Power in Alternating-Current Circuits — Question 1

Question 1

Positive average load power can include intervals of returned energy

A sinusoidal load has rms voltage 120 V, rms current 5.00 A, and lagging power factor .600. Write $v=\sqrt2V\cos\omega t$ and $i=\sqrt2I\cos(\omega t-\phi)$.

1. Derive instantaneous power and its mean.

2. Find its minimum and maximum values.

Question 1 - Solution

Positive average load power can include intervals of returned energy

(a) Method and calculation

$$p=2VI\cos\omega t\cos(\omega t-\phi)=VI[\cos\phi+\cos(2\omega t-\phi)].$$$$\boxed{P=VI\cos\phi=(120)(5)(.600)=360\,W}.$$

(b) Interpretation and further calculation

$$S=VI=600\,VA,\quad p_{\min}=P-S,\quad p_{\max}=P+S.$$$$\boxed{p_{\min}=-240\,W,\quad p_{\max}=960\,W}.$$

Physical check

Negative instantaneous power represents returned stored energy; it does not imply negative average consumption.