Q 12-07-024NEETJEE MainMedium
In a series LCR circuit, $L = 0.5$ H, $C = 8\ \mu$F and $R = 25\ \Omega$. The quality factor at resonance is
Answer: (C) $10$
Concept. $Q = \dfrac{\omega_0L}{R}$ with $\omega_0 = \dfrac{1}{\sqrt{LC}}$.
Step 1. Resonance. $\omega_0 = \dfrac{1}{\sqrt{0.5 \times 8 \times 10^{-6}}} = 500$ rad/s.
Step 2. Quality factor. $Q = \dfrac{500 \times 0.5}{25} = 10$.
Answer. $10$ (Option C).
Solution by Sreeraj P, M.Sc Physics