Skip to main content

Delta load line currents and power, ABC sequence

Short answer

Compute phase currents from Vph/Zph for each delta branch, then line currents via difference. |IA|=45.1 A, |IB|=5.59 A, |IC|=44.5 A. Total power = sum of Iph²×R = 18.8 kW.

A delta connected load of ZAB = 52∠45°Ω, ZBC = 52∠-30°Ω and ZCA = 10∠0°Ω are connected to a 380V, 3 phase ac source. Find the magnitude of the line currents and total power absorbed by loads, when phase sequence is ABC.

Delta load: Z_AB = 52∠45° Ω, Z_BC = 52∠−30° Ω, Z_CA = 10∠0° Ω. V_L = 380 V, sequence ABC.

V_AB = 380∠0°, V_BC = 380∠−120°, V_CA = 380∠120° V (phase voltage = line voltage in delta).

Phase currents

  • I_AB = 380∠0° / 52∠45° = 7.31∠−45° = 5.17 − j5.17 A
  • I_BC = 380∠−120° / 52∠−30° = 7.31∠−90° = 0 − j7.31 A
  • I_CA = 380∠120° / 10∠0° = 38.0∠120° = −19.0 + j32.91 A

Line currents

  • I_A = I_AB − I_CA = 24.17 − j38.08 = 45.1 A (∠−57.6°)
  • I_B = I_BC − I_AB = −5.17 − j2.14 = 5.59 A (∠−157.5°)
  • I_C = I_CA − I_BC = −19.0 + j40.22 = 44.5 A (∠115.3°)

(Sum = 0.)

Total power

  • P_AB = 7.31² × 52 cos 45° = 1964 W
  • P_BC = 7.31² × 52 cos 30° = 2405 W
  • P_CA = 38² × 10 × cos 0° = 14 440 W

Total power = 18 809 W ≈ 18.8 kW

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electrical Engineering chapter 8 questions →