Quick Question about power factor angle

[FrogPad]

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):

pf_{load1}=0.9\,\,\,LEADING

would the angle be negative? like...
\theta_{z\,\,load1}=-\cos^{-1}(0.9)

[Tom Mattson]

Hi there,

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

So when you have \theta_V-\theta_I>0, the current lags the voltage and the power factor is said to be "lagging". When the current leads the voltage then we have \theta_I>\theta_V, which of course implies that the argument of the cosine function, \theta_V-\theta_I, is negative.

[FrogPad]

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: