Sylvia Else wrote:

Anyway, all a test would do is show that the answer is probably correct.
It wouldn't make it any more obvious.

Sylvia.

Sylvia,

Draw yourself a vector diagram. Then with some simple trigonometry it
*should* be more obvious.

Assume a resistive load between two of the three phases, with a load of
1 unit current and 1 unit voltage. The load will thus be 1 unit power.

Each single phase meter will see the in phase voltage as 1 / sqrt(3)
(240/415).

Each single phase meter will see an in phase current of 1 x cos(30).
Remember that cos(30) = 1/2 sqrt(3).

This gives the power measured by each meter as
V*I = 1/sqrt(3) * 1/2 sqrt(3)

The two square roots of three cancel, which, unsurprisingly leaves 1/2.
Thus each meter records 1/2 the power in the load, and you will thus get
billed correctly.

David