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The circuit shown in the figure below will act as an ideal source with respect to terminal A and B when the frequency is ___________(a) Zero(b) 1 rad/s(c) 4 rad/s(d) 16 rad/sThe question was posed to me by my college professor while I was bunking the class.My question comes from Advanced Problems on Resonance in section Resonance & Magnetically Coupled Circuit of Network Theory

Answer»

The CORRECT answer is (c) 4 rad/s

To explain: Z(s) t = \(\frac{(Ls)\frac{1}{Cs}}{Ls + \frac{1}{Cs}} = \frac{Ls}{LCs^2+1} \)

PUTTING s = jω, we GET, Z (jω) = \(\frac{jωL}{1 -LCω^2} \)

For ideal current SOURCE, Z (jω) = ∞

Or, 1 – LCω^2 = 0

Or, ω = \(\frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{\frac{1}{16}×1}}\) = 4 rad/s.



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