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Delta connected coils line current and power

Short answer

Coil impedance Z = 7 + j9.425 = 11.74∠53.4° Ω. Phase current = 400/11.74 = 34.07 A. Line current = √3 × 34.07 = 59.0 A. Total power = 3 × 34.07² × 7 = 24.4 kW.

Three similar coils each of resistance 7Ω and inductance of 0.03 H are connected in Delta to a 400 V, 3 phase, 50 Hz supply. Calculate the line current and the total power consumed.

Each coil: R = 7 Ω, L = 0.03 H, delta connected to V_L = 400 V, 50 Hz.

  • X_L = 2πfL = 314.16 × 0.03 = 9.425 Ω
  • Z_ph = 7 + j9.425 = 11.74∠53.4° Ω
  • Power factor = R/Z = 7/11.74 = 0.596 lagging

Phase current

In delta, V_ph = V_L = 400 V.

I_ph = 400/11.74 = 34.07 A

Line current

I_L = √3 I_ph = 1.732 × 34.07 = 59.0 A

Total power

P = 3 I_ph² R = 3 × 34.07² × 7 = 24 380 W ≈ 24.4 kW

Check: √3 V_L I_L cos φ = 1.732 × 400 × 59.01 × 0.596 = 24.4 kW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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