EE 101
Handout # 28
Prof. Abbas El Gamal
March. 15, 2003
Sample Final Solutions
1.
(a) First we draw the circuit in the phasor representation.
1 V

j
V

j

2
j
j
R
1 A
As we found in problem 2 of HW7,
R
=

Z
th

maximizes the average power
dissipated in
R
. So, we need to find
Z
th
between the two terminals of
R
(after
removing
R
). We turn off all sources and find that
Z
th
= 2
j
Ω
.
Thus
R
= 2Ω.
(b) To find the average power delivered to the 2Ω resistor, we find
V
th
,
i.e.
, the open
circuit voltage between the two terminals of the resistor after taking it out. Here
we can use superposition.
We turn off the two voltage sources and we get
V
1
= 1
×
2
j.
Next we turn on the 1V source only and by voltage divider get
V
2
= 1
×
j
j

2
j
=

1
.
Finally, we turn on the

j
V source to get
V
3
=
j.
Adding up, we get that
V
th
=

1 + 3
j
V
.
Thus the current thr’ the resistor is
I
R
=

1 + 3
j
2 + 2
j
= 0
.
5 +
j
A
,
and the average power dissipated in the resistor is
p
R
avg
=
1
2

I
R

2
R
= 1
.
25W
.
1
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(c) By average power conservation the sum of the average power dissipated in the
circuit elements is zero.
Now since in our circuit the only elements that can
supply/dissipate power are the sources and the resistor (reactive elements don’t
dissipate or supply average power), the total average power supplied by the sources
equals the average power dissipated in the resistor, which is 1
.
25W.
2. First we analyze the circuit for
t <
0, assuming steady state. In this case the transistor
is ON and the inductor is replaced by a short circuit. The equivalent circuit is shown
below. To find
i
b
we use the KVL equation
3

10kΩ
i
b

1

100Ω
×
100
i
b
= 0
.
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 Fall '08
 PingLi
 Average Power, Electrical impedance, Thévenin's theorem, rth

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