I?ve only just read this thread and there are a lot of opinions out there. The first question that would need to be answered is how do you know the load is 2228W? Is that nameplate data? Based on what type of load connection? First of all, if your load is connected directly to only two phases of the 120/208 3-phase system (no connection to neutral at all), that is, the load consists of two connection points, think of it as a single resistor with leads at each end, connected directly across two of the phases, what you have is a simple 1-phase system. The line and phase currents are the same because what goes in one side of the load has to come out of the other and since there is no other branching of the current all of it returns to the source. Thus, alternative 1 is the proper one; 208V appears across the load, divide 2228W by 208V = 10.7A. Furthermore, since the load is resistive (is it a heater?) the voltage and current are in phase, and any discussion that involves the sq.rt of 3 is wrong. This is not a 3? circuit at this point. Also, it would be wrong to bring in any calculations involving 120V because the neutral does not play any role in this circuit. It is a simple resistor connected across 208V.
That said, things get interesting when 3 of these loads (assumed equal) are connected to all three phases, as they might in a high wattage application. The loads can get connected either in ∆ or in Y. In ∆, since each load sees the same situation as above, the total power is 3X the individual power, or 3X2228W=6684W. The current through each load will be the same, that is, 10.7A. However the current through the line is the vector sum (now we do have to worry about the polyphase situation) of the currents through two of the loads. Consider the 3-wire node at any apex of the ∆. These currents are 120? out of phase from each other. Vector addition results in Line current = Load current X sq.rt of 3, or in this case 18.53A. This is the current that will flow in each of the three circuit breakers as a result of this load. Now you will remember that we can calculate the total load by P=(v)(I)(sq. rt of 3), or in this case, 208x18.53xsq.rt of 3=6684W, identical to the result we got above by multiplying each load by 3.
Connecting that same load in Y configuration, however, yields totally different results both for load current as well as load dissipation (which is why I asked on top how you knew that the load dissipation was 2228W.)
Please let me know if you need more.
Heinz R.
