How Does V^2/R Relate to Electrical Power and Efficiency?

[kmarinas86]

(V^2/R) / (VI) = (VI) / (RI^2) = V / RI

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 = ?

 

[berkeman]

Your definitions appear to have some typos and issues mixed in. Can you post links to the sources of the equations you are using? Especially the equations you are using to calculate the Power Factor. Thanks.