Rick,
Please note that there are now two discussions going on. The original poster's discussion of two phase to neutral loads on a three phase wye system (as I understand his post; I could be wrong) and how to calculate the neutral current in this case. The second discussion is what happens when you have a single phase load connected between two phases of a three phase wye system.
For the latter discussion: It is entirely reasonable and correct to consider this to be a single phase load supplied by two voltage sources connected in series. However it is essential when doing the addition to consider the relative phase angle of these two voltage sources. If the two voltage sources are in phase, then you can simply add the two voltages together, and you have the 'split phase' system that you described. But if there is any phase angle between the two sources, then the resulting total voltage will be less than the sum of the individual voltages.
In a conventional 120/208V wye system, the voltage produced by the individual transformer coils are _120V_. I connect these two 120V sources in series, and low and behold I get 208V placed across my single phase load. Since both sources are in series with the load, both sources supply the full load current. This means that the total VA supplied by the two sources is greater than the total W consumed by the load. Entirely irrelevant as far as the load is concerned, but clearly an issue when sizing the sources.
To the original poster: in the example diagram that you just gave, you have 400W at 120V on one leg and 150W at 120V on a separate leg. As you've drawn things, this could either be a single phase split system (240V split to 120V each leg) or a 208V wye system (208V three phase with 120V each leg). In the latter case, use the formula that JBD posted at the top of this thread. First calculate the current in each phase, then plug those number in to the formula. The result is rather less than 4.7A.
-Jon
