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Q. An AC circuit has $R =100 \Omega, C =2 \mu F$ and $L =80 mH ,$ connected in series. The quality factor of the circuit is :

JEE MainJEE Main 2020Alternating Current

Solution:

$Q =\frac{1}{ R } \sqrt{\frac{ L }{ C }}=\frac{1}{100} \sqrt{\frac{80 \times 10^{-3}}{2 \times 10^{-6}}}$
$=\frac{1}{100} \sqrt{40 \times 10^{3}}$
$=\frac{200}{100}=2$