[time-nuts] Modern motherboard with RS232 port

Dave Martindale dave.martindale at gmail.com
Tue Aug 21 16:46:56 UTC 2012

I like to think of it this way:
If you are talking instantaneous measurements, then watts is indeed always
volts * amps.  With a resistive load, the signs of volts and amps are
always the same, and the product of the two is always non-negative.  If you
calculate the average of instantaneous watts over time, you get average

If you have an inductive load, watts is still volts * amps.  But the phase
shift between current and voltage means that the instantaneous power is
sometimes negative, which means that the load is (at that instant)
returning power to the source.  But averaging instantaneous watts, both
positive and negative values, still gives you average power.

The problem comes when we want to calculate watts with devices that only
measure voltage, or only measure current.  With a resistive load, where the
instantaneous power is never negative, you can calculate power by measuring
only voltage, calculating the RMS voltage, and knowing the resistance.  But
that doesn't work for non-resistive loads because the instantaneous current
is no longer proportional to the instantaneous voltage.  If both are still
sinusoidal, knowing the phase shift lets you calculate power.  But that
doesn't work either for arbitrary waveforms.


On Sun, Aug 19, 2012 at 10:32 PM, Chuck Harris <cfharris at erols.com> wrote:

> The long and the short of it is that when AC encounters a reactive
> load, it results in a current that is not in phase with the voltage.
> Power is equal to volts x amps only when the current and voltage are
> in phase.... which can only happen if the load is purely resistive.
> If you hang a perfect capacitor across the power line, or a perfect
> inductor, you will draw lots of current, but no power.
> -Chuck Harris
> Tom Knox wrote:
>> Hi Ed;
>> I may not have had enough coffee yet, but if Volt X Amps = Watts why
>> would there be a difference?
>> Best Wishes;
>> Thomas Knox
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