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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]
    Last edited: Jul 6, 2006
  2. jcsd
  3. Jul 6, 2006 #2
    Sorry for the double post. I do not know how to delete this thread.
  4. Jul 12, 2006 #3
    The phase can vary between pi/2 and -pi/2.
    When phase >= 0, then the load is capacitive because the maximum current comes before the maximum voltage. This is leading.

    When phase < 0, then the load is inductive because the maximum voltage comes before the maximum current. This is lagging.

    power factor is defined as

    power-factor = cos(phase)

    and the idea is to solve for the phase.

    so, if you are given just pf = 0.9 then this is vague because

    0.9 = cos(t) -> t = -0.45, 0.45

    which to choose?

    if pf = 0.9 leading then we know phase >=0 so choose 0.45
    if pf = 0.9 lagging then we know phase < 0 so choose -0.45
  5. Jul 12, 2006 #4
    just to be 100% clear I guess I should have stated

    when phase = 0 then the load is purely resistive because the maximum current happens at the same time as the maximum voltage. Here pf = 1 and the leading and lagging term obviously do not appply.
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