A series LCR circuit is resonating with a sou | vedclass

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(D) The resonant angular frequency of a series $LCR$ circuit is given by $\omega_{R} = \frac{1}{\sqrt{LC}}$.From the given diagrams,the time period $T

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current will lead source voltage after replacing the source

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(D) The resonant angular frequency of a series $LCR$ circuit is given by $\omega_{R} = \frac{1}{\sqrt{LC}}$.From the given diagrams,the time period $T$ of the emf in diagram $-2$ is greater than that in diagram $-1$ $(T_2 > T_1)$. Since angular frequency $\omega = \frac{2\pi}{T}$,it follows that $\omega_2 When the operating frequency $\omega$ is less than the resonant frequency $\omega_R$,the capacitive reactance $X_C = \frac{1}{\omega C}$ is greater than the inductive reactance $X_L = \omega L$.Therefore,the circuit behaves as a capacitive circuit,and the current leads the source voltage.

$A$ series $LCR$ circuit is resonating with a source whose emf varies with time as described in diagram $-1$. If we replace the source by another source whose emf varies with time according to diagram $-2$,then:

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