1 1 10 1 84 3X X

RL C °

FIG. P33.17


252 Chapter 33

P33.20 (a) Z R X XL C= + −( ) = + −( ) =2 2 2 268 0 16 0 101 109. . Ω

X LL = = ( )( ) =ω 100 0 160 16 0. . Ω

XCC = =

( ) ×( ) =−

1 1

100 99 0 101016ω .

Ω

(b) IV

Zmaxmax .

.= = =∆

Ω40 0

0 367V

109A

(c) tan.

..φ = − = − = −X X

RL C 16 0 101

68 01 25:

φ = − = −0 896 51 3. .rad °

Imax .= 0 367 A

ω =100 rad s

φ = − = −0 896 51 3. .rad °

P33.21 X f LL = = ( )( ) =2 2 60 0 0 460 173π π . . Ω

Xf C

C = =( ) ×( ) =

−

1

2

1

2 60 0 21 0 10126

6π π . . Ω

(a) tan .φ = − = − =X X

RL C 173 126

0 314150

Ω ΩΩ

φ = =0 304 17 4. .rad °

(b) Since X XL C> , f is positive; so voltage leads the current .

P33.22 For the source-capacitor circuit, the rms source voltage is ∆V Xs C= ( )25 1. .mA For the circuit

with resistor, ∆V R X Xs C C= ( ) + = ( )15 7 25 12 2. . .mA mA This gives R XC= 1 247. . For the cir-

cuit with ideal inductor, ∆V X X Xs L C C= ( ) − = ( )68 2 25 1. . .mA mA So X X XL C C− = 0 368 0. .

Now for the full circuit

∆V I R X X

X I X

s L C

C C

= + −( )( ) = ( ) +

2 2

225 1 1 247 0 3. . .mA 668

19 3

2X

I

C( )= . mA

P33.23 Xf C

C = =( ) ×( ) = ×

−

1

2

1

2 60 0 20 0 101 33 10

12π π . ..

Hz F88 Ω

Z = ×( ) + ×( ) ≈ ×50 0 10 1 33 10 1 33 103 2 8 2 8. . .Ω Ω Ω

I

V

Zrmsrms

8

V

1.33 10A= =

×= × −∆

Ω5 000

3 77 10 5.

∆V I Rrms body rms body A( ) = = ×( ) ×−3 77 10 50 0 105 3. . VΩ( ) = 1 88.



P33.24 XCC = =

( ) ×( ) =−

1 1

2 50 0 65 0 1049 06ω π . .

. Ω

X LL = = ( ) ×( ) =−ω π2 50 0 185 10 58 13. . Ω

Z R X XL C= + −( ) = ( ) + −( ) =2 2 2 240 0 58 1 49 0 41 0. . . . Ω

I

V

Zmax ..= = =

∆ max A150

41 03 66

(a) ∆V I RR = = ( )( ) =max .3 66 40 146 V

(b) ∆V I XL L= = ( )( ) = =max . . .3 66 58 1 212 5 212 V

(c) ∆V I XC C= = ( )( ) = =max . . .3 66 49 0 179 1 179V V

(d) ∆ ∆V VL C− = − =212 5 179 1 33 4. . . V

P33.25 R = 300 Ω

X LL = = ⎛⎝

⎞⎠ ( ) =−ω π

π2

5000 200 2001s H. Ω

XCC = = ⎛

⎝⎞⎠ ×( )⎡

⎣⎢⎤⎦⎥

=− −−

12

50011 0 101 6

1

ωπ

πs F. 990 9. Ω

Z R X XL C= + −( ) =2 2319 Ω and

φ = −⎛⎝

⎞⎠ =−tan .1 20 0

X X

RL C °

*P33.26 Let Xc represent the initial capacitive reactance. Moving the plates to half their original separation

doubles the capacitance and cuts XCC = 1

ωin half. For the current to double, the total impedance

must be cut in half: Z Z R X X R XX

i f L C LC= + −( ) = + −⎛

⎝⎞⎠2 2

22 2 2

2

, . With XL = R, algebra

then gives

R R X R RX

R RX X

CC

C C

2 2 22

2 2

42

2 2

+ −( ) = + −⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

− + == − +

=

8 4

3

2 2R RX X

X R

C C

C

FIG. P33.24

FIG. P33.25

XL = 200 Ω

XC = 90.9 ΩR = 300 Ω

ZXL – XC = 109 Ωf


254 Chapter 33

Section 33.6 Power in an AC Circuit

P33.27 ω =1 000 rad s, R = 400 ,Ω C = × −5 00 10 6. F, L = 0 500. H

∆Vmax =100 V, ωL = 500 ,Ω 1200

ωC

⎛

⎝⎜

⎞

⎠⎟ = Ω

Z R L

C

IV

= + −⎛

⎝⎜

⎞

⎠⎟ = + =

=

2

2

2 21400 300 500ω

ω Ω

∆max

maxx .Z

= =100

5000 200 A

The average power dissipated in the circuit is P = =⎛⎝⎜

⎞⎠⎟

I RI

Rrms2 .max

2

2

P = ( ) ( ) =0 200

2400 8 00

2..

AWΩ

P33.28 Z R X XL C= + −( )2 2or X X Z RL C−( ) = −2 2

X X

X

L C

L

−( ) = ( ) − ( ) =

= −

75 0 45 0 60 02 2

1

. . .

tan

Ω Ω Ω

φ −−⎛⎝

⎞⎠ = ⎛

⎝⎞⎠ =

=

−X

R

I

C tan.

