"Meindert Sprang" wrote in message ...
"James Hahn" wrote in message
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"Meindert Sprang" wrote in message
...
snip

There are two possible scenarios: one battery wears more quickly than
the
other, eventually reaching a state where the cell voltages reach higher
values due to increasing internal resistance. Thus the charges shuts off
too
early, leaving the better battery not topped up.
The charger does not monitor cell voltages. It monitors the voltage
across
all the cells in series.

That is exactly what I meant and exactly the reason why it can go wrong. So
I would like to invite you to shoot holes in my following argument:

The charge of a battery is the product of current x time.

Both batteries are in series and one load is connected to the set, operating
at 24V. Another load is connected across only one battery, operating at 12V.
So it is evident that one battery is discharged more than the other.

I start to charge the set in series, so the current through both batteries
is exactly the same. Since one battery is discharged more than the other and
the current throug both is the same, one battery must be charged longer that
the other. Exactly how am I going to achive that with the same current
through both batteries?
One battery will reach the full state before the other but is still being
charged with full current because the other battery hasn't reached the
voltage that corresponds with full charge.

Now, if you can come up with an valid argument why this shouldn't damage one
battery, I bow and take my hat off.

Meindert

Duh! the two batteries are in series. The one with the 12v tap cannot become
more discharged than the one without because the battery without the tap
will charge the one with. Or the one with will discharge the one without
until they are both the same. This is the same thing that happens with
individual cells in a 12v battery. With two 12v batteries in a series one
simply has more cells in a series.

CN