Can you this question about wattmeters connected to 3-phase generators?

[Wall-ED]

Two wattmeters are connected to lines A and B of three-phase three-wire ABC system with a line voltage of 208v feeding a balanced Y-connected load. The wattmeter readings were Wa= -200W and Wb=1000W. Find the line current and the load impedance.

 

[Diver300]

The total load is 800 W. As the load is symmetrical, the power is 800 / 3 W in each phase, so 267 W.

The magnitude of the current is the same in each line, and the voltage is 208 * sqrt(3) = 360 V for each wattmeter. The difference in power is due to the difference in phase. One phase is 30 degrees in advance and the other is 30 degrees behind what they would be if they were connected to the neutral point.

If the phase lag is Φ, then the power is V * I * cos(Φ) for each phase
so
V * I * cos(Φ + 30) = -200
V * I * cos(Φ - 30) = 1000

Working that all out:-

cos(Φ + 30)/cos(Φ - 30) = -0.2

(cos Φ * cos 30 - sin Φ sin 30) / (cos Φ * cos 30 + sin Φ sin 30 ) = -0.2

√3/2 cos Φ - 0.5 sin Φ = -0.2 ( √3/2 cos Φ + 0.5 sin Φ)

√3/2 cos Φ - 0.5 sin Φ = - 0.1 √3 cos Φ - 0.1 sin Φ

0.6 √3 cos Φ = 0.4 sin Φ

sin Φ / cos Φ = 0.6 * √3 / 0.4

tan Φ = 2.598

Φ = 68.95 degrees

so the power in each leg is 267 W and the current lags the voltage by 68.95 degrees.
(It could lead by that amount, but I'll assume lagging)

The current is 267/208/cos(68.95) = 3.574 amps

The load impedance in each leg is 208 / 3.574 = 58.2 Ω

That is the magnitude of the impedance. If that is to be split into real and imaginary components, the real impedance is

58.2 * cos (68.95) = 20.9 Ω

The imaginary impedance is 58.2 * sin (68.95) = 54.3 j Ω

So the total impedance is 20.9 + 54.3 j Ω for each leg of the Y load.