On Dec 10, 9:36*pm, "Martin H. Eastburn"
wrote:
We bought a new house once and the contractor put in 130v bulbs.
Not he 115v ones we buy in the store - and run them on 120 or 125v.

Anyway - when we sold the house 11 years later we still had some of
the original bulbs.

Consider : *P=E*I * If E drops - the power drops. *The bulb runs cooler.
* * * * * * *P=E^2/R *or R = E^2/P * 130*130/100 *= 13*13 = 269 ohms hot.
* * * * * * *(rule of thumb 1/10 of hot = cold resistance or 27 ohms for surges).
* * * * * * *I=P/E = 100/130 = .76 amps
* * Now - using the 130 bulb with 269 ohm filament and we run it at 120 :

* * * * * * * P (used) = 120*120/269 * *or 14400/269 = 53.53 watts.
* * * * * * * P=E*I * so I=P/E *I = 53/120 = .44 amps

You assume that the temparature, thus the resistance, of the filament
is the same at 130V as it is at 120V. This is certainly *not* true.
At 120V, the lower filament temperature not only will the bulb use
less power (though less than expected using your calculations) will
make the bulb less efficient (lumens per watt), costing you money too.

lower used wattage, longer life due to the derrating.

Much longer, yes. Bulb life is a function of something like the 16th
power of service voltage. It's still not saving money, unless there
is a cost associated with replacement in addition to the bulb cost.