Thanks Daniel!
On Monday, July 6, 2015, Daniel Young wrote:
> Hey Gary,
>
>
>
> Basically the area of the dropping will lower the current of that module
> (and the current of that string if this is a string inverter). The amount
> of current loss will be the % of the dropping area vs the area of
Hey Gary,
Basically the area of the dropping will lower the current of that module (and
the current of that string if this is a string inverter). The amount of current
loss will be the % of the dropping area vs the area of the cell it landed on.
Voltage won’t be affected by things the scale
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