- #1

- 979

- 1

VA = VI = Apparent power

W = RI^2 = Real power

PF = RI/V = Power factor

(V^2/R) / VA = VA / W = V / RI = 1 / PF

(V^2/R) = VA^2 / W

VA>=W

Therefore:

V^2/R >= VA

!

(V^2 / R) * PF = VA

VA * PF = W

!

What is V^2/R? If you have an large inductor, you could have V^2/R with very little current! What causes the magnetic field?

Is it "IR^2"? I know that equals joule loss. (real power=joule loss)

Efficiency = (? - Joule loss) / ?

Is it "VI"? I know that equals the electrical power input. (electrical power input>=joule loss)

Efficiency = (VI - ? - Joule loss) / VI

Is it "V^2/R"? But this is greater than or equal to electrical power input! (V/R>=I) What does this mean when PF<1 ?

Efficiency = ?