# How is it possible to use a phase as a return path?

On the 3 phase system, it is possible to supply a 380V load using 2 phases, say L1 and L2. I understand that the phases alternate and that one phase acts as a return path for another, but this doesn't seem to always be true when looking at the graph.

Lets say I supply a 380V coil with only the Red and Purple phases, at time 0.1 both will be in the positive half. Now, does this mean that there wont be a return path at time 0.1? What will actually happen to the load at the instant where both phases are either in the positive half or negative half? • All 3 phases are low impedance sources so differential power is available at a high RMS than Y-N Jul 8, 2018 at 17:12
• At time 0.1 what is the voltage across the coil? Jul 8, 2018 at 22:41 Figure 1. Voltage on red phase relative to purple phase. The lower set of arrows shows the resultant 'waveform' samples referenced to a straight X-axis.

Now, does this mean that there wont be a return path at time 0.1?

Have a closer look. The red and purple cross at 1/12 (not 1/10) which is 30° into the purple cycle.

What will actually happen to the load at the instant where both phases are either in the positive half or negative half?

All that matters is where they are relative to each other.

Note also that the phase to phase voltage is 30° ahead of the red phase. simulate this circuit – Schematic created using CircuitLab

Figure 2. The red-purple phasor (green) is 30° ahead of the red phasor.

• Oops! At least I said it was 1/12 of 360°. Can I still get marks for method even if the answer is wrong? Fixed, thanks, Andy. Jul 8, 2018 at 17:52
• +1 for sheer audacity LOL Jul 8, 2018 at 17:59

@Transistor's answer is correct, but I wanted to add a preface to it.

What will actually happen to the load at the instant where both phases are either in the positive half or negative half?

The main confusion seems to be the misconception that the sign of the voltages matter. However, current flows from higher voltage to lower voltage; doesn't matter if they're both positive or negative.

The current does stop at the instant when the voltages are equal, and then flips immediately afterward, just as in the 3-phase case:

Then we can look at the effective voltage difference by subtracting one phase from the other. And this is where @Transistor's answer starts.