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Neutral current and power, unbalanced star

Short answer

Phase voltage 239.6 V; phase currents 23.96 A (0°), 15.97 A (−120°), 11.98 A (120°). Neutral current magnitude is 10.57 A. Total power is 12.44 kW.

For the circuit shown in figure below, calculate the current through the neutral and the total power consumed in the load. [Figure: 415 V balanced supply feeding a four-wire star load with 10 Ω in one phase, 15 Ω in another phase and 20 Ω in the third phase; neutral current IN.]

From the figure: a 415 V balanced supply feeds a four-wire star load with 10 Ω, 15 Ω and 20 Ω resistors in the three phases and a neutral wire carrying I_N. The sequence is not shown, so assume R–Y–B with 10 Ω in R, 15 Ω in Y and 20 Ω in B.

V_ph = 415/√3 = 239.6 V. Take V_R = 239.6∠0°, V_Y = 239.6∠−120°, V_B = 239.6∠+120° V.

Line (phase) currents

  • I_R = 239.6/10 = 23.96∠0° A = 23.96 + j0
  • I_Y = 239.6/15 = 15.97∠−120° A = −7.99 − j13.83
  • I_B = 239.6/20 = 11.98∠+120° A = −5.99 + j10.38

Neutral current

I_N = I_R + I_Y + I_B = (23.96 − 7.99 − 5.99) + j(0 − 13.83 + 10.38) = 9.98 − j3.46 A

|I_N| = 10.57 A (at −19.1°). The neutral current flows because the load is unbalanced.

Total power

All loads are resistive, so P = Σ I²R:

  • Phase R: 23.96² × 10 = 5741 W
  • Phase Y: 15.97² × 15 = 3827 W
  • Phase B: 11.98² × 20 = 2870 W

Total power = 12 438 W ≈ 12.44 kW

Interchanging the phases of the load changes the angle of I_N but not its magnitude, and the total power stays the same, so the answer does not depend on the assumed order.

✓ Answer checked by Claude Opus 5.5 (high effort)

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