Is it possible to use a 60V scale voltmeter as an amperometer to read up to 4A or similar? I tried to connect the two entrance poles with an appropriate resistance, just by trial and error, and found the appropriate value. But the resistance burns after a couple of minutes, with a current as low as 1A. Probably I would need a high-power dissipating resistance, but I have not. May I build such a resistance (low value, like 2 Ohms, but hi power) by cutting a big piece of not much conductive metal out of a thick plate?
Why not invest some money in a high power resistor like this:
http://www.amazon.com/Chassis-Mounted-Aluminum-Wirewound-Resistor/dp/B008MLKDN8/ref=sr_1_14?ie=UTF8&qid=1406703871&sr=8-14&keywords=resistor+2+ohm+100+watt
sure, you're right, that little money is not worth wasting time! So, if, say, R=2 Ohms, and P=100 W, it means that, with a voltage V=100 V as I have to use (for anodization of Aluminium), I can dissipate through that R a current of up to sqrt(P/R)=sqrt(50)=7A, great! And, how about the reached temperature? I don't have much spece where to mount it, far from other components of plastic device sidewalls... Because anodization will have to go on at that high power dissipation for as long as 10+ hrs (overnight). Ok, if the housing is Al, I hope it will not overcome 660 Celsius. Shall I add moire dissipating frames, or a fan? Thank you for your reply!
Well, apparently the resistor has been designed for up to 100 W, hence I would assume that the temperature will stay at non-critical values. I, for one, would mount it on something thermally insulating (ceramic maybe). A fan might help, but I don't think it's necessary. Anyway, you should order at least two resistors ;).
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