Q 12-07-135JEE MainJEE Main 2020 (6 Sep, Shift 1)Easy
An AC circuit has $R = 100\ \Omega$, $C = 2\ \mu\text{F}$ and $L = 80$ mH, connected in series. The quality factor of the circuit is:
Answer: (A) $2$
$$Q = \frac1R\sqrt{\frac LC} = \frac{1}{100}\sqrt{\frac{80\times10^{-3}}{2\times10^{-6}}} = \frac{1}{100}\sqrt{4\times10^4} = \frac{200}{100} = 2$$
Solution by Sreeraj P, M.Sc Physics