.1 60 053 1

45.0

rms

ΩΩ

∆

°

VV

Z

V I

rms

rms rms

V

75.0A= =

= ( ) =

2102 80

Ω

∆

.

cosP φ 2210 2 80 53 1 353V A W( )( ) ( ) =. cos . °

P33.29 (a) P = ( ) = ( ) −( ) =I Vrms rms∆ cos . cos . .φ 9 00 180 37 0 1 29° ××103 W

P = I Rrms2 so 1 29 10 9 003 2. .× = ( ) R and R = 16 0. Ω

(b) tanφ = −X X

RL C becomes tan . :−( ) = −

37 016

°X XL C so X XL C− = −12 0. Ω

P33.30 X LL = = ( )( ) =ω π2 60 0 0 025 0 9 42. . .s H Ω

Z R X XL C= + −( ) = ( ) + ( ) =2 2 2 220 0 9 42 22 1. . .Ω Ω

(a) IV

Zrmsrms V

22.1A= = =

∆Ω

1205 43.

(b) φ = ⎛⎝

⎞⎠ =−tan

.

..1 9 42

20 025 2° so power factor = =cos .φ 0 905

(c) We require φ = 0. Thus, X XL C= : 9 421

2 60 0 1.. s

Ω = ( )−π C

and C = 281 Fµ

(d) P Pb d= or ∆∆

V IV

Rb b bd

rms rmsrms( ) ( ) =

( )cosφ

2

∆ ∆ ΩV R V Id b b brms rms rms( ) = ( ) ( ) = ( )cos .φ 20 0 120 VV A V( )( )( ) =5 43 0 905 109. .



*P33.31 Consider a two-wire transmission line taking in power P

IVrms

rms

.= P∆

Then power loss = =I Rrms2

line

P100

.

Thus,P P

∆VR

rms

⎛⎝⎜

⎞⎠⎟

( ) =2

12100

or RV

1

2

200= ( )∆ rms

P

RA

V1

2

200= =

( )ρ ∆ rms

P or A

r

V= ( )

=( )

π ρ2

4

2002

2

P ∆ rms

and the diameter is 2800

2rV

=( )

ρπ

P ∆

(a) 2800 1 7 10 18 0008

rV

= × ( )( )

−( . Ω∆

m) 20 000 W m

π 22 39 5= ⋅. V m/∆V

(b) The diameter is inversely proportional to the potential difference.

(c) 2r = 39.5 V⋅m/1 500 V = 2.63 cm

(d) ∆V = 39.5 V⋅m/0.003 m = 13.2 kV

*P33.32 (a) XL = ωL = 2p (60/s) 0.1 H = 37.7 Ω

Z = (1002 + 37.72)1/2 = 107 Ω power factor = cosf = 100/107 = 0.936

(b) The power factor cannot in practice be made 1.00. If the inductor were removed or if the generator were replaced with a battery, so that either L = 0 or f = 0, the power factor would be 1, but we would not have a magnetic buzzer.

(c) We want resonance, with f = 0. We insert a capacitor in series with

XL = X

C so 37.7 Ω = 1 s/2pC 60 and C = 70.4 mF

P33.33 One-half the time, the left side of the generator is positive, the top diode conducts, and the bottom diode switches off. The power supply sees resistance

1

2

1

2

1

R RR+⎡

⎣⎢⎤⎦⎥

=−

and the power is∆V

Rrms( )2

.

The other half of the time the right side of the generator is positive, the upper diode is an open circuit, and the lower diode has zero resistance. The equivalent resistance is then

R RR R

Req = + +⎡

⎣⎢⎤⎦⎥

=−1

3

1 7

4

1

and P =( )

=( )∆ ∆V

R

V

Rrms

eq

rms

2 24

7

The overall time average power is: ∆ ∆ ∆V R V R Vrms rms rms

( )⎡⎣ ⎤⎦ + ( )⎡⎣ ⎤⎦ =( )2 2 24 7

2

11

14RR

FIG. P33.31

R1

R1

RL∆Vrms

FIG. P33.33

~∆V

RR R

2R


256 Chapter 33

Section 33.7 Resonance in a Series RLC Circuit

P33.34 (a) fLC

= 1

2π C

f L= =

( ) ×⎛⎝

⎞⎠ =

−

1

4

1

4 10 400 102 2 2 10 2 12π π

A

s Vs

C

As66 33 10 13. × − F

(b) CA

d d=

∈=

∈κ κ0 02

=∈

⎛⎝⎜

⎞⎠⎟

= × ×× ×

− −Cd

κ 0

1 2 13 36 33 10 10. F mm

1 8.85 1108 46 1012

1 2

3−

−⎛⎝⎜

⎞⎠⎟

= ×F

m.

(c) X f LL = = × × × =−2 2 10 400 10 25 110 12π π s Vs A . Ω

P33.35 ω π06 82 99 7 10 6 26 10

1= ×( ) = × =. . rad sLC

CL

= =×( ) ×( ) =

−

1 1

6 26 10 1 40 101 82

02 8 2 6ω . .

. pF

P33.36 L = 20 0. mH, C = × −1 00 10 7. , R = 20 0. ,Ω ∆Vmax = 100 V

(a) The resonant frequency for a series RLC circuit is fLC

= =1

2

13 56

π. .kHz

(b) At resonance, IV

Rmaxmax .= =

∆5 00 A

(c) From Equation 33.38, QL

R= =

ω0 22 4.

