EE 101
Handout # 16
Prof. Abbas El Gamal
Feb 16 2003
Solutions to Sample Midterm Problems
1. To find the power dissipated in the voltage source
v
, we need to find the current
i
flowing into the positive terminal of the voltage source (in the asscosiated reference
direction).
The easiest way to do this is by superposition.
Let’s call
i
1
the current
when the 1A current source is on and the
v
V and 2V voltage sources are off (short
circuits). Then, we have a current divider, and
i
1
= 1
1
1+1
=
1
2
A. Now let
i
2
be the
current when the
v
V voltage source is on and the others are off. In this case we have
a voltage source in series with the equivalent of a 2Ω resistor, and
i
2
=

v
2
A assuming
the appropriate reference direction. Finally let
i
3
be the current when the 2V voltage
source is on and the others are off.
Again, we have a 2V source in series with the
equivalent of a 2Ω resistor, and
i
3
=
2
2
= 1A. The current
i
is the sum of the three
components, thus
i
=
i
1
+
i
2
+
i
3
=
1
2

v
2
+ 1 =
3

v
2
A
.
To find the power, we simply multiply
i
and
v
to get
p
=
i
×
v
=
3

v
2
v
=
3
v

v
2
2
W
.
To find the voltage
v
that maximizes the power, we take the derivative of the power
with respect to voltage, set it to zero, and solve for
v
. Thus
dp
dv
=
3

2
v
2
= 0
⇒
v
=
3
2
V
.
2.
(a) We perform source transformation to convert the voltage source and the 2Ω series
resistor into a 1A current source in parallel with a 2Ω resistor. We then write the
node voltage equations for
n
1
, n
2
, and
n
3
. The fourth equation is
e
2
=
e
3
.
7
/
4

1
0

1
/
4

1
3
/
2
0
0
0
0
2

1
0
1

1
0
e
1
e
2
e
3
e
4
=
1
0
0
0
(b) The power dissipated by the opamp is the current (let’s call it
i
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 Fall '08
 PingLi
 Volt, Thévenin's theorem, Voltage source, rth

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