Three identical coils each [4.2 + j5.6] Ω are connected in star across 415 V, three-phase, 50 Hz, AC supply. Determine (a) the line current (b) the power factor and (c) the total power.

\( Z_{ph} = 4.2 + j5.6 = \sqrt{4.2^2 + 5.6^2}\angle\tan^{-1}(5.6/4.2) = 7\angle53.13^\circ\ \Omega \).

Star: \( V_{ph} = \dfrac{415}{\sqrt3} = 239.6 \) V.

(a) Line current: \( I_L = I_{ph} = \dfrac{239.6}{7} = \) 34.23 A.

(b) Power factor: \( \cos\phi = R/Z = 4.2/7 = \) 0.6 lagging.

(c) Total power: \( P = \sqrt3V_LI_L\cos\phi = \sqrt3\times415\times34.23\times0.6 = 14762 \) W ≈ 14.76 kW (check: 3 × 34.23² × 4.2 = 14.76 kW ✓). (Reactive power = 19.68 kVAR, apparent power = 24.60 kVA.)