b From the graph, as t ,V R 25V,so there is no voltage drop across the inductor, so its internal resistance must be zero... 30.63: a In the R-L circuit the voltage across the resistor
Trang 14
3 2
(0.00285
A)(1.40V)106.12()/(/
b) Terminal a is at a higher potential since the coil pushes current through from b to
aand if replaced by a battery it would have the terminal at a
m)120.0(2
)m1080.4()1800()500(2
2 0 m
π
μ πr
A N μ K K
Trang 230.9: For a long, straight solenoid:
.//
di b a Thus, the current is decreasing
b) From above we have that di(4.00A/s)dt 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
42
0.12()m1000.5(
J)(0.390m)150.0(44
2 2
4 0
2 0
μ
πrU N
A)(80.0
J)1073.1(222
1
2
7 2
I
U L LI
U
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
Trang 330.16: a) free space: (0.0290m ) 3619J.
2
T)(0.560V
)450(2
T)(0.560V
2V
0
2
0 m
U K
2
6 0
2 0 0
2
m1.25T)
(0.600
J)1060.3(22
J)1060.3(2
3
6 0
)A50.2()600(2
b) From Eq (30.10), 7.53J/m
2
)T1035.4(2
3 0
2 3
c) Volume V2rA2 (0.0690m)(3.5010 6 m2)1.5210 6 m3
d) U uV (7.53J/m3)(1.5210 6m3)1.1410 5 J
)m0690.0(2
)m1050.3()600(2
6 2
6 2
L
(3.65 10 H)(2.50A) 1.14 10 J
2
12
V00.60
L
iR dt
di
H50.2
)00.8()A500.0(V00.6A
00
dt
di i
8.00
V00.6)1
(s
200
8.00
V00.6
Trang 4(1000
A0.0259
e1A0.030)
e(1
R Battery L
R
) ( max
10 20
Ri V
i
μs
R / L /
(or, could use V L dt di att20s)
L
c)
30.21: a) i/R(1et/ ), L/R
2 1 2
1 max
max ε/Rsoii /2when(1e / τ) ,ande / τ
i
μs
L t τ
50.0
H)10(1.252)ln(R
ln2and
)(
1et/τ t/τ
μs L
t ln(0.2929)/R 30.7
Trang 530.22: a) 2.13A
H0.115
J)260.0(22
LI U
/ ( 2 2 2
) / (
2
12
12
12
12
1ln)2(120
H115.02
1ln2
21
4
) / ( 2
e R L t
240
V60
)H160.0(2
1ln2
2 / 1 )
/ ( 0
2 /
e I
)V00.6()1
()1
( ( / ) 2 ( / ) 2 ( 8 00 / 2 50 H ) 0
t t
L R t
L
R e
I i
b) 2 2 ( / ) 2 2 (1 ( 8 00 / 2 50 H ) )2
00.8
)V00.6()1
R
ε R
()W50.4( ( 3 20 s 1 )t 2
c) (1 (R/L)t) (R/L)t 2 ( (R/L)t 2 (R/L)t)
R
ε e
L
ε L e
R
ε dt
di iL
Trang 630.26: When switch 1 is closed and switch 2 is open:
.)
