Discuss the power in an $AC$ circuit containing only an inductor.

(N/A) In an $AC$ circuit containing only an inductor,the current lags behind the voltage by a phase angle of $\frac{\pi}{2}$ radians,which corresponds to a time delay of one-fourth of a period,$\frac{T}{4} = \frac{\pi/2}{\omega}$.
The instantaneous voltage is $V = V_m \sin(\omega t)$ and the instantaneous current is $I = I_m \sin(\omega t - \frac{\pi}{2}) = -I_m \cos(\omega t)$.
The instantaneous power $P_L$ supplied to the inductor is given by:
$P_L = IV = [I_m \sin(\omega t - \frac{\pi}{2})] \times [V_m \sin(\omega t)]$
$P_L = -I_m V_m \cos(\omega t) \sin(\omega t)$
Using the identity $\sin(2\theta) = 2\sin\theta \cos\theta$,we get:
$P_L = -\frac{I_m V_m}{2} \sin(2\omega t)$
The average power $\langle P_L \rangle$ over a complete cycle is the average of the instantaneous power over time $T$:
$\langle P_L \rangle = \left\langle -\frac{I_m V_m}{2} \sin(2\omega t) \right\rangle$
Since the average value of $\sin(2\omega t)$ over a complete cycle is zero,we have:
$\langle P_L \rangle = -\frac{I_m V_m}{2} \times 0 = 0$
Thus,the average power supplied to a pure inductor over one complete cycle is zero.