[jaydee]

Originally posted by bigben2k
Jaydee, I think you've been working with waterpumps too long

LOL, no argument there!

Contrary to waterflow, where the resistance (pressure drop) across a block changes with the flow rate, the electrical resistance of the heater will remain (essentially) the same, regardless of voltage. (a poor analogy, I know).

Yeah I understand this. Just like a resistor as pH said.

All in all, the equation V=R * I (V: voltage, I: current (amps) and R: Resistance (in Ohms)) is always true. It's a law, just like many other things in physics.

Math sucks. Let me try and put this to were I can understand it better. Resistance x Current = voltage?

So if I have 10ohms of resistance and 5 amps it equels 50 volts?

So assuming that the heater will put out 150 Watts, at a voltage of 40, using the other equation (also a law) where P=V * I, you can see that the current that the heater will draw is 3.75 amps.

Ok, so to get the amps from 150watts at 40VDC into amps I would divide 150 by 40 to get 3.75amps?

Back to the first equation (V=RI), you can then calculate that the resistance of the heater is ~10.7 Ohms. This does not change (well, just a little bit, maybe up to 15%, but just because it's a heater).

Ok.

So if you reduce the voltage to 12, and applying V=RI, where R is (still) ~10.7 Ohms, you would figure out that it will equal about 1.12 amps. Now knowing the amps, and using P=VI, you can then calculate the power output to be ~13 Watts.

Ok.
But back to your question...

Power supplies usually work in one of two ways: they allow you to control either the voltage (most common) or the current.

If you control the voltage, the resistance determines the current.
If you control the current, the resistance determines the voltage.

So your power supply may do either, but not both. Perhaps the "amp" setting is some kind of failsafe trigger, where if it's exceeded, it shuts down?

All the ones I been looking at have Amperage and Voltage adjustment. I.E.: http://www.alltronics.com/images/ac350.JPG

I belive you maybe correct though that it is just a current limit and not an actual setting. Makes since... This stuff doesn't click when you are ignorant of other important info like above as I am.

It's not so bad.

Theoretically (and someone correct me here), you simply use your voltage measurement's margin of error, in %, and add it to your current measurement error margin (also in %), then add 1 or 2% for secondary heat loss (better to quantify it yourself, but not easy), and you have a relatively accurate measurement of the power that your testbench applies, complete with error margins.

Will have to dig into this later.