/(ln
0
) / ( 0 0
0
0
L R t
i I
t
e I i t L
R I
i
t d L
R i
i d L
R i dt
di iR
dt
di L
14
12 2
6 2
2 2
)H1037.2()1040.5(4
14
3 2
5 2
min 2 2 max
Trang 730.29: a) 2 2π LC 2π (1.50H (6.0010 5F)
ω
π T
H50.1(
s0230.0cos
)C1020.7()cos(
ωt Q
q t
5.4310 4C.Signs on plates are opposite to those at t 0
e) 0.0230 s, ωQsin(ωt)
dt
dq i
H)10(6.00H)(1.50
s0.0230sin
H)10H)(6.00(1.50
C107.20
5 5
C)10(5.432
3 5
2 4 2
Li
U L
Trang 830.30: (a) Energy conservation says U L(max)=U C(max)
A0.871H
1012
F1018V)22.5(
CV2
12
1
3
6 max
2 2
2
at 41 period: (12 10 H)(18 10 F)
2)2
(4
14
LC T
C10150
C T
Trang 930.32: 1917rad/s
)F1020.3(H0850.0(
A1050
max max max
7 2
2
s1917
A1000.5)C1043.4(
LC dt
q d
6.4510 6C
F1060.3
C1050.8
ω
i Q ωQ
J450.0)F1050.2(2
)C1050.1(2
.C1050.1)F1050.2(H400.0()A50.1(
10
2 5 max
2 max
5 10
Q
)F1050.2(H400.0(
11
(must double the frequency since it takes the required value twice per period)
A
1ss
CV
ΩV
CsΩV
CHFH]
q d
We will solve the equation using:
.11
0)cos(
)cos(
1
)
cos(
)sin(
)(
cos
2 2
2
2
2 2
2
LC
ω LC
ωt LC
Q ωt
Q ω q LC
dt
q
d
ωt Q ω dt
q d ωt
ωQ dt
dq ωt
Trang 1030.37: a) cos ( ).
2
12
C
ωt Q
sin2
12
LC
ω C
ωt Q
ωt Q
Lω Li
2
1)(
cos2
C
Q U
cos2
C Q
22)cos(
2
)
sin(
)cos(
2
) 2 / ( )
2 / ( 2
2 2
) 2 / ( )
2 / (
L
R A ω t
ω e
L
R A dt
q d
t ω Ae
ω t
ω e
L
R A dt dq
t L R t
L R
t L R t
L R
ω2Ae (R/ 2L)tcos(ωt)
2
2 2
2
2 2 2
2 2
41
0
12
2
L
R LC ω
LC L
R L
R q LC
q dt
dq L
R dt
q d
At
.4//
12
2
tan2
andcos
0sincos
2and
cos
2
R LC L
R
ω L
R Q
ω L
QR Q
A
A ω A
L
R dt
dq Q
A q
Trang 1130.39: Subbing , , , 1,
C k R b L m q
a) Eq (13.41): 0 Eq.(30.27): 2 0
2 2
dq L
R dt
q d m
kx dt
dx m
b dt
x d
4
1:
)28.30.(
Eq
2 2
2
L
R LC
ω m
b m
C
sF
L
30.41:
LC LC
L R LC LC
L R LC L
R
1126
1146
14
1F)
10(4.60H)(0.285
1)
H285
11
,0
R
0
)95.0(4
11
)41
(95
L
C R LC
L R LC ω
ω
F)10(2.50
(0.0975)H)
4(0.450)
)95.0(1(
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
Trang 1230.44: a) ((0.124A)cos[(240π s)t]
dt
d L dt
di L
ε
ε(0.250H)(0.124A)(240)sin((240 s)t)(23.4V)sin((240π s)t)
b) max 23.4V;i0,since the emf and current are 90out of phase
c) imax 0.124A; 0,since the emf and current are 90 out of phase
22
)(2)
r
dr Nih μ hdr r π
Ni μ hdr
B
b a
b a
b a
)()
/)(1(ln)
/
(
ln
2 0 2
h N μ a
a b a
a b a a b a
b
1
2 2 2 1 0 1
2 2 1 0 1
1 1 0 1
2 2 1
2 2 2
1
r π N N μ l
A N N μ l
IA N μ IA
A N A
A I
N I
2 1 0 2 2
dt
di l
r π N N μ dt
di l
A N μ N dt
d N
1
2 2 2 1 0 2 2
12
1
dt
di l
r π N N μ dt
di M dt
di M
Trang 1330.47: a) Lε/(di/dt)(30.0V)/(4.00 A/s)7.5H.
