Download ECE 2202 Sample Problem capacitors

The circuit shown below has a switch which closed at t = 0. The voltages v1 and v2 were
measured before the switch was closed, and it was found that
v1 (t )  15[V], for t  0, and
v2 (t )  7[V], for t  0.
In addition, for time greater than zero, it was determined that

iR (t )  22e
500 s1  t
 
mA, for t  0.
Explore the energy stored in the capacitors for t < 0, and for t = .
1[k]
C1 =
6[F]
+
v1
t=0
iR(t)
-
+
v2
C2 =
3[F]
-
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
1
2
wSTO ,C1 (0)  C1  v1 (0)  , or
2
1
2
wSTO ,C1 (0)   6 106 [F] 15[V]  675[  J].
2
1
2
wSTO ,C 2 (0)  C2  v2 (0)  , or
2
1
2
wSTO ,C 2 (0)   3 106 [F]  7[V]  73.5[  J].
2
So, the total stored energy, in the combination of the two capacitors, is 748.5[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 v1 and v2 will be equal.
What value will these voltages have? We can solve this by integrating the current iR(t),
with the defining equation for the capacitor, that is

1
v2 () 
iR ( s )ds  v2 (0), and
C2 0

1
v1 () 
iR ( s )ds  v1 (0).
C1 0

500 s 1  s
1
v2 () 
0.022  A  e   ds  7  V   7.667  V  , and
6

3 10  F 0

500 s 1  s
1
 
v1 () 
0.022
A
e
ds  15  V   7.667  V .


6

6 10  F 0
You can check the integrals above to make sure this is correct.
Now, let’s look at the energy after a long time. We get
1
2
wSTO ,C1 ()  C1  vFINAL  , or
2
1
2
wSTO ,C1 ()   6 106 [F]  7.667[V]  176.3[  J].
2
1
2
wSTO ,C 2 ()  C2  vFINAL  , or
2
1
2
wSTO ,C 2 ()   3 106 [F]  7.667[V]  88.17[  J].
2
So, the total stored energy, in the combination of the two capacitors, after a long time, is
264.5[J].
Clearly, there has been a loss of energy. Where did the energy go? It was dissipated in
the resistor. If you were to find the power in the resistor, and integrate it over time, you would
get the same amount of energy dissipated in the resistor, as that which has been lost from the
capacitors.
Now, we note that we had two series capacitors in the original circuit. Let us consider the
possibility of using equivalent circuits here, and what happens when we do so. The equivalent
capacitance for the two series capacitors would be
CEQ 
 6[ F] 3[ F]  2[ F].
C1C2

C1  C2
9[  F]
This equivalent capacitance would have a voltage across it, at t = 0, of {15 – (-7)}[V]. Thus, this
equivalent capacitance would have energy stored in it of
1
2
wSTO ,CEQ (0)  CEQ  22[V] , or
2
1
2
wSTO ,CEQ (0)   2 106 [F]  22[V]  484[  J].
2
This number should look familiar. The total stored energy, in the combination of the two
capacitors, was 748.5[J]. The total stored energy, in the combination of the two capacitors,
after a long time, was 264.5[J]. The difference in energy between these two conditions is
484[J]. Does this seem like a coincidence? It is not. It is exactly what we should have
expected.
This must be the result that must occur, for our equivalent circuits to be equivalent with
respect to the outside world. We found that this much energy was lost in the circuit, and it must
have been lost in the resistor. Therefore, if we find an equivalent circuit with respect to the
resistor, the same thing must happen to the resistor in each case; the same thing must happen in
the actual circuit, and in the equivalent circuit, for the resistor.
However, the behavior inside the equivalent is not the same. In the original circuit, there
was still some energy stored after a long time. In the equivalent, there is no energy stored after a
long time. Equivalent circuits means that the two circuits are equivalent, with respect to the
outside world.
The energy stored in the two series capacitors after a long time, in this case, is a
minimum. You can’t get the last 264.5[J] out of these capacitors, no matter what you do, as
long as they are connected together in series. You can drive the energy in one capacitor to zero,
but in doing so, you will increase the energy in the other. The total will not go below this
minimum energy level. A similar situation exists for parallel inductors.
The circuit shown below has a switch which closed at
t = 0. The voltages v1 and v2 were measured before the
switch was closed, and it was found that
v1 (t )  15[V], for t  0, and
v2 (t )  7[V], for t  0.
In addition, for time greater than zero, it was determined
that

