In the circuit shown, the ratio of the qualit | vedclass

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Question

(A) The quality factor $(Q)$ for a series $LCR$ circuit is given by $Q = \frac{\omega_0 L}{R}$, where $\omega_0 = \frac{1}{\sqrt{LC}}$ is the resonant

Vedclass · Accepted Answer

$\sqrt{\frac{1}{LC}} \frac{L^2}{R^2}$

Vedclass · Answer

(A) The quality factor $(Q)$ for a series $LCR$ circuit is given by $Q = \frac{\omega_0 L}{R}$, where $\omega_0 = \frac{1}{\sqrt{LC}}$ is the resonant angular frequency.Substituting $\omega_0$, we get $Q = \frac{1}{\sqrt{LC}} \cdot \frac{L}{R}$.The band width $(BW)$ is given as $\frac{R}{L}$.We need to find the ratio of the quality factor to the band width:Ratio $= \frac{Q}{BW} = \frac{\frac{1}{\sqrt{LC}} \cdot \frac{L}{R}}{\frac{R}{L}} = \frac{1}{\sqrt{LC}} \cdot \frac{L}{R} \cdot \frac{L}{R} = \sqrt{\frac{1}{LC}} \cdot \frac{L^2}{R^2}$.Thus, the correct option is $A$.

In the circuit shown, the ratio of the quality factor $(Q)$ and the band width $(BW)$ is (Given: band width = $R/L$)

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