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Q#1
Sinusoidal Steady State Analysis
GATE EE 2004
MCQ
+1 mark
-0.33 marks
In the figure the value of Z in Figure, which is most appropriate to cause parallel resonance at 500 Hz, is

125.00 mH
304.20 µF
2.0µF
0.05µF
Explanation:
For parallel resonance to be happen, imaginary part of admittance of parallel network should be zero.
Since inductor is already exists. So, Z should be capacitive
Admittance of parallel network, Y= 
=
For imaginary part of admittance of zero





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