For a series LCR circuit,the I vs omega curve | vedclass

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(C) In a series $LCR$ circuit,the resonance frequency $\omega_{r}$ is defined by the condition $X_{C} = X_{L}$,where $X_{C} = \frac{1}{\omega C}$ and

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$(A)$ and $(C)$ only

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(C) In a series $LCR$ circuit,the resonance frequency $\omega_{r}$ is defined by the condition $X_{C} = X_{L}$,where $X_{C} = \frac{1}{\omega C}$ and $X_{L} = \omega L$.$1$. For $\omega  \omega L$,which means $X_{C} > X_{L}$. Thus,the circuit is mainly capacitive. Statement $(A)$ is correct.$2$. For $\omega > \omega_{r}$,we have $X_{L} > X_{C}$,so the circuit is mainly inductive.$3$. At resonance $\omega = \omega_{r}$,the impedance $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$. Since $X_{L} = X_{C}$,$Z = \sqrt{R^{2} + 0} = R$. Thus,the impedance is equal to the resistance. Statement $(C)$ is correct.$4$. Statement $(D)$ is incorrect because at resonance,the impedance is minimum and equal to $R$,not $0$.Therefore,statements $(A)$ and $(C)$ are correct.

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