dt
di L ε
P R
H1050.3
H1050.3
c) ( ) (3.50 10 3H)(0.680A)( /0.0250s)sin( t/0.0250s )
dt
di L
V230.0)(
))s0180.0)(
s6.125((
sin)V299.0(s)0180.0(
))s6.125sin((
)V299.0()(
1 1
dt
di L dt
di L dt
di dt
di L dt
di
.11
and But
.So
1
2 1
eq 2
eq 1 2
L dt
di L
L dt
di L
L dt
di dt
di L
L dt
di dt
di dt
di
dt
di
Trang 1430.50: a)
2
0 encl
0
πr
i μ B i μ πr B I
i μ BdA
dr π
il μ
22
0 0
N
4)
ln(
22
12
0 2 0
π
li μ i a b π
μ l Li
0
πr
i μ B i μ r π B I
1)2(2
2 0
2 0 0 0
2
dr r
l i rdr
l r
i rdr
l u udV dU
2 0
2
π
l i μ r
dr π
l i μ dU
U
b a
U L Li
U which is the same as in Problem 30.50
22
2 1 0 1 0 1
1 1
1 1
1
r π
A N μ r π
i N μ i
A N i
A N μ r
π
i N μ i
A N i
N
22
2 2 0 2 2 0 2
2 2
2 2
2 2 0
2 1 0
2 2 1 0
r π
A N μ r π
A N μ r
π
A N N μ
0 0 0
2 2 0
ε μ
Trang 1530.54: a) 1860
A1045.6
0.12
f
b)
)/1(ln)
/1(ln)
1
( ( / )
f f
t L R f
i i
Rt L
i i L
Rt e
)s1025.7)(
H50.2(2
12
1
A474.0)1(00.8
V00.6)1(
2 2
1 1
e e
0
) 40 6 ( ) 20 3
e W dt
)1(20.3
)1()W50.4
(
2 1
0
) / (
R
L R
L dt
e U
R L
t L R
00.8
H50.2)W50.4
2)
W50.4
R
L e
R
L R L
(0.168) 0.236J
00.8
H50.2)W50.4
Trang 1630.56: a) 5.00 10 J.
240
V60)H160.0(2
12
12
2 2
2 0
R
Ri dt
di iL dt
dU i L
R dt
di e
)V60( 2 2(240/0.160)(4.0010 .)
/ ( 2
P ( ) 2 2 2 ( / )
)240(2
)H160.0()V60(2
3 2
2 2
0
) / ( 2
which is the same as part (a)
30.57: Multiplying Eq (30.27) by i, yields:
.0
2
12
1R
2 2
2 2
P
C
q dt
d Li dt
d R i dt
dq C
q dt
di Li R i i C
q dt
di Li
i
That is, the rate of energy dissipation throughout the circuit must balance over all of the circuit elements
Trang 1730.58: a) If
24
3cos8
32cos)
t Q q
22
12
2
12
1
2)2(
1)(
1
2 2 2
2
2 2
2 2
Q LC
Q L Li
U
LC
Q Q
Q LC q
Q LC
52
T t
π ωt
L LC
π
)2(
1so
C1000.6
F)1050.2)(
H0600.0(
1000.62
2
4
6 max
2 2
C
Q Li
c) (0.0600H)(1.55 10 A) 7.21 10 J
2
12
Q U
U U
U i
22
434
3J
1080.14
12
2
max
8 max
12
1
.C105.204
3
2 2
max
6
C
q Li
U
Q q
Trang 1830.61: The energy density in the sunspot is /2 6.366 104J /m3.
