An ideal inductor of inductance $L$ is connected to an AC source $V=V_0\sin (\omega t)$. The average power dissipated in the inductor over one complete cycle is equal to which of the following?

Solution

1. The instantaneous power is $P=VI$.
2. For an ideal inductor, the current is $I=-(V_0/\omega L)\cos (\omega t)$.
3. The average power is $P_{avg}=(1/T){\int }_0^T{V}_0\sin (\omega t)[-(V_0/\omega L)\cos (\omega t)]dt$.
4. Since $\sin (\omega t)\cos (\omega t)=(1/2)\sin (2\omega t)$, the integral over a full period is zero.
5. Hence the answer is (C).