iR (t )  22e
500 s1  t
 
mA, for t  0.
Explore the energy stored in the capacitors for t < 0, and
for t = .
1[k]
C1 =
6[F]
+
v1
-
t=0
iR(t)
+
v2
C2 =
3[F]
-
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
1
2
wSTO ,C1 (0)  C1  v1 (0)  , or
2
1
2
wSTO ,C1 (0)   6 106 [F] 15[V]  675[  J].
2
1
2
wSTO ,C 2 (0)  C2  v2 (0)  , or
2
1
2
wSTO ,C 2 (0)   3 106 [F]  7[V]  73.5[  J].
2
So, the total stored energy, in the combination of the two
capacitors, is 748.5[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 v1 and v2 will
be equal.
What value will these voltages have? We can solve
this by integrating the current iR(t), with the defining
equation for the capacitor, that is

1
v2 () 
iR ( s )ds  v2 (0), and

C2 0

1
v1 () 
iR ( s )ds  v1 (0).

C1 0

500 s 1  s
1
 
v2 () 
0.022
A
e
ds  7  V   7.667  V  ,


6

3 10  F 0
and

500 s 1  s
1
v1 () 
0.022  A  e   ds  15  V   7.667  V .
6

6 10  F 0
You can check the integrals above to make sure this is
correct.
Now, let’s look at the energy after a long time. We
get
1
2
wSTO ,C1 ()  C1  vFINAL  , or
2
1
2
wSTO ,C1 ()   6 106 [F]  7.667[V]  176.3[  J].
2
1
2
wSTO ,C 2 ()  C2  vFINAL  , or
2
1
2
wSTO ,C 2 ()   3 106 [F]  7.667[V]  88.17[  J].
2
So, the total stored energy, in the combination of the two
capacitors, after a long time, is 264.5[J].
Clearly, there has been a loss of energy. Where did
the energy go? It was dissipated in the resistor. If you
were to find the power in the resistor, and integrate it over
time, you would get the same amount of energy dissipated
in the resistor, as that which has been lost from the
capacitors.
Now, we note that we had two series capacitors in the
original circuit. Let us consider the possibility of using
equivalent circuits here, and what happens when we do
so. The equivalent capacitance for the two series
capacitors would be
CEQ
6[  F] 3[  F]

C1C2


 2[  F].
C1  C2
9[  F]
This equivalent capacitance would have a voltage across
it, at t = 0, of {15 – (-7)}[V]. Thus, this equivalent
capacitance would have energy stored in it of
1
2
wSTO ,CEQ (0)  CEQ  22[V] , or
2
1
2
wSTO ,CEQ (0)   2 106 [F]  22[V]  484[  J].
2
This number should look familiar. The total stored
energy, in the combination of the two capacitors, was
748.5[J]. The total stored energy, in the combination of
the two capacitors, after a long time, was 264.5[J]. The
difference in energy between these two conditions is
484[J]. Does this seem like a coincidence? It is not. It
is exactly what we should have expected.
This must be the result that must occur, for our
equivalent circuits to be equivalent with respect to the
outside world. We found that this much energy was lost
in the circuit, and it must have been lost in the resistor.
Therefore, if we find an equivalent circuit with respect to
the resistor, the same thing must happen to the resistor in
each case; the same thing must happen in the actual
circuit, and in the equivalent circuit, for the resistor.
However, the behavior inside the equivalent is not
the same. In the original circuit, there was still some
energy stored after a long time. In the equivalent, there is
no energy stored after a long time. Equivalent circuits
means that the two circuits are equivalent, with respect to
the outside world.
The energy stored in the two series capacitors after a
long time, in this case, is a minimum. You can’t get the
last 264.5[J] out of these capacitors, no matter what you
do, as long as they are connected together in series. You
can drive the energy in one capacitor to zero, but in doing
so, you will increase the energy in the other. The total
will not go below this minimum energy level. A similar
situation exists for parallel inductors.