Tài liệu Physics exercises_solution: Chapter 31 docx

31.29: a At resonance, the power factor is equal to one, because the impedance of the circuit is exactly equal to the resistance, so 1... 31.41: a If the original voltage was lagging t

31.1: a) 31.8V.

2

V0.452

)H00.5()srad100(

V0

)H00.5()srad1000(

V0

V0.60

31.4: a)   I Vω C(60.0V)(100rad s)(2.2010 6 F)0.0132A.

ωC

I IX

1202





π πf

X L πfL ωL

L

)F100.4()Hz80(2

12

11

12

12

)F1050.2()Hz60(2

12

11

6



11

ωL ωC X

A)850.0

I C ωC

I V

C C

31.8: 1.63 10 Hz.

)H1050.4()A1060.2(2

)V0.12(2

6 4

V f L Iω

cosV)80.3(

t

t R

11

).)srad012(cos(

V)10.1())srad120(cos(

)250()A1038.4(

))sradcos((120A)

1038.4(1736

))srad120cos((

)V60.7()(cos

3

3

t t

iR

v

t

t X

ωt v

L X ωC ωL X LC ω

31.12: a) Z  R2 (ωL)2  (200)2 ((250rad/s)(0.400H))2 224 b) 0.134A

224

V0



V L

V8.26

V4.13arctan

696

V0

)A0431.0(

V;

62.8)200()A0431.0(

V7.28arctan

31.14: a)

.567)F1000.6()ad/s250(

1)

H400.0()rad/s250()/

)A0529.0(

V

V V

e)

31.15: a)

b) The different voltages are:

.Note

.V85.12,

V60.7,V5.20:ms20At

90250cos(

)V4.13(),cos(250V)

8.26(),26.6cos(250

V)0.30

(

v v v v

v v

t

t v

t v

t v

L R L

R

L R

31.16: a)

b) The different voltage are:

.Note

.V5.27,

V45.2,V1.25:

ms

20

At

)90250cos(

)V7.28(),250cos(

)V62.8(),3.73250cos(

v v

t

t v

t v

t v

C R C

R

C R

31.17: a) Z  R2 (ωL1/ωC)2

2 6

2 ((250rad/s)(0.0400H) 1/ ((250rad/s)(6.00 10 F))))

arctan/

the current

d) V R  IR(0.0499A)(200)9.98V;

;V99.4)H400.0)(

srad250()A0499.0

)A0499.0(

31.18: a)

The different voltages plotted above are:

)

90250cos(

)V3.33()90250cos(

)V

)V98.9(),6.70250cos(

t v

t v

t v

C L

.and

resonance,At

.Hz1132

so

mA61.7/

5.394)

500160

()200()(

C

2 2

2 2

L

C L

ωC ωL R

values R200,L0.400H,andC6.0010 6 F:

a) ω1000rad/s:Z 307, 49.4;

.1.75,

779:

rad/s200

;7.10,

204:

rad/s600

Z ω

b) The current increases at first, then decreases again since

Z

V

I c) The phase angle was calculated in part (a) for all frequencies

31.21: V2 V R2 (V L V C)2

V0.50)V0.90V0.50()V0

)H100.20()Hz1025.1(2

1)

1 3

3

9 3

The impedance of the circuit is

.830)

752()350()

Z

The average power provided by the supply is then

W32.7)1.65cos(

830

)V120()cos(

)

2 rms rms

2 2

R X

X R I

IR

C L

Z

V Z

V

2 rms

2

Z

R Z

V



Z

R Z

V Z

V

P av

2 rms

2 rms cos 

W

5.43)0.75()105(

)V0.80(

2

2 2

R Z

R



.8.45)698.0

(

cos

698

0

344

240

F)1030.7()Hz400(2

1)

H120.0()Hz400(2)

For pure capacitors and inductors there is no average energy flow

31.26: a) The power factor equals:

.181.0))H20.5()s/rad60)2(((

)360(

)360()

(

cos

2 2

R

R Z

R



b)

.W62.2)181.0())H(5.20s)/rad60)2(((

)360(

)V240(2

1cos2

1

2 2

2 2

V

31.27: a) At the resonance frequency, Z R.

