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33

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 Rectifiers and Filters

Alternating Current Circuits

ANSWERS TO QUESTIONS

Q33.1 If the current is positive half the time and negative half thetime, the average current can be zero. The rms current is notzero. By squaring all of the values of the current, they allbecome positive. The average (mean) of these positive values isalso positive, as is the square root of the average.

Q33.2 ∆∆

VV

avg =max

2, ∆

∆V

Vrms =

max

2

Q33.3 AC ammeters and voltmeters read rms values. With anoscilloscope you can read a maximum voltage, or test whetherthe average is zero.

Q33.4 Suppose the voltage across an inductor varies sinusoidally.Then the current in the inductor will have its instantaneous

peak positive value 14

cycle after the voltage peaks. The voltage

is zero and going positive 14

cycle (90°) before the current is

zero and going positive.

Q33.5 If it is run directly from the electric line, a fluorescent light tube can dim considerably twice in everycycle of the AC current that drives it. Looking at one sinusoidal cycle, the voltage passes throughzero twice. We don’t notice the flickering due to a phenomenon called retinal imaging. We do notnotice that the lights turn on and off since our retinas continue to send information to our brainsafter the light has turned off. For example, most TV screens refresh at between 60 to 75 times persecond, yet we do not see the evening news flickering. Home video cameras record information atfrequencies as low as 30 frames per second, yet we still see them as continuous action. A vividdisplay of retinal imaging is that persistent purple spot you see after someone has taken a picture ofyou with a flash camera.

Q33.6 The capacitive reactance is proportional to the inverse of the frequency. At higher and higherfrequencies, the capacitive reactance approaches zero, making a capacitor behave like a wire. As thefrequency goes to zero, the capacitive reactance approaches infinity—the resistance of an opencircuit.

Q33.7 The second letter in each word stands for the circuit element. For an inductor L, the emf ε leads thecurrent I—thus ELI. For a capacitor C, the current leads the voltage across the device. In a circuit inwhich the capacitive reactance is larger than the inductive reactance, the current leads the sourceemf—thus ICE.

265

266 Alternating Current Circuits

Q33.8 The voltages are not added in a scalar form, but in a vector form, as shown in the phasor diagramsthroughout the chapter. Kirchhoff’s loop rule is true at any instant, but the voltages across differentcircuit elements are not simultaneously at their maximum values. Do not forget that an inductor caninduce an emf in itself and that the voltage across it is 90° ahead of the current in the circuit in phase.

Q33.9 In an RLC series circuit, the phase angle depends on the source frequency. At very low frequencythe capacitor dominates the impedance and the phase angle is near –90°. The phase angle is zero atthe resonance frequency, where the inductive and capacitive reactances are equal. At very highfrequencies φ approaches + °90 .

Q33.10 − °≤ ≤ °90 90φ . The extremes are reached when there is no significant resistance in the circuit.

Q33.11 The resistance remains unchanged, the inductive resistance doubles, and the capacitive reactance isreduced by one half.

Q33.12 The power factor, as seen in equation 33.29, is the cosine of the phase angle between the current andapplied voltage. Maximum power will be delivered if ∆V and I are in phase. If ∆V and I are 90° outof phase, the source voltage drives a net current of zero in each cycle and the average power is zero.

Q33.13 The person is doing work at a rate of P = Fv cosθ . One can consider the emf as the “force” thatmoves the charges through the circuit, and the current as the “speed” of the moving charges. Thecosθ factor measures the effectiveness of the cause in producing the effect. Theta is an angle in realspace for the vacuum cleaner and phi is the analogous angle of phase difference between the emfand the current in the circuit.

Q33.14 As mentioned in Question 33.5, lights that are powered by alternating current flicker or get slightlybrighter and dimmer at twice the frequency of the AC power source. Even if you tried using twobanks of lights, one driven by AC 180° of phase from the other, you would not have a stable lightsource, but one that exhibits a “ripple” in intensity.

Q33.15 In 1881, an assassin shot President James Garfield. The bullet was lost in his body. AlexanderGraham Bell invented the metal detector in an effort to save the President’s life. The coil is preservedin the Smithsonian Institution. The detector was thrown off by metal springs in Garfield’s mattress, anew invention itself. Surgeons went hunting for the bullet in the wrong place and Garfield died.

Q33.16 As seen in Example 33.8, it is far more economical to transmit at high voltage than at low voltage, asthe I R2 loss on the transmission line is significantly lower. Transmitting power at high voltagepermits the use of step-down transformers to make “low” voltages and high currents available to theend user.

Q33.17 Insulation and safety limit the voltage of a transmission line. For an underground cable, thethickness and dielectric strength of the insulation between the conductors determines the maximumvoltage that can be applied, just as with a capacitor. For an overhead line on towers, the designermust consider electrical breakdown of the surrounding air, possible accidents, sparking across theinsulating supports, ozone production, and inducing voltages in cars, fences, and the roof gutters ofnearby houses. Nuisance effects include noise, electrical noise, and a prankster lighting a hand-heldfluorescent tube under the line.

Q33.18 No. A voltage is only induced in the secondary coil if the flux through the core changes in time.

Chapter 33 267

Q33.19 This person needs to consider the difference between the power delivered by a power plant andI R2 losses in transmission lines. At lower voltages, transmission lines must carry higher currents totransmit the same power, as seen in Example 33.8. The high transmitted current at low voltageactually results in more internal energy production than a lower current at high voltage. In his

formula ∆V

Ra f2

, the ∆V does not represent the line voltage but the potential difference between the

ends of one conductor. This is very small when the current is small.

Q33.20 The Q factor determines the selectivity of the radio receiver. For example, a receiver with a very lowQ factor will respond to a wide range of frequencies and might pick up several adjacent radiostations at the same time. To discriminate between 102.5 MHz and 102.7 MHz requires a high-Qcircuit. Typically, lowering the resistance in the circuit is the way to get a higher quality resonance.

Q33.21 Both coils are wrapped around the same core so that nearly all of the magnetic flux created by theprimary passes through the secondary coil, and thus induces current in the secondary when thecurrent in the primary changes.

Q33.22 The frequency of a DC signal is zero, making the capacitive reactance at DC infinite. The capacitorthen acts as an open switch. An AC signal has a non-zero frequency, and thus the capacitivereactance is finite, allowing a signal to pass from Circuit A to Circuit B.

