Q 12-07-116JEE MainJEE Main 2021 (24 Feb, Shift 1)Easy
A resonance circuit having inductance and resistance $2\times10^{-4}$ H and $6.28\ \Omega$ respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is ______. ($\pi = 3.14$)
Numerical value type. Enter your answer.
Answer: 2000
$$Q = \frac{\omega L}{R} = \frac{2\pi\times10^7\times2\times10^{-4}}{6.28} = \frac{6.28\times2\times10^3}{6.28} = 2000$$
Solution by Sreeraj P, M.Sc Physics