V1290

;2582/

)(/1

V1290

;2582/

/1(

V150b)

V150Ω)(300A)500.0(

C

L L L

R

IX

V

C L ωC X

IX V C

L LC L

ωL

X

IR V

IR IZ

V

c) cos 2 ,since andcos 1atresonance

2

1 2

2 (ωL ωC)

R

V I

1

2 2

I Thus, the amplitude of the voltage across the inductor is V  ωL I( )(0.300A)(50.0rad/s)(9.00H)135V

31.29: a) At resonance, the power factor is equal to one, because the impedance of the

circuit is exactly equal to the resistance, so 1

Z R

b) Average power:   75W

150

V1502

rms 2

31.31: a) At resonance:

0.400H 6.00 10 F

11

6

ω

0 645.5rad s103Hz b)

,V2.212

V0.302

rms rms rms rms

1

R

V V I

V V

A106.0

.V4.27H400.0srad5.645A106.0

2 6

0

rms 3

0 rms 2

V C

ω

I V

L ω I V

d) If the resistance is changed, that has no affect upon the resonance frequency:

Hz103s

rad5.645

V120

I

V Z R

31.33: a) 10

12

1202

V12.0rms

R

V I

c) P av IrmsVrms 2.40A12.0V28.8W

500W

28.8

V

120 2 rms

12000

.500

.5

2 2

N

N I

00.8

108

12 3 2

1 2

1 2

2

1 2

N N

N R R

40

1V0.601

2 1

ωL R

ωC

R ω

R L R

1A

0385.0779

V30

.779F

1000.6srad2001H400.0srad200200

rms

2 6 2

and,V5.20V18.2V7.22

,V7.22F1000.6srad200

A0272.0

,V18.2H400.0srad200A0272.0

2 0 30 rms 5 2

3 4

6

rms rms

3

rms rms

V

V

ωC

I X I

V

ωL I X I

V

C L

b) If ω1000rad s, using the same steps as above in part

(a): Z 307,V1 13.8V,V2 27.6V,V3 11.5V,V4 16.1V,V5 21.2V

ω

π t t ω

π t ω

π t π n

ωt

2

32

21when

1

,222

sin23sinsin

cos

t t

t t

t

I ω

I π

π ω

I ωt ω

I dt ωt I

I π

ω I

ω

I t t

31.40: a)     2 120Hz0.332

250

π ω

XL L ωL

X L

b) 2 2 400  2 250 2 472 ,cos

Z

R X

rms rms

R Z

V

av

31.41: a) If the original voltage was lagging the circuit current, the addition of an

inductor will help it “catch up,” since a pure LR circuit would have the voltage leading This will increase the power factor, because it is largest when the current and voltage are in phase

b) Since the voltage is lagging, the impedance is dominated by a capacitive element so

we need an inductor such that X L  X0,where X0 is the original capacitively dominated reactance (this could include inductors, but the capacitors “win”)

.41

.6.412

.4360

2.430

.60720.0720.0

2 2

2 2 0

2 2

X L ωL X

X

R Z X

X R Z

Z R

C C

L

C

31.42: 80.0 2 2 2 50.0 2

A 00 3 V 240 rms

I R P

4 3

2 2

I

X X X

X R

V;

10667

b) In part (a) we found I = 0.211 mA

c) X L  X C and R = 0 gives that the source and inductor voltages are in phase;

the voltage across the capacitor lags the source and inductor voltages by 180

2

2 2

2

2 1

L

X

X C

ω C

ω L

ω L ω

19

13

1

2 3

3

3 1

L

X

X X

C ω C

ω

L ω L ω

capacitor’s reactance is greater than that of the inductor

c) Since X L  X C at ω1, that is the resonance frequency

31.45: 2 2 2 2 2  2

Z

V L ω R I V V

V

V

2 2

ω

ωR ωC

R

R V

V ωC

I V

ωC

ωC ωC

ωC V

V

s

V I

12

1

2 2

2 2

2

ωC ωL

R

R V R

Z

V R

c) The average power and the current amplitude are both greatest when the

denominator is smallest, which occurs for .

LC

ω C ω L

2 6 2

P av

 ω , ,  .

ω ,

ω

2 2

2

00000022000

31.48: a)

R

L Vω Z

L Vω Iωω

I ωC

31.49: a)

4

12

12

12

12

12

rms 2

122

12

12

12

rms 2

R

LV ωC

ωL R

V L

LI

2 2

2 2

2 2

14

14

14

14

14

14

1

2 2

2

2 2

2 2 2

2 2

ωC ωL

R C ω

V ωC

ωL R

C ω

V C

4and

1

R

LV U

U LC

31.50: a) Since the voltage drop between any two points must always be equal, the

parallel LRC circuit must have equal potential drops over the capacitor, inductor and resistor, so v R v L v C v Also, the sum of currents entering any junction must equal the current leaving the junction Therefore, the sum of the currents in the branches must equal the current through the source: ii R i L i C

i C  leads the voltage by 90

c) From the diagram,

2 2

V C Vω R

V I

I I

11

ωL

ωC R

Z Z

V I

2 2

11

V C Vω I

L ω C ω LC

0

0 0

0 0

11

V

P av

2 2

rms cos 

  at resonance where R < Z so power is a maximum.

c) At ωω0, I and V are in phase, so the phase angle is zero, which is the same as a

series resonance

31.52: a) 0.778A.