SOLUTIONS TO PROBLEMS

Section 33.1 AC Sources

Section 33.2 Resistors in an AC Circuit

P33.1 ∆ ∆ ∆v t V t V t t ta f b g b g a f a f a f= = = =max sin sin sin sinω ω π2 200 2 2 100 283 628rms V

P33.2 ∆Vrms V2

V= =170

120

(a) P = → = =∆

ΩV

RRrms V

W

b g a f2 212075 0

193.

(b) R = =120100

1442 V

W

a fΩ

P33.3 Each meter reads the rms value.

∆

∆Ω

V

IVR

rms

rmsrms

V2

V

V24.0

A

= =

= = =

10070 7

70 72 95

.

..

FIG. P33.3


P33.4 (a) ∆ ∆v V tR = max sinω

∆ ∆v VR = 0 250. maxb g, so sin .ω t = 0 250 , or ω t = −sin .1 0 250a f.The smallest angle for which this is true is ω t = 0 253. rad Thus, if t = 0 010 0. s ,

ω = =0 253

25 3.

. rad

0.010 0 s rad s .

(b) The second time when ∆ ∆v VR = 0 250. maxb g, ω t = −sin .1 0 250a f again. For this occurrence,ω πt = − =0 253 2 89. . rad rad (to understand why this is true, recall the identitysin sinπ θ θ− =a f from trigonometry). Thus,

t = =2 89

0 114.

. rad

25.3 rad s s .

P33.5 i I tR = max sinω becomes 0 600 0 007 00. sin .= ωb g .Thus, 0 007 00 0 600 0 6441. sin . .b g a fω = =−

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

P33.6 P = I Vrms rms∆b g and ∆Vrms V= 120 for each bulb (parallel circuit), so:

I IV1 2

1 1501 25= = = =

P∆ rms

W120 V

A. , and RVI

R11

2120

96 0= = = =∆

Ωrms V1.25 A

.

IV3

3 1000 833= = =

P∆ rms

W120 V

A. , and RVI3

3

120144= = =

∆Ωrms V

0.833 A .

P33.7 ∆Vmax .= 15 0 V and Rtotal = + =8 20 10 4 18 6. . .Ω Ω Ω

IV

RI

I R

maxmax .

.

.. .

= = = =

= =FHG

IKJ =

∆Ω

Ω

totalrms

speaker rms2

speaker

V18.6

A

A2

W

15 00 806 2

0 80610 4 3 38

2

P a f

Section 33.3 Inductors in an AC Circuit

P33.8 For Imax .= 80 0 mA , Irms mA2

mA= =80 0

56 6.

.

XVI

X fL LX

f

L

LL

b g

a f

min

.

.

= = =

= → = ≥ ≥

rms

rms

V0.056 6 A

2 20.0

H

50 0884

22

8847 03

Ω

Ωπ

π π

Chapter 33 269

P33.9 (a) XVIL = = =∆

Ωmax

max ..

1007 50

13 3

LXL= = = =ω π

13 32 50 0

0 042 4 42 4.

.. .a f H mH

(b) XVIL = = =∆

Ωmax

max ..

1002 50

40 0

ω = =×

=−XL

L 40 042 4 10

9423.

. rad s

P33.10 At 50.0 Hz, X LX

LL= =

FHG

IKJ = =2 50 0 2 50 0

2 60 050 060 0

54 0 45 060 0π ππ

. ..

.

.. .. Hz Hz

Hz Hza f a f a f a fΩ Ω

IVX

V

XL Lmax

max

..= = = =

∆ ∆

Ω

2 2 10045 0

3 14rms V

Ab g a f

.

P33.11 i tV

LtL a f

a f a fb gb ge j

= −FHGIKJ =

−

× −

∆ max sin. sin . .

. .ωω

π π π

π2

80 0 65 0 0 015 5 2

65 0 70 0 10 3

V

rad s H

i tL a f a f a f= =5 60 1 59 5 60. sin . . A rad A

P33.12 ω π π= = =2 2 60 0 377f . s rad sb gX L

IVX

I I

i t I t

U Li

L

L

= = ⋅ =

= = =

= = =

= = ⋅FHG

IKJ = °=

= = ⋅ =

ω

ωπ

377 0 020 0 7 54

12015 9

2 2 15 9 22 5

22 52 60 0

22 5 120 19 5

12

12

0 020 0 19 5 3 802 2

s V s A

V7.54

A

A A

As

1 s180

A A

V s A A J

rmsrms

rms

b gb g

a fa f a f a f a f

b ga f

. .

.

. .

sin . sin.

. sin .

. . .

max

max

Ω

∆Ω

P33.13 LN

IB=

Φ where ΦB is the flux through each turn. N LI

X V

XBL L

LΦ

∆, max max

, max= =

ωd i

NV

fBL

Φ∆

, .. max

, rms 2 V s2

T C mN s

N mJ

JV C

T m= =⋅ ⋅ ⋅

⋅FHG

IKJ

⋅FHGIKJ ⋅FHGIKJ = ⋅

2

2120

60 00 450

d ia fπ π

.


Section 33.4 Capacitors in an AC Circuit

P33.14 (a) Xf CC =

12π

: 1

2 22 0 10175

6π f . ×<

−e j Ω

1

2 22 0 10 1756π . ×<

−e ja ff f > 41 3. Hz

(b) XCC ∝1

, so X X4412

22a f a f= : XC < 87 5. Ω

P33.15 I IV

XV f C

Cmax = = =2

22 2rms

rmsrms

∆∆

b g b g π

(a) Imax . .= × =−2 120 2 60 0 2 20 10 1416 V s C V mAa f b ge jπ

(b) Imax . .= × =−2 240 2 50 0 2 20 10 2356 V s F mAa f b ge jπ

P33.16 Q C V C V C Vmax max= = =∆ ∆ ∆b g b g b g2 2rms rms

P33.17 I V Cmax max . . .= = × =− −∆b g a fa fe je jω π48 0 2 90 0 3 70 10 1001 6 V s F mA

P33.18 XCC = =

×=

−

1 1

2 60 0 1 00 102 65

3ω π . ..

s C V b ge jΩ

v t V tC a f = ∆ max sinω , to be zero at t = 0

iVX

tCC

= + = + °LNM

OQP= °+ ° = −

−

−∆

Ωmax sin

.sin . . sin . .ω φ πb g a f a f a f2 120

2 652

60180

90 0 64 0 120 90 0 32 01

1

V

s s

A A

Section 33.5 The RLC Series Circuit

P33.19 (a) X LL = = × =−ω π2 50 0 400 10 1263.a fe j Ω

XC

Z R X X

V I Z

C

L C

= =×

=

= + − = + − =

= = × =

−

−

1 1

2 50 0 4 43 10719

500 126 719 776

250 10 776 194

6

2 2 2 2

3

ω π . .

max max

a fe jb g a fe ja f

V

Ω

Ω

∆ FIG. P33.19

(b) φ =−F

HGIKJ =

−FHG

IKJ = − °− −tan tan .1 1 126 719

50049 9

X XR

L C . Thus, the Current leads the voltage.

