Consider the applied emf
$$
\mathrm{E}=\mathrm{E}_0 \sin (\omega t)
$$
current developed is
$$
\mathrm{I}=\mathrm{I}_0 \sin (\omega \mathrm{t} \pm \phi)
$$
Instantaneous power output of the AC source
$$
\begin{aligned}
\mathrm{P} &=\mathrm{EI}=\left(\mathrm{E}_0 \sin \omega \mathrm{t}\right) \quad\left[\mathrm{I}_0 \sin (\omega \mathrm{t} \pm \phi)\right] \\
&=\mathrm{E}_0 \mathrm{I}_0 \sin \omega \mathrm{t} \cdot \sin (\omega \mathrm{t}+\phi)
\end{aligned}
$$
On solving ,
So, $\mathrm{P}=\frac{\mathrm{E}_0 \mathrm{I}_0}{2}[\cos \phi-\cos (2 \omega \mathrm{t}+\phi)]$
Average power $\mathrm{P}_{\mathrm{av}}=\frac{\mathrm{V}_0}{\sqrt{2}} \frac{\mathrm{I}_0}{\sqrt{2}} \cos \phi$
$$
=\frac{\mathrm{V}_0 \mathrm{I}_0}{2} \cos \phi
$$
or, $\quad=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{ms}} \cos \phi$ where $\phi$ is the phase difference.
From Eq. (i) when $\cos \phi < \cos (2 \omega \mathrm{t}+\phi) \quad\left(\therefore \mathrm{P}_{\mathrm{av}} < 0\right)$ So, the instantaneous power output of an AC source can be negative
From Eq. (ii), $\mathrm{P}_{\mathrm{av}}>0$
$\therefore \quad \cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}>0 \quad$ (Always positive)
So, $\mathrm{R}, \mathrm{Z}$ the reactance are always positive. Thus, the average power output of an AC source cannot be negative.
Hence, $\mathrm{P}_{\mathrm{av}}$ never be negative.