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Trying to understand the effect of reactive on true power

  1. Jan 23, 2010 #1
    If apparent power is 720 volt amps and the true power is 624 watts which tells u that the reactive is 360 rva, why is it that the true power is not 50% of the apparent power when the reactive is 50% of the apparent?
    I would think that if you have 720va and then 360 rva that the true power should then be 360watts.
    That since half the power is coming back, then the true power(err the result of apparent - reactive) would be cut in half like two differing voltages in a DC circuit which cancels out 12v-7v=5v.
     
  2. jcsd
  3. Jan 23, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi foolios ! Welcome to PF! :smile:
    From the PF Library on voltage …​

    [tex]P_{apparent}\ =\ V_{rms}I_{rms}\ =\ |{P_{complex}|\ =\ \sqrt{P_{average}^2+ Q_{average}^2}[/tex]

    where … [itex]Q[/itex] is the reactive or imaginary power (involving no net transfer of energy), and [itex]V_{rms}\text{ and }I_{rms}[/itex] are the root-mean-square voltage and current, [itex]V_{peak}/\sqrt{2}\text{ and }I_{peak}/\sqrt{2}[/itex].

    So Papparent = √(3602 + 6242) = 720. :wink:
     
  4. Feb 19, 2010 #3
    An alternative way to think of reactive power is in terms of transmission lines. Imaginary power is the portion of incoming power that is reflected back to the generator rather than transmitted and then dissipated in the resistive load as heat.
     
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