Chapter 30
-
Upload
anon-158680 -
Category
Documents
-
view
425 -
download
3
description
Transcript of 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
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
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
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)
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
Rε
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.
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
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
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
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
VΩ
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
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
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.
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
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
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
rπ
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
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)).
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.
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
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:
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
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
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.
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.
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
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
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.
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.
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
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
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
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
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
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
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.