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ECE 201 Spring 2010
Homework 33 Solutions
Problem 30
(a)
Y
in
(
jω
)=
1
R
+
1
jωL
=0
.
05

j
0
.
25
ω
Z
in
(
jω
)=
1
Y
in
(
jω
)
=
j
20
ω
5+
jω
(b)
I
L
=
I
in
Z
in
(
jω
)
jωL
= 10
×
10(
i
+
j
)
j
20
=5
√
2
e

jπ/
4
⇒
i
L
(
t
) = 10 cos(5
t

π/
4)
mA
Problem 40
Let the impedances of
R
1
,C
and
R
2
,L
combinations be
Z
1
and
Z
2
respec
tively.
Z
1
=
±
1
500
+
j
(400
×
5
×
10

6
)
²

1
= 250(1

j
)
Z
2
=
³
1
100
+
1
j
(400
×
0
.
125)
´

1
= 20(1 + 2
j
)
1
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View Full DocumentLet the phasor current through the source be
I
.
I
=
V
270

j
210
⇒
V
C
=
I
×
250(1

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 Spring '08
 Chunghorng,R

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