(d) ∆V X I LIL L, max max .max kV= = =ω0 2 24

P33.37 The resonance frequency is ω0

1=LC

. Thus, if ω ω= 2 0 ,

X LLC

LL

CL = = ⎛⎝⎜

⎞⎠⎟ =ω 2

2 and XC

LC

C

L

CC = = =1

2

1

2ω

Z R X X RL

CL C= + −( ) = + ⎛⎝

⎞⎠

2 2 2 2 25. so IV

Z

V

R L Crms

rms rms= =+ ( )

∆ ∆2 2 25.

and the energy delivered in one period is E t= P ∆ :

EV R

R L C

V RC

R C=

( )+ ( )

⎛⎝

⎞⎠ =

( )∆ ∆rms rms

2

2

2

22 25

2

.

πω ++ ( ) =

( )+2 25

4

4 9 00

2

2. .LLC

V RC LC

R C Lπ

π ∆ rms

With the values specifi ed for this circuit, this gives:

E =( ) ( ) ×( ) ×− −4 50 0 10 0 100 10 10 0 102 6 3 2π . . .V FΩ 33 1 2

2 64 10 0 100 10 9 00 10 0 10

H

F

( )( ) ×( ) + ×− −. . .Ω 33 242

HmJ( ) =



P33.38 The resonance frequency is ω0

1=LC

. Thus, if ω ω= 2 0 ,

X LLC

LL

CL = = ⎛⎝⎜

⎞⎠⎟ =ω 2

2 and XC

LC

C

L

CC = = =1

2

1

2ω

Then Z R X X RL

CL C= + −( ) = + ⎛⎝

⎞⎠

2 2 2 2 25. so IV

Z

V

R L Crms

rms rms= =+ ( )

∆ ∆2 2 25.

and the energy delivered in one period is

E tV R

R L C

V R= =

( )+ ( )

⎛⎝

⎞⎠ =

( )P ∆∆ ∆rms rms

2

2

2

2 25

2

.

πω

CC

R C LLC

V RC LC

R C L2

2

22 25

4

4 9 00+ ( ) =( )

+. .π

π ∆ rms

P33.39 For the circuit of Problem 20, ω0 3 6

1 1

160 10 99 0 10251= =

×( ) ×( )=

− −LC H Frad s

.

QL

R= =

( ) ×( )=

−ω0

3251 160 10

68 00 591

rad s H

..

Ω

For the circuit of Problem 21, QL

R

L

R LC R

L

C= = = = ×

×

−

−

ω03

6

1 1

150

460 10

10

H

21.0Ω FF= 0 987.

The circuit of Problem 21 has a sharper resonance.

Section 33.8 The Transformer and Power Transmission

P33.40 (a) ∆V2

1

13120 9 23,rms V V= ( ) = .

(b) ∆ ∆V I V I1 1 2 2, , , ,rms rms rms rms=

120 0 350 9 23 2 V A V ,rms( )( ) = ( ). . I

I2

42 04 55,

..rms

W

9.23 V A= = for a transformer with no energy loss.

(c) P = 42 0. W from part (b).

P33.41 ∆ ∆VN

NVout in V( ) = ( ) = ⎛

⎝⎞⎠ ( )

max max2

1

2 000

350170 == 971 V

∆Vout rms

VV( ) = ( )

=971

2687

P33.42 (a) ∆ ∆VN

NV2

2

11,rms ,rms( ) = ( ) N2

2 200 80

1101 600=

( )( )= windings

(b) I V I V1 1 2,rms ,rms ,rms 2,rms∆ ∆( ) = ( ) I1,rms A= ( )( ) =1 50 2 200

11030 0

..

(c) 0 950 1 2. I V I V,rms 1,rms ,rms 2,rms∆ ∆( ) = ( ) I1

1 20 2 200

110 0 95025 3,

.

..rms A= ( )( )

( )=


258 Chapter 33

P33.43 (a) R = ×( ) ×( ) =−4 50 10 6 44 10 2904 5. .M mΩ Ω and IVrms

rms5

W

5.00 10 VA= = ×

×=P

∆5 00 10

10 06.

.

Ploss rms2 A kW= = ( ) ( ) =I R 10 0 290 29 02. .Ω

(b) PPloss = ×

×= × −2 90 10

5 00 105 80 10

4

63.

..

(c) It is impossible to transmit so much power at such low voltage. Maximum power transfer occurs when load resistance equals the line resistance of 290 Ω, and is

4 50 10

2 2 29017 5

3 2.

.×( )

⋅ ( ) =V

kW,Ω

far below the required 5 000 kW.

Section 33.9 Rectifi ers and Filters

P33.44 (a) Input power = 8 W

Useful output power = = ( ) =I V∆ 0 3 2 7. .A 9 V W

effi ciency = = = =useful output

total input

W

8 W

2 70 34 34

.. %

(b) Total input power = Total output power

8 2 7

5 3

W W wasted power

wasted power W

= +

=

.

.

(c) E t= = ( )( )⎛⎝

⎞⎠

⎛⎝

⎞⎠P ∆ 8 31

86 400W 6 d

s

1 d

1 J

1 Ws== ×

×⎛⎝⎜

⎞⎠⎟ =1 29 10 4 88. $ .J

$0.135

3.6 10 J6

P33.45 (a) The input voltage is ∆V IZ I R X I RC

Cin = = + = +⎛⎝⎜

⎞⎠⎟2 2 2

21

ω. The output voltage is

∆V IRout = . The gain ratio is ∆∆V

V

IR

I R C

R

R C

out

in

=+ ( )

=+ ( )2 2 2 2

1 1ω ω.

