inductance and resistance $2 imes 10^{-4} mathrm{H}$ and $6.28 Omega$ respectively oscillates at $10 mathrm{MHz}$
Question:

A resonance circuit having inductance and resistance $2 \times 10^{-4} \mathrm{H}$ and $6.28 \Omega$ respectively oscillates at $10 \mathrm{MHz}$ frequency. The value of quality factor of this resonator is__________

$[\pi=3.14]$

Solution:

Given: $\mathrm{L}=2 \times 10^{-4} \mathrm{H}$

$\mathrm{R}=6.28 \Omega$

$\mathrm{f}=10 \mathrm{MHz}=10^{7} \mathrm{~Hz}$

Since quality factor,

$\mathrm{Q}=\omega_{0} \frac{\mathrm{L}}{\mathrm{R}}=2 \pi \mathrm{f} \frac{\mathrm{L}}{\mathrm{R}}$

$\therefore \mathrm{Q}=2 \pi \times 10^{7} \times \frac{2 \times 10^{-4}}{6.28}$

$Q=2 \times 10^{3}=2000$

$\therefore$ Ans. is 2000