In the circuit shown in the figure, the ratio | vedclass

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(A) Given: Resistance $R = 10 \, \Omega$, Inductance $L = 3.0 \, 	ext{H}$, Capacitance $C = 27 \, \mu	ext{F} = 27 	imes 10^{-6} \, 	ext{F}$.The re

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$10$

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(A) Given: Resistance $R = 10 \, \Omega$, Inductance $L = 3.0 \, 	ext{H}$, Capacitance $C = 27 \, \mu	ext{F} = 27 	imes 10^{-6} \, 	ext{F}$.The resonant angular frequency is $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{3 	imes 27 	imes 10^{-6}}} = \frac{1}{\sqrt{81 	imes 10^{-6}}} = \frac{1}{9 	imes 10^{-3}} = \frac{1000}{9} \, 	ext{rad/s}$.The bandwidth is $\Delta \omega = \frac{R}{L} = \frac{10}{3} \, 	ext{rad/s}$.The quality factor is $Q = \frac{\omega_0}{\Delta \omega} = \frac{1000/9}{10/3} = \frac{1000}{9} 	imes \frac{3}{10} = \frac{100}{3}$.We need to find the ratio of the quality factor to the bandwidth: $\frac{Q}{\Delta \omega} = \frac{100/3}{10/3} = \frac{100}{10} = 10 \, 	ext{s}$.

In the circuit shown in the figure, the ratio of the quality factor $(Q)$ to the bandwidth $(\Delta \omega)$ is $.............$ s.

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