30.22: a)
A13.2
H0.115
J)260.0(2
2
2
1
2
L
U
ILIU
V.256)(120A)13.2(
IR
b)
2
0
)/(222)/(
2
1
2
1
2
1
2
1
2
1
and
LIUeLiLiUIei
tLRtLR
s.1032.3
2
1
ln
)2(120
H115.0
2
1
ln
2
2
1
4
)/(2
R
L
t
e
tLR
30.23: a)
A.250.0
240
V60
0
R
I
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)
.s1062.4
2
1
ln
)240(
)H160.0(
2
1
ln
2
1
4
2/1
)/(
0
2/1
R
L
te
I
i
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
ee
R
eIiP
).1()W50.4(
)s20.3(
1
t
eP
b)
2)H50.2/00.8(
2
2)/(
2
2
)1(
00.8
)V00.6(
)1(
ttLR
R
ee
R
ε
RiP
.)1()W50.4(
2)s20.3(
1
t
R
eP
c)
)()1(
)/(2)/(
2
)/()/(
tLRtLRtLRtLR
L
ee
R
ε
e
L
ε
Le
R
ε
dt
di
iLP
).()W50.4(
)s40.6()s20.3(
11
tt
L
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
0
0
LRt
i
I
t
eIit
L
R
Ii
td
L
R
i
id
L
R
i
dt
di
iR
dt
di
L
30.27: Units of
s/)s(/H/ RL
units of time.
30.28: a)
πf
LC
ω 2
1
.H1037.2
)F1018.4()106.1(4
1
4
1
3
1226222
Cf
L
b)
F.1067.3
)H1037.2()1040.5(4
1
4
1
11
3252
min
22
max
Lf
C
30.29: a)
)F1000.6()H50.1(22
2
5
πLCπ
ω
π
T
s.rad105,s0596.0 ω
b)
.C1020.7)V0.12)(F1000.6(
45
CVQ
c)
.J1032.4)V0.12)(F1000.6(
2
1
2
1
3252
0
CVU
d)
.0)cos(,0At
ωtQQqt
)F1000.6)(H50.1(
s0230.0
cos)C1020.7()cos(s,0230.0
5
4
ωtQqt
C.1043.5
4
Signs on plates are opposite to those at
.0
t
e)
)sin(,s0230.0 ωtωQ
dt
dq
it
A.0.0499
H)10(6.00H)(1.50
s0.0230
sin
H)10H)(6.00(1.50
C107.20
55
4
i
Positive charge flowing away from plate which had positive charge at
.0
t
f) Capacitor:
.J102.46
F)102(6.00
C)10(5.43
2
3
5
242
C
q
U
C
Inductor:
.J101.87A)(0.0499H)50.1(
2
1
2
1
322
LiU
L
30.30: (a) Energy conservation says (max)=(max)
CL
UU
A0.871
H1012
F1018
V)22.5(
CV
2
1
2
1
3
6
max
22
max
LCVi
Li
The charge on the capacitor is zero because all the energy is in the inductor.
(b)
LCT
2
2
at
41
period:
F)10(18H)10(12
2
)2(
4
1
4
1
63
LCT
s1030.7
4
at 43 period:
s1019.2)s1030.7(3
4
3
34
T
(c) CCVq
405)V(22.5F)18(
0
30.31:
F0.30
V
10
29
.
4
C10150
3
9
V
Q
C
For an L-C circuit,
LCω 1
and
LCωπT
22
mH601.0
)2(
2
C
T
L
30.32:
rad/s1917
)F1020.3()H0850.0(
1
6
ω
a)
C1043.4
srad1917
A1050.8
7
4
max
maxmaxmax
ω
i
Q
ωQi
b) From Eq. 31.26
2
1
4
2722
s1917
A1000.5
)C1043.4(
LCiQq
.C1058.3
7
30.33: a)
)sA80.2()F1060.3()H640.0(0
1
6
2
2
dt
di
LCqq
LC
dt
qd
C.1045.6
6
b)
.V36.2
F
10
60
.
