Question

Show that the current leads the voltage in phase by $\pi /2$ in an $AC$ circuit containing an ideal capacitor.

Answer 1

Given that,

Phase = $\frac{\pi }{2}$

If,

$V=V_0\sin \omega t$

We know that,

$q=CV$

$q=C{V}_0\sin \omega t$

Now, the current

$I=\frac{dq}{dt}$

$I=\frac{d\left(C{V}_0\sin \omega t\right)}{dt}$

$I=\omega C{V}_0\cos \omega t$

$I=\omega C{V}_0\sin (\omega t+\frac{\pi }{2})$

Hence, the current is $\omega C{V}_0\sin (\omega t+\frac{\pi }{2})$