Question

In the circuit shown in the figure, the ratio of the quality factor and the band width is__ s.

Answer 1

Answer: 10

$\Delta \omega =\frac{R}{L}$
$Q=\frac{{\omega }_0}{\Delta \omega }={\omega }_0\frac{L}{R}$
${\omega }_0=\frac{1}{\sqrt{3\times 27\times 10^{-6}}}=\frac{1}{9\times 10^{-3}}$
$\frac{Q}{\Delta \omega }=\frac{{\omega }_0\frac{R}{R}}{\frac{R}{L}}={\omega }_0\frac{L^2}{R^2}=\sqrt{\frac{1}{LC}}\frac{L^2}{R^2}$
$=\frac{1}{9\times 10^{-3}}\times \frac{9}{100}=10s$