3
C1050.8
6
6
C
q
30.34: a)
.
max
max
maxmaxmax
LCi
ω
i
Q
ωQi
J450.0
)F1050.2(2
)C1050.1(
2
.C1050.1)F1050.2()H400.0()A50.1(
10
25
max
2
max
510
max
C
Q
U
Q
b)
14
10
s1018.3
)F1050.2()H400.0(
11
2
2
2
LC
f
(must double the frequency since it takes the required value twice per period).
30.35:
.ssA
A
1
s
s
C
V
Ω
V
C
s
Ω
V
C
HFH][
222
sLCLC
30.36: Equation (30.20) is
.0
1
2
2
q
LC
dt
qd
We will solve the equation using:
.
11
0)cos()cos(
1
).cos()sin()(cos
22
2
2
2
2
2
LC
ω
LC
ωt
LC
Q
ωtQωq
LCdt
qd
ωtQω
dt
qd
ωtωQ
dt
dq
ωtQq
30.37: a)
.
)(cos
2
1
2
1
222
C
ωtQ
C
q
U
C
.
1
since,
)(sin
2
1
)(sin
2
1
2
1
2
22
2222
LC
ω
C
ωtQ
ωtQLωLiU
L
b)
)(sin
2
1
)(cos
2
1
2222
2
Total
ωtQLωωt
C
Q
UUU
LC
)(sin
1
2
1
)(cos
2
1
222
2
ωtQ
LC
L
ωt
C
Q
))(sin)((cos
2
1
22
2
ωtωt
C
Q
Total
2
2
1
U
C
Q
is a constant.
30.38: a)
)cos(
)2/(
tωAeq
tLR
)sin(
2
2)cos(
2
).sin()cos(
2
)2/()2/(
2
2
2
)2/()2/(
tωe
L
R
A
ωtωe
L
R
A
dt
qd
t
ωAeωtωe
L
R
A
dt
dq
tLRtLR
tLRtLR
).cos(
)2/(2
tωAeω
tLR
2
2
2
2
2
2
2
2
2
4
1
0
1
22
L
R
LC
ω
LCL
R
L
R
q
LC
q
dt
dq
L
R
dt
qd
b)
:0,,0
dt
dq
iQqtAt
.
4//12
2
tan
2
and
cos
0sincos
2
andcos
22
LRLCL
R
ωL
R
Q
ω
L
QRQ
A
A
ωA
L
R
dt
dq
QAq
30.39: Subbing
,
1
,,,
C
kRbLmqx
we find:
a) Eq. (13.41):
.0:)27.30.(Eq0
2
2
2
2
LC
q
dt
dq
L
R
dt
qd
m
kx
dt
dx
m
b
dt
xd
b) Eq. (13.43):
.
4
1
:)28.30.(Eq
4
2
2
2
2
L
R
LC
ω
m
b
m
k
ω
c) Eq. (13.42):
).cos(:)28.30.(Eq)cos(
)2/()2/(
tωAeqtωAex
tLRtmb
30.40:
.
A
V
VC
s
F
H
2
C
L
C
L
30.41:
LCLC
LR
LCLC
LR
LCL
R
LC
6
11
2
6
11
4
6
1
4
1
22
2
2
2
.4.45
F)10(4.60H)6(0.285
1
F)10(4.60H)(0.285
1
)H285.0(2
44
R
30.42: a) When
s.rad298
F)10(2.50H)(0.450
11
,0
5
0
LC
ωR
b) We want
2
222
0
)95.0(
4
1
1
)41(
95.0
L
CR
LC
LRLC
ω
ω
.8.83
F)10(2.50
(0.0975)H)4(0.450
))95.0(1(
4
5
2
C
L
R
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(( tπ
dt
d
L
dt
di
L
ε
).)s240((sin)V4.23())s240sin(()240()A124.0()H250.0( tπtε
b)
,0;V4.23
max
i
since the emf and current are
90
out of phase.
c)
,0;A124.0
max
i
since the emf and current are
90
out of phase.
