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Re: [vox-tech] [OT] Electrical Engineering Question
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Re: [vox-tech] [OT] Electrical Engineering Question

ach, one more thing.  i can't believe i forgot this (but the truth is,
my department knows i hate experimental physics, and has given me
lectures and discussions for the past few years instead of running

light bulbs are highly non-ohmic (linear) devices.  the power they
consume is highly dependent on the temperature of the filament.  in
other words, the power they consume depends on how long they've been
left on.   for the life of me, i can't remember whether they consume
more or less power as they heat up.  for your sake, i hope it's less.

250W is some kind of average; i don't know how they get that number, but
i'm sure they have some kind of procedure like "leave the lightbulb on
for an hour and divide by 3600" or something like that.

from what i remember about the physics 9C labs, 25% sounds pretty
reasonable to me.


begin Jeff Newmiller <jdnewmil@dcn.davis.ca.us> 
> On Fri, 31 Jan 2003, Rod Roark wrote:
> > I got my first electric bill at the new house; looks too
> > high.  So I decided to do an experiment.
> > 
> > Outside the house is an electric meter.  It reads KWH
> > accumulated on 5 dials, and has a horizontal platter that
> > appears to spin about 100 revolutions per KWH (anyone know
> > if this is exactly true for a standard meter?).
> > 
> > So I figure that means 10 watt-hours per rev, or 36,000
> > watt-seconds per rev.
> > 
> > I timed one revolution with most things in the house turned
> > off.  45 seconds.  Then I turned on a 250W light bulb and
> > timed it again.  32 seconds.  So:
> > 
> >   36,000 watt-secs / 45 secs =  800 watts
> >   36,000 watt-secs / 32 secs = 1125 watts
> > 
> >   1125 - 800 = 325 watts -- for a 250W bulb.
> > 
> > How come?  Should I complain to PG&E, or is there some
> > gotcha that I'm missing?
> Go look on your meter face for a number labelled Kh.  A typical value is
> 7.2.  Units are Watt-hours per revolution.
> Count the number of seconds (S) it takes for the disk to revolve once. (I
> usually time two or three turns and average. Total measurement intervals
> between 40 and 90 seconds will usually give good results (though if it
> spins three times you will need to remember to divide the total interval
> by three before using the answer).
> Power = Kh * 3600 / S
> The only difference between your equation and mine is your use of Kh=10.
> Assuming you have a regular meter like mine, the calculation should yield
> 810W - 576W = 234W, which could be reasonable. I don't know what kind of
> light bulb you used... torchiers output is often adjustable. Beware of
> "hidden loads" too... a refrigerator kicking off or on toward the end of
> your measurement interval can reduce or increase respectively your
> apparent load.
> For those looking to check these answers: the energy meter is almost
> certainly more accurate than whatever test equipment the average computer
> geek is likely to have laying around for measuring energy consumption
> (energy meters have a tested accuracy of <0.3%).  P=V*I always for
> simultaneous/instantaneous measurements, but the root-mean-square readings
> typically reported by DVMs ONLY hold for _sinusiodal waveforms that are in
> phase_.  Those conditions are often not valid in the real world, so beware
> of getting out your DVM to double-check your energy meter. :)

First they ignore you, then they laugh at you, then they fight you,
then you win. -- Gandhi, being prophetic about Linux.

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