Quote:
Originally Posted by rallypanam
I don't understand this.
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Batteries die a slow, corrosive death....they basically rust* internally over time and recharge cycles.
*They actually form a sulphate layer...not exactly an oxide layer.....(I'm not a chemist..)
The chemistry (lead acid, alkaline, Ni-Mh, Li-ion, etc) determines the cell voltage....that's galvanic potential.
http://www.corrosionist.com/galvanic_co ... _chart.htm[youtube_id][/youtube_id]
When a battery ages, the plates corrode and form sulphate layers. Measuring an old near dead battery with a voltmeter does not draw any current (an ideal voltmeter has infinite resistance), so the small remaining non-corroded surface area gives the cell the correct voltage reading for no load.
However, if you put a significant load on the battery, a fresh battery has lots of clean plate surface area to allow zillions ions/electrons to flow producing lots of current with very little voltage drop.
In an old, crusty battery, the corroded plates have very little conductive surface area left, and simply cannot provide the zillions of ions/electrons needed to produce meaningful current. In this case, the voltage drops (sags) because there is just not enough ions flowing to keep the cell at it's design voltage.
Here is a simple mechanical analogy:
Lay out a garden hose and cap the end with a pressure gauge, turn on the faucet. The gauge reads 50 psi. Remove the gauge and lots of water flows.
Repeat the above but step on the hose until it's just got a tiny bit of area the remains open under your foot. Cap the hose with the pressure gauge and turn on the faucet.....you'll still see 50psi. Since no water is flowing the pressure is the same throughout the entire hose.
Remove the gauge with your foot still on the hose and you will get a mouse-fart's (there's that phrase again!
) worth of flow, since you are stepping on the hose.
Pressure=voltage and current=flow.
So, with a dead or almost dead battery, the cell voltages will be normal or close to normal under no load, but when you try to pull current out of it, there is just not enough conductive surface area left to provide sufficient current.