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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