ECE 201 Spring 2010
Homework 13 Solutions
Problem 9
(a)
2R and 6R in series gives 8R. 8R and 8R in parallel gives 4R. Thus 12R is in
series with
V
s
in the simplified circuit. Thus the Thevenin voltage is given
by
V
oc
=
V
s
12
R
×
1
2
×
6
R
=
V
s
4
=
30
V
To find the Thevenin resistance, we short circuit
V
s
. Thus 8R and 8R are
in parallel, which gives 4R, which in turn is in series with 2R. This gives 6R
and 6R in parallel. Thus
R
th
= 3
R
= 900
Ω
.
(b)
Using the Thevenin equivalent circuit to simplify our calculations,
P
R
L
=
30
900 +
R
L
2
R
L
=
0
.
1875
W,
0
.
24
W,
0
.
244898
W
(
R
L
= 300
Ω
,
600
Ω
,
1200
Ω
)
The Thevenin equivalent circuit analysis allows us to analyze the circuit
without the load once and then plug in the various load resistor values to
compute the relevant quantities.
However, using earlier techniques would
require 3 separate analyses to compute these values.
Hence the use of a
Thevenin equivalent reduces the e
ff
ort needed to obtain the answers.
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 Spring '08
 Chunghorng,R
 Thévenin's theorem, rth

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