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