(b) As ω → 0, 1

ωC→ ∞ and

∆∆V

Vout

in

.→ 0

As ω → ∞, 1

0ωC

→ and∆∆V

V

R

Rout

in

.→ = 1

(c) 1

2 12 2=

+ ( )R

R Cω

RC

R22 2

214+ =

ω ω 2 2

2

1

3C

R= ω π= =2

1

3f

RC f

RC= 1

2 3π



P33.46 (a) The input voltage is ∆V IZ I R X I R CCin = = + = + ( )2 2 2 21 ω . The output voltage is

∆V IXI

CCout = =

ω. The gain ratio is

∆∆V

V

I C

I R C

C

R C

out

in

.=+ ( )

=+ ( )

ω

ω

ω

ω2 2 2 21

1

1

(b) As ω → 0, 1

ωC→ ∞ and R becomes negligible in comparison. Then

∆∆V

V

C

Cout

in

→ =1

11

ωω

.

As ω → ∞, 10

ωC→ and

∆∆V

Vout

in

→ 0 .

(c) 1

2

1

12 2=

+ ( )ω

ω

C

R C R

C C2

2

2 2

1 4+⎛⎝⎜

⎞⎠⎟ =

ω ω R C2 2 2 3ω = ω π= =2

3f

RC

fRC

= 3

2π

P33.47 For this RC high-pass fi lter, ∆∆V

V

R

R XC

out

in

=+2 2

.

(a) When∆∆V

Vout

in

= 0 500. ,

then 0 5000 500

2 2

..

0.500

ΩΩ( ) +

=XC

or XC = 0 866. .Ω

If this occurs at f = 300 Hz, the capacitance is

C

f XC

= =( )( )

= × =−

1

2

1

2 300 0 866

6 13 10 64

π π Hz

F

.

.

Ω

113 Fµ

(b) With this capacitance and a frequency of 600 Hz,

XC =( ) ×( ) =−

1

2 600 6 13 100 4334π Hz F.. Ω

∆∆

ΩΩ Ω

V

V

R

R XC

out

in 0.500=

+=

( ) +2 2 2

0 500

0 433

.

.(( )=

20 756.

FIG. P33.47


260 Chapter 33

P33.48 For the fi lter circuit, ∆∆V

V

X

R XC

C

out

in

=+2 2

.

(a) At f = 600 Hz, Xf C

C = =( ) ×( ) = ×

−

1

2

1

2 600 8 00 103 32 10

9

4

π π Hz F

.. ΩΩ

and ∆∆

Ω

Ω Ω

V

Vout

in 90.0= ×

( ) + ×( )3 32 10

3 32 10

4

2 4 2

.

.≈≈ 1 00.

(b) At f = 600 kHz, Xf C

C = =×( ) ×( ) =

−

1

2

1

2 600 10 8 00 1033 2

3 9π π Hz F

.. ΩΩ

and ∆∆

ΩΩ Ω

V

Vout

in 90.0 .2=

( ) + ( )=33 2

330 346

2 2

..

P33.49 ∆∆vv

out

in

=+ −( )

R

R X XL C2 2

(a) At 200 Hz: 1

4

8 00

8 00 400 1 400

2

2 2= ( )( ) + −[ ]

.

.

ΩΩ π πL C

At 4 000 Hz: 8 00 8 0001

8 0004 8 002

2

2. .Ω Ω( ) + −⎡⎣⎢

⎤⎦⎥ = ( )π

πL

C

At the low frequency, X XL C− < 0. This reduces to 4001

40013 9π

πL

C− = − . Ω [1]

For the high frequency half-voltage point, 8 0001

800013 9π

πL

C− = + . Ω [2]

Solving Equations (1) and (2) simultaneously gives C = 54 6. Fµ and L = 580 Hµ

(b) When X XL C= , ∆∆

∆∆

vv

vv

out

in

out

in

=⎛⎝⎜

⎞⎠⎟

=max

.1 00

(c) X XL C= requires fLC0 4 5

1

2

1

2 5 80 10 5 46 10894= =

×( ) ×( )=

− −π π . .H FHHz

(d) At 200 Hz,∆∆vv

out

in

= =R

Z

1

2and X XC L> ,

so the phasor diagram is as shown:

φ = −

⎛⎝⎜

⎞⎠⎟ = −

⎛⎝⎜

⎞⎠⎟− −cos cos1 1 1

2

R

Zso

∆ ∆v vout in leads by 60.0° .

At f0 , X XL C= so

∆ ∆v vout inand have a phase difference of 0°° .

At 4 000 Hz,∆∆vv

out

in

= =R

Z

1

2and X XL C− > 0

Thus, φ = ⎛⎝

⎞⎠ =−cos .1 1

260 0°

or ∆ ∆v vout in lags by 60.0°

FIG. P33.49(d)

or

or

∆Vout

∆Vin

f

∆Vout

∆Vin

fR

ZXL – XC f

R

ZXL – XC

f

FIG. P33.49(a)

continued on next page



(e) At 200 Hz and at 4 kHz,

P =( )

=( )( )

=( ) ( )∆ ∆ ∆v v vout,rms in,rms

2 21 2 1 2 1 2

R Riin,max V

W⎡⎣ ⎤⎦ = ( )

( ) =2

210 0

8 8 001 56

R

.

..

Ω

At f0 , P =( )

=( )

=( )⎡⎣∆ ∆ ∆v v vout,rms in,rms in,max

2 21 2

R R

⎤⎤⎦ = ( )( ) =

2210 0

2 8 006

R

.

.

V.25 W

Ω

(f ) We take QL

R

f L

R= = =

( ) ×( )−ω π π

0 0

42 2 894 5 80 10

8 00

Hz H

.

. Ω== 0 408.

