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A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is _____________(a) 25(b) 50(c) 100(d) 200I got this question by my college director while I was bunking the class.Origin of the question is Dot Convention in Magnetically Coupled Circuits topic in division Resonance & Magnetically Coupled Circuit of Network Theory

Answer» CORRECT choice is (b) 50

Easiest explanation: Q = \(\frac{f_0}{BW}\)

And f0 = 1/2π (LC)^0.5

BW = R/L

Or, Q = \(\frac{1}{R}(\frac{L}{C})^{0.5}\)

When R, L and C are DOUBLED, Q’ = 50.


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