P33.20 ωω

ωLC LC

= → = =× ×

= ×− −

1 1 1

57 0 10 57 0 101 75 10

6 6

4

. ..

e je j rad s

f = =ωπ2

2 79. kHz

Chapter 33 271

P33.21 (a) X LL = = × =− −ω π2 50 0 250 10 78 51 3. . s H e je j Ω

(b) XCC = = × =− − −1

2 50 0 2 00 10 1 591 61

ωπ . . . s F ke je j Ω

(c) Z R X XL C= + − =2 2 1 52b g . kΩ

(d) IVZmaxmax= =

×=

∆Ω

210138

V1.52 10

mA3

(e) φ =−L

NMOQP = − = − °− −tan tan . .1 1 10 1 84 3

X XR

L C a f

P33.22 (a) Z R X XL C= + − = + − =2 2 2 268 0 16 0 101 109b g a f. . Ω

X L

XC

L

C

= = =

= =×

=−

ω

ω

100 0 160 16 01 1

100 99 0 10101

6

a fa f

a fe j

. .

.

Ω

Ω

(b) IVZmaxmax .

.= = =∆

Ω40 0

0 367 V

109 A

(c) tan.

..φ =

−=

−= −

X XR

L C 16 0 10168 0

1 25 :

φ = − = − °0 896 51 3. . rad

Imax .= 0 367 A ω = 100 rad s φ = − = − °0 896 51 3. . rad

P33.23 X f LL = = =2 2 60 0 0 460 173π π . .a fa f Ω

Xf CC = =

×=

−

12

1

2 60 0 21 0 10126

6π π . .a fe j Ω

(a) tan .φ =−

=−

=X X

RL C 173 126

0 314 150 Ω Ω

Ωφ = = °0 304 17 4. . rad

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

*P33.24 For the source-capacitor circuit, the rms source voltage is ∆V Xs C= 25 1. mAa f . For the circuit with

resistor, ∆V R X Xs C C= + =15 7 25 12 2. . mA mAa f a f . This gives R XC= 1 247. . For the circuit with idealinductor, ∆V X X Xs L C C= − =68 2 25 1. . mA mAa f a f . So X X XL C C− = 0 368 0. . Now for the full circuit

∆V I R X X

X I X X

I

s L C

C C C

= + −

= +

=

2 2

2 225 1 1 247 0 368

19 3

b ga f b g b g. . .

.

mA

mA


P33.25 Xf CC = =

×= ×

−

12

1

2 60 0 20 0 101 33 10

128

π π . ..

Hz F a fe jΩ

Z

IVZ

V I R

= × + × ≈ ×

= =×

= ×

= = × × =

−

−

50 0 10 1 33 10 1 33 10

5 0003 77 10

3 77 10 50 0 10 1 88

3 2 8 2 8

5

5 3

. . .

.

. . .

V1.33 10

A

A V

rmsrms

8

rms body rms body

Ω Ω Ω

∆Ω

∆ Ω

e j e j

b g e je j

P33.26 XCC = =

×=

−

1 1

2 50 0 65 0 1049 0

6ω π . ..a fe j

Ω

X L

Z R X X

IVZ

L

L C

= = × =

= + − = + − =

= = =

−ω π2 50 0 185 10 58 1

40 0 58 1 49 0 41 0

15041 0

3 66

3

2 2 2 2

. .

. . . .

..max

a fe jb g a f a f

Amax

Ω

Ω

∆

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

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

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

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

FIG. P33.26

P33.27 R = 300 Ω

X LL = = FHGIKJ =−ω π

π2

5000 200 2001 s H .a f Ω

XCC = = F

HGIKJ ×

LNM

OQP =− −−

12

50011 0 10 90 91 6

1

ωπ

π s F . .e j Ω

Z R X XL C= + − =2 2 319b g Ω and

φ =−F

HGIKJ = °−tan .1 20 0

X XR

L C

XL = 200 Ω

XC = 90.9 ΩR = 300 Ω

φ

ZXL - XC = 109 Ω

FIG. P33.27

*P33.28 Let Xc represent the initial capacitive reactance. Moving the plates to half their original separation

doubles the capacitance and cuts XCC =

1ω

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

must be cut in half: Z Zi f= 2 , R X X R XX

L C LC2 2 2

2

22

+ − = + −FHGIKJb g ,

R R X R RX

R RX X R RX X

X R

CC

C C C C

C

2 2 22

2 2 2 2

42

2 2 8 4

3

+ − = + −FHGIKJ

FHG

IKJ

− + = − +

=

b g

Chapter 33 273

P33.29 (a) XL = = ×2 100 20 5 1 29 104π Hz H a fa f. . Ω

ZVI

X X Z R

X XC

C

L C

L C

= = =

− = − = −

− = × − = ± =

∆Ω

Ω Ω

Ω Ω

rms

rms

V4.00 A

HznF or 124 nF

20050 0

50 0 35 0

1 29 101

2 10035 7 123

2 2 2 2 2

4

.

. .

. .

b g b g b g

b gπ

(b) ∆ ΩV I XL L, . . . rms rms A kV= = × =4 00 1 29 10 51 54a fe jNotice that this is a very large voltage!

FIG. P33.29

Section 33.6 Power in an AC Circuit

P33.30 X LL = = =ω 1 000 0 050 0 50 0s H b gb g. . Ω

XC

Z R X X

Z

C

L C

= = × =

= + −

= + − =

− −11 000 50 0 10 20 0

40 0 50 0 20 0 50 0

6 1

2 2

2 2

ωs F

b ge j

b ga f a f

. .

. . . .

