[RE-wrenches] Strings and series of batteries with reverse return linkup

[Antony Tersol]

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)
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