Chapter 30

33
30.1: a) V, 0.270 /s) A 830 ( H) 10 25 . 3 ( ) / ( 4 1 2 dt di M and isconstant. b) Ifthe second coilhasthe sam e changing current, then the induced voltage isthe sam e and V. 270 . 0 1 30.2: Fora toroidalsolenoid, . 2 / and , / 1 1 0 B 1 2 2 2 r A i N i N M B So, . 2 / 2 1 0 r N AN M 30.3: a) H. 1.96 A) (6.52 / Wb) 0320 . 0 ( ) 400 ( / 1 2 2 i N M B b) W hen Wb. 10 7.11 (700) / H) (1.96 A) 54 . 2 ( / A, 54 . 2 3 1 2 B 2 1 N M i i 30.4: a) H. 10 6.82 s) / A 0.242 ( / V 10 65 . 1 ) / ( / 3 3 2 dt di M b) A, 20 . 1 , 25 1 2 i N Wb. 10 3.27 25 / H) 10 (6.82 A) 20 . 1 ( / 4 3 2 1 2 N M i B c) mV. 2.45 s) / A (0.360 H) 10 82 . 6 ( / and s / A 360 . 0 / 3 2 1 2 dt Mdi dt di 30.5: Ωs. 1 A)s / (V 1 AC)s / (J 1 A / J 1 A / Nm 1 A / Tm 1 A / Wb 1 H 1 2 2 2 30.6: Fora toroidalsolenoid, ). / ( / / dt di i N L B So solving for N w e have: turn 238 s) / A (0.0260 Wb) (0.00285 A) (1.40 V) 10 6 . 12 ( ) / ( / 3 dt di i N B 30.7: a) V. 10 4.68 s) / A (0.0180 H) 260 . 0 ( ) / ( 3 1 dt di L b) Term inal a isata higherpotentialsince the coil pushescurrentthrough from b to a and ifreplaced by a battery itw ould have the term inalat . a 30.8: a) H. 130 . 0 m) 120 . 0 ( 2 ) m 10 80 . 4 ( ) 1800 ( ) 500 ( 2 / 2 5 2 0 2 0 m m π μ πr A N μ K K L b) W ithoutthe m aterial, H. 10 2.60 H) 130 . 0 ( 500 1 1 4 m m K L K L

description

12th edition physic

Transcript of Chapter 30

Page 1: Chapter 30

30.1: a) V,0.270/s)A830(H)1025.3()/( 412 dtdiM and is constant.

b) If the second coil has the same changing current, then the induced voltage is the same and V.270.01 30.2: For a toroidal solenoid, .2/and,/ 110B12 22

rAiNiNM B So, .2/210 rNANM 30.3: a) H.1.96A)(6.52/Wb)0320.0()400(/ 12 2

iNM B

b) When Wb.107.11(700)/H)(1.96A)54.2(/A,54.2 312B2 1

NMii 30.4: a) H.106.82s)/A0.242(/V1065.1)/(/ 33

2 dtdiM

b) A,20.1,25 12 iN

Wb.103.27

25/H)10(6.82A)20.1(/4

3212

NMiB

c) mV.2.45s)/A(0.360H)1082.6(/ands/A360.0/ 3212 dtMdidtdi

30.5: Ωs.1A)s/(V1AC)s/(J1A/J1A/Nm1A/Tm1A/Wb1H1 222

30.6: For a toroidal solenoid, )./(// dtdiiNL B So solving for N we have:

turns.238s)/A(0.0260Wb)(0.00285

A)(1.40V)106.12()/(/3

dtdiiN B

30.7: a) V.104.68s)/A (0.0180H)260.0()/( 31

dtdiL b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a battery it would have the terminal at .a

30.8: a) H.130.0m)120.0(2

)m1080.4()1800()500(2/252

020mm

πμ

πrANμKKL

b) Without the material, H.102.60H)130.0(50011 4

mm

KLK

L

Page 2: Chapter 30

30.9: For a long, straight solenoid: .//and/ 2

00 lANμLlNiAμiNL BB 30.10: a) Note that points a and b are reversed from that of figure 30.6. Thus, according to Equation 30.8, s./A00.4H0.260

V04.1 L

VVdtdi ab Thus, the current is decreasing.

b) From above we have that .)s/A00.4( dtdi After integrating both sides of this expression with respect to ,t we obtain

A.4.00s)(2.00A/s)(4.00A)0.12(s)/A00.4( iti 30.11: a) H.0.250A/s)(0.0640/V)0160.0()/(/ dtdiL b) Wb.104.50(400)/H)(0.250A)720.0(/ 4 NiLB

