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Capacitor for single-phase motor pf correction

Short answer

Compute active power P = 440×40×0.7 = 12320 W. Then Qc = P (tan(arccos 0.7) – tan(arccos 0.9)) = 6602 VAR. C = Qc/(2πf V²) = 6602/(314.16×440²) = 108.5 μF.

A single phase motor takes a current of 40A at pf 0.7 lagging from a 440V, 50HZ supply. What value must a shunting capacitor have to raise the power factor to 0.9 lagging.

Given V = 440 V, f = 50 Hz, I = 40 A, cos φ₁ = 0.7 lagging, target cos φ₂ = 0.9 lagging.

Motor power

  • Active power P = V I cos φ₁ = 440 × 40 × 0.7 = 12 320 W
  • φ₁ = cos⁻¹ 0.7 = 45.57°, tan φ₁ = 1.0202
  • φ₂ = cos⁻¹ 0.9 = 25.84°, tan φ₂ = 0.4843

Reactive power

  • Q₁ = P tan φ₁ = 12 320 × 1.0202 = 12 569 VAR
  • Q₂ = P tan φ₂ = 12 320 × 0.4843 = 5 967 VAR

The capacitor must supply Q_c = Q₁ − Q₂ = 6 602 VAR.

Capacitance

Q_c = V²/X_C = ωCV²

C = Q_c / (ωV²) = 6602 / (314.16 × 440²) = 6602 / (6.0820 × 10⁷)

C = 108.5 μF

The supply current after correction is P/(V × 0.9) = 12 320/(440 × 0.9) = 31.1 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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