Solution 14.96
L
R
1
2L
2L
2
L
1
CsR1
LCRsRsL
sC
1
+
++
⋅
i
i2
2
1
RV
Z
Z
V+
=
where
=
+
22
2
RZ
Z
L
Ri
V1
Vo
−
Z2
Z1
Therefore,
o
V
)LCRsRsL(R
2L
2
LL
++
Solution 14.97
i
1i2L
L
1
2L
L
o
sC1RsC1R
R
sC1R
RV
Z
Z
VV ++
⋅
+
=
+
=
L
Ri
V1
Vo
Z2
Z1
Solution 14.98
π=−π=−π=ω−ω= 44)432454(2)ff(2B
1212
Solution 14.99
Cf2
1
C
1
Xcπ
=
ω
=
×
103
300
X
-4
L
Solution 14.100
1
Solution 14.101
1
Solution 14.102
(a) When
0R
s
=
and
∞=
L
R
, we have a low–pass filter.
(b) We obtain
Th
R
across the capacitor.
Solution 14.103
Cj1||RR
R
)(
12
2
i
o
ω+
==ω
V
V
H
,
ω= js
Solution 14.104
The schematic is shown below. We click Analysis/Setup/AC Sweep and enter Total