30.12: a) J.540.02/A)(0.300H)0.12(21 22 LIU

b) W.2.16)180(A)300.0( 22 RIP c) No. Magnetic energy and thermal energy are independent. As long as the current is constant, constant.U

30.13: πr

AlNμLIU42

1 2202

turns.2850A)0.12()m1000.5(

J)(0.390 m)150.0(44224

02

0

μπ

AIμπrUN

30.14: a) J.101.73h)/s3600h/day(24W)200( 7PtU

b) H.5406A)(80.0

J)1073.1(2221

2

7

22

IULLIU

30.15: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability is used in place of .0

Page 3: Chapter 30

30.16: a) free space: J.3619)m0290.0(2

T)(0.560V2

V 3

0

2

0

2

BuU

b) material with J.04.8)m0290.0()450(2T)(0.560V

2V450 3

0

2

0m

2

m K

BuUK

30.17: a) 32

60

20

0

2

m1.25T)(0.600

J)1060.3(22Volume

2

B

UBVolUu .

b) T.9.11T4.141m)(0.400

J)1060.3(22 23

6002

B

VolU

B

30.18: a) .mT35.4)m0690.0(2

)A50.2()600(2

00

r

NIB

b) From Eq. (30.10), .m/J53.72

)T1035.4(2

3

0

23

0

2

Bu

c) Volume .m1052.1)m1050.3()m0690.0(22V 3626 rA d) .J1014.1)m1052.1()m/J53.7( 5363 uVU

e) .H1065.3)m0690.0(2

)m1050.3()600(2

6262

02

0

r

ANL

J1014.1)A50.2()H1065.3(21

21 5262 LIU same as (d).

30.19: a) .s/A40.2H50.2V00.60When.

dtdii

LiR

dtdi

b) When .A/s800.0H50.2

)00.8()A500.0(V00.6A00.1

dtdii

c) At A.413.0)1(8.00

V00.6)1(s200.0 s)(0.250H)50.2/00.8()/(

eeRεit tLR

d) As A.750.08.00

V00.6

Rit

Page 4: Chapter 30

30.20: (a) mA,30A030.01000V30

max i long after closing the switch. )b

V4.0V26V30εV26A)(0.0259Ω)(1000

A0.0259

e1A0.030)e(1

RBatteryL

R

)(max

1020

VVRiV

ii μsμs

R/L/t

(or, could use s)20at tLV dt

diL

c)

30.21: a) RLeRi t /),1(/ / 2

121

maxmax and,)(1when2/so/ τ/tτ/t eeiiRεi

μsLtτ/t 3.1750.0

H)10(1.252)ln(Rln2and)(ln

3

21

b) max2

21

max2

21 ; LiULiU

2/when maxmax21 iiUU

2929.02/11eso211 // τtτte μsLt 30.7R/ln(0.2929)

Page 5: Chapter 30

30.22: a) A13.2H0.115J)260.0(22

21 2

LUILIU

V.256)(120A)13.2( IR

b)

2

0)/(222)/(

21

21

21

21

21and LIUeLiLiUIei tLRtLR

s.1032.3

21ln

)2(120H115.0

21ln

2

21

4

)/(2

RLt

e tLR

30.23: a) A.250.0240

V600

RI

b) A.137.0A)250.0( )s10(4.00H)0.160/(240)/(0

4

eeIi tLR

c) ,V9.32)240()A137.0( iRVV abcb and c is at the higher potential.

d) .s1062.421ln

)240()H160.0(

21ln

21 4

2/1)/(

0

2/1

RLte

Ii tLR

30.24: a) .V60and00At bcab vvt b) .0andV60As bcab vvt c) .V0.24V0.36V0.60andV0.36A150.0When bcab viRvi

30.25: a) )1(00.8

)V00.6()1()1( )H50.2/00.8(2

)/(2

)/(0

ttLRtLR eeR

eIiP

).1()W50.4( )s20.3( 1 teP

b) 2)H50.2/00.8(2

2)/(2

2 )1(00.8

)V00.6()1( ttLRR ee

RεRiP

.)1()W50.4( 2)s20.3( 1 tR eP

c) )()1( )/(2)/(2

)/()/( tLRtLRtLRtLRL ee

Rεe

LεLe

dtdiiLP

).()W50.4( )s40.6()s20.3( 11 ttL eeP

d) Note that if we expand the exponential in part (b), then parts (b) and (c) add to give part (a), and the total power delivered is dissipated in the resistor and inductor.

Page 6: Chapter 30

30.26: When switch 1 is closed and switch 2 is open:

.)/(ln

0

)/(00

00

LRt

i

I

t

eIitLRIi

tdLR

iid

LRi

dtdiiR

dtdiL

30.27: Units of s/)s(/H/ RL units of time.