Additional Problems

P33.50 The equation for ∆v( )t during the fi rst period (usingy mx b= + ) is:

∆∆

∆

∆ ∆

v

v v

tV t

TV

Tt

( ) = ( ) −

( )⎡⎣

⎤⎦ = ( )⎡

2

12

maxmax

avg⎣⎣ ⎤⎦ = ( ) −

⎡⎣⎢

⎤⎦⎥∫ ∫2

0

2 2

0

21dt

V

T Tt dt

T T∆ max

∆∆

v( )⎡⎣ ⎤⎦ =( ) ⎛

⎝⎞⎠

−[ ]=

=2

2 3

02

2 1

3avg

V

T

T t T

t

t T

max ==( )

+( ) − −( )⎡⎣ ⎤⎦ =( )

=

∆ ∆

∆ ∆

V V

V

max max

2

3 3

2

61 1

3

rms vv( )⎡⎣ ⎤⎦ =( )

=2

2

3 3avg

max∆ ∆V Vmax

*P33.51 (a) Z 2 = R2 + (XL – X

C)2 760 2 = 4002 + (700 − X

C)2 417 600 = (700 − X

C)2

There are two values for the square root. We can have 646.2 = 700 − XC or −646.2 = 700 − X

C.

XC can be 53.8 Ω or it can be 1.35 kΩ.

(b) If we were below resonance, with inductive reactance 700 Ω and capacitive reactance 1.35 kΩ, raising the frequency would increase the power. We must be above resonance,

with inductive reactance 700 Ω and capacitive reactance 53.8 Ω.

(c) 7602 = 2002 + (700 − XC)2 537 600 = (700 − X

C)2 Here +733 = 700 − X

C has no solution

so we must have −733.2 = 700 − XC and X

C = 1.43 kΩ .

FIG. P33.50


262 Chapter 33

P33.52 The angular frequency is ω π= =2 60 377s s. When S is open, R, L, and C are in series with the source:

R X XV

IL Cs2 2

2 2201 19+ −( ) = ⎛

⎝⎞⎠ = ⎛

⎝⎞⎠ =∆ V

0.183 A. 44 104× 2Ω (1)

When S is in position 1, a parallel combination of two R’s presents equivalent resistanceR

2, in

series with L and C:

RX XL C2

204 504 10

22

23⎛

⎝⎞⎠ + −( ) = ⎛

⎝⎞⎠ = ×V

0.298 A. 2Ω (2)

When S is in position 2, the current by passes the inductor. R and C are in series with the source:

R XC2 2

2420

2 131 10+ = ⎛⎝

⎞⎠ = ×V

0.137 A2. Ω (3)

Take equation (1) minus equation (2):

3

47 440 102 3R = ×. 2Ω R = 99 6. Ω

Only the positive root is physical. We have shown than only one resistance value is possible. Now equation (3) gives

XC

C = × − ( )⎡⎣

⎤⎦ = =2 131 10 99 6 106 7

14 2 1 2

. . . Ω Ωω

Only the positive root is physical and only

one capacitance is possible.

C X CC= ( ) = ( )⎡⎣ ⎤⎦ = × =− − −ω 1 1 5377 106 7 2 49 10s F. .Ω

Now equation (1) gives

X X

X

L C

L

− = ± × − ( )⎡⎣

⎤⎦ = ±1 194 10 99 6 44 994 2 1 2

. . . Ω Ω

== + = =

= =

106 7 44 99 61 74. . . or 151.7 Ω Ω Ω Ω ω

ω

L

LXL 00 164. H or 0.402 H = L

Two values for self-inductance are possible.

P33.53 ω0 6

11 1

0 050 0 5 00 102 000= =

( ) ×( )=

−

−

LC . .H Fs

so the operating angular frequency of the circuit is

ω ω= = −0 1

21 000 s

Using Equation 33.37, P =( )

+ −( )∆V R

R L

rms

2 2

2 2 2 202 2

ωω ω ω

P =( ) ( )( )

( ) ( ) + (400 8 00 1 000

8 00 1 000 0 050 0

2 2

2 2

.

. . )) −( ) ×⎡⎣ ⎤⎦=

2 6 21 00 4 00 10

56 7. .

. WFIG. P33.53



*P33.54 (a) At the resonance frequency XL and X

C are equal. The certain frequency must be higher

than the resonance frequency for the inductive reactance to be the greater.

(b) It is possible to determine the values for L and C, because we have three independent equations in the three unknowns L, C, and the unknown angular frequency w. The equations are

2 0002 = 1/LC 12 = wL and 8 = 1/wC

(c) We eliminate w = 12/L to have 8 wC = 1 = 8(12/L)C = 96C/L so L = 96C

Then 4 000 000 = 1/96 C 2 so C = 51.0 mF and L = 4.90 mH

*P33.55 The lowest-frequency standing-wave state is NAN. The distance between the clamps we

represent as d d= =NN .λ2

The speed of transverse waves on the string is v = = =fT

f dλµ

2 .

The magnetic force on the wire oscillates at 60 Hz, so the wire will oscillate in resonance at 60 Hz.

Td

0 01960 4

2 2

. kg ms= ( ) T d= ⋅( )274 2kg m s2

Any values of T and d related according to this expression will work, including

if m Nd T= =0 200 10 9. . . We did not need to use the value of the current and magnetic

fi eld. If we assume the subsection of wire in the fi eld is 2 cm wide, we can fi nd the rms value of the magnetic force:

F I B TB = = ( )( )( ) = sin . . sin .θ 9 0 02 0 015 3 90 2 7A m ° 55 mN

So a small force can produce an oscillation of noticeable amplitude if internal friction is small.

