In the circuit shown,L = 1 mu H,C = 1 mu F,an | vedclass

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(D) The impedance of the $LCR$ series circuit is given by $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.[$A$] The current is in phase with the

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$A, B, D$

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(D) The impedance of the $LCR$ series circuit is given by $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.[$A$] The current is in phase with the voltage at resonance,where $\omega L = \frac{1}{\omega C}$,which gives $\omega = \frac{1}{\sqrt{LC}}$. This frequency is independent of $R$. Thus,statement [$A$] is correct.[$B$] As $\omega 	o 0$,the capacitive reactance $X_C = \frac{1}{\omega C} 	o \infty$. Thus,the impedance $Z 	o \infty$ and the current $I = \frac{V_0}{Z} 	o 0$. Thus,statement [$B$] is correct.[$C$] At high frequencies $(\omega \gg \frac{1}{\sqrt{LC}} = 10^6 	ext{ rad } s^{-1})$,the inductive reactance $X_L = \omega L$ dominates over the capacitive reactance $X_C = \frac{1}{\omega C}$. The circuit behaves like an inductor,not a capacitor. Thus,statement [$C$] is incorrect.[$D$] Resonance occurs at $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-6} 	imes 10^{-6}}} = 10^6 	ext{ rad } s^{-1}$. At this frequency,the current is in phase with the voltage. Thus,statement [$D$] is correct.Therefore,statements [$A$],[$B$],and [$D$] are correct.

List-$I$	List-$II$
$A$. $AC$ generator	$I$. Detects the presence of current in the circuit
$B$. Galvanometer	$II$. Converts mechanical energy into electrical energy
$C$. Transformer	$III$. Works on the principle of resonance in $AC$ circuit
$D$. Metal detector	$IV$. Changes an alternating voltage for smaller or greater value

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