Obtain the formula for current when an $AC$ voltage is applied to a capacitor and draw the graphs of $V$ and $I$ versus $\omega t$.

(N/A) pure capacitor connected in an $AC$ circuit is shown in the figure. The capacitor is connected to an $AC$ voltage source $V = V_{m} \sin \omega t$.
When a capacitor is connected to a voltage source in a $DC$ circuit,current flows only for the short time required to charge the capacitor.
As charge accumulates on the capacitor plates,the voltage across them increases,opposing the current.
When the capacitor is fully charged,the current in the circuit falls to zero.
When the capacitor is connected to an $AC$ source,it limits or regulates the current but does not completely prevent the flow of charge.
The capacitor is alternately charged and discharged as the current reverses each half cycle.
Let $q$ be the charge on the capacitor at any time $t$. The instantaneous voltage $V$ across the capacitor is $V = \frac{q}{C}$,where $C$ is the capacitance.
From Kirchhoff's loop rule:
$V_{m} \sin \omega t - \frac{q}{C} = 0$
$\therefore V_{m} \sin \omega t = \frac{q}{C}$
$\therefore q = C V_{m} \sin \omega t$
Now,the current $I = \frac{dq}{dt}$:
$I = \frac{d}{dt} [C V_{m} \sin \omega t]$
$I = C V_{m} \omega \cos \omega t$
$I = \frac{V_{m}}{1 / \omega C} \cos \omega t$
Using $\cos \omega t = \sin \left(\omega t + \frac{\pi}{2}\right)$ and defining $I_{m} = \frac{V_{m}}{1 / \omega C} = \omega C V_{m}$:
$I = I_{m} \sin \left(\omega t + \frac{\pi}{2}\right)$
This shows that the current leads the voltage by a phase angle of $\frac{\pi}{2}$.