A pure capacitor connected in $AC$ circuit is shown in figure. Capacitor is connected with ac voltage $\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin \omega t$.

When a capacitor is connected to a voltage source in a dc circuit, current will flow 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 $\mathrm{V}$ across the capacitor is $\mathrm{V}=\frac{q}{\mathrm{C}}$ where $\mathrm{C}$ is the capacitance of capacitor.

From the Kirchhoff's loop rule,

$\mathrm{V}_{\mathrm{m}} \sin \omega t-\frac{q}{\mathrm{C}}=0$

$\therefore \mathrm{V}_{\mathrm{m}} \sin \omega t=\frac{q}{\mathrm{C}}$

$\therefore\mathrm{CV}_{\mathrm{m}} \sin \omega t=q$

$\text { Now } \mathrm{I}=\frac{d q}{d t}$

$\therefore \mathrm{I}=\frac{d}{d t}\left[\mathrm{~V}_{\mathrm{m}} \mathrm{C} \sin \omega t\right]$

$\therefore \mathrm{I}=\mathrm{V}_{\mathrm{m}} \mathrm{C} \omega \cos \omega t$

$\therefore \mathrm{I}=\frac{\mathrm{V}_{\mathrm{m}}}{1 / \omega \mathrm{C}} \cdot \cos \omega t$

Now putting $\cos \omega t=\sin \left(\omega t+\frac{\pi}{2}\right)$

$\therefore \quad \mathrm{I}=\mathrm{I}_{\mathrm{m}} \sin \left(\omega t+\frac{\pi}{2}\right)$

$\ldots$ $(2)$