MWBC and very low power factors

The principle of an MWBC is that two hots can share a neutral if they're on opposite phases, since the current on the neutral will cancel instead of adding. But isn't this only true for loads with reasonable power factors? Consider if one phase of an MWBC had a load that was nearly purely capacitive, and the other phase had a load that was nearly purely inductive. The voltages are 180 degrees out of phase, but the capacitive load's current is pushed almost 90 degrees ahead of its voltage, and the inductive load's current is pushed almost 90 degrees behind its voltage. Wouldn't this situation result in the currents adding up after all, and potentially overloading the neutral wire? Are MWBCs considered safe just because loads with such low power factors are nearly nonexistent in practice, or is there something else that prevents this from being a problem?

Even with your power factor shift it is not that large over all -- most motors have a power factor of 80 or better. The vast majority of loads are resistive or inductive not only in residential but industrial also. Yes, there are some high capacitive loads, but they are rare. Harmonics actually are more of a concern for multiwire branch circuits on 3-phase systems as the harmonics can cause the neutral to become overloaded. I have measured this but rarely found it to be at a hazard level.

An example of this would be 23 amps measured on a #12 neutral. Yes, that is overloaded but there is a huge safety factor. You can see this by seeing that a controls cabinet wiring 12 awg can handle 60 amps beyond the enclosure 120 amps inside (now that’s crazy). To see these values with your own eyes NEC table 430.72.B. We try and limit the harmonic overloading -- in many cases it is only at startup when it’s the worst then it drops down to normal levels.

I have never found a problem with multiwire branch circuits being overloaded on the neutral and prior to GFCIs being required everywhere I used to use them all the time. So yes, in theory you could overload the neutral but with the safety factors built into code reality is you won’t overload split phase. Harmonic overload is very possible on 3-phase but not normally a hazard on branch circuits.

Theoretically -- and with determination, practically -- yes, you are absolutely correct that you can overload the neutral with reactive loads. The mechanism is exactly as you describe.

Let's find the values for a worst-case scenario, where your neutral current is twice as large as both your phase currents. Assuming both your phases only have reactive loads on them, one inductive (with "L" Henries), one capacitive ("C" Farads), you can do some complex math to arrive at:

L * C = 1 / w^2

...where w is omega in rad/s and is equal to 2 * pi * f, and where f is frequency in Hz -- say 60 Hz for North America. In other words:

L * C = 1 / (2 * pi * 60)^2 = ~7.04E-6 s^2 rad^-2

Your job at this point is to find inductor and capacitor values that fit the equation and put phase currents between 50% and 100% of the main breaker trip level.

To find current levels, know that inductor impedance magnitude is |Z_L| = 2 * pi * f * L. For capacitor impedance it's |Z_C| = 1/(2 * pi * f * C). Take your voltage magnitude (110 or 120 V) and divide by your impedance magnitude to find current magnitude. Again, keep this between 50-100% of your main breaker's trip level.

Once you have your L and C values, you will have to create that particular inductance and capacitance on each of the two legs in the house by utilizing all the small 15 A circuits in a way that will not overload each circuit nor the reactive devices themselves. This will likely involve distributed parallel inductors on the first phases's circuits and likewise with capacitors on the opposite phase, in an attempt to create the overall L and C values. Remember, inductance adds in series and capacitance in parallel. Depending on the sizing of reactive devices practically available, you may also need to throw in some resistors to really hone in on whatever current magnitude you're shooting for.

If you do the above, then your main neutral current will be twice as big as both your phase currents and between 100 and 200% of its current rating. If you manage to load both phases to 100% then your neutral will be at 200% and would likely melt its insulation at the very least, if not start a fire.

Your answer is absolutely "yes", but it's so impractical for most people to accomplish and would never happen in any practical situation that there is no need to protect against it. There are easier ways to burn down your house!

As the other comment mentions, the most practical way neutrals are overloaded is with nonlinear (i.e. harmonic or noise-injecting) loads; however, it's still the same mechanism at play. In the case of harmonics, it's possible that a certain frequency band of current is inductive on one leg and capacitive on the other, or in the case of noise that the randomness of phase (i.e. randomly inductive and randomly capacitive) causes the currents to add going into the neutral.

• Wire current ratings are set based upon worst-case heat dissipation scenarios. If one had a long wire with six inches of fiberglass insulation wrapped around it, the amount of current required to make the insulation overheat would be far less than in anything resembling a typical insulation (bear in mind that because copper is a good conductor of heat, a short stretch of cable that's buried in insulation could conduct heat to the nearby parts). If the neutral wire generates 4x normal heat while hot1 and hot2 generate 1x, the cable as a whole would only produce 6x/3, i.e. 2x, and... Commented Feb 7 at 16:25
• ...unless it's buried in mountains of insulation along it's entire length it would seem unlikely to pose an actual danger even under the extreme scenario posited. Commented Feb 7 at 16:27
• It doesn't really make any sense to think of harmonics in terms of "inductive" or "capactive". The problem with harmonics is that they don't always cancel in the same way the fundamental does. Commented Feb 7 at 22:00
• For the fundamental we have "sin(ωt) + sin(ωt + ⅔π) + sin(ωt - ⅔π) = 0" but for the third harmonic we have "sin(3ωt) + sin(3(ωt + ⅔π)) + sin(3(ωt - ⅔π)) = 3sin(3ωt)" Commented Feb 7 at 22:06