At 18:51 4/3/2000, BRENT CHRISTENSEN wrote:
>My impression has been that when running the charge gets "soaked up"
>(excuse my use of layman's terms) by the battery that needs it (i.e. has
>the lowest voltage).  Is this not the case?

Pretty close.  Since the batteries are connected in parallel they will
always have very nearly the same voltage, which in this case will be the
charging voltage; but more current goes to the one with lower charge as its
internal resistance (for charging, not discharging) is lower.

>   My limited experience with electrical issues has been that in most
> cases electricity behaves just like water, and will follow the path of
> least resistance,

Like water it will follow all paths that do not have infinite resistance,
but the amount is inversely proportional to the resistance.

>  as well as flowing into the "most empty bucket"

Yes, if you stipulate that the emptiness of the bucket is represented by
the EMF (electromotive force, i.e. voltage) exerted by the electrons trying
to escape from the bucket.  An emptier bucket has proportionately less
voltage opposing current flow into it, and will accept proportionately more
charging current.  This is a very simple electrical equation if the bucket
is a capacitor; if it's a battery the energy storage is by means of an
electrochemical reaction and explaining it gets a lot more complicated.

Note that once you remove the charging current, if you leave the batteries
connected together the one with higher voltage will feed the other until
they are equal -- that's why you don't parallel an old and a new battery,
or a large and a small one, except for charging purposes.

:)
david

David Beierl - Providence, RI
  http://pws.prserv.net/synergy/Vanagon/
'84 Westy "Dutiful Passage"
'85 GL "Poor Relation"