Correct power factor formua

My instructor asked me why have I written the power factor formula as

If $\varphi$ is the phase angle between the current and voltage, then the power factor is equal to the cosine of the angle, $cos\,\varphi$: $$|P| = |S|\,\cdot\,cos\,\varphi$$

with his reasoning that $P$ must always be positive with the exception when the resistance is negative, so I should not have the absolute value there.

Now I cannot remember the reason why, but I am pretty sure I have found it somewhere, but since I did a poor job with references, now I cannot find the origin of this formula.

I have found it on wikipedia.

Is this formula wrong? If yes, how can it be written correctly in this form. If no, can you forward me to some more reading on this? Many thanks.

• P=S⋅cosφ would account for regenerative breaking of a motor (a.k.a. a generator). – jippie Feb 29 '16 at 19:14
• Hello @jippie, could You please elaborate further? I am still confused. Thank You. – delmadord Feb 29 '16 at 19:21
• P>0 means power is sent into the load, like a motor pushing a cart forward. If P<0 then the motor is actally braking and with that generating power, probably delivering power to the grid. A regular elevator consumes power to pull the carriage up, whereas a somewhat modern elevator will deliver power back to the grid while the carriage goes down and the motor is just used for braking. Similar with electric and hybrid cars. P<0 indicates power being generated. – jippie Feb 29 '16 at 19:26
• This fact is clear to me. In Your first comment, I got confused by the word breaking instead of braking (silly me). Does this all apply for the absolute values in formula too, please? – delmadord Feb 29 '16 at 19:32
• Whoops silly me. Using the formula without the absolute value for P makes it easy to identify whether a motor is in generator mode or in motor mode, provided you have all the signs and phasors correct. – jippie Feb 29 '16 at 19:44

I would say that the most general formula is:

$$P=Re\{\underline{S}\}=Re\{\underline{V}·\underline{I}^{\star}\}=S\cos{\theta}$$

where I'm using underline to identify phasors.

S is the module of $\underline{S}$ and it is always positive (it is simply the voltage amplitude times the current amplitude).

$\theta$ is the angle between voltage and current. If this angle is bigger than 90º it means that the load is in reality providing active power to the source and not the other way around.

Therefore P can be positive or negative.

The real power (P) should not be an absolute value. Let's look at the math. Power factor is defined as the ratio of real power (P) to apparent power (S). But S is a complex number, so a simple ratio is meaningless. We need to use the magnitude of S:

$$pf = \frac {P} {|S|}$$

If we only have S, we can use it to get a formula for P. P is the real part of S:

$$P = Re\{S\} = |S| \cos \varphi$$

where $\varphi$ is the argument of S -- the angle between the complex vector S and the positive real axis. If S has a negative real component, then P should be negative. And indeed, $\cos \varphi$ is negative when $-90^\circ < \varphi < 90^\circ$.

This is a bit confusing because power triangles are almost always drawn with positive real power, which makes it look like the angle is between P and S. But it's not -- the angle is a property of the complex number S alone. So a negative power factor tells you that the load is supplying power to the generator. If you don't care about that, you can take the absolute value of the cosine to keep the power factor positive:

$$|pf| = \frac {|P|} {|S|} = |\cos \varphi|$$

UPDATE: The Wikipedia article is incorrect. It contradicts itself: The formula contains the absolute value of P, but then immediately says the power factor can be negative.

• Thank you for the answer. After some more investigation, I have found that the formula containing absolute values is on the wikipedia. Since you are proposing that the absolute value should not be present on P, should the wikipedia be changed too, please? – delmadord Mar 1 '16 at 7:43
• Yeah, I think Wikipedia is wrong in this case. I updated my answer. – Adam Haun Mar 1 '16 at 13:49
• I have found another source, where the absolute value is on the cosine. en.wikiversity.org/wiki/Power_factor – delmadord Mar 7 '16 at 13:35
• They're defining PF to be between 0 and 1, so they have to have the absolute value there. They're also ignoring cases where the load generates power. That's fine as long as they're consistent. – Adam Haun Mar 7 '16 at 15:14