Ω

Ω FIG. P33.30

(a) IVZrmsrms V

50.0 = =

∆

Ωb g 100

I

X XR

L C

rms A

40.0

=

=−F

HGIKJ

= = °

2 00

30 036 9

.

arctan

arctan.

.

φ

φ ΩΩ

(b) P = = °=∆V Irms rms V 2.00 A Wb g a fcos cos .φ 100 36 9 160

(c) PR I R= = =rms2 A W2 00 40 0 1602. .a f Ω

P33.31 ω = 1 000 rad s , R = 400 Ω , C = × −5 00 10 6. F , L = 0 500. H

∆Vmax = 100 V , ω L = 500 Ω ,1

200ωCFHGIKJ = Ω

Z R LC

IVZ

= + −FHG

IKJ = + =

= = =

22

2 21400 300 500

100500

0 200

ωω

A

Ω

∆max

max .

The average power dissipated in the circuit is P = =FHGIKJI R

IRrms

2 max2

2.

P = =0 200

2400 8 00

2..

A W

a f a fΩ


P33.32 Z R X XL C= + −2 2b g or X X Z RL C− = −b g 2 2

X X

X XR

IVZ

V I

L C

L C

− = − =

=−F

HGIKJ =

FHG

IKJ = °

= = =

= = ° =

− −

b g a f a f

b g a fa f a f

75 0 45 0 60 0

60 053 1

2102 80

210 2 80 53 1 353

2 2

1 1

. . .

tan tan.

.

.

cos . cos .

45.0

V75.0

A

V A W

rmsrms

rms rms

Ω Ω Ω

ΩΩ

∆Ω

∆

φ

φP

P33.33 (a) P = = − ° = ×I Vrms rms W∆b g a f a fcos . cos . .φ 9 00 180 37 0 1 29 103

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

(b) tanφ =−X XR

L C becomes tan .− ° =−

37 016

a f X XL C : so X XL C− = −12 0. Ω .

P33.34 X LL = = =ω π2 60 0 0 025 0 9 42. . .s H b gb g Ω

Z R X XL C= + − = + =2 2 2 220 0 9 42 22 1b g a f a f. . . Ω Ω

(a) IVZrmsrms V

22.1 A= = =

∆Ω

1205 43.

(b) φ = FHGIKJ = °−tan

..

.1 9 4220 0

25 2 so power factor = =cos .φ 0 905 .

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

2 60 0 1.

.

sΩ =

−π e jCand C = 281 Fµ .

(d) P Pb d= or ∆∆

V IV

Rb b bd

rms rmsrmsb g b g b g

cosφ =2

∆ ∆ ΩV R V Id b b brms rms rms V A Vb g b g b g a fa fa fa f= = =cos . . .φ 20 0 120 5 43 0 905 109

P33.35 Consider a two-wire transmission line:

IVrms

rms=

P∆

and power loss = =I Rrms2

lineP

100.

Thus, P P

∆VR

rms

FHG

IKJ =

2

12100

b g or RV

1

2

200=

∆ rmsb gP

Rd

AV

1

2

200= =ρ ∆ rmsb g

Por A

r d

V= =π ρ2

42002

2

a fb g

P

∆ rms

and the diameter is 2800

2rd

V=

ρ

π

P

∆a f .

R1

R1

RL∆Vrms

FIG. P33.35

Chapter 33 275

P33.36 One-half the time, the left side of the generator is positive, the topdiode conducts, and the bottom diode switches off. The powersupply sees resistance

12

12

1

R RR+LNM

OQP =−

and the power is ∆V

Rrmsb g2

.

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 zeroresistance. The equivalent resistance is then

R RR R

Req = + +LNM

OQP =−1

31 7

4

1

and P = =∆ ∆V

RV

Rrms

eq

rmsb g b g2 247

.

~V∆

RR R

R2

FIG. P33.36

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

R

rms rms rmsb g b g b g2 2 24 7

211

14

+= .

Section 33.7 Resonance in a Series RLC Circuit

P33.37 ω π06 82 99 7 10 6 26 10

1= × = × =. .e j rad s

LC

CL

= =× ×

=−

1 1

6 26 10 1 40 101 82

02 8 2 6ω . .

.e j e j

pF

P33.38 At resonance, 1

22

ππ

f Cf L= and

1

22π f L

Cb g

= .

The range of values for C is 46 5. pF to 419 pF .

*P33.39 (a) fLC

=1

2π

Cf L

= =×

FHGIKJ = ×

−

−14

1

4 10 400 106 33 102 2 2 10 2 12

13

π π

A

s Vs

CAs

Fe j

.

(b) CA

d d=

∈=

∈κ κ0 02

=∈FHGIKJ =

× ×× ×

FHG

IKJ = ×

− −

−−Cd

κ 0

1 2 13 3

12

1 236 33 10 10

108 46 10

..

F mm1 8.85 F

m

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


P33.40 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

21

3 56π

. kHz .

(b) At resonance, IVRmaxmax .= =

∆5 00 A .

(c) From Equation 33.38, QL

R= =ω 0 22 4. .

(d) ∆V X I LIL L, max max . max kV= = =ω 0 2 24

P33.41 The resonance frequency is ω 01

=LC

. Thus, if ω ω= 2 0 ,

X LLC

LLCL = =

FHGIKJ =ω

22 and X

CLCC

LCC = = =

12

12ω

Z R X X RLCL C= + − = + FHGIKJ

2 2 2 2 25b g . so IVZ

V

R L Crms

rms rms= =+

∆ ∆2 2 25. b g

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

EV R

R L C

V RC

R C LLC

V RC LC

R C L=

+FHGIKJ = +

=+

∆ ∆ ∆rms rms rmsb gb g

b g e j b g2

2

2

2

2

22 252

2 25

4

4 9 00. . .πω

ππ

.

With the values specified for this circuit, this gives:

E =× ×

× + ×=

− −

− −

4 50 0 10 0 100 10 10 0 10

4 10 0 100 10 9 00 10 0 10242

2 6 3 2 3 1 2

2 6 3

π . . .

. . .

V F H

F H mJ

a f a fe j e ja f e j e j

Ω

Ω.

P33.42 The resonance frequency is ω 01

=LC

. Thus, if ω ω= 2 0 ,

X LLC

LLCL = =

FHGIKJ =ω

22 and X

CLCC

LCC = = =

12

12ω

.

Then Z R X X RLCL C= + − = + FHGIKJ

2 2 2 2 25b g . so IVZ

V

R L Crms

rms rms= =+

∆ ∆2 2 25. b g

and the energy delivered in one period is

E tV R

R L C

V RC

R C LLC

V RC LC

R C L= =

+FHGIKJ = +

=+

P∆∆ ∆ ∆rms rms rmsb gb g

b g e j b g2

2

2

2

2

22 252

2 25

4

4 9 00. . .πω

ππ

.

Chapter 33 277

P33.43 For the circuit of Problem 22, ω 03 6

1 1

160 10 99 0 10251= =

× ×=

− −LC H F rad s

e je j.

QL

R= =

×=

−ω 0

3251 160 10

68 00 591

rad s H

b ge j.

.Ω

.

For the circuit of Problem 23, QL

RL

R LC RLC

= = = =××

=−

−ω 0

3

61 1

150460 10

100 987

H

21.0 FΩ. .

The circuit of Problem 23 has a sharper resonance.

Section 33.8 The Transformer and Power Transmission

P33.44 (a) ∆V21

13120 9 23, rms V V= =a f .

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

120 0 350 9 23 2 V A V , rmsa fa f a f. .= I

I242 0

4 55,.

. rms W

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

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

P33.45 ∆ ∆VNN

Vout in V Vb g b g a fmax max= = FHG

IKJ =2

1

2 000350

170 971

∆Vout rms

V Vb g a f

= =971

2687

P33.46 (a) ∆ ∆VNN

V22

11, rms , rmsd i d i= N2

2 200 80

1101 600= =

b ga f windings

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

11030 0

..

a fb g

(c) 0 950 1 2. I V I V, rms 1, rms , rms 2, rms∆ ∆d i d i= I11 20 2 200

110 0 95025 3,

.

.. rms A= =

a fb ga f


P33.47 The rms voltage across the transformer primary is

NN

V1

2∆ 2, rmsd i

so the source voltage is ∆ ∆V I RNN

Vs s, rms , rms , rms= +11

22d i .

The secondary current is ∆V

RL

2, rmsd i, so the primary current is

NN

V

RI

L

2

1

21

∆ , rms, rms

d i= .

FIG. P33.47

Then ∆∆ ∆

VN V R

N R

N V

Nss

L, rms

, rms , rms= +

2 2

1

1 2

2

d i d i

and RN R

N VV

N V

NsL

s= −FHGG

IKJJ= −

FHG

IKJ =

1

2 2

1 2

2

5 50 02 25 0

80 05 25 0

287 5

∆∆

∆ ΩΩ

, rms, rms

rms V

V V

d id i a f

a fa f, .

..

.. .

P33.48 (a) ∆ ∆VNN

V22

11, rms , rms= d i N

N

V

V2

1

2

1

310 0 1083 3= =

×=

∆

∆,

,

.. rms

rms

V120 V

(b) I V I V2 2 0 900, rms , rms 1, rms 1, rms∆ ∆d i d i= .

I2, rms V V

24.0 V10 0 10 0 900

1201203. .× = F

HGIKJe j a f

ΩI2, rms mA= 54 0.

(c) ZV

I22

2

310 0 10185= =

×=

∆Ω, rms

, rms

V0.054 A

k.

P33.49 (a) R = × × =−4 50 10 6 44 10 2904 5. . M m Ω Ωe je j and IVrms

rms5 W

5.00 10 V A= =

××

=P

∆5 00 10

10 06.

.

Ploss rms2 A kW= = =I R 10 0 290 29 02. .a f a fΩ

(b)PPloss =

××

= × −2 90 105 00 10

5 80 104

63.

..

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

4 50 10

2 2 29017 5

3 2.

.×

⋅=

V

kW

e ja fΩ far below the required 5 000 kW.

Chapter 33 279

Section 33.9 Rectifiers and Filters

*P33.50 (a) Input power = 8 W

Useful output power = = =I V∆ 0 3 2 7. . A 9 V Wa fefficiency = = = =

useful outputtotal input

W8 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= = FHG

IKJFHGIKJ = ×

×

FHG

IKJ =P ∆ 8 31

86 4001 29 10 88 W 6 d

s1 d

1 J1 Ws

J$0.135

3.6 10 J6a fa f . $4.

*P33.51 (a) The input voltage is ∆V IZ I R X I RCCin = = + = +FHGIKJ

2 2 22

1ω

. The output voltage is

∆V IRout = . The gain ratio is ∆∆VV

IR

I R C

R

R C

out

in=

+=

+2 2 2 21 1ω ωb g b g.

(b) As ω → 0 , 1

ωC→∞ and

∆∆VV

out

in→ 0

As ω →∞ , 1

0ωC

→ and ∆∆VV

RR

out

in→ = 1

(c)12 12 2=

+

R

R Cωb g

RC

R22 2

214+ =

ωω 2 2

21

3C

R= ω π= =2

13

fRC

fRC

=1

2 3π

P33.52 (a) The input voltage is ∆V IZ I R X I R CCin = = + = +2 2 2 21 ωb g . The output voltage is

∆V IXICCout = =

ω. The gain ratio is

∆∆VV

I C

I R C

C

R C

out

in=

+=

+

ω

ω

ω

ω2 2 2 21

1

1b g b g.

(b) As ω → 0 , 1

ωC→∞ and R becomes negligible in comparison. Then

∆∆VV

CC

out

in→ =

11

1ωω

. As

ω →∞ , 1

0ωC

→ and ∆∆VV

out

in→ 0 .

(c)12

1

12 2=

+

ω

ω

C

R Cb gR

C C2

2

2 21 4

+FHGIKJ =

ω ωR C2 2 2 3ω = ω π= =2

3f

RC

fRC

=3

2π


P33.53 For this RC high-pass filter, ∆∆VV

R

R XC

out

in=

+2 2.

(a) When ∆∆VV

out

in= 0 500. ,

then 0 500

0 5002 2

..

0.500

Ω

Ωa f +=

XC

or XC = 0 866. Ω .

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

Cf XC

= = = × =−12

12 300 0 866

6 13 10 6134

π πµ

Hz F Fa fa f.

.Ω

.

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

XC =×

=−

1

2 600 6 13 100 433

4π Hz F a fe j.

. Ω

∆∆

Ω

Ω Ω

VV

R

R XC

out

in

0.500 =

+=

+=

2 2 2 2

0 500

0 4330 756

.

..

a f a f.

