Question:
An AC circuit has $R=100 \Omega, C=2 \mu \mathrm{F}$ and $L=80 \mathrm{mH}$, connected in series. The quality factor of the circuit is :
Correct Option: 1
Solution:
(1) Quality factor,
$Q=\frac{1}{R} \sqrt{\frac{L}{C}}=\frac{1}{100} \sqrt{\frac{80 \times 10^{-3}}{2 \times 10^{-6}}}$
$=\frac{1}{100} \sqrt{40 \times 10^{3}}=\frac{200}{100}=2$