Watts vs Volt-Amps - huh ??

OK lets start over.

It would seem that large inductive loads like motors and compressors with synchronous motors are going to look inductive and it is these thing that represent the largest loads and would dominate the trusses in my carpenter's square. IS the power company going to charge me more if I don't have a flat roof?

I might have a few small switch mode DC devices, but I don't think that is going to change the pitch on my roof much is it?

In the USA, residential power is not metered by PF, so there is no real point in correcting for it.

I think that you understand the issue quite well Mike, but what you are saying is exactly wrong.

VoltAmps (VA) is just the simple product of the RMS voltage times the RMS current, with no allowance for phase angle. It is useful for both AC and DC.
The Watts figure (W) takes into account the phase difference between voltage and current (zero for DC or for an AC resistive load). and is always less than or equal to the VA number.

One reason that VA is needed for AC work is that it tells you exactly how much current the conductors must be able to carry, even when the actual net power going over the same wires is close to zero.

PF by definition (for sine waves anyway) is just Watts/VA, and so is a number less than or equal to 1.

(That same information came right after Mike's original post back in 2012, but since the original was quoted here I am also repeating the correction.)

Electronic Switcher PF

When you have an inductive load such as a motor, the current waveforms will be pretty much a sine wave;
since they lag the voltage sine, you have power factor of less than one. A cap load has a sine current leading
the voltage, so in those cases a cap load bank may also be connected to cancel the out of phase inductive
current and bring the PF to near 1. That works because the dynamic sum of sine waves (of the same frequency)
is another sine wave.

When non linear elements such as diodes or switched transistors are used, current waveforms are created
which are more like abbreviated pulses than sine waves. Creating the inverse of these waveforms is not
practical, so they cause a nasty power factor. Caps are only really effective against sines.

Adding more filtering in load inputs helps smooth out input current, at some cost. In recent years there
has been recognition of the electronic switcher PF issue. Some switchers now include input circuits that
cause the input current waveform to approach an in phase sine. These add cost; how widely these are
deployed at this time, industrial or consumer, I can't say, but try reading the spec labels. Bruce Roe

Utilities do not charge residential customers for VARS. Only large commercial and industrial users have to pay more for VARS.

may be of interest. One snippet from that exchange:

-- snip --
Utilities do care about power factor to residential buildings, but hassling them about it is more trouble than it's worth due to the relatively light load compared to serious industrial customers.

Instead, the utilities hassle your legislators. It is logistically easier and much more cost effective for them to get laws passed to require certain power factors from devices commonly used by residential customers than to try to enforce or charge for low power factors from those customers directly. In the EU, for example, any electronic device drawing more than some specified amount of power (used to be around 70 W if I remember right, but I think the threshold has been lowered recently) must have a power factor above some limit to get CE certification. More and more electronic devices start with switching power supplies that do active power factor control, partially as a result of such laws emerging worldwide.
-- snip --

Completely useless information.

Answer this GURU DAN. It is not a trick question. First year 2nd day electrical student test question.

What is the total PF of a Desktop consuming .120 Kva with a PF of .83, plus a space heater consuming 2 Kva at a PF = 1.

Once you figure out the answer, you will then know the article is nonsense and useless. Give you a hint the PF = .9905 or 1 which is as good as it gets. Very Simple high school algebra 101. PF = True Power / Apparent Power, so 2100 watts / 2120 Kva = .9905 PF.

ahj?

With respect to an electrical inspection AHJ = Authority Having Jurisdiction.