L
=
L
d I
dt
,
at all times, so
V
L
→
0 as
t
→ ∞
, and we can
replace the inductor in the figure by a straight
wire. When the current now comes from
R
1
to the junction where
I
1
splits into
I
2
and
I
,
there is no resistance in the “
I
” path but a
nonzero resistance
R
2
in the other. Naturally,
the current takes the path with no resistance.
Since we do pass through
R
1
in any case, the
equivalent resistance is now
R
1
. At
t
=
∞
I
=
E
R
1
and
I
2
= 0
Alternate Solution:
An equation of the
form
L
dI
dt
−
R I
− E
= 0
has a solution which is composed of a ho
mogeneous solution and a particular solution.
The particular solution corresponds to the
first term in the
I
in the problem statement;
the homogeneous corresponds to the second
term. The homogeneous solution is therefore
the only one of relevance to the time con
stant
τ
x
, and we can disregard any batteries
–
Faraday’s
Law
and
AC
circuits
–
yeazell
–
(58010)
8
E
. This provides a simple solution to Part 3:
Just ignore the battery and find the equiva
lent resistance connected to
L
, in this case
020(part2of2)10.0points
4.
I
2
=
E
R
2
Explanation:
See previous explanation.
019(part1of2)10.0points
An
LC
circuit is shown in the figure below.
The 27 pF capacitor is initially charged by the
6 V battery when
S
is at position
a
. Then
S
is thrown to position
b
so that the capacitor
is shorted across the 12 mH inductor.
12 mH
27 pF
6 V
S
b
a
What is the maximum value for the oscil
lating current assuming no resistance in the
circuit?
What is the maximum energy stored in the
magnetic field of the inductor?
Correct answer: 4
.
86
×
10

10
J.
Explanation:
U
max,ind
=
U
max,cap
=
Q
2
max
2
C
=
1
2
C V
2
=
1
2
(2
.
7
×
10

11
F) (6 V)
2
=
4
.
86
×
10

10
J
.
021
10.0points
The
emf
E
drives the circuit shown below
at angular frequency
ω.
E
C
R
L
–
Faraday’s
Law
and
AC
circuits
–
yeazell
–
(58010)