Inductance
3015
(b)
2
(
/
)
2
2(
/
)
R L t
R L t
L
di
R
dU
di
i
e
i
iL
Ri
e
R
dt
L
dt
dt
R
−
−
=
⇒
= −
⇒
=
= −
=
E
E
.
4
2
2(240/0.160)(4.00 10
)
(60 V)
4.52 W.
240
L
dU
e
dt
−
−
×
= −
= −
Ω
(c)
In the resistor,
4
2
2
2
2(
/
)
2(240/0.160)(4.00 10
)
(60 V)
4.52 W
240
R L t
R
R
dU
P
i R
e
e
dt
R
−
−
−
×
=
=
=
=
=
Ω
E
.
(d)
2
2
2(
/
)
( )
R L t
R
P
t
i R
e
R
−
=
=
E
.
2
2
2
2(
/
)
3
2
0
(60 V) (0.160 H)
5.00
10
J,
2
2(240
)
R L t
R
L
U
e
R
R
R
∞
−
−
=
=
=
=
×
Ω
∫
E
E
which is the same as
part (a).
E
VALUATE
:
During the decay of the current all the electrical energy originally stored in the inductor is dissipated
in the resistor.
30.55.
I
DENTIFY
and
S
ET
U
P
:
Follow the procedure specified in the problem.
2
1
2
Li
is the energy stored in the inductor
and
2
/2
q
C
is the energy stored in the capacitor. The equation is
0.
di
q
iR
L
dt
C
−
−
−
=
E
XECUTE
:
Multiplying by °
i
gives
2
0.
di
qi
i R
Li
dt
C
+
+
=
(
)
(
)
2
2
1
1
1
2
2
2
2
,
L
d
d
d
di
di
U
Li
L
i
L
i
Li
dt
dt
dt
dt
dt
⎛
⎞
=
=
=
=
⎜
⎟
⎝
⎠
the
second term.
2
2
1
1
(
)
(2 )
,
2
2
2
C
d
d
q
d
dq
qi
U
q
q
dt
dt
C
C dt
C
dt
C
⎛
⎞
=
=
=
=
⎜
⎟
⎝
⎠
the third term.
2
,
R
i R
P
=
the rate at which
electrical energy is dissipated in the resistance.
,
L
L
d
U
P
dt
=
the rate at which the amount of energy stored in the
inductor is changing.
,
C
C
d
U
P
dt
=
the rate at which the amount of energy stored in the capacitor is changing.
E
VALUATE
:
The equation says that
0;
R
L
C
P
P
P
+
+
=
the net rate of change of energy in the circuit is zero. Note
that at any given time one of
or
C
L
P
P
is negative. If the current and
L
U
are increasing the charge on the capacitor
and
C
U
are decreasing, and vice versa.
30.56.
I
DENTIFY
:
The energy stored in a capacitor is
2
1
2
C
U
Cv
=
. The energy stored in an inductor is
2
1
2
L
U
Li
=
.
Energy conservation requires that the total stored energy be constant.
S
ET
U
P
:
The current is a maximum when the charge on the capacitor is zero and the energy stored in the
capacitor is zero.
E
XECUTE
:
(a)
Initially
16.0 V
v
=
and
0
i
=
.
0
L
U
=
and
2
6
2
4
1
1
2
2
(5.00
10
F)(16.0 V)
6.40
10
J
C
U
Cv
−
−
=
=
×
=
×
.
The total energy stored is 0.640 mJ .
(b)
The current is maximum when
0
q
=
and
0
C
U
=
.
4
6.40
10
J
C
L
U
U
−
+
=
×
so
4
6.40
10
J
L
U
−
=
×
.
2
4
1
max
2
6.40
10
J
Li
−
=
×
and
4
max
3
2(6.40
10
J)
0.584 A
3.75
10
H
i
−
−
×
=
=
×
.
E
VALUATE
:
The maximum charge on the capacitor is
80.0 C
Q
CV
μ
=
=
.
30.57.
I
DENTIFY
and
S
ET
U
P
:
Use
2
1
2
C
C
U
CV
=
(energy stored in a capacitor) to solve for
C
. Then use Eq.(30.22) and
2
f
ω
π
=
to solve for the
L
that gives the desired current oscillation frequency.
E
XECUTE
:
2
2
2
1
2
12.0 V;
so
2
/
2(0.0160 J)/(12.0 V)
222 F
C
C
C
C
C
V
U
CV
C
U
V
μ
=
=
=
=
=
2
1
1
so
(2
)
2
f
L
f
C
LC
π
π
=
=
3500 Hz gives
9.31
H
f
L
μ
=
=
E
VALUATE
:
f
is in Hz and
ω
is in rad/s; we must be careful not to confuse the two.
30.58.
I
DENTIFY
:
Apply energy conservation to the circuit.
S
ET
U
P
:
For a capacitor
/
V
q C
=
and
2
/2
U
q
C
=
. For an inductor
2
1
2
U
Li
=
E
XECUTE
:
(a)
6
max
4
6.00
10
C
0.0240 V.
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 Spring '06
 Buchler
 Physics, Inductance, Energy, Inductor, IMAX, Limax

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