Salvo wrote:ktmrfs wrote:
Post #5
For series, cable power loss is I^2R, or V*I or V^2/R. To satisfy Salvo I’ll use V*I to calculate the power loss.
You haven't understood the previous discussion. I don't care how you calculate cable power loss. That's never been an issue.
Note that the parallel case with double the current has 4x the power loss, not double the power loss. (11.6W vs. 2.9W)
Abain, that's never been the issue. If you double current, cable power loss will be 4 times has high; no matter what configuration you use.
The intelligent way to figure out what size cable is needed is to use a 1% or 2% voltage drop rule of thumb. Whatever percent voltage drop is used, battery charging current will drop by same amount, not 4 times that amount.
Sal