FIG. P33.53

P33.54 For the filter circuit, ∆∆VV

X

R XC

C

out

in=

+2 2.

(a) At f = 600 Hz, Xf CC = =

×= ×

−

12

1

2 600 8 00 103 32 10

94

π π Hz F a fe j.

. Ω

and∆∆

Ω

Ω Ω

VV

out

in

90.0 =

×

+ ×≈

3 32 10

3 32 101 00

4

2 4 2

.

..

a f e j.

(b) At f = 600 kHz , Xf CC = =

× ×=

−

12

1

2 600 10 8 00 1033 2

3 9π π Hz F

e je j.. Ω

and∆∆

Ω

Ω Ω

VV

out

in

90.0 .2 =

+=

33 2

330 346

2 2

..

a f a f.

Chapter 33 281

P33.55∆∆VV

R

R X XL C

out

in=

+ −2 2b g

(a) At 200 Hz:14

8 00

8 00 400 1 400

2

2 2=+ −

.

.

Ω

Ω

a fa f π πL C

.

At 4 000 Hz: 8 00 8 0001

8 0004 8 002

22. . Ω Ωa f a f+ −

LNM

OQP

=ππ

LC

.FIG. P33.55(a)

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

8 00013 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=FHGIKJ =

max

.1 00 .

(c) X XL C= requires fLC0

4 5

12

1

2 5 80 10 5 46 10894= =

× ×=

− −π π . . H F Hz

e je j.

(d) At 200 Hz, ∆∆VV

RZ

out

in= =

12

and X XC L> ,

so the phasor diagram is as shown:

φ = − FHGIKJ = −

FHGIKJ

− −cos cos1 1 12

RZ

so

∆ ∆V Vout in leads by 60.0° .

At f0 , X XL C= so

R

ZXL - XC

φ or φ∆Vout

∆Vin

R

ZXL - XC φ

orφ ∆Vout

∆Vin

FIG. P33.55(d)∆ ∆V Vout in and have a phase difference of 0° .

At 4 000 Hz, ∆∆VV

RZ

out

in= =

12

and X XL C− > 0 .

Thus, φ = FHGIKJ = °−cos .1 1

260 0

or ∆ ∆V Vout in lags by 60.0° .

(e) At 200 Hz and at 4 kHz,

P = = = = =∆ ∆ ∆

Ω

V

R

V

R

V

Rout, rms in, rms in, max V

W

d i b gd i b g b g a fa f

2 2 2 21 2 1 2 1 2 10 08 8 00

1 56..

. .

At f0 , P = = = = =∆ ∆ ∆

Ω

V

R

V

R

V

Rout, rms in, rms in, max V

.25 W

d i d i b g a fa f

2 2 2 21 2 10 02 8 00

6..

.

(f) We take: QL

Rf L

R= = =

×=

−ω π π

0 04

2 2 894 5 80 10

8 000 408

Hz H

a fe j.

..

Ω.


Additional Problems

P33.56 The equation for ∆v ta f during the first period (using y mx b= + ) is:

∆∆

∆

∆ ∆∆

v tV tT

V

vT

v t dtV

T Tt dt

T T

a f b g

a f a f b g= −

= = −LNMOQPz z

2

1 212 2

0

2 2

0

maxmax

max

ave FIG. P33.56

∆∆ ∆ ∆

∆ ∆∆ ∆

vV

TT t T V V

V vV V

t

t T

a f b g b g a f a f b g

a f b g

22 3

0

23 3

2

22

2

2 1

3 61 1

3

3 3

ave

rmsave

max

= FHGIKJ

−= + − − =

= = =

=

=

max max max

max

P33.57 ω 06

11 1

0 050 0 5 00 102 000= =

×=

−

−

LC . . H F s

b ge jso the operating angular frequency of the circuit is

ωω

= = −0 1

21 000 s .

Using Equation 33.37, P =+ −

∆V R

R L

rmsb ge j

2 2

2 2 2 202 2

ω

ω ω ω

P =+ − ×

=400 8 00 1 000

8 00 1 000 0 050 0 1 00 4 00 1056 7

2 2

2 2 2 6 2

a f a fb ga f b g b g a f

.

. . . .. W .

Q ≈ 12 5.b g

FIG. P33.57

*P33.58 The angular frequency is ω π= =2 60 377s s . When S is open, R, L, and C are in series with thesource:

R X XVIL C

s2 22 2

4201 194 10+ − = FHG

IKJ = FHG

IKJ = ×b g ∆

Ω V

0.183 A 2. . (1)

When S is in position 1, a parallel combination of two R’s presents equivalent resistance R2

, in series

with L and C:

RX XL C2

204 504 10

22

23F

HGIKJ + − = FHG

IKJ = ×b g V

0.298 A 2. Ω . (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+ = FHGIKJ = ×

V0.137 A

2. Ω . (3)

Take equation (1) minus equation (2):

34

7 440 102 3R = ×. 2Ω R = 99 6. Ω

continued on next page

Chapter 33 283

(only the positive root is physical.) Now equation (3) gives

XCC = × − = =2 131 10 99 6 106 7

14 2 1 2. . .a f Ω Ω

ω (only the positive root is physical.)

C X CC= = = × =− − −ωb g b g1 1 5377 106 7 2 49 10s F. .Ω .

Now equation (1) gives

X X

X L

LX

L

L C

L

L

− = ± × − = ±

= + = =

= = =

1 194 10 99 6 44 99

106 7 44 99 61 74

0 164

4 2 1 2. . .

. . .

.

a f

or 151.7

H or 0.402 H

Ω Ω

Ω Ω Ω Ω ω

ω

P33.59 The resistance of the circuit is RVI

= = =∆

Ω12 0

19 0.

. V

0.630 A .

The impedance of the circuit is ZVI

= = =∆

Ωrms

rms

V0.570 A

24 0

42 1.

. .

Z R L

L Z R

2 2 2 2

2 2 2 21 1377

42 1 19 0 99 6

= +

= − = − =

ω

ω. . .a f a f mH

*P33.60 The lowest-frequency standing-wave state is NAN. The distance between the clamps we represent

as L d= =NNλ2

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

f L= = =λµ

2 . The magnetic

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

TL

0 01960 4

2 2

. kg ms= b g T L= 274 2 kg ms2e j

Any values of T and L related according to this expression will work, includingif m NL T= =0 200 10 9. . . We did not need to use the value of the current and magnetic field. If

we assume the subsection of wire in the field is 2 cm wide, we can find the rms value of themagnetic force:

F I B TB = = °=sin . . sin .θ 9 0 02 0 015 3 90 2 75 A m mNa fa fb g .