*P33.56 φ ω ω= −⎛⎝

⎞⎠

−tan/1 1L C

Rchanges from −90° for w = 0 to 0

at the resonance frequency to +90° as w goes to infi nity. The slope of the graph is df/dw :

d

d L C

R

RL

C

φω ω ω ω

=+ −⎛

⎝⎜

⎞⎠⎟

− −⎛⎝⎜

⎞⎠⎟

1

11

1 11

12 2

/( ) At resonance

we have w0L = 1/w

0C and LC = 1/w

02.

Substituting, the slope at the resonance point is

d

d RL

CLC

L

R

Qφω ωω0

1

1 0

1 1 2 22

0

=+

+⎛⎝

⎞⎠ = =

P33.57 (a) When ωL is very large, the bottom branch carries negligible current. Also, 1

ωCwill be

negligible compared to 200 Ω and45 0

225. V

200mA

Ω= fl ows in the power supply and the

top branch.

(b) Now 1

ωC→ ∞ and ωL → 0 so the generator and bottom branch carry 450 mA .

FIG. P33.56

ww0

f

90°

−90°


264 Chapter 33 264 Chapter 33

P33.58 (a) With both switches closed, the current goes only through generator and resistor.

i tV

Rt( ) =

∆ max cosω

(b) P =( )1

2

2∆V

Rmax

(c) i tV

R Lt

L

R( ) =

++ ⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

−∆ max cos tan2 2 2

1

ωω ω

(d) For 011 0 0= =

− ( )⎛⎝⎜

⎞⎠⎟

−φω ω

tanL C

R

We require ωω0

0

1L

C= , so C

L= 1

02ω

(e) At this resonance frequency, Z R=

(f ) U C V CI XC C= ( ) =1

2

1

22 2 2∆

U CI X CV

R C

VCmax max

max max= =( )

=(1

2

1

2

12 2

2

202 2

∆ ∆ω

))2

22

L

R

(g) U LI LV

Rmax maxmax= =

( )1

2

1

22

2

2

∆

(h) Now ω ω= =22

0LC

.

So φω ω

=− ( )⎛

⎝⎜⎞⎠⎟

=− ( )⎛

⎝⎜⎞− −tan tan1 11 2 1 2L C

R

L C L C

R ⎠⎠⎟=

⎛⎝⎜

⎞⎠⎟

−tan .1 3

2R

L

C

(i) Now ωω

LC

= 1

2

1 ω ω

= =1

2 20

LC

P33.59 (a) IV

RR, .rms

rms V

80.0 A= = =∆

Ω100

1 25

(b) The total current will lag the applied voltage as seen in the phasordiagram at the right.

IV

XL

L

,rmsrms V

2 60.0 s H= = ( )( )

=−

∆ 100

0 2001

1π ...33 A

Thus, the phase angle is: φ =⎛

⎝⎜⎜

⎞

⎠⎟⎟ =− −tan tan

.,1 1 1 33I

IL

R

rms

,rms

A

1.25 A.

⎛⎝⎜

⎞⎠⎟ = 46 7. °

FIG. P33.58

FIG. P33.59

IR

IL

∆V

I

f



P33.60 Suppose each of the 20 000 people uses an average power of 500 W. (This means 12 kWh per day, or $36 per 30 days at 10c| per kWh.) Suppose the transmission line is at 20 kV. Then

IVrms

rms

W

VA= =

( )( )P∆

20 000 500

20 000103~

If the transmission line had been at 200 kV, the current would be only ~ .102 A

P33.61 R = 200 Ω, L = 663 mH, C = 26 5. F,µ ω = −377 1s , ∆Vmax .= 50 0 V

ω L = 250 Ω,1

100ωC

⎛

⎝⎜

⎞

⎠⎟ = Ω, Z R X XL C= + −( ) =2 2

250 Ω

(a) IV

Zmaxmax .

.= = =∆

Ω50 0

0 200V

250A

φ = −⎛⎝

⎞⎠ =−tan .1 36 8

X X

RL C ° (∆V leads I )

(b) ∆V I RR,max V= =max .40 0 at φ = 0°

(c) ∆VI

CC ,

max .max V= =ω

20 0 at φ = −90 0. ° (I leads ∆V )

(d) ∆V I LL , max . max V= =ω 50 0 at φ = +90 0. ° (∆V leads I )

P33.62 L = 2 00. H, C = × −10 0 10 6. F, R = 10 0. ,Ω ∆v t t( ) = ( )100sinω

(a) The resonant frequencyω0 produces the maximum current and thus the maximum power delivery to the resistor.

ω0 6

1 1

2 00 10 0 10224= =

( ) ×( )=

−LC . .rad s

(b) P =( )

= ( )( ) =

∆V

Rmax

.

2 2

2

100

2 10 0500 W

(c) IV

Z

V

R L Crms

rms rms= =+ − ( )( )

∆ ∆2 2

1ω ω

and IV

Rrmsrms( ) =

max

∆

I R I Rrms2

rms2= ( )1

2 max or

∆ ∆V

ZR

V

RRrms rms( )

=( )2

2

2

2

1

2

This occurs where Z R2 22= : R LC

R2

2

212+ −

⎛⎝⎜

⎞⎠⎟

=ωω

ω ω ω4 2 2 2 2 2 22 1 0L C L C R C− − + = or L C LC R C2 2 4 2 2 22 1 0ω ω− +( ) + =

2 00 10 0 10 2 2 00 10 0 102 6 2 4 6. . . .( ) ×( )⎡⎣

⎤⎦ − ( ) ×(− −ω )) + ( ) ×( )⎡

⎣⎤⎦ + =−10 0 10 0 10 1 02 6 2 2. . ω

Solving this quadratic equation, we fi nd that ω 2 51130= , or 48 894

ω1 48 894 221= = rad s and ω2 51130 226= = rad s


266 Chapter 33

P33.63 (a) From Equation 33.41, N

N

V

V1

2

1

2

= ∆∆

.