30.28: a) πfLC

ω 21

.H1037.2)F1018.4()106.1(4

14

1 31226222

CfL

b) F.1067.3)H1037.2()1040.5(4

14

1 113252

min22max

LfC

Page 7: Chapter 30

30.29: a) )F1000.6()H50.1(222 5 πLCπωπT

s.rad105,s0596.0 ω b) .C1020.7)V0.12)(F1000.6( 45 CVQ

c) .J1032.4)V0.12)(F1000.6(21

21 3252

0 CVU

d) .0)cos(,0At ωtQQqt

)F1000.6)(H50.1(s0230.0cos)C1020.7()cos(s,0230.0

5

4ωtQqt

C.1043.5 4 Signs on plates are opposite to those at .0t

e) )sin(,s0230.0 ωtωQdtdqit

A.0.0499H)10(6.00H)(1.50

s0.0230sinH)10H)(6.00(1.50

C107.2055

4

i

Positive charge flowing away from plate which had positive charge at .0t

f) Capacitor: .J102.46F)102(6.00

C)10(5.432

35

242

CqU C

Inductor: .J101.87A)(0.0499H)50.1(21

21 322 LiU L

Page 8: Chapter 30

30.30: (a) Energy conservation says (max) = (max) CL UU

A0.871

H1012F1018V)22.5(

CV21

21

3

6

max

22max

LCVi

Li

The charge on the capacitor is zero because all the energy is in the inductor. (b)

LCT 22

at 41 period: F)10(18H)10(122

)2(41

41 63

LCT

s1030.7 4

at 43 period: s1019.2)s1030.7(343 34 T

(c) CCVq 405)V(22.5F)18(0

30.31: F0.30V1029.4C10150

3

9

VQC

For an L-C circuit, LCω 1 and LCωπT 22

mH601.0)2( 2

C

TL

Page 9: Chapter 30

30.32: rad/s1917)F1020.3()H0850.0(

16

ω

a) C1043.4srad1917A1050.8 7

4max

maxmaxmax

ω

iQωQi

b) From Eq. 31.26 2

1

42722

s1917A1000.5)C1043.4(

LCiQq

.C1058.3 7

30.33: a) )sA80.2()F1060.3()H640.0(01 62

2

dtdiLCqq

LCdtqd

C.1045.6 6

b) .V36.2F1060.3C1050.8

6

6

Cq

30.34: a) .maxmax

maxmaxmax LCiω

iQωQi

J450.0

)F1050.2(2)C1050.1(

2

.C1050.1)F1050.2()H400.0()A50.1(

10

25max

2

max

510max

CQU

Q

b) 14

10s1018.3

)F1050.2()H400.0(

11222

LCf

(must double the frequency since it takes the required value twice per period).

30.35: .ssAA1s

sC

VCsΩ

VCHFH][ 222 sLCLC

30.36: Equation (30.20) is .012

2

qLCdt

qd We will solve the equation using:

.110)cos()cos(1

).cos()sin()(cos

222

2

22

2

LCω

LCωt

LCQωtQωq

LCdtqd

ωtQωdt

qdωtωQdtdqωtQq

Page 10: Chapter 30

30.37: a) .)(cos21

21 222

CωtQ

CqUC

.1since,)(sin21)(sin

21

21 2

222222

LCω

CωtQωtQLωLiU L

b) )(sin21)(cos

21 2222

2

Total ωtQLωωtCQUUU LC

)(sin121)(cos

21 222

2

ωtQ

LCLωt

CQ

))(sin)((cos21 22

2

ωtωtCQ

Total

2

21 U

CQ

is a constant.

30.38: a) )cos()2/( tωAeq tLR

)sin(

22)cos(

2

).sin()cos(2

)2/()2/(2

2

2

)2/()2/(

tωeL

RAωtωeL

RAdt

qd

tωAeωtωeL

RAdtdq

tLRtLR

tLRtLR

).cos()2/(2 tωAeω tLR

2

22

2

22

2

2

2

41

0122

LR

LCω

LCLR

LRq

LCq

dtdq

LR

dtqd

b) :0,,0 dtdqiQqtAt

.4//12

2tan

2and

cos

0sincos2

andcos

22 LRLCL

RωL

RQωL

QRQA

AωAL

RdtdqQAq

Page 11: Chapter 30

30.39: Subbing ,1,,,C

kRbLmqx we find:

a) Eq. (13.41): .0:)27.30.(Eq0 2

2

2

2

LCq

dtdq

LR

dtqd

mkx

dtdx

mb

dtxd

b) Eq. (13.43): .4

1:)28.30.(Eq4 2

2

2

2

LR

LCω

mb

mkω

c) Eq. (13.42): ).cos(:)28.30.(Eq)cos( )2/()2/( tωAeqtωAex tLRtmb

30.40: .A

VVCs

FH 2

CL

CL

30.41: LCLC

LRLCLC

LRLCL

RLC 6

1126

1146

14

1 222

22

. 4.45F)10(4.60H)6(0.285

1F)10(4.60H)(0.285

1)H285.0(2 44

R

30.42: a) When s.rad298F)10(2.50H)(0.450

11,050

LCωR

b) We want 2222

0

)95.0(4

11

)41(95.0

LCR

LCLRLC

ωω

.8.83F)10(2.50

(0.0975)H)4(0.450))95.0(1(45

2

CLR

30.43: a)

b) Since the voltage is determined by the derivative of the current, the V versus t graph is indeed proportional to the derivative of the current graph.

Page 12: Chapter 30

30.44: a) ]s)240cos[()A124.0(( tπdtdL

dtdiLε

).)s240((sin)V4.23())s240sin(()240()A124.0()H250.0( tπtε

b) ,0;V4.23max i since the emf and current are 90 out of phase. c) ,0;A124.0max i since the emf and current are 90 out of phase.

30.45: a) ).(ln22

)(2

)( 000 abNihμrdrNihμhdr

rπNiμhdrB

b

a

b

a

b

aB

b) ).(ln2

20 ab

πhNμ

iNL B

c) .2

L2

)()/)(1(ln)/(ln2

02

2

a

abπ

hNμa

aba

abaabab

30.46: a) .1

22210

1

2210

1

110

1

22

1

22212 l

rπNNμl

ANNμl

IANμIA

ANAA

IN

INM BB

b) .1

1

222101

1

210222

2

dtdi

lrπNNμ

dtdi

lANμN

dtd

Nε B

c) .2

1

2221022

121 dtdi

lrπNNμ

dtdiM

dtdiMε

Page 13: Chapter 30

30.47: a) .H5.7)/A00.4/()V0.30()//( sdtdiεLdtdiLε

b) .Wb360)s0.12)(V0.30(

fif tεdtdε

c) W.1440/s)A00.4)(A0.48)(H50.7( dtdiLiPL

.0104.0W138240)0.60()A0.48( 22R

R

L

PPRiP

30.48: a) ))s0250.0/cos()A680.0(()H1050.3( 3 tdtd

dtdiL

.V299.0s0250.0

)A680.0)(H1050.3( 3max

b) .Wb1095.5400

)A680.0)(H1050.3( 63

maxmax

N

LiB

c) .)s0250.0/(sin)s0250.0/)(A680.0)(H1050.3()( 3 tdtdiLt

V230.0)(

))s0180.0)(s6.125((sin)V299.0(s)0180.0())s6.125sin(()V299.0()(

1

1

t

tt

30.49: a) Series: ,eq2

21

1 dtdiL

dtdi

Ldtdi

L

but iii 21 for series components so eq2121 thus, LLL

dtdi

dtdi

dtdi

b) Parallel: Now . where, 21eq2

21

1 iiidtdiL

dtdi

Ldtdi

L

.11

and But.So

1

21eq

2

eq

1

eq

2

eq2

2

eq121

LLL

dtdi

LL

dtdi

LL

dtdi

dtdi

LL

dtdi

dtdi

LL

dtdi

dtdi

dtdi

dtdi

Page 14: Chapter 30

30.50: a) .2

2 00encl0 πr

iμBiμπrBIμdiB

b) .2

0 ldrπr

iμBdAd B

c) b

a

b

aBB ).ab(

πilμ

rdr

πilμd ln

2200

d) ).ln(2

0 abπ

μli

NL B

e) ).ln(4

)ln(22

121 2

0202 abπliμiab

πμlLiU

30.51: a) .2

2 00encl0 πr

iμBiμrπBIμd lB

b) .4

)2(22

1)2(2

20

2

0

00

2

drrli

rdrlri

rdrluudVdUBu

c) )./(ln44

20

20 ab

πliμ

rdr

πliμdUU

b

a

b

a

d) ),/(ln2

221 0

22 ab

πμl

iULLiU which is the same as in Problem 30.50.

