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

  1. May 6, 2008 #1
    (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 = ????
     
  2. jcsd
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