400

V311

;311

A0.672arctan

31.53: a)

ωL

V I C;

Vω I

At high frequency, the induced emf in the inductor resists the violent changes and passes little current The capacitor never gets a chance to fill up so passes charge freely

)1050.0)(

H0.2(

11

ω

2

) ωL

v C (Vω R

(1000s

100VF)

1050.0)(

sV)(1000100

(200

V

1 6

31.54: a) Note that as  and 1 0.

C ω L

ω ,

ω Thus, at high frequencies the current through R is nearly zero and the power dissipated by the circuit is1

kW

44.10

.40

V)240

b) Now we let ω0, and so ω L0and 1 

C

ω Thus, at low frequencies the

current through R is nearly zero and the power dissipated by the circuit is2

.kW960.00

.60

V)240

31.55: Connect the source, capacitor, resistor, and inductor in series.

31.56: a)

.6.20)560.0)(

7.36(cos

7.36W)

220(

)560.0(V)120(coscos

2 2

rms

2 rms

P

V Z Z

V P

av av

b) Z  R2 X L2 X L  Z2R2  (36.7)2(20.6)2 30.4.But at0



 this is resonance, so the inductive and capacitive reactances equal each other So:

.1005.1)Hz)(30.40

.50(2

12

11

F π

X f π ωX

C ωC

X

C C

V)120

31.57: a) tan X X Rtan

R

X X

C L C



.102)54tan(

)180(

350     



)180(

)W140(rms

R I

av

rms rms

31.58: a) For ω800rad s,Z  R2 (ωL1 ωC)2

V

155H)s)(2.00rad

800)(

A0971.0(V

V

243F)10s)(5.0rad

(800

A0971.01

V

V.48.6)A)(5000971

.0(V

A0971.01030

V100

1030F)))

100.5(rad/s)((8001H))(2.0rad/s800((

)500(

7

2 7 2

V I Z



b) Repeating exactly the same calculations as above for

V.400V;

100A;

0.2000.;

;500Z

:srad

;6.48

;0971.0

;9.60

;1030:

srad

31.59: a) 0.75A.

480

V360

C

X

V I IX V

A0.75

) 80 ( ) 160 (

2 2

X

R Z X

X

d) If 0 then 1 X ωL.Forus,X 341 if ω ω0

ωC X

1094.12H)[

100.1(

11

2 6 6

2 0

ω

.126.0

Hz)(2.8610

(94.02

1H)

10Hz)(1.0010

0.94(2991

99

199

11

100

2100

11

2)

(01.0)(

6 6

6

2 2

2 2

2 2

2

2 1

R

ωC) L

(ω ωC) L

(ω R

ωC) / L (ω R

R

R

V ωC)

L (ω R

R V ω

P ω

This answer is very sensitive to the capacitance so you may have to carry the first part

of the problem out to more significant figures

31.61: The average current is zero because the current is symmetrical above and below

the axis We must calculate the rms-current:

26

63

44

2

0

2 0

2 0

2 0 2

2 0

2 0 2

0

3 2

2 0 2

2

2 2 0 2

0

I I I

I τ τ I I

τ I t

τ

I (t)dt I τ

t I (t) I τ

31.62: a) 786rad s.

F)10H)(9.0080

.1(

11

b) Z  R2 (ω L1 ω C)2

A

200.0300

V60

.300F)))

10s)(9.00rad

786((

1H)s)(1.80rad

786((

)300

(

rms rms

2 7 2

2)

(

04

21

4)1(1

(2

1

2 2

rms

2 rms 2

2 2

2

2

2 rms

2 rms 2

2 2 2

2

2 rms

2 rms 2

2 2 2

rms rms

rms

0 0

0 0

V C

L R ω L

ω

I

V R C

L C ω

L

ω

I

V ωC

ωL R

ωC) ωL

R

V Z

V I

I

Substituting in the values for this problem, the equation becomes: (ω2)2(3.24)

.01023.1)10

R ω

ω1 2 28rad sec.(iii) 3, rms0 20A, 1 2 288

Width gets smaller as R gets smaller; Irms0 gets larger as R gets smaller.

