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In the circuit shown the ratio of quality factor and the bandwidth is
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$10 \mathrm{~s}$
$\begin{aligned} & \text { Bandwidth }=2 \Delta \omega=\frac{\mathrm{R}}{\mathrm{L}} \\ & \text { Q factor }=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \\ & \frac{\mathrm{Q} \text { factor }}{\text { Band width }}=\frac{\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}}{\frac{\mathrm{R}}{\mathrm{L}}}=\frac{1}{\mathrm{R}^2} \frac{\mathrm{L}^{3 / 2}}{\sqrt{\mathrm{C}}} \\ & \frac{\mathrm{Q}_{\text {pacs; }}}{\text { Bandwidth }}=\frac{3 \sqrt{3}}{100 \sqrt{27 \times 10^{-6}}}=\frac{3 \sqrt{3}}{100 \times 3 \sqrt{3} \times 10^{-3}}=10 \mathrm{~s}\end{aligned}$
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