I was viewing this from an "imbalance perspective". where if connected to a 1 ph multiwire branch ckt, vectorly the neutral amps would be 0, would it not?
To be clear, in my picture there are six amps flowing from A to Neutral, and five amps flowing from B to neutral, so in a single phase system the neutral current would be one amp. In a 208Y/120 system, there would be 5.56 amps flowing on the neutral.
Think of the three circles in my drawing as being three teams of people, say 100 people in each team. The "A" team has 50 individuals who want to run to "C", and 50 who want to run to "B". When the load establishes a path between the "A" team and the neutral, then the teams go where they want to go, using the neutral connection as a path, or bridge.
Coincidentally, the path from "B" to neutral is available at the same time in the picture.
Due to the resistance of the load (the width of the path), only six individuals are walking between "A" and "N". Only five are walking between "B" and "N".
In the picture, only the members of "A" team that want to go to "B" move, because the ones that want to go to "C" are waiting for their path to do so.
This is how I picture it, as goofy as it may sound.