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Capacitor for power factor correction of lamp

Short answer

Lamp: S=180VA, P=80W, cos φ₁=0.444 lag, Q₁=161.2VAR. For unity pf, Qc=161.2VAR, C=8.91µF. For 0.95 lag, Qc=134.95VAR, C=7.46µF. Capacitor in parallel.

A fluorescent lamp takes a current of 0.75A when connected across a 240V, 50Hz a.c supply. The power consumed by the lamp is 80 watt. Calculate the value of the capacitance to be connected in parallel with the lamp to improve the power factor to (i) unity (ii) 0.95 lagging.

Given V = 240 V, f = 50 Hz, I = 0.75 A, P = 80 W.

Lamp data

  • Apparent power S = VI = 240 × 0.75 = 180 VA
  • Power factor cos φ₁ = P/S = 80/180 = 0.444 lagging (φ₁ = 63.6°)
  • Reactive power Q₁ = √(S² − P²) = √(180² − 80²) = 161.2 VAR

A capacitor in parallel must supply Q_c = Q₁ − Q₂ so that the net reactive power Q₂ gives the required power factor. Then C = Q_c / (ωV²), with ω = 2π × 50 = 314.16 rad/s and ωV² = 314.16 × 240² = 1.8096 × 10⁷.

(i) Unity power factor

Q₂ = 0, so Q_c = 161.2 VAR.

C = 161.2 / (1.8096 × 10⁷) = 8.91 × 10⁻⁶ F = 8.91 μF

(ii) 0.95 lagging

tan φ₂ = tan(cos⁻¹ 0.95) = 0.3287

Q₂ = P tan φ₂ = 80 × 0.3287 = 26.3 VAR

Q_c = 161.2 − 26.3 = 134.95 VAR

C = 134.95 / (1.8096 × 10⁷) = 7.46 × 10⁻⁶ F = 7.46 μF

Answer: 8.91 μF for unity, 7.46 μF for 0.95 lagging. The supply current falls from 0.75 A to 0.333 A (unity) or 0.351 A (0.95).

✓ Answer checked by Claude Opus 5.5 (high effort)

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