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

P33.61 (a) When ω L is very large, the bottom branch carries negligible current. Also, 1

ωC will be

negligible compared to 200 Ω and 45 0

225. V

200 mA

Ω= flows in the power supply and the

top branch.

(b) Now 1

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


P33.62 (a) With both switches closed, the current goes only through generatorand resistor.

i tVR

ta f = ∆ max cosω

(b) P =12

2∆V

Rmaxb g

(c) i tV

R Lt

LR

a f =+

+ FHGIKJ

LNM

OQP

∆ max cos arctan2 2 2ω

ωω

FIG. P33.62

(d) For 010 0= =

−FHG

IKJφ

ω ωarctan

L C

Rb g

.

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= =12

12

2 2 2∆b g

U CI X CV

R C

V L

RCmax maxmax max= = =

12

12

12

2 22

202 2

2

2

∆ ∆b g b gω

(g) U LI LV

Rmax maxmax= =

12

12

22

2

∆b g

(h) Now ω ω= =22

0 LC.

So φω ω

=−F

HGIKJ =

−FHG

IKJ=

FHG

IKJarctan arctan arctan

L C

R

L C L C

R RLC

1 2 1 2 32

b g b g.

(i) Now ωω

LC

=12

1ω

ω= =

12 2

0

LC.

P33.63 (a) IVRR , . rmsrms V

80.0 A= = =

∆Ω

1001 25

(b) The total current will lag the applied voltage as seen in the phasor

diagram at the right.

IVXL

L, rms

rms V

2 60.0 s H A= = =

−

∆ 100

0 2001 33

1π e ja f..

Thus, the phase angle is: φ =FHG

IKJ =

FHG

IKJ = °− −tan tan

..,1 1 1 33

46 7I

IL

R

rms

, rms

A1.25 A

.

φ

IR

IL

∆V

I

FIG. P33.63

Chapter 33 285

P33.64 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 10¢ per kWh). Suppose the transmission line is at 20 kV. Then

IVrms

rms

W

V A= =

P∆

20 000 500

20 000103b ga f

~ .

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

P33.65 R = 200 Ω , L = 663 mH , C = 26 5. Fµ , ω = −377 1 s , ∆Vmax .= 50 0 V

ω L = 250 Ω , 1

100ωCFHGIKJ = Ω , Z R X XL C= + − =2 2 250b g Ω

(a) IVZmaxmax .

.= = =∆

Ω50 0

0 200 V

250 A

φ =−F

HGIKJ = °−tan .1 36 8

X XR

L 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)

P33.66 L = 2 00. H , C = × −10 0 10 6. F , R = 10 0. Ω , ∆v t ta f b g= 100 sinω

(a) The resonant frequency ω 0 produces the maximum current and thus the maximum powerdelivery to the resistor.

ω 06

1 1

2 00 10 0 10224= =

×=

−LC . .a fe j rad s

(b) P = = =∆V

Rmax

.b g a f

a f2 2

2100

2 10 0500 W

(c) IVZ

V

R L Crms

rms rms= =+ −

∆ ∆

2 21ω ωb gd i

and IVRrmsrmsb gmax

=∆

I R I Rrms2

rms2=

12 e jmax

or∆ ∆V

ZR

V

RRrms rmsb g b g2

2

2

212

= .

This occurs where Z R2 22= : R LC

R22

212+ −

FHG

IKJ =ω

ω

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− + + =e j2 00 10 0 10 2 2 00 10 0 10 10 0 10 0 10 1 02 6 2 4 6 2 6 2 2. . . . . .a f e j a fe j a f e j×LNM

OQP − × + ×LNM

OQP + =− − −ω ω .

Solving this quadratic equation, we find that ω 2 51 130= , or 48 894

ω 1 48 894 221= = rad s and ω 2 51 130 226= = rad s .


P33.67 (a) From Equation 33.41,NN

VV

1

2

1

2=∆∆

.

Let input impedance ZVI1

1

1=∆

and the output impedance ZVI2

2

2=∆

so thatNN

Z IZ I

1

2

1 1

2 2= . But from Eq. 33.42,

II

VV

NN

1

2

2

1

2

1= =∆∆

.

So, combining with the previous result we have NN

ZZ

1

2

1

2= .

(b)NN

ZZ

1

2

1

2

8 0008 00

31 6= = =.

.

P33.68 P I RVZ

R= = FHGIKJrms

2 rms∆ 2

, so 250120

40 02

2 W V

=a f a f

Z. Ω : Z R L

C= + −

FHG

IKJ

22

1ω

ω

250120 40 0

40 0 2 0 185 1 2 65 0 10

2

2 62=

+ − × −

a f a fa f a f e j

.

. . .π πf fand 250

576 000

1 600 1 162 4 2 448 5

2

2 2 2=+ −

f

f f. .e j

12 304

1 600 1 351 1 5 692 3 5 995 300

2

2 4 2=+ − +

ff f f. .

so 1 351 1 6 396 3 5 995 300 04 2. .f f− + =

f 226 396 3 6 396 3 4 1 351 1 5 995 300

2 1 351 13 446 5 1 287 4=

± −=

. . .

.. .

b g b gb gb g or

f = 58 7. Hz or 35.9 Hz

P33.69 IVRR =

∆ rms ; IV

LL =∆ rms

ω; I

V

CC = −

∆ rms

ωb g 1

(a) I I I I VR

CLR C Lrms rms= + − = F

HGIKJ + −FHG

IKJ

2 22

21 1b g ∆ ω

ω

(b) tanφ =−

= −LNM

OQPFHG

IKJ

I II

VX X V R

C L

R C L∆

∆rmsrms

1 1 1

tanφ = −LNM

OQP

RX XC L

1 1FIG. P33.69

Chapter 33 287

P33.70 (a) I VR

CLrms rms= + −

FHG

IKJ∆

1 12

2

ωω

∆ ∆V Vrms rms→ b gmax when ω

ωC

L=

1

fLC

= =× ×

=− −

12

1

2 200 10 0 150 10919

3 6π π H F Hz

.e j

(b) IVRR = = =

∆Ω

rms V80.0

A120

1 50.

