Maximum diversity help

[schoe]

Hi all I'm struggling a bit with the calculations for max diversity on lighting circuits using the 66% of total current demand.

Could someone help me work these out

Three circuits

17 x 50 watts = 850 divided 230 x 66 - 243
10 x 50 watts = 500 divided 230 x 66 - 143
5 x 50 watts = 250 divided 230 x 66 - 72

Can anyone tell me if these calculations are right as they look to high to me, and if not how do you work them out. Are the figures 243, 143,72 amps?

Thanks as always chris

 

[Rampage]

Using the first one as an example:
17x50 watts = 850 watts,
850/230 = 3.695 Amps
3.695 Amps x 0.66 (66% diversity) = 2.439 Amps after diversity.

You were multiplying by 66 for the diversity, it should be 0.66, as that is 66%, 0.5 is 50%, 0.25 is 25% etc.

Work the rest out yourself as its the best way to learn

 

[Deleted member 26818]

I take it that these are down lighters, not bayonet?

 

[schoe]

Yep down lighters