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Remind me...current flow on single ended 3 phase loads

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hrc

Electrical
Nov 8, 2001
104
I am pretty sure we discussed this before....but I was trying to run thru my notes and did a search but didnt find what I was looking for.

Given a 208wye system, leg to leg connections to a load (single phase load). Each load draws 14A. So, if I have 3 loads, balanced across the legs (AB,AC,BC) each LEG will have a total current draw that is additive so 28A would be the draw on each leg. Correct?
 
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Use kVA. 208 V x 14 A = 2912 VA per phase. x 3 phase = 8,736 VA.

8.74 kVA at 208 V: I = 8736/(208 x 1.732) = 24.24 A.

Remember youare adding currents vectorially.
 
OK..that makes sense. I've done some measuring but at lower currents and the delta difference is harder to see there, but makes sense.

Thanks again guys, appreciate the help
 
Just thinking....what about unbalanced loads?

say 12A on one leg, 14A on another and 16 on the third?

A 208*12=2496
B 208*14=2912
C 208*16=3328

So AB = (2496/(208*1.732)) + (2912/(208.1.732)) = 15A

??
 
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