(V^2/R) / (VI) = (VI) / (RI^2) = V / RI(adsbygoogle = window.adsbygoogle || []).push({});

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 = ????

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# (v^2/r) >= V*i

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums - The Fusion of Science and Community**