Physics - EMI/AC Question with Solution | TestHub

The instantaneous e.m.f. is $E=E_0\sin (\omega t)$. The current in a pure inductor lags the e.m.f. by $\pi /2$, so $I=I_0\sin (\omega t-\pi /2)=-I_0\cos (\omega t)$.

Instantaneous power $P=EI=E_0{I}_0\sin (\omega t)(-\cos (\omega t))=-\frac{E_0{I}_0}{2}(2\sin (\omega t)\cos (\omega t))=-\frac{E_0{I}_0}{2}\sin (2\omega t)$.

Thus, the angular frequency of the instantaneous power is $2\omega$.