Question

1) Can the instantaneous power output of an ac source ever be negative?

2) ) Can the average power output of an $AC$ source be negative?

Answer 1

(i) Formula used: $P=V_0{I}_0\sin \omega t\sin (\omega t+\varphi )$

Applied voltage in AC circuit containing,

$V=V_0\sin \left(\omega t\right)$

Hence current developed is
$I=I_0\sin (\omega t\pm \varphi )$

Instantaneous power output of the AC source

$P=V_I=\left(V_0\sin \omega t\right)\left(I_0\sin \right(\omega t+\varphi )$

= $V_0{I}_0\sin \omega t\sin \sin (\omega t+\varphi )$

=$V_0{I}_0\sin \omega t[\sin \omega tcos\varphi +\cos \omega t\sin \varphi ]$

=$V_0{I}_0[si{n}^2\omega t\cos \varphi +\sin \omega tcos\omega tsin\varphi$

=$V_0{I}_0[\frac{(1-\cos 2\omega t)}{2}\cos \varphi +\frac{1}{2}\sin 2\omega tsin\varphi$


=$\frac{V_0{I}_0}{2}[\cos \varphi -\cos 2\omega t\cos \varphi +\sin 2\omega t\sin \varphi ]$


=$\frac{V_0{I}_0}{2}[cos\varphi -(cos2\omega tcos\varphi -sin2\omega tsin\varphi ]$

$P=\frac{V_0{I}_0}{2}[\cos \varphi -\cos (2\omega t+\varphi \left)\right]$

Clearly, from the equation, Taken phase angle ,$\varphi \pm ve$
When $\cos <\cos (2\omega t+\varphi )$ ,the instantaneous power output of an AC source will be negative.

Final answer: Yes, the instantaneous power of an $AC$ source can be negative.

(ii)

Formula used:

$P_{avg}=\frac{V_{rms}^2}{Z}\cos \left(\varphi \right)$ , $P_{av}=\frac{V_0}{\sqrt{2}}\frac{I_0}{\sqrt{2}}cos\varphi$

Apply average power in LCR circuit,

$P_{avg}=\frac{V_{rms}^2}{Z}\cos \left(\varphi \right)$

Here, $\cos \left(\varphi \right)=\frac{R}{Z}\ge 0$

As $RandZ$ are always non-negative, $\cos \left(\varphi \right)$ is always non-negative. Hence, average power cannot be negative.

Final Answer: No.