A single-phase load draws 10 kW at 0.8 power factor lagging from a 230 V, 50 Hz supply. What capacitance must be connected in parallel to raise the power factor to 0.95 lagging? By how much does the supply current fall?
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The capacitor supplies the difference in reactive power: Q_c = P(tan φ₁ − tan φ₂).
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Worked answer.
The real power P = 10 kW does not change when a capacitor is added. Only the reactive power drawn from the supply changes.
Step 1: Reactive power before correction. cos φ₁ = 0.8, so sin φ₁ = 0.6 and tan φ₁ = 0.75.
Q₁ = P tan φ₁ = 10 × 0.75 = 7.5 kVAR.
Step 2: Reactive power after correction. cos φ₂ = 0.95, so φ₂ ≈ 18.19° and tan φ₂ ≈ 0.3287.
Q₂ = 10 × 0.3287 ≈ 3.287 kVAR.
Step 3: Capacitor kVAR. Q_c = Q₁ − Q₂ = 7.5 − 3.287 ≈ 4.21 kVAR.
Step 4: Capacitance. For a capacitor across voltage V, Q_c = V² ωC, so
C = Q_c / (2πf V²) = 4213 / (2π × 50 × 230²) = 4213 / (314.16 × 52,900) ≈ 2.535 × 10⁻⁴ F ≈ 254 µF.
Step 5: Supply current. Before: I₁ = P / (V cos φ₁) = 10,000 / (230 × 0.8) ≈ 54.3 A. After: I₂ = 10,000 / (230 × 0.95) ≈ 45.8 A. The current falls by about 8.6 A (about 16%).
Check. Apparent power after correction is √(10² + 3.287²) ≈ 10.53 kVA, and 10.53 kVA / 230 V ≈ 45.8 A, which matches Step 5.
Common mistake. Using kVA instead of kW as the base for tan φ, or forgetting that P stays constant. Another frequent slip is using ω = 50 instead of 2π × 50.
Likely follow-up. "Why not correct all the way to unity?" The last few percent need disproportionately more kVAR, and over-correction at light load can make the power factor leading and raise the voltage. Utilities typically target around 0.95 to 0.99.
Original UniCircle question. Last reviewed 28 Oct 2026.