Originally Posted by
onehundred80
Surely the wattage is determined by the resistance of the filament. (E x E) ÷ R = W
Long lasting filaments have a higher resistance and less light output.
I'm not too sure of what I say, maybe you can enlighten me.
What we need is Sparkie, he knows what wires will turn him into smoke.
Enlighten you? On Ohm's law for power in a DC circuit when voltage and resistance is known? Naw nothing to 'enlighten' you about there. What you should be doing is taking the Ohm's law for DC circuits for finding the UNKNOWN (resistance) and applying it to filaments that exist for automotive bulb use. Lets see, as a reference only (as system voltage varies with load and supply) lets use 12.5v at 100 watt output, we come up with approx 1.56 ohms. Now lets find out how much a filament would cost in this application in the gaseous environment necessary to not blow out with any reasonable time span. I don't think there is a known filament with this specification, but if someone who has these elcheapo bulbs would take a quality DVOM and report back here the bulbs resistance, we'll see how much actual power in watts these bulbs produce at the specified voltages applied.