In article ,
wrote:
George E. Cawthon wrote:

...watts = 0.4 x volts x amps, for some undercounter fluorescents.

Do fluorescent lights have a power factor?

Sure... 0.4 in the case above.

But, that said, if the undercounter light fixture say 15W, then that
is how much is used regardless of power factor

Yes.

and w=va still applies. or am I wrong...

You am wrong. W = PFxVxA.

Nick


QUESTION about this "power factor":

(it has been *so long* since I understood any
of that stuff, that I've forgotten all but a
few words describing it.)

With DC, the pf is 1.0?

With AC, I'm not sure what it is.

Of course there's the "rms" stuff, trying to
get an average value of a sine-wave.

The pf, I recall from *ages* (decades) ago, had something to
do with the voltage and current "waves" getting out of sync
with each other, due to a coil or a capacitor (one shifting
in one direction, one in the other).

I recall something about having to use trig to get
the pf, maybe it was the sine or cosine of the degrees
of lead or lag?

If so, then since those functions range between plus and
minus one, then maybe the pf I dimly recall is the
reciprocal of yours?

Anyway -- here's my question:

How do you get a substantial pf for a fluorscent (sp?)
light? (Huge electric motor, I understand how.)

And a pf for an incandescent light, that would involve
no phase shift at all?

Obvously, I could use some mental fill-in!

Thanks!

David