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RLC Circuits and Resonance - Technical MCQs

Q1:

What is the Q (Quality factor) of a series circuit that resonates at 6 kHz, has equal reactance of 4 kilo-ohms each, and a resistor value of 50 ohms?

A 0.001

B 50

C 80

D 4.0

ANS:C - 80

Quality Factor=(0.159)/f*R*C.

XL=XC.
Xc=1/2* π * f * c.
By using this formula we get the value of C=6631.45pF(6631.45x10^-12),
Substitute the value of C, f and R to the formula of quality factor.
We get Q=79.922 approximately 80.