A series LCR circuit with inductance L = 10,H | vedclass

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Question

(A) Given: $L = 10\,H$,$C = 10 	imes 10^{-6}\,F$,$R = 50\,\Omega$,and $V = 200 \sin(100t)$.Comparing with $V = V_{m} \sin(\omega t)$,we get angular f

Vedclass · Accepted Answer

$
u_{0} = 
u = \frac{50}{\pi}\,Hz$

Vedclass · Answer

(A) Given: $L = 10\,H$,$C = 10 	imes 10^{-6}\,F$,$R = 50\,\Omega$,and $V = 200 \sin(100t)$.Comparing with $V = V_{m} \sin(\omega t)$,we get angular frequency $\omega = 100\,rad/s$.The frequency of the $AC$ source is $
u = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi}\,Hz$.The resonant frequency $
u_{0}$ of the $LCR$ circuit is given by $
u_{0} = \frac{1}{2\pi\sqrt{LC}}$.Substituting the values: $
u_{0} = \frac{1}{2\pi\sqrt{10 	imes 10 	imes 10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-4}}} = \frac{1}{2\pi 	imes 10^{-2}} = \frac{100}{2\pi} = \frac{50}{\pi}\,Hz$.Thus,$
u_{0} = 
u = \frac{50}{\pi}\,Hz$.

$A$ series $LCR$ circuit with inductance $L = 10\,H$,capacitance $C = 10\,\mu F$,and resistance $R = 50\,\Omega$ is connected to an $AC$ source of voltage $V = 200 \sin(100t)\,V$. If the resonant frequency of the $LCR$ circuit is $\nu_{0}$ and the frequency of the $AC$ source is $\nu$,then:

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