I have always assumed, through school and otherwise, that the neutral of a single phase 120/240v will always carry the unbalance of the two line currents. For instance if line 1 was carrying 5 amps and line 2 was carrying 10 amps the neutral current would be 5 amps. If lines 1 and 2 were both carrying 10 amps the neutral would be carrying 0 amps.

However I had been asked a question that really made me think about this. What if line 1 was carrying a pure capacitive load drawing 5 amps and line 2 had a pure inductive load drawing 5 amps. Would the neutral current still be 0 amps?

What about if line 1 had a pure inductive load drawing 5 amps and line 2 had a pure resistive load drawing 5 amps. Would the neutral current here be 0 amps as well?

Thanks for all information you can provide.

Dave
The current in the neutral will be 0 amps if the currents are equal and opposite polarity as is the case when both loads,line 1 and 2 hot to neutral,are exactly the same(resistive or capacitive/inductive or a like combination).
If the load on one side is reactive and the other side is resistive the currents do not cancel to zero in the neutral because the current leads the applied voltage by 90deg for a capacitive load and lags the applied voltage by 90deg for an inductive load.Fortunately not all loads are purely reactive but have some resistive component which does allow a portion of the current to balance out.
This also brings up the subject of harmonics created by some loads(computer switching power supplies)that do not get canceled in the neutral because there is no equal and opposite phase(180deg) current from the other side.
The answer to case 1&2 is the neutral current will not be zero for the reasons stated above.
Chris

Chris,
There are no documented reports of excessive neutral current in the neutral of 120/240 volt single phase systems with harmonic loads. These currents, do for the most part, cancel in 120/240 single phase systems. The only neutral problems caused by nonlinear loads are on 120/208 and 277/480 volt 3 phase wye systems. As a matter of fact the easiest and cheapest way to eliminate the neutral problems caused by nonlinear loads is to use 120/240 single phase systems in place of the more commonly used 120/208 volt wye systems.
Don

Taking the neutral as a reference point, the voltage on hot legs L1 and L2 will be 180 degrees out of phase.

Assume pure capacitance on the L1 side and pure inductance on the L2 side. Current in L1 will lead voltage by 90 degrees, while current in L2 will lag the voltage by 90 degrees. Because the two voltages are 180 deg. different to begin with, that puts both currents in phase and therefore they must add in the neutral.

If one leg stays purely reactive while the other is pure resistance, then the L1 and L2 currents will be 90 degrees out of phase with each other, and thus the neutral current will be reduced.

Don,
I did not mean to imply that there would be excessive current in the neutral because of non linear loads on one side.I was trying to make the point(not very well I guess)that the neutral is the reference point for many measurements and could be corrupted by these non linear loads.
Paul,
Good point about the current doubling(as compared to the line current) in the neutral when the load is capacitive on one side and inductive on the other because of the voltage polarity and the load phase lead/lag relationship.
Chris

Let me open up this discussion a bit further.
Think I'm missing something here. How does one of the basic transformation formulas, Ip*Np=Is*Ns, applies to single-phase transformer with 120/240 V output?
I guess that secondary side has two windings wraped around the core in opposite directions?

If that were the case the two 120 volt voltages would cancel. In distribution transformers the 240 volt secondary is simply center-tapped.
Control transformers have two separate 120 volt windings that can be connected in parallel for 120 volt output, or series for 240.

Ed

Dan2,

Remember, when applying the formulas for transformers, you must keep the references points equal. If you are refering to line-line currents on the primary then you will have to use the line-line currents on the secondary.
The addition of a tap (a mid-point one is a neutral) on a single winding secondary does not change the formulas.

A reconnectable two winding secondary does not change the formulas either, but you now have two values that must be added (normal series connection) or subtracted (i.e. buck connections) depending on the desired output.

Dave posted theoretical question where load is PURE, thus no problem with harmonics and resistive fraction of load.
I agree with "pauluk" with add inductive current and capacitive current for total 10A, but add capacitive 5A and resistive 5A is as sum of two phasors shifted 90 deg. (diagonal of square) and total current is 7.07A shifted 45 deg.
Andre