Let input impedance ZV

I11

1

= ∆ and the output impedance Z

V

I22

2

= ∆

so that N

N

Z I

Z I1

2

1 1

2 2

= But from Eq. 33.42, I

I

V

V

N

N1

2

2

1

2

1

= =∆∆

So, combining with the previous result we haveN

N

Z

Z1

2

1

2

= .

(b) N

N

Z

Z1

2

1

2

8 000

8 0031 6= = =

..

P33.64 P = = ⎛⎝

⎞⎠I R

V

ZRrms

2 rms∆ 2

, so 250120

40 02

2 W

V = ( ) ( )

Z. :Ω Z R L

C= + −

⎛⎝⎜

⎞⎠⎟

2

21ω

ω

250120 40 0

40 0 2 0 185 1 2 65 0

2

2= ( ) ( )

( ) + ( ) −

.

. . .π πf f ××( )⎡⎣ ⎤⎦⎡⎣

⎤⎦

−10 62

and

250576 000

1 600 1 162 4 2 448 5

2

2 2 2=+ −( )

f

f f. .

12 304

1 600 1 3511 5 692 3 5 995 300

2

2 4 2=+ − +

f

f f f. . so 1 3511 6 396 3 5 995 300 04 2. .f f− + =

f 2

26 396 3 6 396 3 4 1 3511 5 995 300

2 1 35=

± ( ) − ( )( ). . .

. 1113 446 5 1 287 4( ) = . .or

f = 58 7. Hz or 35.9 Hz There are two answers because we could be above resonance or

below resonance.

P33.65 IV

RR =∆ rms ; I

V

LL =∆ rms

ω; I

V

CC =

( )−∆ rms

ω 1

(a) I I I I VR

CL

R C Lrms rms= + −( ) =⎛⎝⎜

⎞⎠⎟+ −

⎛⎝⎜

⎞2 2

2

1 1∆ ωω ⎠⎠

⎟2

(b) tanφ = − = −⎡

⎣⎢

⎤

⎦⎥⎛⎝⎜

⎞⎠

I I

IV

X X V RC L

R C L

∆∆rms

rms

1 1 1⎟⎟

tanφ = −⎡

⎣⎢

⎤

⎦⎥R

X XC L

1 1

FIG. P33.65



*P33.66 An RLC series circuit, containing a 35.0-Ω resistor, a 205-mH inductor, a capacitor, and a power supply with rms voltage 200 V and frequency 100 Hz, carries rms current 4.00 A. Find the capacitance of the capacitor.

We solve for C 2500 = 352 + (129 − 1/628C)2 1275 = (129 − 1/628C)2

There are two possibilities: 35.7 = 129 − 1/628C and −35.7 = 129 − 1/628C 1/628C = 93.1 or 1/628C = 164.5

C = either 17.1 mF or 9.67 mF

P33.67 (a) X XL C= = 1884 Ω when f = 2 000 Hz

LX

fL= = =

2

1 8840 150

π π

4 000 rad s H

Ω. and

Cf XC

=( )

=( )( )

=1

2

1

4 000 1 88442 2

π π rad s nF

Ω.

X fL = ( )2 0 150π . H Xf

C =( ) ×( )−

1

2 4 22 10 8π . F

Z X XL C= ( ) + −( )40 0 2 2. Ω

(b) Impedance, Ω

FIG. P33.66

f X X ZL C (Hz)

300

600

800

1 000

1 500

2 00

( ) ( ) ( )Ω Ω Ω

00

3 000

4 000

6 000

10 000

283

565

754

942

1 410

1 880

2 830

33 770

5 650

9 420

12 600

6 280

4 710

3 770

2 510

1 880

1 260

9442

628

377

1 2300

5 720

3 960

2 830

1100

40

1 570

2 830

5 0200

9 040

FIG. P33.67(b)


268 Chapter 33

P33.68 ω061

1 00 10= = ×LC

. rad s

For each angular frequency, we fi nd

Z R L C= + −( )2 21ω ω

then IZ

= 1 00. V

and P = ( )I 2 1 00. Ω

The full width at half maximum is:

∆ ∆

∆

f

f

= =−( )

= × −

ωπ

ωπ2

1 000 5 0 999 5

2

1 00 10

2

0

3 1

. .

. s

ππ= 159 Hz

while

R

L2

1 001593π π

=×( ) =−

.

2 1.00 10 HHz

Ω

*P33.69 (a) We can use sin sin sin cosA BA B A B+ = +⎛

⎝⎞⎠ −⎛

⎝⎞⎠2

2 2 2 2to fi nd the sum of the two sine

functions to be

E E t

E E1 2

1 2

24 0 4 5 35 0 35 0+ = ( ) +( )+

. sin . . cos .cm ° °

== ( ) +( )19 7 4 5 35 0. sin . .cm t °

Thus, the total wave has amplitude 19 7. cm and has a constant phase difference of

35 0. ° from the fi rst wave.

(b) In units of cm, the resultant phasor is

y y y i iR = + = ( )+ ° +1 2 12 0 12 0 70 0 12 0. ˆ . cos . ˆ . sin 770 0

16 1 11 3

16 1 11 32 2

. ˆ

. ˆ . ˆ

. .

j

i j

y

( )= +

= ( ) + ( )R aat cm at 35.0tan

.