30.52: a) ,22

210110

1

1

1

11

1

rπANμ

rπiNμ

iAN

iN

L B

ANμrπiNμ

iAN

iN

L B

22

220220

2

2

2

22

2

b) .222 21

220

210

2

2102 LLrπ

ANμrπ

ANμrπ

ANNμM

30.53: EεμEμεBμ

BEεuu EB 002

000

220

22

T.1017.2)V/m650( 600

με

Page 15: Chapter 30

30.54: a) .1860A1045.6

0.123

ViVR

f

b) )/1(ln

)/1(ln)1( )/(

ff

tLRf ii

RtLiiLRteii

H.963.0))45.6/86.4(1(ln

)s1025.7)(1860( 4

L

30.55: a) After one time constant has passed:

.J281.0)A474.0)(H50.2(

21

21

A474.0)1(00.8

V00.6)1(

22

11

LiU

eeR

i

Or, using Problem (30.25(c)):

.)()50.4(7/3

0

)40.6()20.3( dteeWdtPU ttL

J281.040.6

)1(20.3

)1()W50.4(21

ee

b)

)1()W50.4()1()W50.4( 1

/

0

)/(tot e

RL

RLdteU

RLtLR

J517.000.8

H50.2)W50.4( 1tot

eU

c) dteeU tLRRL

tLRR )21()W50.4( )/(2

/

0

)/(

)1(

2)1(2)W50.4( 21 e

RLe

RL

RL

J.236.0)168.0(00.8

H50.2)W50.4(

RU

The energy dissipated over the inductor (part (a)), plus the energy lost over the resistor (part (c)), sums to the total energy output (part (b)).

Page 16: Chapter 30

30.56: a) J.1000.5240

V60)H160.0(21

21

21 3

222

0

RLLiU

b) tLRLtLR eR

RidtdiiL

dtdU

iLR

dtdie

Ri )/(2

22)/(

W.52.4240

)60( )1000.4)(160.0/240(22

4

eV

dtdU L

c) In the resistor:

.W52.4240

)V60( .)1000.4)(160.0/240(22

)/(22

2 4

ee

RεRi

dtdU tLRR

d) tLRR e

RεRitP )/(2

22)(

J,1000.5)240(2

)H160.0()V60(2

32

22

0

)/(22

RL

Rεe

RεU tLR

R

which is the same as part (a).

30.57: Multiplying Eq. (30.27) by i, yields:

.021

21R

22222

CLR PPPCq

dtdLi

dtdRi

dtdq

Cq

dtdiLiRii

Cq

dtdiLii

That is, the rate of energy dissipation throughout the circuit must balance over all of the circuit elements.

Page 17: Chapter 30

30.58: a) If 24

3cos8

32cos)cos(8

3 QQTT

QtQqTt

.222

122

121

2)2(1)(1

2222

22222

BE UC

qC

QLC

QLLiU

LCQQQ

LCqQ

LCi

b) The two energies are next equal when .8

58

52

TtπωtQq

30.59: μFVUCCVUV CCCCC 222)V0.12/()J0160.0(2/2 so ;V0.12 22221

Cπf

LLCπ

f 2)2(1 so

21

H31.9givesHz3500 μLf

30.60: a) .V0240.0F1050.2C1000.6

4

6

max

CQV

b) A1055.1F)1050.2)(H0600.0(

1000.622

1 3

4

6

max

22max

LCQi

CQLi

c) .J1021.7)A1055.1)(H0600.0(21

21 8232

maxmax LiU

d) If C

qC

QUUUUii CL 22

43

43J1080.1

41

21 2

2

max8

maxmax

times.allfor

21

21

.C105.2043

22

max

6

CqLiU

Qq

Page 18: Chapter 30

30.61: The energy density in the sunspot is ./mJ10366.62/ 340

2 μBuB

The total energy stored in the sunspot is .VuU BB

The mass of the material in the sunspot is .ρVm

;21 so 2

BB UmvUK VuVvρ B2

21

The volume divides out, and /sm102/2 4 ρuv B

30.62: (a) The voltage behaves the same as the current. Since , iVR the scope must be across the 150 resistor.

(b) From the graph, as ,V25, RVt so there is no voltage drop across the inductor, so its internal resistance must be zero.

)1( /max

rtR eVV

when .63.0)1(, max1

max VVVt eR From the graph, when τtVV ms5.0 ,V16 63.0 max

H075.0)150()ms5.0(ms5.0/ LRL

(c) Scope across the inductor:

Page 19: Chapter 30

30.63: a) In the R-L circuit the voltage across the resistor starts at zero and increases to the battery voltage. The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the voltage drops, so the oscilloscope is across the solenoid.

b) At t the current in the circuit approaches its final, constant value. The voltage doesn’t go to zero because the solenoid has some resistance .LR The final voltage across the solenoid is ,LIR where I is the final current in the circuit.

c) The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current is zero and there is no voltage drop across any of the resistance in the circuit.

d) As 0, LIRIRεt

V50ε and from the graph V15LRI (the final voltage across the inductor), so

A3.5/RV)35(andV35 IRI

e) 4.3A)(3.5/V)15(so V,15 LRRI L

LL ViRVε where,0 includes the voltage across the resistance of the solenoid.

