A balanced three-phase load draws a line current of 20 A from a 415 V (line-to-line) supply at a power factor of 0.8 lagging. Find the real power, apparent power and reactive power.
Show the explanation
Worked answer.
- Apparent power S = √3 × V_L × I_L = 1.732 × 415 × 20 ≈ 14,376 VA ≈ 14.38 kVA.
- Real power P = S × pf = 14,376 × 0.8 ≈ 11,501 W ≈ 11.5 kW.
- sin φ = √(1 − 0.8²) = 0.6, so reactive power Q = S × 0.6 ≈ 8,626 var ≈ 8.63 kvar (lagging, i.e. inductive).
Check: P² + Q² = S² (11.50² + 8.63² ≈ 14.38²). Common mistake: forgetting √3 or mixing phase and line voltage.
Original UniCircle question. Last reviewed 21 Oct 2026.