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David Combs

In article ,
hr(bob) wrote:
....

- harvested from Usenet

For a power factor of one, the voltage and current are in exact
agreement/phase for the peaks and nulls. If you apply a voltage to a
pure resistor, they are in phase and the power used is the product of
the voltage and current. If you apply a voltage to an inductor/motor/
transformer, the current lags behind the voltage somewhat and the
power into the device is not the product of the voltage times the
current, but the product of the voltage times the current times the
power factor. If you apply a voltage to a capacitor/condensor, the
current leads the voltage somewhat. Again, the actual power into the
device is the product of the voltage times the curent times the power
factor. If you have a very large inductance. the current will lag far
behind the voltage, and the actual power used in the device is small.
But, the power company still has to be able to provide the maximum
current to the device, but does not get much $$ since the power
atually consumed is much less. That is why the power companies do not
like inductive or capacitive loads. Compact fluorescent lights tend
to look like capacitive loads because they have a large inrush current
that leads the voltage and so altho the power companies will save some
power when these are adopted over a widespeard area, they will still
have to be able to provide the peak current that these lamps draw.
Fortunately, that peak curent is still well below what a resistive
incandescent light that puts out the same lumens uses.

Hope this helps.

A not only much delayed but probably stupid question:

If you phase-shifted volt-amp pair, and multiplied them together
at each time-instant, and addeded them up (ie integrated the
product dt)?, would that give the same result as multiplying
the peak voltage and peak current by the pf?

Thanks,

David

(I gotta go read that wikipedia article!)