2 Phase Power

P = 2(220V*22A) = 9680W

Question for you. What is the amperage on the neutral?

Thanks for your answer, Waregle. Current should be both neither 22A nor 44A (if there were 3 phases and 3 equal loads then curr = 0). I always have had difficulties with these phasor calculations.

In=(1/√(Ia²+Ib²))*100

Ampers in the neutral should be about 31A.

Then ,

P = 1.41*220*31?

I always thought we multiply the square root of 2 with line voltage and phase amperage (just like the sq root of 3 for 3 phase load) - now it seems we should multiply it with phase voltage and neutral amperage?

Since these currrents are 120 degrees apart, the neutral current must be added using vectors.

In = Ia @0 degrees + Ib @ 120 degrees.

In = 22 +J0 + (-22 cos 120 + 22 sin 120)

In =22 + 0 - 22 x .50 + 22 x 0.866 = 22 amps at 60 degrees.

Are we talking about phases at 120° or 90°

Did not know it is so complicated. Thank you for answers. But, in any way, IS the power consumed 9680 W ?

The voltage give appear to be a wye connection.

Wye with neutral - 2 shoulders of which have each a resistor 10 Ohm. Phase to phase 380 V, phase to neutral 220 V.