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ECE 53a
FINAL EXAM Solutions
Pamela Cosman
3/17/08
Problem 1: We begin by considering the circuit before time t=0, when it is in DC steady state with
the switch open. The inductor can be replaced by a short circuit. So, for
t <
0, the voltage across
the inductor is zero. The circuit reduces to a 9A current source with two resistors in parallel, so
the current through the inductor can be found by current division:
i
L
(
t
= 0
−
) =
9
×
12
12 + 6
= 6
A
=
i
L
(
t
= 0
+
)
At time 0, the switch closes, e±ectively shorting out everything to the left of it. So the circuit at
this point can be treated as just a 3H inductor and a 6Ω resistor. The time constant is
τ
=
L
R
=
3
6
= 0
.
5
So the current will decay as
i
L
(
t
) =
i
L
(
t
= 0
+
)
e
−
t/τ
= 6
e
−
2
t
We are asked for the voltage however. The initial condition on the voltage can be found by con
sidering that the initial current of 6A goes (from left to right) across the 6Ω resistor, producing a
voltage drop of 36V. So
V
L
(
t
) =
V
L
(
t
= 0
+
)
e
−
t/τ
=

36
e
−
2
t
If the switch did not ²ip again, then this would be the answer for all
t >
0. But because the switch
does ²ip again, this expression is valid only for 0
< t <
1. We evaluate the situation at
t
= 1.
i
L
(
t
= 1) = 6
e
−
2
≈
0
.
812
At this point, the switch opens again, and the situation at time
t
= 0
+
is this:
9A
12
Ω
6
Ω

V
+
L
3H
(t)
I
1
+

V
R
+
I
2

V
2
where
I
1
= 0.812 and
I
2
can be found from KCL to be:
I
2
= 9 +
I
1
= 9

0
.
812 = 8
.
188
A
which means that, using KVL:
V
2
=
V
R
+
V
L
where
V
2
= 12
×
I
2
= 12
×
8
.
2 = 98
.
256
V
V
R
=

6
×
I
1
= 6
×
0
.
812 = 4
.
872
V
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View Full Documentwhich leads to
V
L
(
t
= 1
+
) =
V
2

V
R
= 98
.
256

4
.
87 = 93
.
386
V
The time constant of the circuit, with the switch open is:
τ
=
L
12 + 6
=
3
18
=
1
6
As time goes towards inFnity, the circuit will settle into a new DC steady state, in which the
inductor again acts like a short circuit, so the voltage across it must go to zero. So for time after 1
sec:
V
L
(
t
) = 93
.
386
e
−
6(
t
−
1)
Problem 2: Two of the four missing values should have been easy. These are worth 2 points each.
•
When the load resistance is zero, there can be no voltage sustained across it. The voltage is zero.
•
When the load resistance is inFnite, there can be no current through it. The current is zero.
•
The middle line is easy too. This is also worth 2 points. By Ohm’s Law, if the voltage across a
200Ω resistor is 10 V, then the current through it is
I
L
=
V
L
R
L
=
10
200
= 0
.
05
A
•
The only tricky item is the voltage across the inFnite resistance. This is worth 4 points. Here we
need to invoke a Thevenin or Norton equivalent model for the unknown circuit.
I will use a Norton model. The short circuit current is given as
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 Spring '08
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