1) As the resistor heats up, its resistance increases from 1 ohm.

2) Your ammeter might have a finite resistance when trying to make this type of measurement.

All of these non-idealistic become insignificant at lower currents. I think rather than a bench supply, I think you should try the experiment with a 1k or higher resistor.

]]>Then, from Kirchoff’s law, you know that the voltage across the mystery resistor and the voltage across the real resistor have to add up to 9v. So the voltage across the mystery resistor must be 3v, when the current through it is .6 amps, so the internal resistance is 3v/.6A = 5 ohms. Once you calculate the internal resistance, try other load resistors, and see if your results are consistent. But be careful of short-circuiting the battery! They can and do explode! ]]>