On Sun, 11 Sep 2005 02:08:43 +0000, David Combs wrote:

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?

PF is kinda meangless with DC, since P=VA.

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.

Not average. When Volts is multiplied by amps (power) there is a squared
term in there. RMS == Root Means Square (root of the mean square), so
when you calculate power, the squared term is taken into account. Average
voltage isn't all that useful; the average for sine wave is zero. ;-)

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).

Or a non-linear load adding harmonic content to the waveforms.

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?

Cosine of the angle between them, but that really only works for
pure sine waves.

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?

The fluorescent is a non-linear load. It chops the current waveform,
adding harmonics to it. The above formula isn't very useful in this case.

(Huge electric motor, I understand how.)

A "huge electric motor" should have a PF very close to one if it's heavily
loaded. If it's creating no mechanical work, then it will have a worse
PF.

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

The phase shift will be small, but it will only conduct current during
part of the cycle, chpping the current waveform. Since the current isn't
being used during the entire waveform, the power uring this part of the
cycle isn't useful.

Obvously, I could use some mental fill-in!

Thanks!

--
Keith