Hello. Does anyone know the equation I can use to find the load on an individual leg, assuming we're using 208v and the load is unbalanced? Hypothetically, say AB=50, BC=40, and CA=60. How can I find out how much I'm pulling on A,B, and C? Thank you.
This is true ONLY for switchmode power supplies.With the exception of some variation in powerfactor, wattage does not change with voltage, so just add everything in watts and divide by your legs.
QUESTION CLARIFICATION:Hello. Does anyone know the equation I can use to find the load on an individual leg, assuming we're using 208v and the load is unbalanced? Hypothetically, say AB=50, BC=40, and CA=60. How can I find out how much I'm pulling on A,B, and C? Thank you.
Dang, @JD couldn't you have found that in English, or at least a language all (some) of us could understand?Here's the long answer (prepare for headache): http://www.electrical-engineering-assignment.com/12-unbalanced-delta-connected-load
Applying @YesItWillWork 's formulae (isn't Excel the best thing since sliced bread ever?),... say AB=50, BC=40, and CA=60. ...
I would agree. Since we are dealing with amps, not watts, any PF effect is already baked into the amprage. So, unless we are on a cruise ship dealing with delta wired dimmers and 208/240 volt lamps (whatever!) we should all be able to put this in the knowledge bank and move on!I'm writing late at night, but I'm not convinced we need to constrain the maths to purely resistive loads.
If we broaden the constraint to all loads needing to share a common power factor, regardless of whether that is 1 (a resistive load), or something less than 1, then because we have a consistent voltage / current phase relationship on all loads, the 2 pi /3 shift between phases holds, and thus the formulas presented above hold true...
I'm writing late at night, but I'm not convinced we need to constrain the maths to purely resistive loads.
If we broaden the constraint to all loads needing to share a common power factor, regardless of whether that is 1 (a resistive load), or something less than 1, then because we have a consistent voltage / current phase relationship on all loads, the 2 pi /3 shift between phases holds, and thus the formulas presented above hold true...
It is true that the three legs would have to have the same type of power factor distortion for the formula to hold. A lower than 100 power factor can have several causes, for example, a magnetic ballast would have a classic phase lag in ampacity, whereas a standard phase-chopped dimmer would score its low power factor in a different way and vary all over the place depending on the dimmer setting, and lastly, a computer power supply would score its power factor due to the poor distribution caused by the fact that a load only exists when the diodes are in forward conduction (Line voltage greater than supply cap charge.)
In our line of work, if the loads were ALL magnetic ballasts, or ALL electronic ballasts, the formula (in my opinion) would hold. If you were to mix and match, the results would vary.
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