A single‑phase AC load consumes P=5 kW at V=400 V_rms, f=50 Hz, and has a lagging power factor of cosφ=0.6. A capacitor is connected in parallel with the load to bring the power factor to unity. Calculate the required capacitance.

A single‑phase AC load consumes $P = 5$ kW at $V = 400$ V $_{rms}$ , $f = 50$ Hz, and has a lagging power factor of $cos ϕ = 0.6$ . A capacitor is connected in parallel with the load to bring the power factor to unity. Calculate the required capacitance.

hard 3 marks Class 12 · Physics · Alternating Current

Answer

133 μF

The correct answer is 133 μF — Capacitance: $C = \frac{1}{2 π f X _{C}} = \frac{1}{2 π \times 50 \times 24.0} \approx 1.33 \times 1 0^{- 4}$ F, i.e. $133$ µF.

Step-by-step solution

Determine the power factor angle: $ϕ = cos^{- 1} (0.6) \approx 53.1 3^{\circ}$ .
Compute reactive power of the load: $Q_{l o a d} = P tan ϕ = 5000 \times tan (53.1 3^{\circ}) \approx 6665$ VAR.
Required capacitor reactance magnitude: $X_{C} = \frac{V ^{2}}{Q _{l o a d}} = \frac{40 0 ^{2}}{6665} \approx 24.0$ Ω.
Capacitance: $C = \frac{1}{2 π f X _{C}} = \frac{1}{2 π \times 50 \times 24.0} \approx 1.33 \times 1 0^{- 4}$ F, i.e. $133$ µF.

Common mistake: Using

V 2

instead of

V_{r m s}

for reactive power

Concept tested: Power factor correction with capacitor

Published 17 September 2026 · How this solution was prepared and checked

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A single‑phase AC load consumes P=5 kW at V=400 Vrms​, f=50 Hz, and has a lagging power factor of cosϕ=0.6. A capacitor is connected in parallel with the load to bring the power factor to unity. Calculate the required capacitance.

Step-by-step solution

A single‑phase AC load consumes $P = 5$ kW at $V = 400$ V $_{rms}$ , $f = 50$ Hz, and has a lagging power factor of $cos ϕ = 0.6$ . A capacitor is connected in parallel with the load to bring the power factor to unity. Calculate the required capacitance.