[Physics] What does a positive gradient on a graph of V plotted against I mean in terms of EMF and internal resistance

electric-currentelectrical-resistanceelectricityexperimental-physicsvoltage

According to the equation $V = E-Ir$, the gradient of a graph of $V$ against $I$ should be $-r$ (internal resistance) and the Y intercept should be the EMF. Am I right?

In an experiment I used a constant wind speed to drive a small motor, keeping everything constant and changing the resistance in the circuit using a variable resistor, measuring the output in volts. How is it that the graph shows a positive gradient (and therefore a negative internal resistance ?).

The equation of the line is $0.0732934 + 0.0396359x$.

http://cl.ly/image/0b3i393E0E0G

Best Answer

Negative resistance is not uncommon. You see it in arc lamps, too. For your motor (used as a generator), I would guess that it is most efficient at higher current, possibly having field coils and not permanent magnets.

The voltage from a generator comes from moving a wire through a magnetic field.

With field coils in series, the magnetic field is generated by the current flowing through the (generator).

There is probably a small residual permanent field in the iron, so you get some voltage even when there is no current.

I bet if you could run up to higher currents, eventually you would get max power from the generator, and start seeing voltage go down again.

I am going on this being a 'universal' motor you have salvaged from somewhere. Read about universal or series wound motors at the motor wiki:http://en.wikipedia.org/wiki/Electric_motor

I found a discussion group with one sensible (to my mind) answer amongst the cruft (see engineertony's reply) here: http://cr4.globalspec.com/thread/77573/How-to-Make-a-Generator-from-a-Universal-Motor

My own take is, go for it! It kinda works, it looks like you can get real power from it, though the voltage is all over the map, you will want to regulate it.