The given load is star connected load.
L = 0.02 H
Inductive reactance X = 2 × π × 50 × 0.02 ≈ 6.28Ω
R = 10 Ω
Impedance, Zph = 10 + j6.28 = 11.81∠32.13°
Phase voltage, \(V_{ph} = \frac{440}{\sqrt 3} = 254.03 V\)
Line current
\(I_{ph} = I_L = \frac {V_ph}{z_ph} = \frac {254.03}{11.81} = 21.51 A\)
Power in three-phase is given by
\(P = \sqrt 3 \times 440 \times \frac{440}{\sqrt 3 \times 11.8} \times \cos 32.12 = 13 .89 k W\)