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)

and current developed is;

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))

= E_{0} I_{0} sinωt . \sin (ωt + ϕ)

= \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

= V_{rms} I_{rms} \cos ϕ . . . (ii)

Where

ϕ

is the phase difference.

Clearly, from Eq. (i):

When,

cosϕ < \cos (2 ωt + ϕ)

\Rightarrow P < 0

Because;

\cos ϕ = \frac{R}{Z} > 0

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