(N/A) Let the applied $EMF$ be $E = E_m \sin \omega t$ and the current be $I = I_m \sin(\omega t \pm \phi)$.
The instantaneous power $P$ is given by $P = EI = E_m I_m \sin(\omega t) \sin(\omega t \pm \phi)$.
Using the trigonometric identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we get:
$P = \frac{E_m I_m}{2} [\cos(\phi) - \cos(2\omega t \pm \phi)]$.
Since the term $\cos(2\omega t \pm \phi)$ oscillates between $-1$ and $1$,the instantaneous power $P$ can be negative when $\cos(2\omega t \pm \phi) > \cos(\phi)$. This occurs during parts of the cycle when the current and voltage have opposite signs,indicating energy is being returned to the source.
The average power over a complete cycle is $P_{avg} = \frac{E_m I_m}{2} \cos \phi = V_{rms} I_{rms} \cos \phi$. Since $V_{rms}$,$I_{rms}$,and $\cos \phi$ (power factor) are generally positive for passive circuits,the average power cannot be negative. It represents the net energy consumed by the circuit per unit time.