30.45: a)
).(ln
22
)(
2
)(
000
ab
Nih
μ
r
dr
Nih
μ
hdr
r
π
Niμ
hdrB
b
a
b
a
b
a
B
b)
).(ln
2
2
0
ab
π
hNμ
i
N
L
B
c)
.
2
L
2
)(
)/)(1(ln)/(ln
2
0
2
2
a
ab
π
hNμ
a
ab
a
ab
aabab
30.46: a)
.
1
2
2210
1
2210
1
110
1
22
1
222
12
l
r
πNNμ
l
ANN
μ
l
IAN
μ
IA
AN
A
A
I
N
I
N
M
BB
b)
.
1
1
2
2210
1
1
210
222
2
dt
di
l
r
πNNμ
dt
di
l
AN
μ
N
dt
d
N
ε
B
c)
.
2
1
2
2210
22
121
dt
di
l
r
πNNμ
dt
di
M
dt
di
M
ε
30.47: a)
.H5.7)/A00.4/()V0.30()//( sdtdiεL
dt
di
L
ε
b)
.Wb360)s0.12)(V0.30(
fif
tε
dt
d
ε
c)
W.1440/s)A00.4)(A0.48)(H50.7(
dt
di
LiP
L
.0104.0W138240)0.60()A0.48(
22
R
R
L
P
P
RiP
30.48: a)
))s0250.0/cos()A680.0(()H1050.3(
3
t
dt
d
dt
di
L
.V299.0
s0250.0
)A680.0)(H1050.3(
3
max
b)
.Wb1095.5
400
)A680.0)(H1050.3(
6
3
max
max
N
Li
B
c)
.)s0250.0/(sin)s0250.0/)(A680.0)(H1050.3()(
3
t
dt
di
Lt
V230.0)(
))s0180.0)(s6.125((sin)V299.0(s)0180.0(
))s6.125sin(()V299.0()(
1
1
t
tt
30.49: a) Series:
,
eq
2
2
1
1
dt
di
L
dt
di
L
dt
di
L
but
iii
21
for series components so
eq21
21
thus, LLL
dt
di
dt
di
dt
di
b) Parallel: Now
. where,
21eq
2
2
1
1
iii
dt
di
L
dt
di
L
dt
di
L
.
11
and But.So
1
21
eq
2
eq
1
eq
2
eq
2
2
eq
121
LL
L
dt
di
L
L
dt
di
L
L
dt
di
dt
di
L
L
dt
di
dt
di
L
L
dt
di
dt
di
dt
di
dt
di
30.50: a)
.
2
2
0
0encl0
πr
iμ
BiμπrBIμdiB
b)
.
2
0
ldr
πr
iμ
BdAd
B
c)
b
a
b
a
BB
).ab(
π
ilμ
r
dr
π
ilμ
d ln
22
00
d)
).ln(
2
0
ab
π
μ
l
i
N
L
B
e)
).ln(
4
)ln(
2
2
1
2
1
2
0
2
0
2
ab
π
liμ
iab
π
μ
lLiU
30.51: a)
.
2
2
0
0encl0
πr
iμ
BiμrπBIμd
lB
b)
.
4
)2(
22
1
)2(
2
2
0
2
0
00
2
dr
r
li
rdrl
r
i
rdrluudVdU
B
u
c)
)./(ln
44
2
0
2
0
ab
π
liμ
r
dr
π
liμ
dUU
b
a
b
a
d)
),/(ln
2
2
2
1
0
2
2
ab
π
μ
l
i
U
LLiU
which is the same as in Problem 30.50.
30.52: a)
,
22
2
10110
1
1
1
1
1
1
rπ
ANμ
rπ
iNμ
i
AN
i
N
L
B
rπ
ANμ
rπ
iNμ
i
AN
i
N
L
B
22
2
20220
2
2
2
2
2
2
b)
.
222
21
2
20
2
10
2
210
2
LL
r
π
ANμ
rπ
ANμ
rπ
ANNμ
M
30.53:
EεμEμεB
μ
B
Eε
uu
EB 00
2
00
0
2
2
0
22
T.1017.2)V/m650(
6
00
με
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