The circuit shown below has a switch which closed at t = 0.
The voltages
v
1
and
v
2
were
measured before the switch was closed, and it was found that
1
( )
15[V], for
0, and
v t
t
=
<
2
( )
7[V], for
0.
v t
t
= 
<
In addition, for time greater than zero, it was determined that
(
29
[
]
1
500 s
( )
22
mA , for
0.
t
R
i t
e
t


=
Explore the energy stored in the capacitors for t < 0, and for t =
∞
.
1[k
Ω
]
+

v
1
C
1
=
6[
μ
F]
C
2
=
3[
μ
F]
t
= 0
+

v
2
i
R
(t)
Solution:
For
t
≤
0, we have no current through the capacitors, and therefore no change in the
voltages across the capacitors.
Thus, we can write
(
29
(
29
(
29
2
, 1
1
1
2
6
, 1
1
(0)
(0)
, or
2
1
(0)
6 10 [F] 15[V]
675[ J].
2
STO C
STO C
w
C v
w
μ

=
=
×
=
(
29
(
29
(
29
2
, 2
2
2
2
6
, 2
1
(0)
(0)
, or
2
1
(0)
3 10 [F]
7[V]
73.5[ J].
2
STO C
STO C
w
C v
w

=
=
×

=
So, the total stored energy, in the combination of the two capacitors, is 748.5[
μ
J].
Now, once the switch closes, current begins to flow.
This makes sense, since when the
switch closes, we now have 22[V] across the resistor, so current must flow.
If you think about
it, it makes sense that the current will decrease with time.
As the charges move, the voltages
across the capacitors will increase and decrease, so that the voltage across the resistor will
decrease.
When the current decreases to zero, the two voltages
v
1
and
v
2
will be equal.
What value will these voltages have?
We can solve this by integrating the current
i
R
(t)
,
with the defining equation for the capacitor, that is