> > Oh no, here we go again with voodoo physics and math! > > I'm sorry, but that was downright insulting. Getting upset by Olin's (or anyone else's) rudeness is pointless and unproductive. The best approach is to take the generally sound technical advice, discard any (in Olin's case, usually but not always small, ) voodoo and gibberish, ignore all downright insults and general insults and proceed with life. Don't let 5 minutes of Olin's rudeness (or anyone else's) ruin a whole day of your time. (Not that I am yet fully adept at taking my own advice :-) ). Let's try a new approach at the same task. A Tungsten filament bulb operated at a higher voltage will have more current flow but not proportionately so as Tungsten bulbs have a positive temperature coefficient and therefore a higher resistance at higher powers. This effect is very significant - the inrush current of a cold bulb is many times that of one at operating temperature. An Aside: The only other real contender for bulb filaments is Carbon with a negative tempco - very hard to make them last). Affects during run up to temperature will complicate this - if the time on is not large relative to time to a stable temperature the "Physics" will change somewhat. Cooling time between cycles will also have some effect on peak temperatures reached if ton is not large compared to thermal time constant.). I would expect that thermal tc would be very similar up and down but maybe not (as eg gas transfer effects in bulb may operate differently )(seems unlikely but good to think of how things may affect results) Continuing. Power while operating is V^2/R. Energy is Power x time. Energy is what we care about over a whole cycle. If resistance at 12v is R then say resistance at 24 V = kR Let dc be fraction of time on. At 12v dc = 1 At 24v dc <1 The original info said that the current was the same in both cases. Here we may well descend into arguments about whether this was average current, RMS current etc. We can assume that it was NOT peak current. I'll assume it was AVERAGE current as I suspect this is what they would measure. As far as your car battery is concerned, average current is what it cares about (within reason) for energy purposes. 9Deep discussions on battery charging, cell chemistries, gas formation and consequent plate impedances etc left for another lifetime). Energy 12v in a unit time = V^2/res x dc E @ 12v = 12^2/R x 1 = 144/R watt-seconds For the 24v AVERAGE current to be equal to the 12v average current then if k = 1 Iavg = Ion x dc - 1 Ion = V/res= 24/res = 24 /(kR) - 2 Iavg = 12/R (as equal to 12v case) -3 1,2,3 gives 12/R = 24/kR x dc - 4 4 rearranged dc = 12k/24 (as one might expect) ie the higher the bulb resistance the longer the bulb must be on at 24v to make the current equal the current at 12v E @ 24v = V^2/res x dc in unit time = 24^2/kR x 12k/24 (k cancels, so why did we bother :-) ) = 288/R watt seconds ie the energy per unit time is double. This is an entirely unsurprising result as we deemed that both had the same average current and as P= V x I then doubling the voltage will double the power for fixed average current. Will the bulb be brighter? You bet! Will the bulb be hotter? You bet! Will the bulb last less long? Safe bet :-( Why do it this way at all? Why not just provide a somewhat higher voltage and run the bulb in a linear 100% on mode - hotter, brighter, whiter, more moth-near-the-flame-ier ? I don't know Mayhaps some of the mechanical aspects serendipitously happen to make things better. Maybe the gas cooling (such as it is, in the bulb. (Bulbs are typically filled with Argon but I have no idea what % of energy leaves filament by convection). All this ignores spectral efficiency of the human eye etc which others have mentioned. Also moving energy out of infra-red into visible. Also one may wish to suggest it's NOT energy that counts. Or ... Now I just know someone's going to criticise me and ruin my day :-) RM -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.