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Star system with three loads, neutral current

Short answer

Line currents: I_R = 9.24 A ∠0°, I_Y = 20.99 A ∠-100°, I_B = 15.40 A ∠110°. Total power = 10.2 kW. Neutral current = 6.2 A (phasor sum of three line currents).

A three phase star connected system with line voltage 400 V is connected to three loads: 25∠0°, 11∠-20° and 15∠10° (also connected in star). Find the line to line current, total power and current in the neutral of the system.

Taking the three loads Z_R = 25∠0° Ω, Z_Y = 11∠−20° Ω, Z_B = 15∠10° Ω as star connected with a neutral wire, in RYB sequence. The loads are unbalanced.

V_ph = 400/√3 = 230.9 V. Take V_R = 230.9∠0°, V_Y = 230.9∠−120°, V_B = 230.9∠+120° V.

Line currents

  • I_R = 230.9∠0° / 25∠0° = 9.24 A ∠0°
  • I_Y = 230.9∠−120° / 11∠−20° = 20.99 A ∠−100°
  • I_B = 230.9∠120° / 15∠10° = 15.40 A ∠+110°

In rectangular form: I_R = 9.24 + j0; I_Y = −3.65 − j20.67; I_B = −5.27 + j14.47 A.

Neutral current

I_N = I_R + I_Y + I_B = (9.24 − 3.65 − 5.27) + j(0 − 20.67 + 14.47) = 0.32 − j6.20 A

|I_N| = 6.2 A (at about −87°)

Total power

P = Σ I²R_load, with R = Z cos θ of each load:

  • Phase R: 9.238² × 25 × cos 0° = 2133 W
  • Phase Y: 20.995² × 11 × cos(−20°) = 4556 W
  • Phase B: 15.396² × 15 × cos 10° = 3502 W

Total P = 10 191 W ≈ 10.2 kW

Answer: line currents 9.24 A, 20.99 A, 15.40 A; total power about 10.19 kW; neutral current about 6.2 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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