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!


It's a 3 x 208V, 200A source (√3 x 208 x 200 = 72kVA) with a phase voltage (line to neutral) of 120V (208/√3).

The total load would be 72kW with 60 lamps rated 120V, 1200W (3 x 20 lamps connected line to neutral) or 60 lamps rated 208V, 1200W (3 x 20 lamps connected line to line).

The total connected load in either case would be 72kW.

The load could be a mix of 120V and 208V lamps, provided they are equally shared between the phases (balanced load). The load could be unbalanced, provided 24kW per phase is not exceeded (lamps could even be lit between a single line and neutral or between 2 lines).

The error in your calculations stems from not considering 2 equal currents from each line, with a phase difference between them, when the 208V lamps are connected line to line. Those currents would be (1200 x 20)/208=115.4A. The resultant of those currents would be (115.4*√3)=200A.

  • \$\begingroup\$ Okay, that makes sense, thank you. If I were to load only two legs with 20 208v lamps line to line, then the effective current would be 100A on A and B legs, with 0A on leg C? \$\endgroup\$ – Peter C Jun 7 '20 at 13:17
  • \$\begingroup\$ Anytime, Peter! 115.4A to be exact. \$\endgroup\$ – vu2nan Jun 7 '20 at 13:58
  • \$\begingroup\$ Hi Peter, I would be glad should my answer be accepted. Thank you. \$\endgroup\$ – vu2nan Jun 26 '20 at 6:20

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