How do remaining two phases act as return path in three phase system

Since they are out of phase one against the other then there is a potential difference between them and the one that is on the higher potential is enclosed in the lower one.

This is my assumption, so I'm wondering if it is true and if it is not how the rest of the phases at a certain point become current return path.

With the three phases being out-of-phase with each other by 120°, there can never be all three phases at the same voltage. So if, at any given instant, L1 and L2 are at the same voltage, then L2 and L3 will be different, and L3 and L1 will be different.

Three-phase loads normally consist of three elements - heaters, motor windings or whatever. So there will always be a current flowing through at least two of them.

A current will always flow if you connect a load between two points that are at a different voltage. It doesn't matter whether you choose to call them "phases" or "neutral", the electrons don't care.

Since they are out of phase one against the other then there is a potential difference between them and the one that is on the higher potential is enclosed in the lower one.

We don't talk about phases enclosing each other.

I find the easiest way to t think about it is to look at the instantaneous voltages at any time on the 3-phase diagram, see which phase voltage is highest and then to realise that current must flow from that phase to the other two.

Figure 1. A three-phase recitifier.

The three-phase rectifier of Figure 1 provides a good example.

• As shown, D5 and D4 are conducting so section '45' of the voltage vs time graph is relevant.
• In this we can see that the green phase is the highest so it supplies the current to the load, R.
• What isn't so clear from the graph is that the return current will be through the most negative of the other two phases. For the first half of that time period that will be the red phase but then the blue will take over.