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Unbalanced delta loads currents and power

Short answer

Phase currents: I_AB=3.00 A, I_BC=4.00 A, I_CA=2.00 A. Line currents: 4.78 A, 1.04 A, 5.82 A. Total active power absorbed is 1.08 kW.

Three loads (31+j59) Ω, (30-j40)Ω and (80+j60)Ω are connected in delta to a 3 phase, 200 V supply. Find the phase currents, line currents and total power absorbed.

Loads connected in delta: Z_AB = 31 + j59 Ω, Z_BC = 30 − j40 Ω, Z_CA = 80 + j60 Ω. V_L = 200 V, sequence assumed ABC, with the loads in the order AB, BC, CA.

Phase voltages equal line voltages: V_AB = 200∠0°, V_BC = 200∠−120°, V_CA = 200∠120° V.

Phase currents

  • |Z_AB| = 66.65 Ω, so I_AB = 200/66.65 = 3.00 A ∠−62.3° = 1.40 − j2.66
  • |Z_BC| = 50 Ω, so I_BC = 200/50 = 4.00 A ∠−66.9° = 1.57 − j3.68
  • |Z_CA| = 100 Ω, so I_CA = 200/100 = 2.00 A ∠83.1° = 0.24 + j1.99

Line currents (by KCL at the junctions)

  • I_A = I_AB − I_CA = 1.16 − j4.65 = 4.78 A ∠−76.0°
  • I_B = I_BC − I_AB = 0.17 − j1.02 = 1.04 A ∠−80.3°
  • I_C = I_CA − I_BC = −1.33 + j5.67 = 5.82 A ∠103.2°

(Their sum is zero.)

Total power

  • P_AB = 3.00² × 31 = 279.2 W
  • P_BC = 4.00² × 30 = 480 W
  • P_CA = 2.00² × 80 = 320 W

Total power = 1079 W ≈ 1.08 kW

✓ Answer checked by Claude Opus 5.5 (high effort)

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