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(D) In an $AC$ circuit, the instantaneous current $I = I_0 \sin(\omega t + \phi)$ and voltage $V = V_0 \sin(\omega t)$.When the current is zero, $\sin

Vedclass · Accepted Answer

$A, B$ and $D$ only

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(D) In an $AC$ circuit, the instantaneous current $I = I_0 \sin(\omega t + \phi)$ and voltage $V = V_0 \sin(\omega t)$.When the current is zero, $\sin(\omega t + \phi) = 0$, which implies $\omega t + \phi = 0$ or $\pi$.When the voltage is maximum, $\sin(\omega t) = 1$, which implies $\omega t = \frac{\pi}{2}$.Substituting $\omega t = \frac{\pi}{2}$ into the current equation: $\sin(\frac{\pi}{2} + \phi) = 0$, which means $\cos(\phi) = 0$, so $\phi = \pm \frac{\pi}{2}$.This phase difference of $\frac{\pi}{2}$ occurs in a pure inductor $(\phi = -\frac{\pi}{2})$, a pure capacitor $(\phi = +\frac{\pi}{2})$, or an $LC$ circuit at resonance or specific conditions where the net reactance is zero (if current is zero and voltage is non-zero, it implies pure reactive behavior).Therefore, options $A, B$, and $D$ are correct.

List-$I$	List-$II$
$A$. Purely capacitive circuit	$I$. $I$ leads $V$ by $90^{\circ}$
$B$. Purely inductive circuit	$II$. $I$ and $V$ are in phase
$C$. $LCR$ series at resonance	$III$. $V$ leads $I$ by angle $\theta$
$D$. $LCR$ series circuit	$IV$. $V$ leads $I$ by $90^{\circ}$

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