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Neutral current and power, 380 V unbalanced load

Short answer

Phase voltage 219.4 V; phase currents 21.94 A, 14.63 A, 7.31 A. Neutral current magnitude is 12.67 A. Total power consumed is 9.63 kW.

For the circuit shown below, calculate the current through the neutral and total power of the circuit. [Figure: 380 V balanced source, star connected; load impedances 10 Ω, 30 Ω and 15 Ω in the three phases; neutral current IN.]

From the figure: a 380 V balanced three-phase star source feeds a star load with resistors of 10 Ω, 30 Ω and 15 Ω, and a neutral wire carrying I_N. Assume R–Y–B sequence with R = 10 Ω, Y = 15 Ω, B = 30 Ω. (The magnitude of I_N and the total power are the same whichever way the three resistors are assigned to the phases.)

V_ph = 380/√3 = 219.4 V. V_R = 219.4∠0°, V_Y = 219.4∠−120°, V_B = 219.4∠120° V.

Currents

  • I_R = 219.4/10 = 21.94∠0° A = 21.94 + j0
  • I_Y = 219.4/15 = 14.63∠−120° A = −7.31 − j12.67
  • I_B = 219.4/30 = 7.31∠120° A = −3.66 + j6.33

Neutral current

I_N = I_R + I_Y + I_B = (21.94 − 7.31 − 3.66) + j(−12.67 + 6.33) = 10.97 − j6.33 A

|I_N| = 12.67 A (∠−30°). This magnitude is the same whichever order the phases are assumed in.

Total power

The load is purely resistive, so P = Σ I²R:

  • 21.94² × 10 = 4813 W
  • 14.63² × 15 = 3209 W
  • 7.31² × 30 = 1604 W

Total power = 9627 W ≈ 9.63 kW

✓ Answer checked by Claude Opus 5.5 (high effort)

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