Antony Tersol wrote:
Assume 3 parallel battery stings, with identical batteries of internal
resistance r.
Assume identical wire leads of resistance R.
Then when the batteries have the same state of charge and their
internal resistances are equal, one can solve for the current in each
string.
I1 = I3
I2 = I1 r / (r+R)
where I2 is charging current thru the middle string.
The amount that the middle current is reduced is a function of the
relative sizes of the wire and battery resistances.
For R << r, I2 --> I1.
For r << R, I2 --> 0.
With a buss bar arrangement, I1 = I2 = I3 = V/(2R + r)
One thing this does not take into account though, is the inductance of
all the wires brought
on by the wire lengths and their loop area, which in a usual inverter
with ripple current, will also limit the current
through the system... That will also have some undesired consequences
with the voltages
seen at the inverter terminals due to resonances, etc.
In ~can~ be significant.
boB
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