Does fuse amperage required change with voltage

Things can get a little complicated, but the simple answer is NO

Complicated because with lights, their resistance increases with heat, that is, lights draw more current when cold, than when lit. But in the range that "is" "when they are lit" they don't change MUCH, so we can assume they stay the same.

So you will have to have enough "fudge factor" to cover the "power up surge" when you first turn them on, as they will draw more for a short time.

Google up "ohms law."

Simply, we assume the lamp resistance "stays the same" when at operating brilliance.

Since R (resistance) is now assumed to be a (pretty much) constant, this means that if voltage goes UP, then current goes UP as well

Ohms law is I = E / R, which means

Current in amps = Voltage divided by Resistance

So if E (Voltage) goes up, I (Current) must go up

http://hvacwebtech.com/images/pie.gif

Hmmm... something's not making sense to me. I agree with everything you said until you got to the part that if E (voltage) goes up I (current) goes up too.

Current and Voltage are inversely proportional so since wattage=current x voltage given the same wattage if current goes up voltage goes down and vice-versa. What am I missing?