31.63: a)

R

V Z

Z

.)11

2 2

ωL

)2

(

1Thus

1so

1

2 0

2 0

4 0 2 2

2 4

0 2 2 2

0

ω ω ω ω

ω L C

ω ω

L

C LC

212

1Δ

2

1

0

2 0 2

0

2 0 2 0

2

0

4 0 2 2

L ω

ω

ω L ω)

ω (ω

ω L

2 0

2 0

2 0

2 0

2 0

12

11

1)2

(

1

ω

ω ω

ω ω

ω ω

ω ω

ω ω ω ω

0

ω

ω LC

ω



Putting this together gives

84

22

11

0

3 2 2 2 2

0 0

2 0 2 2

ω

ω L ω L ω

ω ω

ω ω

L ωC

so,4

ω L R Z ω L ω

R

V Z

V I

      

L

R L

R R

L

R

4

34

34

2 2

2 2

L

R ω

4

3

C

L R

LC L

L

R ω

ω

ω     As R increases so does the width.

F)10H)(0.40050

.2(

1A;

815

120

6 0

secrad04.1s,

rad1000A

80(ii)

sec

rad4.10H50.2

V Z

V

R

V I

ωC

0 max

L R

V C Rω

V X

V L ω R

V X I

2

12

12

1

2

2 2

2 2

C max

V L C

L R

V C CV

2

12

1

2

2 2

max

V L LI

V C

ω

L ω R

V Z

V

I

4

92

2

2 2

49

2

2 2

0 max

C

L R

V C

L C

L R

V C

ω IX

4

9

0 max

C

L R

V C L C

L R

V L

ω IX

22

1

2

2 2

C max max

C

L R

LV CV

12

1

2

2 2

max

C

L R

LV LI

U L







31.66: ω2ω0.

4

9)

2

12

2 0 0

2

C

L R

V C

ω L ω R

V Z

4

92

1

2 2

0 max

C

L R

V C L C

L R

V C

ω IX

49

2

2 2

0 max

C

L R

V C

L C

L R

V L ω IX

2

1

2

2 2

max max

C

L R

LV CV

2

1

2

2 2

max

C

L R

LV LI

I V dt p T R P

2

1][2))2cos(

1(2

1)(

2

1)sin(

)cos(

ωLI dt

di Li

)sin(

2

2

ωt I

V ωt ωt

I V i v i C

q C

q dt

d dt

2

1)(cos2 ωt V I ωt V I ωt I

V p p p

))

sin(

)sin(

)cos(

)(

cos(ωt V ωt V ωt V ωt I

31.68: a) V R maximum when 0 1

LC ω

ω V

2 2

2 2

2

2 2

))1((

)1)(

1()

1(0

)10

ωC ωL

R

C ω L C ω L L Vω ωC

ωL R

VL

ωC ωL

R

L Vω dω

d dω

12

11

21

)1

()

1(

2 2

2 2 2

2 2 2

2

2

2 4 2

2 2 2

C R LC ω

C R LC ω

C ω C

L C ω

R

C ω L

ω ωC ωL

12

)1

()

1(

))1((

)1)(

1()1(0

)1(0

2

2 2

2 2

2

2

2 4 2

2 2 2

2 3 2 2

2 2

2 2

2

2 2

L

R LC ω

L ω C

L L ω

R

C ω L

ω ωC

ωL

R

ωC ωL

R C

C ω L C ω L V ωC ωL

R C ω

V

ωC ωL

R ωC

V dω

d dω

31.69: a) From the current phasors we know that

A

400.0500

V200

.500F)

10s)(1.25rad

1000(

1H)

s)(0.50rad

1000()400

(

)1(

2 6 2

2 2

(10001

H)s)(0.500rad

1000(

.500)

300()400(

300400

F)10s)(1.25rad

(1000

1H)

)(0.50srad1000(400

2 2

6 cpx

25

68)

300(400

V200cpx

256Re(

)Im(

tan

) cpx (



I I

f) (400 ) (128 96 )V

25

68cpx

10s)(1.25rad

1000(

125

68V

V

)160201(H)s)(0.500rad

1000(25

68

6 cpx

cpx cpx

i

i i C

I i

i

i i ωL iI

V

Lcpx

L

g) Vcpx V Rcpx V Lcpx V Lcpx (12896i)V (120160i)V

(192256i)V200V