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) $10$
$\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{4 \times 10^{-6}}} = 500$ rad/s. $Q = \dfrac{\omega_0L}{R} = \dfrac{250}{25} = 10$.
Solution by Sreeraj P, M.Sc Physics