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ECE 201 Spring 2010
Homework 40 Solutions
Problem 30
(a)
Z
th
= (
R
1

R
2
)

j/ωC
= (20

j
10) Ω
V
oc
=
V
s
R
2
R
1
+
R
2
= 20
V
rms
For maximum power transfer,
Z
L
=
Z
*
th
= (20 +
j
10) Ω.
(b)
P
avg
=
V
2
oc
4
R
L
= 5
W
Problem 37
(a)
Here we cannot apply the maximum power transfer theorem because
R
L
is
±xed. The expression for average power across
R
L
is given by
P
avg
=
V
2
s
R
L
(
R
+
R
L
)
2
+ (
ωL

1
/ωC
)
2
Clearly, the average power is maximum when the denominator is minimum,
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This note was uploaded on 11/06/2010 for the course ECE 201 taught by Professor All during the Spring '08 term at Purdue University.
 Spring '08
 ALL

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