Ncert Exercises & Solutions

Can the instantaneous power output of an AC source ever be negative?
Can the average power output be negative?

Hint: The power dissipated depends on the power factor.

Step 1: Let the applied emf;

E = E_{0} \sin (ωt)

$E = E_{0} \sin (ωt)$

and current developed is;

I = I_{0} \sin (ωt \pm ϕ)

$I = I_{0} \sin (ωt \pm ϕ)$

Instantaneous power output of the AC source;

P = EI = (E_{0} sinωt) (I_{0} \sin (ωt \pm ϕ))

$P = EI = (E_{0} sinωt) (I_{0} \sin (ωt \pm ϕ))$

= E_{0} I_{0} sinωt . \sin (ωt + ϕ)

$= E_{0} I_{0} sinωt . \sin (ωt + ϕ)$

= \frac{E_{0} I_{0}}{2} [cosϕ - \cos (2 ωt + ϕ)] . . . (i)

$= \frac{E_{0} I_{0}}{2} [cosϕ - \cos (2 ωt + ϕ)] . . . (i)$

Step 2: Average power;

P_{av} = \frac{V_{0}}{\sqrt{2}} \frac{I_{0}}{\sqrt{2}} \cos ϕ

$P_{av} = \frac{V_{0}}{\sqrt{2}} \frac{I_{0}}{\sqrt{2}} \cos ϕ$

= V_{rms} I_{rms} \cos ϕ . . . (ii)

$= V_{rms} I_{rms} \cos ϕ . . . (ii)$

Where

ϕ

$ϕ$ is the phase difference.

Clearly, from Eq. (i):

When,

cosϕ < \cos (2 ωt + ϕ)

$cosϕ < \cos (2 ωt + ϕ)$

\Rightarrow P < 0

$\Rightarrow P < 0$

Because;

\cos ϕ = \frac{R}{Z} > 0

$\cos ϕ = \frac{R}{Z} > 0$

No, the average power output of an AC source cannot be negative.