Obtain an equation for the current when an $AC$ voltage is applied to an inductor and draw a graph of $V$ and $I$ versus time.

(N/A) The figure shows an $AC$ source connected to an inductor of inductance $L$. The inductor has negligible resistance,so the circuit is a purely inductive $AC$ circuit.
Let the voltage across the source be $V = V_m \sin \omega t$.
Using Kirchhoff's loop rule,$V - L \frac{dI}{dt} = 0$,where $-L \frac{dI}{dt}$ is the self-induced $emf$.
Therefore,$V = L \frac{dI}{dt}$,which implies $\frac{dI}{dt} = \frac{V}{L}$.
Substituting $V = V_m \sin \omega t$,we get $\frac{dI}{dt} = \frac{V_m}{L} \sin \omega t$.
Integrating with respect to time $t$,we get $I = \int \frac{V_m}{L} \sin \omega t \, dt = -\frac{V_m}{L \omega} \cos \omega t + C$.
Since the current is purely sinusoidal,the integration constant $C$ must be zero.
Thus,$I = -\frac{V_m}{L \omega} \cos \omega t = \frac{V_m}{\omega L} \sin(\omega t - \frac{\pi}{2})$.
Defining $I_m = \frac{V_m}{\omega L}$,we have $I = I_m \sin(\omega t - \frac{\pi}{2})$.
This shows that the current lags behind the voltage by a phase angle of $\frac{\pi}{2}$.