V9.27,whenso,3.14,10V,50

)]1(1[ so ),1(,

tot

/

tot

/

tot

l

τtL

tL

VτtRRε

eRRεVe

RεiiRεV

From the graph, LV has this value when t = 3.0 ms (read approximately from the graph), so .mH43)3.14()ms0.3(Thenms.0.3/ tot LRLτ

Page 20: Chapter 30

30.64: (a) Initially the inductor blocks current through it, so the simplified equivalent circuit is

A333.0150

V50

Rεi

0,A333.0resistor)50withparallelin(inductorV7.16

it.throughflowscurrentnosince0V16.7A)333.0()50(V33.3A)333.0()100(

231

42

3

4

1

AAAVV

VVV

(b) Long after S is closed, steady state is reached, so the inductor has no potential drop across it. Simplified circuit becomes

A230.050

V5.11

A153.075

V5.11A,385.0

V11.5V38.5V500;V5.38)A385.0()100(

A385.0130

V50/

3

21

43

21

i

ii

VVVV

Rεi

Page 21: Chapter 30

30.65: a) Just after the switch is closed the voltage 5V across the capacitor is zero and there is also no current through the inductor, so ,.0 54323 VVVVV and since

2435 and,0and0 VVVV are also zero. 04 V means 3V reads zero.

1V then must equal 40.0 V, and this means the current read by 1A is A.0.800)(50.0/V)0.40(

A.800.0 so 0but, 14321432 AAAAAAAA

A;800.041 AA all other ammeters read zero.

V0.401 V and all other voltmeters read zero.

b) After a long time the capacitor is fully charged so .04 A The current through the inductor isn’t changing, so .02 V The currents can be calculated from the equivalent circuit that replaces the inductor by a short-circuit.:

V0.24)0.50(A480.0readsA;480.0)33.83(/V)0.40(

1

1

IVAI

The voltage across each parallel branch is V16.0V4.02V0.40

V0.16,0 5432 VVVV

.thatNotezero.readsA.320.0readsmeansV0.16.A160.0readsmeansV0.16

132

43423

AAAAAVAV

C192)V(16.0F)0.12(soV0.16)c 5 CVQV

d) At t = 0 and .0, 2 Vt As the current in this branch increases from zero to 0.160 A the voltage 2V reflects the rate of change of current.

Page 22: Chapter 30

30.66: (a) Initially the capacitor behaves like a short circuit and the inductor like an open circuit. The simplified circuit becomes

A500.0150

V75

Ri

0A,500.0V0.50parallel)(in

V50.0A)50.0()100(,0V25.0A)50.0()50(

231

4

43

1

AAAVV

VVRiV

2

(b) Long after S is closed, capacitor stops all current. Circuit becomes

V0.753 V and all other meters read zero.

(c) nC,5630V)(75nF)75( CVq long after S is closed.

Page 23: Chapter 30

30.67: a) Just after the switch is closed there is no current through either inductor and they act like breaks in the circuit. The current is the same through the 15.0and0.40 resistors and is equal to A;455.0A.455.0)0.150.40(V)0.25( 41 AA

.032 AA

b) After a long time the currents are constant, there is no voltage across either inductor, and each inductor can be treated as a short-circuit . The circuit is equivalent to:

A585.0)73.42(V)0.25( I

A.0.585reads1A The voltage across each parallel branch is )(40.0A)(0.585V0.25

A.107.0)0.15(V)60.1(reads A.160.0)0.10V)60.1(readsA.320.0)0.5(V)60.1(readsV.1.60 432

AAA

30.68: (a) ,s50.0sincems40.025mH10 τRLτ steady state has been reached, for

all practical purposes.

A00.225V50 Ri The upper limit of the energy that the capacitor can get is the energy stored in the

inductor initially.

C1090.0)F1020()H1010()A00.2(21

2363

max

0max20

2max

Q

LCiQLiC

QUU LC

(b) Eventually all the energy in the inductor is dissipated as heat in the resistor.

