# Quick Question about power factor angle

1. Jul 6, 2006

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)

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

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

Last edited: Jul 6, 2006
2. Jul 6, 2006

Sorry for the double post. I do not know how to delete this thread.

3. Jul 12, 2006

### xw3850

Basically,
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.

Now,
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

4. Jul 12, 2006

### xw3850

actually,
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.