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Quick Question about power factor angle

  1. Jul 6, 2006 #1
    I just took an exam, and one of the questions had a load with a power factor that was leading. ALL of our examples have dealt with LAGGING power factors, so I was unsure about determining the angle.

    So for example)
    Find the power factor angle of the following load (load 1):

    [tex] pf_{load1}=0.9\,\,\,LEADING [/tex]

    would the angle be negative? like...
    [tex] \theta_{z\,\,load1}=-\cos^{-1}(0.9) [/tex]
  2. jcsd
  3. Jul 13, 2006 #2

    Tom Mattson

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    Gold Member

    Hi there,

    The answer to your question is "yes". The power factor is defined as: [itex]pf=\cos(\theta_V-\theta_I)[/itex]. The terms "leading" and "lagging" pertain to the relationship that the current has with respect to the voltage.

    So when you have [itex]\theta_V-\theta_I>0[/itex], the current lags the voltage and the power factor is said to be "lagging". When the current leads the voltage then we have [itex]\theta_I>\theta_V[/itex], which of course implies that the argument of the cosine function, [itex]\theta_V-\theta_I[/itex], is negative.
  4. Jul 14, 2006 #3
    Thank you for the thorough reply. I really didn't want to just memorize that I should toss a negative sign in there. Thanks for the background :smile:
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