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Homework Help: Reactive Power Apparent Power Issue

  1. Aug 8, 2011 #1
    This is not a HW question.

    Assume you have a sinusoidally varying voltage, such that current lags by some phase angle phi

    Then you can write the power as a function of time as


    You can break this up via trig identity into the following

    Eq1: p(t)=2*Vrms*Irms*( [1+cos(2*omega*t)]*cos(phi) + sin(2*omega*t)*sin(phi) )

    If you average over 1 period, then you obtain that the average REAL power is

    P=Vrms*Irms*cos(phi) This is average real power

    The peak real power would be twice this value since (take t=0 in Eq1:)
    real power is associated with cos(phi) term.

    The average reactive power is zero, it bounces back and forth between inductive and capacitive loads (B and E fields).

    But the peak reactive power is defined as Q=Irms*Vrm*sin(phi).


    I see often that the Apparent power S = P + i*Q. Unfortunately it is never throuroughly defined. Why would would be adding the Average value of real power and Peak value of Reactice power and call it the apparent power.

    Am I misunderstanding this? Shouldn't we be adding the PEAK real power and PEAK reactive power and call this the apparent power.

    Also, other relations are P=(I^2)R Q=(I^2)X S=(I^2)Z


    Am i correct in saying that the reason why it is defined this way is so that "you" could say that the phase angle is given by

    Phi = arctan(Q/P).

    I'm just a little weired out by combining the Average Real power with Peak Reactive power.

    Last edited: Aug 8, 2011
  2. jcsd
  3. Aug 8, 2011 #2


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    Science Advisor
    Homework Helper

    hi azaharak! :smile:
    no, that's the complex power

    the apparent power is VrmsIrms … it's voltage times current
    so far as i know, neither complex power nor apparent power have any precise physical significance

    (but you need to know them because they come up in exam questions! :wink:)

    from the pf library on https://www.physicsforums.com/library.php?do=view_item&itemid=303"


    Power = work per time = voltage times charge per time = voltage times current:

    [tex]P = VI =\ V_{max}I_{max}\cos(\omega t + \phi/2)\cos(\omega t - \phi/2)[/tex]
    [tex]=\ V_{max}I_{max}(\cos\phi + \cos2\omega t)/2[/tex]
    [tex]=\ V_{rms}I_{rms}(\cos\phi + \cos2\omega t)[/tex]
    [tex]=\ (V_{rms}^2/|Z|)(\cos\phi + \cos2\omega t)[/tex]​

    So (instantaneous) power is the constant part, [itex]P_{av} = V_{rms}I_{rms}\cos\phi[/itex] (the average power), plus a component varying with double the circuit frequency, [itex]V_{rms}I_{rms}\cos2\omega t[/itex] (so a graph of the whole power is a sine wave shifted by a ratio [itex]\cos\phi[/itex] above the x-axis).

    Apparent power, reactive power, and complex power, are convenient mathematical definitions with no precise physical significance: apparent power is the (constant) product of the r.m.s. voltage and current, [itex]S=V_{r.m.s.}I_{r.m.s.}[/itex]: it is what we would expect the average power to be if we knew nothing about reactance! :rolleyes:

    Similarly reactive power is defined as [itex]Q=S\sin\phi=P_{av}\tan\phi[/itex], and complex power is defined as [itex]Se^{j\phi}=P_{av}+jQ[/itex].
    Last edited by a moderator: Apr 26, 2017
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