J. Clarke wrote:
Dan Coby wrote:
J. Clarke wrote:
....
For a resistive load P=E^2/R. If the lamp is rated for 100 watts at
130 volts then it will dissipate (120^2/130^2)*100 watts at 120v or
about 85 watts if its resistance remains constant.
The assumption that the resistance will remain constant is a bad one.
....
... The actual power at 120 volts will be somewhere
between the 85 watts that you calculated and the 100 watts that it
would dissipate at 130 volts.

The standard number given is V^1.6. But the point is that a 100 watt bulb
only draws 100 watts at the design voltage, it's not 100 watts at all
voltages as Brainiac claims.

I hadn't looked up specific numbers; I only used the fact that the power
actually used is what controls the operating cost and that bulbs are
rated for their power consumption at the stated voltage. Hence, the
variability between an ideal 120V and our typical higher voltage that is
still rarely as high as 130V will cause the power consumption to be less
than it would otherwise be albeit w/ a loss of lumens altho I really
don't think it's terribly noticeable unless the lighting already was
marginal.

Anyway, assuming the 1.6 exponent, the reduction factor would be 0.88
instead of 0.85 according to my trusty HP-97. In reality, altho I've
never monitored it for a period of time (altho come to think I do have
sufficient test gear I could; just never thought of doing so as doesn't
really make any difference as it is what it is and will continue to be
so) I'd guess our average would be around 124/125 based on the numbers I
generally have noted when did measure. So, would be less than that in
practice but it _won't_ be 1.0.

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