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A one port active network has an input admittance Y, the magnitude of which is shown in figure as a function of frequency. The circuit is resistive or capacitive in different frequency ranges.
Complete the following table:
Frequency Range | Type of Impedance | Value (Ω/H/F) |
10000 rad/sec < ω | A | P |
10 rad/sec < ω < 1000 rad/sec | B | Q |
A= Capacitive, P=C=10μF and B=Inductive Q=L=0.01H
A= Capacitive, P=C=100μF and B=Inductive Q=L=0.1H
A=Inductive P=L=0.01H and B=Capacitive Q=C=100μF
A= Capacitive, P=C=100μF and B=Inductive Q=L=0.01H
The plot of admittance Vs frequency is shown below,
The admittance of various circuit elements is,
In dB the admittance is given by,
As the magnitude of admittance is constant between
, the impedance would be resistive.
The admittance has a negative slope between
, the impedance is inductive and its value at 
is 20dB
L = 0.01 H
The admittance has a positive slope between
, the impedance is capacitive and its value at 
is 0dB
C = 100μF