Q.An inductor of inductance 200 mH is connected to an A.C. source of peak e.m.f. 220 V and frequency 50 Hz. Calculate the peak current in the circuit.

For a purely inductive circuit (negligible resistance) connected to $e=e_0\sin\omega t$, integrating the self-induced-emf relation $L\,di/dt=e$ (from Kirchhoff's voltage law, since the applied emf must at every instant exactly balance the induced back-emf $e'=-L\,di/dt$) gives $i=i_0\sin\left(\omega t-\dfrac{\pi}{2}\right)$, with peak current $i_0=e_0/(\omega L)$ -- the current LAGS the emf by a phase angle of $\pi/2$ ($90^\circ$); equivalently, the voltage across the inductor is said to lead the current by $90^\circ$. …