A series LCR circuit containing 5.0 , H induc | vedclass

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(C) Given: $L = 5.0 \, H$,$C = 80 \, \mu F = 80 	imes 10^{-6} \, F$,$R = 40 \, \Omega$.First,calculate the resonant angular frequency $\omega_0$:$\om

Vedclass · Accepted Answer

$46 \, rad/s$ and $54 \, rad/s$

Vedclass · Answer

(C) Given: $L = 5.0 \, H$,$C = 80 \, \mu F = 80 	imes 10^{-6} \, F$,$R = 40 \, \Omega$.First,calculate the resonant angular frequency $\omega_0$:$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0 	imes 80 	imes 10^{-6}}} = \frac{1}{\sqrt{400 	imes 10^{-6}}} = \frac{1}{0.02} = 50 \, rad/s$.Next,calculate the bandwidth $\Delta \omega$:$\Delta \omega = \frac{R}{L} = \frac{40}{5.0} = 8 \, rad/s$.The frequencies at which the power is half the maximum power (half-power frequencies) are given by $\omega = \omega_0 \pm \frac{\Delta \omega}{2}$.$\omega_1 = \omega_0 - \frac{\Delta \omega}{2} = 50 - \frac{8}{2} = 50 - 4 = 46 \, rad/s$.$\omega_2 = \omega_0 + \frac{\Delta \omega}{2} = 50 + \frac{8}{2} = 50 + 4 = 54 \, rad/s$.Thus,the angular frequencies are $46 \, rad/s$ and $54 \, rad/s$.

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