Yes, that works, too. I should have realized in advance when I did my calculation that the resistors cannot change the phase relationships so the result can be calculated more easily, without complex numbers. My method would work for general impedances but that's not needed here. As for 220Vac, that is a standard (well, it's more like 230V) for North America, but that's for single phase power. 208Vac is the closest for 3-phase. I think the original poster meant that he has heaters which are intended for 220Vac and he is going to run them from 208. That is a fairly standard thing to do. Most industrial buildings obtain 120V for normal outlets by connecting them in a wye configuration with the center connected to neutral and the phases at 208V relative to each other. Note that the tolerances on three phase power seem (in my experience) to be wider than typical 120V circuits. I've seen 480V be as high as 510 and as low as 460 (about 6% off). Under normal circumstances, I've never seen 120V be more than 125V or less than 115V (4% off). Also, the OP may know this but all power distribution voltages and currents (residential or industrial) are stated as RMS values. This means that a 120V circuit has 120V RMS from hot to neutral, which is actually 170V peak from hot to neutral. Sean On Fri, Sep 4, 2009 at 12:36 PM, Dwayne Reid wrote: > At 07:29 AM 9/3/2009, Aaron wrote: >>I have 3 resistance heaters. =A0One is rated for 880 watts at 220 VAC and >>the other two are each rated for 400 watts at 220 VAC. =A0I am trying to >>calculate the theoretical currents for each phase if I wire the heaters >>in a delta configuration across a 3 phase supply. =A0Can anybody point he= lp? > > Easy. > > Lets call the 3 phase supply lines A, B, C. > > Find the current through each heater when connected across the > supply. =A0Call those currents Iab, Iac, Ibc where Iab is the current > of the load connected from lines A to B. > > Note that 220Vac is NOT a standard voltage in North America - its > 208Vac. =A0That will reduce the heater current (and output power) from > your rated 220Vac. =A0You are going to have to figure out the > resistance of each heater (Eg. R=3D (220V)^2 / 800W), then figure out > the current at 208Vac. =A0Or: find the current at 220V, then divide > that by the ratio of 220V / 208V. =A0Whichever is easier for you. > > Add Iab to Iac, then divide that result by 2, then multiply that by > SqRoot (3) (about 1.73). =A0That amount is the current in conductor > A. =A0In other words, (Iab + Iac) * 0.866 . > Add Iab to Ibc, then divide that result by 2, then multiply that by > SqRoot (3) (about 1.73). =A0That amount is the current in conductor > B. =A0In other words, (Iab + Ibc) * 0.866 . > Add Iac to Ibc, then divide that result by 2, then multiply that by > SqRoot (3) (about 1.73). =A0That amount is the current in conductor > C. =A0In other words, (Iac + Ibc) * 0.866 . > > I hope this helps. > > dwayne > > -- > Dwayne Reid =A0 > Trinity Electronics Systems Ltd =A0 =A0Edmonton, AB, CANADA > (780) 489-3199 voice =A0 =A0 =A0 =A0 =A0(780) 487-6397 fax > www.trinity-electronics.com > Custom Electronics Design and Manufacturing > > -- > http://www.piclist.com PIC/SX FAQ & list archive > View/change your membership options at > http://mailman.mit.edu/mailman/listinfo/piclist > -- = http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist