A single‑phase AC load consumes P=4 kW at V=400 V rms, f=50 Hz, with lagging power factor cos $\varphi$ = 0.65. A capacitor is connected in parallel with the load to bring PF to unity. What capacitance should be used?

Step-by-step solution

1. Find the power factor angle $\varphi$ from cos$\varphi$=0.65.$\text{n}\varphi =\arccos$(0.65)$\approx 49$.5$°$.
2. Compute the tangent: tan$\varphi =\sin \varphi /\cos \varphi =\sqrt{1-0.65^2}$/0.65$\approx 1$.169.
3. Calculate reactive power Q=P$\tan \varphi$=4000\,W$\times 1$.169$\approx 4676$\,var.
4. Determine angular frequency $\omega$=2$\pi$ f=2$\pi \times 50$=314.16\,$\text{rad/s}$.
5. Use the relation for capacitor reactive power Q_c=$V^2\omega$ C to cancel Q: set Q_c=Q.
6. Solve for C: C=Q/($V^2\omega$)=4676/($400^2\times 314$.16)$\approx 9$.3$\times 10^{-5}$\,F.
7. Convert to microfarads: 9.3$\times 10^{-5}$\,F=93\,$\mu \text{F}$.

Concept tested: Power factor correction with capacitor

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked