I have a probably obvious question about powering single phase loads from two legs of a 3 phase supply and how this impacts current draw calculations. This is something that is commonly done in my job (entertainment lighting) but in thinking about it today I have run into some confusion. The system I am working with is commonly 208v/120v 3 phase in Wye configuration.
208v * 200A * 1.732[sqrt(3)] gives me about 72,000 VA theoretical power output of the service. If I were to connect 200A of load on each leg to neutral (120v) I would be pulling 72,000 VA. 200A * 120v * 3 phases = 72,000. Firstly, is my thinking here correct? I understand not to think of this system as delivering 600A.
Now on to the real question. Say I am trying to figure out how many lights (dual voltage for 120v or 208v), that draw a resistive load of 1200w, I can power from the service. At 120v, this load is 10A. So therefore on this system I should be able to power 200A / 10A * 3 lights, or 60 lights. Now, the way I have been calculating and have been taught to calculate the load for 208v is by simply using ohm's law. At 208v, these should draw 5.77A. However, This only lets me power 200A / 5.77A lights, or 34.6 lights. If I instead calculate the current as 1200w / 208v / 1.732, I get 3.33A per light, which works out as 200A / 3.33A = 60 lights. Is it correct to apply the 1.732 figure for the difference in phase angle when using an unbalanced wye configuration and powering these effectively single phase loads from two phases?
Thank you in advance!