Just what is the internal resistance of a 9V battery?

The classic way of measuring this is very simple in principle. First measure the open circuit voltage and then the short circuit current. The internal resistance is then simply V/I. In practice there are complications.

The open circuit voltage is simple and is easily handled by any meter. The current is more difficult. The first problem is that it really isn't a short circuit current because the meter is actually measuring the voltage across its internal shunt: a very small value resistor. The resistance of this shunt is naturally not zero.

In addition it is likely that the internal resistance is not constant. It will increase as the battery ages and as it is used. Also the resistance may vary with the amount of current drawn. Accurately characterizing the internal resistance would be tricky.

So instead I wimped out. I consulted Figure 15. of the Briody paper. This shows the voltage on a 9V battery when a N28B electric match is connected. Knowing the nominal resistance of the electric match it is then simple to compute the internal resistance of the battery.

The voltage across the N28B is 5.06V and its nominal resistance is 1.6 Ω so the current is 5.06V/1.6Ω = 3.16A. The voltage drop in the battery is 9.625V - 5.06V or 4.57V. Since the current is the same the internal resistance is 4.57V/3.16A or 1.45Ω.

I rounded this up to 1.5 Ω for simplicity.

I could have pulled out the 1 Ω 25 Watt 1% resistor that I own and measured the voltage across it but this seemed simpler and accurate enough.