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Sample Problems – ECE 2300
1. 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]
-
2. The switch shown had been open for a long time, then closed at t = 0, and
opened again at 50[s].
a) Find iX(0-).
b) Find iX(0+).
c) Find vX(0-).
d) Find vX(0+).
t = 50[s]
1.5[k]
7.2[k]
3.3[k]
t=0
560[]
+
8[mH]
5.6[k]
+
vX
-
-
+
iS1 =
3.7[mA]
iX
vS1 =
16[k]iX
iS2 =
22[mA]
vS2 =
17[V]
3. For the circuit shown, the switch had been in position a for a long time
before moving to position b at t = 0. The voltage vX before t = 0 was
constant, and equal to -7.34[V].
a) Find vW(0-).
b) Find vW(0+).
c) Find iW(0-).
d) Find iW(0+).
15[]
+
vW
t=0
b
6.8[F]
a
iW
+
-
12.3[V]
4.3[F]
3.2[F]
vX
+
-
5.2[A]
1. 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 1): 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
When we integrate these, we find that
v2     7.667[V] and
v1     7.667[V].
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.
2.
a)
b)
c)
d)
Solutions:
336[A]
-56.8[A]
0
2.201[V]
3. Solutions:
a) -4.96[V]
b) -4.96[V]
c) 0
d) 5.2[A]