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Star alternator supplying delta load (8+j6)

Short answer

Load branch current = 23 A. Power consumed = 12.7 kW. Power factor = 0.8 lagging. Reactive power = 9.52 kVAR.

A star connected alternator supplies a delta connected load. The impedance of the load branch is (8 + J6)Ω. The line voltage is 230 volt. Determine: i) Current in the load branch ii) Power consumed by the load iii) Power factor of the load iv) Reactive power of the load

The alternator is star connected, but the load is delta connected, so the line voltage across each load branch is V_L = 230 V (phase voltage of load = line voltage).

Z_ph = 8 + j6 = 10∠36.87° Ω

i) Current in the load branch

I_ph = V_L/Z = 230/10 = 23 A (line current = √3 × 23 = 39.8 A)

iii) Power factor

cos φ = R/Z = 8/10 = 0.8 lagging

ii) Power consumed

P = 3 I_ph² R = 3 × 23² × 8 = 12 696 W ≈ 12.7 kW

Check: √3 V_L I_L cos φ = 1.732 × 230 × 39.84 × 0.8 = 12.7 kW.

iv) Reactive power

Q = 3 I_ph² X = 3 × 23² × 6 = 9522 VAR ≈ 9.52 kVAR (lagging)

Apparent power S = √(12 696² + 9522²) = 15.87 kVA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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