What will reverse polarity in an Energizer AAA, unpowered, dead battery, circuit?

Question

Circuit is a three-LED reading lamp. Batteries were left in for "a long time, but it worked last time I used it" according to Mom. Unknown if switch was 0 or 1. All three batteries were inserted correctly. Spring,-|+,post; in device. Tested one battery with a Sperry BT500. Black to Neg, Red to Pos. Needle jumped in the WRONG direction. Reversed leads. Needle swung to middle of "REPLACE" range on analog scale, and slowly started dropping further from the "GOOD" range of the scale.

I had searched on Startpage for "Energizer "AAA" reversed polarity", and the first link in the list was "Possible causes for AAA battery to explode".

Mine didn't explode, there were three in the circuit (correctly), no IR, no charging capability.

Answer 1

If one battery has less capacity than the others then it will drop to zero faster and may reverse since the current through the load is going through the battery in reverse.

In other words, it's possible that two good batteries reverse charged a weak one.

This would be particularly likely if Mom replaced two of the three batteries and left an old one in there (perhaps because that was all that was at hand).

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

ESR means Effective Series Resistance which causes Δ voltage / Δ current = ESR

Δ is greek symbol for D as in delta as in difference.

Assumptions.

You want to learn Kirchoff voltage laws (KVL) where all voltages in a loop ( going in same direction) add up to 0.

Note that in my assertion, dead batteries rise in ESR and therefore cannot dissipate much power even if reverse voltage.

This is how we roughly model all batteries where C and ESR vary with Ah and V.