This is excerpted from another thread, and as I read it, I think I stumbled onto a way to measure the internal resistance of a battery...

Quick test: A battery of 4 AA NIMH Duracell 2650 mAh cells into 1Ω + two clipleads gave 3.5 A- plenty for one Igniter. 12V should do even slightly better; would still add a 3Ω resistor to protect relay. # 22 twin lead is about .8Ω for 25 ft.

With a perfect voltage supply, he should have gotten 6A, but only got 3.5A. So total circuit resistance is 1.7Ω, the other .7Ω attributed to the batteries. 0.7Ω/4 = 175mΩ per cell.

Is that correct? Is that how one tests for battery internal resistance?

Is there a better to do it, when the battery may be capable of several hundred amps and/or may explode if you use too low resistance? Imagine testing 48V 5AH 130C hobby lipo pack - scary, and where do you get a 30KW resistor to test it? There has to be a better way.

 

This document says a AA NiMh battery has "less than 50 milliohms" of internal resistance.
I haven't measured it, and I expect some off-brands won't do as well, but they sure are better than alkaline batteries in a photoflash camera because they can dump serious current for short periods.

 

A common way to measure the battery internal impedance is to measure the output voltage both open circuit and with a known load. From that you can determine the internal resistance with a little algebra for two resistors in series (one of which has the known value). You use a resistor value low enough to give reasonable measurement error (probably at least 10 times the anticipated battery internal resistance), but high enough to avoid seriously overloading the battery.

 

http://data.energizer.com/PDFs/BatteryIR.pdf

 

I screwed up strantor's figures a bit by not including battery box which has 4 spring contacts worth .25Ω's each; not a fixed figure because as battery cells are added some spring turns are shorted giving a smaller figure.