Trick, the A,B,C are the currents in phase A, Phase B, and Phase C.

So for a split(single)phase service, the nuetral current is the unbalnced current flowing between phase A and phase B.

So if 30 amps were flowing on phase A, and 15 amps were flowing on Phase B, the nuetral current would be: Yes it is a minus...

A-B=N 30-15=15

On a wye service, if we had 150amps on A, 110amps on B, and 75amps on C.

N= Sq root of (A^+B^+C^)-(AB+AC+CB) N=Sq root of (150^+110^+75^)-((150*110)+(150*75)+(75*110) N=Sq root of (22500+12100+5625)-((16500)+(11250)+(8250)) N=Sq root of (40225)-(36000) N=Sq root of (4225) N=65 amps

How does that look?

[This message has been edited by Dnkldorf (edited 07-28-2006).]

All the above assumes that all loads are pure resistors, like incandescent light bulbs. If we talk about a computer server farm fed by 3 phase 120V, things get more complicated. Computer power supplies draw large spikes of current at the very peak of voltage of the phase they are connected to. When the voltage on phase A peaks, computers on phase A draw their spikes of current and return them to the neutral. Later, 120 degrees of the 60Hz cycle, phase B peaks, and computers on phase B draw their current spikes, and return them to the neutral. Likewise, 120 degrees later for phase C. No overlap of these current spikes (in terms of time). Spikes last about 5% of the cycle time. Note that the neutral will see 3 times as much spike current than any one phase (assuming equal phase loading). Thus you'd need a thicker wire for the neutral.

[This message has been edited by wa2ise (edited 07-30-2006).]