If V = V_m sin ωt is the voltage across the Capacitor, then the current is: 

In a purely capacitive circuit, the current leads the voltage by 90° because first, the current flows through the two plates of the capacitor, where the charge is stored, after that the charge accumulates at the plates of the capacitor and causes an establishment of a voltage difference.

Let, V = Vm sin ωt

We know, \(I = C\frac{{dV}}{{dt}}\)

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

I = ωC Vm cos ωt = Vm ωC sin(ωt + 90°)

I = Im sin \(\left(\omega +\frac{\pi}{2}\right)\)

From the phase diagram, it is clear that the current leads the voltage by 90° (i.e. one-fourth of one cycle).

Note:

• In a purely resistive circuit, the current is in phase with the voltage.
• In a purely inductive circuit, the current lags the voltage by 90°.