IV

LL = = =−

∆ rms V

374 s H A

ω120

0 2001 60

1e ja f..

I V CC = = × =− −∆ rms V s F mAωb g a fe je j120 374 0 150 10 6 731 6. .

(c) I I I IR C Lrms A= + − = + − =2 2 2 21 50 0 006 73 1 60 2 19b g a f b g. . . .

(d) φ =−L

NMOQP=

−LNM

OQP = − °− −tan tan

. ..

.1 1 0 006 73 1 601 50

46 7I I

IC L

R

The current is lagging the voltage .

FIG. P33.70

P33.71 (a) X XL C= = 1 884 Ω when f = 2 000 Hz

LX

fL= = =

21 884

0 150π π

4 000 rad s

HΩ

. and

Cf XC

= = =1

21

4 000 1 88442 2

π πb g b gb g rad s nF

Ω.

X fL = 2 0 150π . Ha f Xf

C =× −

1

2 4 22 10 8πb ge j. F

Z X XL C= + −40 0 2 2. Ωa f b g

f X X ZL C (Hz) 300600800

1 0001 5002 0003 0004 0006 000

10 000

283565754942

1 4101 8802 8303 7705 6509 420

12 6006 2804 7103 7702 5101 8801 260

942628377

1 23005 7203 9602 8301 100

401 5702 8305 0209 040

( ) ( ) ( )Ω Ω Ω

(b) Impedence, Ω

FIG. P33.71(b)


P33.72 ω 061

1 00 10= = ×LC

. rad s

For each angular frequency, we find

Z R L C= + −2 21ω ωb g

then IZ

=1 00. V

and P = I 2 1 00. Ωa f .The full width at half maximum is:

∆∆

∆

f

f

= =−

=×

=−

ωπ

ωπ

π

2

1 000 5 0 999 5

21 00 10

2159

0

3 1

. .

.

b g

s Hz

while

RL2

1 00159

3π π=

×=

−

.

2 1.00 10 H Hz

Ω

e j.

ωω

ωω0

2LC

Z P I R 1

W

0.99900.99910.99930.99950.99970.99991.00001.00011.00031.00051.00071.00091.0010

999.0 999.1 999.3 999.5 999.7 999.910001000.11000.31000.51000.71000.91001

1001.01000.91000.71000.51000.31000.11000.0 999.9 999.7 999.5 999.3 999.1 999.0

2.242.061.721.411.171.021.001.021.171.411.722.062.24

0.199840.235690.337680.499870.735240.961531.000000.961540.735350.500120.337990.236010.20016

Ω Ω Ωa f a f a f a f=

1.0

0.8

0.6

0.4

0.2

0.00.996 0.998 1 1.002 1.004

I R2

(W)

ω/ω 0

FIG. P33.72

P33.73∆∆VV

R

R C

R

R f C

out

in=

+=

+2 2 2 21 1 2ω πb g b g

(a)∆∆VV

out

in=

12

when 1

3ωC

R= .

Hence, fRC

= = =ωπ π2

12 3

1 84. kHz .

∆Vin R

C

∆Vout

FIG. P33.73

continued on next page

Chapter 33 289

(b) Log Gain versus Log Frequency

–4

–3

–2

–1

0

0 1 2 3 4 5 6

Log f

Log ∆Vout/∆Vin

FIG. P33.73(b)

ANSWERS TO EVEN PROBLEMS

P33.2 (a) 193 Ω ; (b) 144 Ω P33.32 353 W

P33.34 (a) 5 43. A; (b) 0 905. ; (c) 281 Fµ ; (d) 109 VP33.4 (a) 25.3 rad/s; (b) 0.114 s

P33.3611

14

2∆VRrmsb gP33.6 1 25. A and 96 0. Ω for bulbs 1 and 2;

0 833. A and 144 Ω for bulb 3

P33.8 7 03. H or more P33.38 46 5. pF to 419 pF

P33.10 3 14. A P33.40 (a) 3 56. kHz; (b) 5 00. A ; (c) 22 4. ;(d) 2 24. kV

P33.12 3.80 J

P33.424

4 9

2

2

π ∆V RC LC

R C Lrmsb g

+P33.14 (a) greater than 41 3. Hz ;

(b) less than 87 5. Ω

P33.44 (a) 9 23. V ; (b) 4 55. A ; (c) 42 0. WP33.16 2C V∆ rmsb g

P33.46 (a) 1 600 turns ; (b) 30 0. A ; (c) 25 3. AP33.18 –32.0 A

P33.48 (a) 83 3. ; (b) 54 0. mA; (c) 185 kΩP33.20 2 79. kHz

P33.50 (a) 0.34; (b) 5.3 W; (c) $4.8P33.22 (a) 109 Ω ; (b) 0 367. A ; (c) Imax .= 0 367 A ,

ω = 100 rad s, φ = −0 896. rad P33.52 (a) see the solution; (b) 1; 0; (c) 3

2π RCP33.24 19.3 mA

P33.54 (a) 1.00; (b) 0.346P33.26 (a) 146 V ; (b) 212 V ; (c) 179 V ; (d) 33 4. V

P33.56 see the solutionP33.28 X RC = 3

P33.58 R = 99 6. Ω , C = 24 9. Fµ , L = 164 mH or402 mHP33.30 (a) 2 00. A; (b) 160 W; (c) see the solution


P33.60 L = 0 200. m and T = 10 9. N , or any valuesrelated by T L= 274 2 kg ms2e j

P33.64 ~103 A

P33.66 (a) 224 rad s ; (b) 500 W ;(c) 221 rad s and 226 rad s

P33.62 (a) i tVR

ta f = ∆ max cosω ; (b) P =∆V

Rmaxb g22

;P33.68 either 58 7. Hz or 35.9 Hz

(c) i tV

R Lt

LR

a f =+

+ FHGIKJ

LNM

OQP

−∆ max cos tan2 2 2

1

ωω

ω;

P33.70 (a) 919 Hz ;(b) IR = 1 50. A, IL = 1 60. A , IC = 6 73. mA ;

(d) CL

=1

02ω

; (e) Z R= ; (f) ∆V L

Rmaxb g2

22; (c) 2 19. A ; (d) − °46 7. ; current lagging

(g) ∆V L

Rmaxb g2

22; (h) tan− F

HGIKJ

1 32R

LC

;P33.72 see the solution

(i) 1

2LC

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