Solution 9.59
For the network in Fig. 9.66, find Zin. Let ω = 100 rad/s.
Figure 9.66
For Prob. 9.59.
Solution
10 Ω
Solution 9.60
Solution 9.61
All of the impedances are in parallel.
Solution 9.62
20j)102)(1010(jLjmH2-33=××=ω→
50)50)(01( =°∠=V
−
-j100 Ω
Solution 9.63
First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with
the corresponding delta.
10 Ω
10 Ω
–j16 Ω
Now all we need to do is to combine impedances.
8 Ω
–j12 Ω
–j16 Ω
Solution 9.64
Find ZT and V in the circuit shown in Fig. 9.71. Let the value of the inductance be j20 Ω.
Figure 9.71
For Prob. 9.64.
Solution
ZT
−
Solution 9.65
)4j3(||)6j4(2
Solution 9.66
For the circuit in Fig. 9.73, calculate ZT and Vab.
Figure 9.73
For Prob. 9.66.
Solution
170
)10j40)(5j20(
+−
III j12
2j8
10j40
+
=
+
=
Vab
I2
I1
I
21ab 10j20– IIV +=
Solution 9.67
(a)
20j)1020)(10(jLjmH20 -33=×=ω→
(b)
10j)1010)(10(jLjmH10
-33
=×=ω→
50j–
)1020)(10(j
1
Cj
1
F20
6–3
=
×
=
ω
→µ
2060||30 =
Solution 9.68
4j–
1
j3
1
2j5
1
eq +
+
+
−
=Y
Solution 9.69
)2j1(
4
1
2j–
1
4
11
o
+=+=
Y
)6.0j8.0()333.0j()1(
6.0j8.0
1
3j–
1
1
11
o
+++=
−
++=
′
Y
Solution 9.70
Make a delta–to-wye transformation as shown in the figure below.
9j7
5j15
)10j15)(10(
15j1010j5
)15j10)(10j–(
an −=
+
−
=
++−
+
=Z
)5j8(||)2( cnbnaneq −+++= ZZZZ
c
b
a
Solution 9.71
We apply a wye–to–delta transformation.
j1
2j
2j2
2j
4j2j2
ab
−=
+
=
+−
=Z
)j1)(4j(
−
1
1
11
+
+=
j4 Ω
Solution 9.72
Transform the delta connections to wye connections as shown below.
6j–18j–||
9j– =
,
)20)(20(
)10)(20(
)10)(20(
b
a
R3
Solution 9.73
Transform the delta connection to a wye connection as in Fig. (a) and then transform the
wye connection to a delta connection as in Fig. (b).
=++++++ ))(2())(4()4)(2( 313221 ZZZZZZ
6.9j4.46
+
)88.61583.7)(906(
°∠°∠
b
a
R1
R2
R3
)07.7911.9)(9012(
°∠°∠
Solution 9.74
One such RL circuit is shown below.
We now want to show that this circuit will produce a 90° phase shift.
20j20–
)20j20)(20j(
+
+
This shows that the output leads the input by 90°.
20 Ω
20 Ω
V
Z
Solution 9.75
Since
)90tsin()tcos( °+ω=ω
, we need a phase shift circuit that will cause the output to