An AC source of emf $V(t) = V_0 sinωt$ is put across a pure capacitor. The value of angular frequency of instantaneous power is:

$2\omega $

The correct answer is option (3) : $2\omega $

$V(t)=V_0sin \omega t$

In a capacitor $I(t)=I_0sin \left(\omega t +\frac{\pi}{2}\right)$

$=I_0cot \omega t $

$P_{in\, f}=V(t) I(t)$

$=V_0sin \omega t I_0 cos \omega t$

$=\frac{V_0I_0}{3}sin 2 \omega t$

Angular frequency $=2 \omega $