Question

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) 200

I 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 1

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.