..− ⎛

⎝⎜

⎞⎠⎟ =1 11 3

16 119 7 °

The answers are identical.

ωω

ωω0

2LC

Z I R 1

W

0.9990

0.9991

0

Ω Ω Ω( ) ( ) ( ) = ( )P

..9993

0.9995

0.9997

0.9999

1.0000

1.0001

1.0003

1..0005

1.0007

1.0009

1.0010

999.0

999.1

999..3

999.5

999.7

999.9

1000

1000.1

1000.3

10000.5

1000.7

1000.9

1001

1001.0

1000.9

1000.7

1000.55

1000.3

1000.1

1000.0

999.9

999.7

999.5

9999.3

999.1

999.0

2.24

2.06

1.72

1.41

1.17

1.022

1.00

1.02

1.17

1.41

1.72

2.06

2.24

0.19984

0.235699

0.33768

0.49987

0.73524

0.96153

1.00000

0.961544

0.73535

0.50012

0.33799

0.23601

0.20016

continued on next page

FIG. P33.68

1.0

0.8

0.6

0.4

0.2

0.00.996 0.998 1 1.002 1.004

I2R(W)

w/w0

FIG. P33.69(b)

y1

y2

70.0°

yR

αx

y w t



(c)

y i jR = +

+

12 0 70 0 12 0 70 0

15 5 80

. cos . ˆ . sin . ˆ

. cos

° °

.. ˆ . sin . ˆ

. cos ˆ . sin

0 15 5 80 0

17 0 160 17 0

° °

°

i j

i

−

+ + 1160° j

y i jR = − + =9 18 1 83 9 36. ˆ . ˆ . cm at 169°

The wave function of the total wave is

y x tR = ( ) − +( )9 36 15 4 5 169. sin . .cm °

ANSWERS TO EVEN PROBLEMS

P33.2 (a)193 Ω (b) 144 Ω

P33.4 (a) 25.3 rad/s (b) 0.114 s

P33.6 3.38 W

P33.8 3 14. A

P33.10 3.80 J

P33.12 (a) greater than 41 3. Hz (b) less than 87 5. Ω

P33.14 2C V∆ rms( )

P33.16 –32.0 A

P33.18 2 79. kHz

P33.20 (a) 109 Ω (b) 0 367. A (c) Imax .= 0 367 A, ω = 100 rad s, φ = −0 896. rad

P33.22 19.3 mA

P33.24 (a) 146 V (b) 212 V (c) 179 V (d) 33 4. V

P33.26 Cutting the plate separation in half doubles the capacitance and cuts in half the capacitive reactance to X

C/2. The new impedance must be half as large as the old impedance for the new

current to be doubled. For the new impedance we then have

(R2 + [R − XC/2]2)1/2 = 0.5(R2 + [R − X

C]2)1/2. Solving yields X

C = 3R.

P33.28 353 W

P33.30 (a) 5 43. A (b) 0 905. (c) 281 Fµ (d) 109 V

P33.32 (a) 0.936 (b) Not in practice. If the inductor were removed or if the generator were replaced with a battery, so that either L = 0 or f = 0, the power factor would be 1, but we would not have a magnetic buzzer. (c) 70.4 mF

P33.34 (a) 633 fF (b) 8.46 mm (c) 25.1 Ω

FIG. P33.69(c)

y1

y2yR

x

ykx - w t

y3


270 Chapter 33

P33.36 (a) 3 56. kHz (b) 5 00. A (c) 22 4. (d) 2 24. kV

P33.38 4

4 9

2

2

π ∆V RC LC

R C Lrms( )

+P33.40 (a) 9 23. V (b) 4 55. A (c) 42 0. W

P33.42 (a) 1 600 turns (b) 30 0. A (c) 25 3. A

P33.44 (a) 0.34 (b) 5.3 W (c) $4.8

P33.46 (a) See the solution. (b) 1; 0 (c) 3

2π RC

P33.48 (a) 1.00 (b) 0.346

P33.50 See the solution.

P33.52 Only one value for R and only one value for C are possible. Two values for L are possible. R = 99.6 Ω, C = 24.9 mF, and L = 164 mH or 402 mH

P33.54 (a) Higher. At the resonance frequency XL = X

C. As the frequency increases, X

L goes up and

XC goes down. (b) It is. We have three independent equations in the three unknowns L, C, and

the certain f. (c) L = 4.90 mH and C = 51.0 mF


P33.58 (a) i tV

Rt( ) =

∆ max cosω (b) P =( )∆V

Rmax

2

2 (c) i t

V

R Lt

L

R( ) =

++ ⎛

⎝⎞⎠

⎡⎣⎢

⎤⎦⎥

−∆ max cos tan2 2 2

1

ωω ω

(d) CL

= 1

02ω

(e) Z R= (f)∆V L

Rmax( )2

22 (g)

∆V L

Rmax( )2

22 (h) tan− ⎛

⎝⎜⎞⎠⎟

1 3

2R

L

C (i) 1

2LC

P33.60 ~103 A

P33.62 (a) 224 rad s (b) 500 W (c) 221 rad s and 226 rad s

P33.64 The frequency could be either 58.7 Hz or 35.9 Hz. We can be either above or below resonance.

P33.66 An RLC series circuit, containing a 35.0-Ω resistor, a 205-mH inductor, a capacitor, and a power supply with rms voltage 200 V and frequency 100 Hz, carries rms current 4.00 A. Find the capacitance of the capacitor. Answer: It could be either 17.1 mF or 9.67 mF.



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