J100.2

)A00.2()H1010(21

21

2

2320

LiUU LR

Page 24: Chapter 30

30.69: a) At ,0t all the current passes through the resistor ,1R so the voltage abv is the total voltage of 60.0 V. b) Point a is at a higher potential than point .b c) V0.60cdv since there is no current through .2R d) Point c is at a higher potential than point .b e) After a long time, the switch is opened, and the inductor initially maintains the

current of .A40.225.0

V0.60

22

R

iR Therefore the potential between a and b is

.V0.96)0.40()A40.2(R1 ivab f) Point b is at a higher potential than point a. g) V156)2540()A40.2()( 21 RRivcd h) Point d is at a higher potential than point c. 30.70: a) Switch is closed, then at some later time:

.V0.15)A/s0.50()H300.0(A/s0.50 dtdiLv

dtdi

cd

The top circuit loop: 60.0 .A50.10.40

V0.60V 111

iRi

The bottom loop: A.80.10.25

V45.00V0.15V60 222

iRi

b) After a long time: ,A40.20.25

V0.602

i and immediately when the switch is

opened, the inductor maintains this current, so .A40.221 ii

Page 25: Chapter 30

30.71: a) Immediately after 1S is closed, ,V0.36and,0,00 cbac vvi since the inductor stops the current flow.

b) After a long time, 0i A180.015050V0.36

0

RR ,

.V0.27V00.9V0.36and,V00.9)50()A18.0(00 cbac vRiv

c) ),1()A180.0()()1()( )s50()(

total

1total ttLRetie

Rti

.3)V00.9(1)V00.9(V0.36)()(v

and1)V00.9()()()50()50(

0

)s50(0

11

1

tstscb

tac

eeRtiεt

eRtitv

Below are the graphs of current and voltage found above.

Page 26: Chapter 30

30.72: a) Immediately after 2S is closed, the inductor maintains the current A180.0i through .R The Kirchoff’s Rules around the outside of the circuit yield:

.0andV0.36)V50()A72.0(,A072.0

Ω50V36

0)50()150()18.0()150()18.0(V0.36

0

000

cbac

L

vvi

iRiiRεε

b) After a long time, ,V0.36acv and .0cbv Thus

,A720.050

V0.36

00

Rεi

A720.0and,0 2 sR ii

c) and,)A180.0()()(,A720.0 )s5.12()(

total0

1 tR

tLRR etie

Rεtii

tts eeAti )s5.12()s5.12( 11

2 4)A180.0()180.0()A720.0()(

Below are the graphs of currents found above.

Page 27: Chapter 30

30.73: a) Just after the switch is closed there is no current in the inductors. There is no current in the resistors so there is no voltage drop across either resistor. A reads zero and V reads 20.0 V. b) After a long time the currents are no longer changing, there is no voltage across the inductors, and the inductors can be replaced by short-circuits. The circuit becomes equivalent to

A267.0)0.75()V0.20( aI

The voltage between points a and b is zero, so the voltmeter reads zero. c) Use the results of problem 30.49 to combine the inductor network into its equivalent:

0.75R is the equivalent resistance.

V0.9VV0.20so0V0.20V0.11)0.75)(A147.0(

A147.0so,ms115.0,0.75,0.20ms144.0)(75.0mH)8.10(/with),1)((says Eq.(30.14)

RLLR

R

τt

VVViRV

itRVεRLτeRεi

Page 28: Chapter 30

30.74: (a) Steady state: A600.0125

V0.75

Ri

(b) Equivalent circuit:

F6.14

F351

F2511

s

s

CC

Energy conservation: 20

2

21

2Li

Cq

)F106.14()H1020()A600.0( 630

LCiq C41024.3

s1049.8)F106.14()H1020(

2

2)2(

41

41

463

t

LCLCTt

Page 29: Chapter 30

30.75: a) Using Kirchhoff’s Rules: and,01

111 RεiRiε

).1(0 )(

2222

2 2 tLReR

iRidtdiL

b) After a long time, .andstill,2

21

1 Rεi

Rεi

c) After the switch is opened, ,))((21

21

2

tLRReRεii and the current drops off.

d) A 40-W light bulb implies .360W40V)(120 22

P

VR If the switch is opened,

and the current is to fall from 0.600A to 0.150 A in 0.0800 s, then: )s0800.0)(H0.22)360(())((

