Example 4.16 A parallel circuit is shown in Figure 4.24.
Calculate the value of R in the circuit for which the circuit will
reasonate.
Solution: Equivalent impedance Z
ZZ
ZZ
=
+
12
12
=
+-
++-
=
+-
++-
()
()
()
()
()
()
()
()
Rjj
Rjj
Rjj
Rj
61
04
61
04
61
04
1064
or
Z
RjjRj
Rj
Rj
=
+-
+-
+++-
()()[(
)]
[()][(
)]
6104102
1021
02
=
+++-+-+-+
+
[()()()]()()
()
()
10241026046041021024
10
RRRjRRjR
R
22
4+
At resonance, the imaginary part of Z will be zero.
Therefore,
j(60 - 4R)(R+ 10) - j2(10R+ 24) = 0
or 60R- 4R
2
+ 600 - 40R- 20R- 48 = 0
or R
2
= 138
or R= 11.75 Ω
4.6.4 Locus Diagram
The magnitude ...
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