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Star three impedances four-wire system

Short answer

Line currents: 23.09 A (−53.13°), 23.09 A (−156.87°), 46.19 A (66.87°). Active, reactive powers: 13.87 kW, 16.0 kVAR. Neutral current magnitude 18.4 A.

Three impedances of (6+j8)Ω, (8+j6)Ω, (3+j4)Ω, are connected in star to a 3-phase, 4 wire system for which the line voltage is 400V. Find the line currents, and active, reactive and apparent power per phase. Also find the current through neutral wire.

Star connected load Z_R = 6 + j8 = 10∠53.13° Ω, Z_Y = 8 + j6 = 10∠36.87° Ω, Z_B = 3 + j4 = 5∠53.13° Ω. V_L = 400 V, four-wire, RYB sequence assumed.

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

Line currents

  • I_R = 230.9∠0° / 10∠53.13° = 23.09 A ∠−53.13° = 13.86 − j18.48
  • I_Y = 230.9∠−120° / 10∠36.87° = 23.09 A ∠−156.87° = −21.24 − j9.07
  • I_B = 230.9∠120° / 5∠53.13° = 46.19 A ∠66.87° = 18.14 + j42.48

Neutral current

I_N = I_R + I_Y + I_B = (13.86 − 21.24 + 18.14) + j(−18.48 − 9.07 + 42.48) = 10.76 + j14.93 A

|I_N| = 18.4 A (∠54.2°)

Power per phase

PhaseI (A)P = I²R (W)Q = I²X (VAR)S = I²Z (VA)
R23.09320042675333
Y23.09426732005333
B46.196400853310 667
Total13 86716 00021 333 (sum)

Total active power 13.87 kW, total reactive power 16.0 kVAR (all lagging), total apparent power √(13 867² + 16 000²) = 21.17 kVA (the arithmetic sum is 21.33 kVA).

✓ Answer checked by Claude Opus 5.5 (high effort)

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