You might want to try measuring the voltage with a light
load, call this v1, then again with a load, call this v2.
If you assume v1 is the open circuit voltage, then v2
is the loaded voltage, and the difference appears across
the internal resistance. Knowing the current with the
heavier load and you can calc the Ri.
A better way is to apply a low ac signal and measure
phase and amplitude, then from this info calculate
the resistance and theoretical capacitance.
In other words, you look at the battery as being a
big capacitor with an unknown series resistance and an
unknown capacitance and try to calculate both from an
ac current perturbation.
I've used standard line frequency 60Hz in the past
to compare rechargeables.
MR. Al's and gwbaltzell suggested very good approaches. My humble approach for a ballpark number is to put 2 identical batteries 'head-to-head' and measure the total resistance between the two negatives with my ohmmeter. This requires identical voltages on each battery and I've only done it with analog meters with high input impedences (one megaohm per volt or more, IIRC).
It's been a while since I've done that - now I mainly use MrAl's first method with two meters, one for voltage and one for current.
[ QUOTE ] IlluminatingBikr said:
[ QUOTE ] evan9162 said:
AA NiMH should give about 25 milliOhms of resistance.
[/ QUOTE ]
So they can supply 60 amps? [img]/ubbthreads/images/graemlins/confused.gif[/img]
[/ QUOTE ]
Yes, but not for long! Remember resistance increases as the cells discharge. Most manufacturers rate their product in terms of sustained drain. This is normally as a multiple of the 5 hour capacity. Typical ranges are 2C to 5C. The actual capacity is significantly derated at these drain levels.
Even cells/batteries drained (preferrable to 1.0V and not below 0.8 V per cell) can still deliver a high current and pose a burn/fire hazard if shorted!