2221 )A600.0(A150.0)A600.0( RtLRR eei

.V7.12)2.21)(A600.0(

0.21360)00.4ln(0800.0

H0.22

22

22

Riε

RRs

e) Before the switch is opened, A0354.0360

V7.12

10

Ri

Page 30: Chapter 30

30.76: Series: .eq2

121

212

21

1 dtdiL

dtdi

Mdtdi

Mdtdi

Ldtdi

L

But .and 211221

21 MMMdtdi

dtdi

dtdiiii

.andwith

,and

haveWe:Parallel

.2or

,)2(So

211221

121

22

eq2

121

1

21eq

eq21

MMMdtdi

dtdi

dtdi

dtdiL

dtdiM

dtdiL

dtdiL

dtdiM

dtdiL

MLLLdtdiL

dtdiMLL

eq

To simplify the algebra let . and ,, 21

dtdiC

dtdiB

dtdiA

So .,, eq2eq1 CBACLMABLCLMBAL Now solve for .oftermsinand CBA

.)2(

)()()2(

0)()()(0)()()(

.using0)()(

21

1121

211

21

21

CLLM

LMBCLMBLLM

BLMBMLCMLBLMBCML

BCABLMAML

But ,)2(

)2()2(

)(

21

121

21

1 CLLM

LMLLMLLM

CLMCBCA

or .2 21

2 CLLM

LMA

Substitute A in B back into original equation.

So CLCLLM

LMMLLMCLML

eq21

1

21

21

)2()(

2)(

.2 eq

21

212

CLCLLM

LLM

Finally, MLL

MLLL

221

221

eq

Page 31: Chapter 30

30.77: a) Using Kirchhoff’s Rules on the top and bottom branches of the circuit:

).1(

)00

).1(0

)/1(

0

)/1(2

20 22

)1(

22

22

2222

)(

11

111

22

2

1

tCRt

tCRt

tCR

tLR

eεCCeRRεtdiq

eRεi

CiR

dtdi

CqRiε

eRεi

dtdiLRiε

b) .A1060.95000

V0.48,0)1()0( 30

22

0

11

eRεie

Rεi

c) As .0,A92.10.25

V0.48)1()(:2

211

1

eRεi

Rεe

Rεit

A good definition of a “long time” is many time constants later.

d) .)1()1( )1(

2

1)()1(

2

)(

121

21211 tCRtLRtCRtLR eRRee

Re

Rii

Expanding the exponentials like :findwe,!32

132

xxxe x

22

2

2

122

11

21

21

CRt

RCt

RRt

LRt

LR

,)(2

122

2

11

RRtO

CRR

LRt

if we have assumed that .1t Therefore:

.s106.1)F100.2()5000(H0.8

)F100.2)(5000)(H0.8(

)1()1(11

352

5

22

22

22

t

CRLCLR

CRLRt

e) At .A104.9)1(25

V48)1(:s1057.1 3)825()(

11

3 1

ttLR eeR

it

f) We want to know when the current is half its final value. We note that the current 2i is very small to begin with, and just gets smaller, so we ignore it and find:

).1)(A92.1()1(A960.0 )()(

1121

11 tLRtLR eeR

ii

s22.0)5.0ln(25

H0.8)5.0ln(500.01

)( 1

RLte tLR

Page 32: Chapter 30

30.78: a) Using Kirchoff’s Rules on the left and right branches:

Left: .)(0)( 121

121 ε

dtdiLiiR

dtdiLRiiε

Right: .)(0)( 221

221 ε

CqiiR

CqRiiε

b) Initially, with the switch just closed, .0and,0 221 qRεii

c) The substitution of the solutions into the circuit equations to show that they satisfy the equations is a somewhat tedious exercise in bookkeeping that is left to the reader. We will show that the initial conditions are satisfied:

.0)])0[cos(1()0()]cos()sin()2[(1()(

0)0sin()sin(,0At

11

1

2

RεiωtωtωRCe

Rεti

ωRεωte

ωRεqt

βt

βt

d) When does 2i first equal zero? rad/s625)2(

112

RCLCω

.s10256.1rad/s625

785.00.7851.00)arctan(

1.00.F)1000.2)(400)(rad/s625(22)tan(

01)tan()2()]cos()sin()2([0)(

3

6

112

tωt

ωRCωt

ωtωRCωtωtωRCeR

ti bt

Page 33: Chapter 30

30.79: a) )())(( 00AirAir dW

WNiKμ

WdDW

NiμABABBA LLB

])[(0 KddDNiμ

.and,where,

])[(

20

200

0

0

0000

20

DNKμLDNμLDLLLL

d

dD

LLL

DdL

DdLLKddDNμ

iNL

ff

ff

B

b) Using 1 mK we can find the inductance for any height .10

DdLL m

__________________________________________________________________ Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury 4Dd 0.63024 H 0.63000 H 2Dd 0.63048 H 0.62999 H 43Dd 0.63072 H 0.62999 H Dd 0.63096 H 0.62998 H __________________________________________________________________ Where are used the values .5

m3

2 109.2 (Hg)and1052.1)O( m d) The volume gauge is much better for the liquid oxygen than the mercury because there is an easily detectable spread of values for the liquid oxygen, but not for the mercury.