1) Is the 1.8 volt difference worth worrying about? i.e. I calculate the 12v bulb will theoretically be putting out 66 watts instead of the 50 watts it is rated at; will it burn out in short order or just reduce it's life some ? (I can buy 4 of the cheap bulbs and still break even). I also know that the resistance can increase with an increase in temperature, which a bulb filament certainly does. So is a simple ohms law calculation even valid for a light bulb? I guess I need to think about the increased heat in the machine also. Which brings me to my next question.

2) If I use a simple voltage dropping resistor to drop the 1.8 volts is the following simple ohms law calculation correct? Or is there another formula that takes into account the heated filament in the bulb?

P=IE or P/E=I .... 50w / 12v = 4.1666 amps

E=IR or E/I=R .... 1.8v / 4.1666a = .432 ohm resistor

3) Let's make that three questions. Where do hobbyist find parts? RadioShack is a joke, and the online website's have good prices until you factor in the $7/8 shipping with a $30 minimum order for $2 or $3 worth of parts.

As well as the .43 ohm resistor for the bulb, I have an old Simpson Meter and need a 1% tolerance 11.5 ohm resistor to repair it. I've called around and all of the repair shops in the area have closed up; seems no body fixes anything any more!

Thanks in advance

Chris