0
B μ
u B
The total energy stored in the sunspot is U B u B V
The mass of the material in the sunspot is m ρV
;2
The volume divides out, and v 2u B /ρ 2104m/s
across the 150resistor
(b) From the graph, as t ,V R 25V,so there is no voltage drop across the inductor, so its internal resistance must be zero
)1
max
r t
t R e From the graph, when
τ t
V
V 0.63 max 16V, 0.5ms
H075.0)150()ms5.0(ms
5.0
L
(c) Scope across the inductor:
Trang 1930.63: 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 R The final voltage across L.the solenoid is IR where I is the final current in the circuit L,
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 t,εIRIR L 0
V50
ε and from the graph I R L 15V (the final voltage across the inductor), so
A3.5/RV)35(andV
whenso,3.14,
10V,
50
)]
1(1
[so
),1
(,
tot
/ tot
/ tot
t L
V τ t R
R ε
e R
R ε V e
R
ε i iR ε
From the graph, V has this value when t = 3.0 ms (read approximately from the L
graph), so τ L/Rtot 3.0ms.ThenL(3.0ms (14.3)43mH
Trang 2030.64: (a) Initially the inductor blocks current through it, so the simplified equivalent circuit is
A333.0150
0,
A333.0
resistor)50
withparallelin
(inductorV
7.16
it
throughflows
currentno
since0
V16.7A)333.0(50(
V33.3A)
333.0(100(
2 3
1
4 2
A
V V
V5.11
A153.075
V5.11A,
385.0
V11.5V38.5V
50
0
;V5.38)A385.0(100(
A385.0130
V50/
3
2 1
4 3
2 1
V V
V V
R ε i
Trang 2130.65: a) Just after the switch is closed the voltage V5 across the capacitor is zero and there
is also no current through the inductor, so V3 0.V2 V3 V4 V5, and since
2 4
0but
1 4 3
A
A;
800.0
4
A all other ammeters read zero
V0.40
1
V and all other voltmeters read zero
b) After a long time the capacitor is fully charged so A4 0 The current through the inductor isn’t changing, so V2 0 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
The voltage across each parallel branch is 40.0V24.0V16.0V
V0.16,
V
.that
Note
zero.readsA
320.0readsmeans
V0.16
A160.0readsmeans
4 3
4 2
3
A A A
A A
V A
V0.16
)
d) At t = 0 and t,V2 0 As the current in this branch increases from zero to
0.160 A the voltage V reflects the rate of change of current.2
Trang 2230.66: (a) Initially the capacitor behaves like a short circuit and the inductor like an open
circuit The simplified circuit becomes
A500.0150
V75
500.0
V0.50parallel)(in
V50.0A)50.0(100(,0
V25.0A)
50.0(50(
2 3
1 4
4 3
A
V V
V V
Ri V 2
(b) Long after S is closed, capacitor stops all current Circuit becomes
V0.75
3
V and all other meters read zero
(c) q CV (75nF)(75V)5630nC, long after S is closed
Trang 2330.67: 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 40.0and15.0resistors and is equal to (25.0V) (40.015.0)0.455A.A1 A4 0.455A;
320.0)0.5(V)60.1(reads
all practical purposes
A00.225V
R
i 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(
2
12
3 6
3 max
0 max
2 0
2 max
Q U
(b) Eventually all the energy in the inductor is dissipated as heat in the resistor.
J100.2
)A00.2()H1010(2
12
1
2
2 3
2 0
Trang 2430.69: a) At t 0,all the current passes through the resistor R so the voltage 1, v is the ab
total voltage of 60.0 V
b) Point a is at a higher potential than point b c) v cd 60.0Vsince there is no current through R2
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 2.40A
25.0
V0.60
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/s
0
dt
di L v dt
di
cd
The top circuit loop: 60.0 1.50A
