In the circuit shown the ratio of quality factor and the bandwidth is

In the circuit shown the ratio of quality factor and the bandwidth is
  1. $10 \mathrm{~s}$
  2. $8 \mathrm{~s}$
  3. $6 \mathrm{~s}$
  4. $4 \mathrm{~s}$

Solution

$\begin{aligned} & \text { Bandwidth }=2 \Delta \omega=\frac{\mathrm{R}}{\mathrm{L}} \\ & \text { Q factor }=\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}} \\ & \frac{\mathrm{Q} \text { factor }}{\text { Band width }}=\frac{\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}}{\frac{\mathrm{R}}{\mathrm{L}}}=\frac{1}{\mathrm{R}^2} \frac{\mathrm{L}^{3 / 2}}{\sqrt{\mathrm{C}}} \\ & \frac{\mathrm{Q}_{\text {pacs; }}}{\text { Bandwidth }}=\frac{3 \sqrt{3}}{100 \sqrt{27 \times 10^{-6}}}=\frac{3 \sqrt{3}}{100 \times 3 \sqrt{3} \times 10^{-3}}=10 \mathrm{~s}\end{aligned}$ *

Asked in: MHT CET 2023 (10 May Shift 1)

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