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

[boB Gudgel]

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