0.40
V0.60
V45.00
V0.15V
V0.60
Trang 2530.71: a) Immediately after S is closed, 1 i0 0,v ac 0,andv cb 36.0V, since the inductor stops the current flow.
b) After a long time, i0 0.180A
15050
V0.36
,
.V0.27V00.9V0.36and
,V00.9)50()A18.0(
i e
V00.9(V0.36)
()
(
v
and1
)V00.9()(
)
(
) 50 ( )
50 ( 0
) s 50 ( 0
1 1
1
t s t
s cb
t ac
e e
R t i ε
t
e R
t i
Trang 2630.72: a) Immediately after S is closed, the inductor maintains the current 2 i0.180Athrough R The Kirchoff’s Rules around the outside of the circuit yield:
.0and
V0.36)V50()A72.0(,A072.0Ω50
V36
0)50()150()18.0()150()18.0(V0.36
0
0 0
L
v v
i
i R
i iR ε
1t R
t L R
R
ε t i
Trang 2730.73: 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
a
I
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
0V
0
20
V0.11)0.75)(
A147.0(
A147.0so,ms115.0,0.75,
0
20
ms144.0)(75.0mH)8.10(/with
),1
)(
(saysEq.(30.14)
L R R
τ t
V V
V
iR
V
i t
R V
ε
R L τ e
R ε i
Trang 2830.74: (a) Steady state: 0.600A
125
V0
F6.14
F35
1F25
11
3.2410 4C
s1049.8)F106.14()H1020(2
2)2
(4
141
4 6
T t
Trang 2930.75: a) Using Kirchhoff’s Rules: 0 ,and
1 1 1
ε i R
i
)
1(
2 2 2
2
R i R
i dt
ε i R
ε i
i and the current drops off
d) A 40-W light bulb implies 360
W40
V)(120 2 2
and the current is to fall from 0.600A to 0.150 A in 0.0800 s,
then: ( ) ) ( 360 ) 22 0 H )( 0 0800 s )
2 (0.600A)e R1 R2 L t 0.150A (0.600A)e R2
.V7.12)2.21)(
A600.0(
0.21360
)00.4ln(
0800
0
H0.22
2 2
2 2
R R s
e) Before the switch is opened, 0.0354A
360
V7.12
Trang 3030.76: Series: 2 eq
12
1 21
2 2
1 1
dt
di L dt
di M dt
di M dt
di L dt
di dt
di i
i
.and
with
,and
,)
2
(
So
21 12 2
1
1 21
2
2
eq
2 12
1 1
2 1
eq
eq 2
1
M M M dt
di dt
di M dt
di
L
dt
di L dt
di M dt
di L
M L
L
L
dt
di L dt
di M L
di B dt
(
)(
)(
)2
(
0)(
)(
)(
0)(
)()(
.using
0)(
)(
2 1
1 1
2 1
2 1
1
2 1
2 1
C L L M
L M B
C L M B L L M
B L M B M L C M L
B L M B C M L
B C A B
L M A M L
(
)2
()2
(
)(
2 1
1 2
1 2
1
L L M
L M L L M L
L M
C L M C
M
L M
L M M L
L
M
C L
M
L
eq 2
1
1 2
1
2
1
)2
(
)(
2
)(
2
C L C L L M
L L M
L
M L L L
2
2 1
2 2 1
Trang 3130.77: a) Using Kirchhoff’s Rules on the top and bottom branches of the circuit:
)
1(
)0
0
)
1(0
) / 1 (
0
) / 1 ( 2 2
0 2
2
) 1 ( 2 2
2 2 2 2
2
2
) ( 1
1
1 1
1
2 2
2 1
t C R
t t C R t
t C R
t L R
e εC Ce
R R
ε t
d
i
q
e R
ε i C
i R dt
di C
ε i dt
0)1()
0
2 2 0
1 1
R
ε i
0.25
V0.48)
1()(:
2
2 1
ε e
R
ε i
1 ( 2
) ( 1
2
t C R t
L R t
C R t
L
R
R e
e R
e R i
Expanding the exponentials like ,wefind:
!321
3 2
2 1 1
2
12
1
C R
t RC
t R
R t
(
2
1 2
2 2
1 1
R
R t
O C R
R L
8
)F100.2)(
5000)(
H0
8
(
)1()1(
11
3 5
2
5
2 2
2 2
2 2
C LR C
R L
(:
s1057
1 1
A92.1()1
(A
960
1 1
t L R t
L
e R i
25
H0.8)5.0ln(
500.0
e R L t
Trang 3230.78: a) Using Kirchoff’s Rules on the left and right branches:
2 1
1 2
dt
di L i i R dt
di L R i i
2 1
2 2
C
q i i R C
q R i i
b) Initially, with the switch just closed, 1 0, 2 andq2 0
R
ε i i
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[(
)sin(
ωt ωt
ωRC e
ωR
ε q
t
βt βt
d) When does i first equal zero? 2 625rad/s
)2(
11
2
RC LC
ω
.s10256.1rad/s625
785.00.785
1.00)arctan(
1.00
F)1000.2)(
400)(
rad/s625(22
)
tan(
01)tan(
)2()]
cos(
)sin(
)2([0
)
(
3 6
1 1
ωRC ωt
ωt ωRC
ωt ωt
ωRC e
R t
Trang 33Ni μ A B A B
B
])
[(
.and
,where
,
])[(
2 0
2 0 0 0
0
0 0
0 0
2 0
D N Kμ L D
N μ L D
L L
L L d
d D
L L L D
d L D
d L L Kd d D N μ i
N L
f f
f f