A single‑phase load consumes $P=3\text{kW}$ at $V=230\text{V rms}$, $f=50\text{Hz}$ and has a lagging power factor of $\cos \varphi =0.70$. To improve the power factor to unity, a capacitor is connected in parallel with the load. What capacitance should be used?\n\nA) $120\mu \text{F}$\nB) $184\mu \text{F}$\nC) $250\mu \text{F}$\nD) $320\mu \text{F}$

Step-by-step solution

1. Apparent power $S=\frac{P}{\cos \varphi }=\frac{3000\text{W}}{0.70}=4285.7\text{VA}$.
2. Reactive power supplied by the load: $Q_{\text{load}}=P\tan \varphi$.
3. Calculate $\tan \varphi =\frac{\sqrt{1-{\cos }^2\varphi }}{\cos \varphi }=\frac{\sqrt{1-0.49}}{0.70}=\frac{0.714}{0.70}=1.020$.
4. Hence $Q_{\text{load}}=3000\times 1.020=3060\text{VAR}$ (lagging).
5. To obtain unity power factor the capacitor must supply an equal leading reactive power: $Q_C=3060\text{VAR}$.
6. For a capacitor $Q_C=V^2\omega C$, where $\omega =2\pi f=2\pi \times 50=314.16\text{rad/s}$.
7. Solve for $C$: $C=\frac{Q_C}{V^2\omega }=\frac{3060}{(230{)}^2\times 314.16}$.
8. Compute denominator: $230^2=52900$, $52900\times 314.16\approx 1.6619\times 10^7$.
9. Thus $C=\frac{3060}{1.6619\times 10^7}\approx 1.84\times 10^{-4}\text{F}$.
10. Convert to microfarads: $1.84\times 10^{-4}\text{F}=184\mu \text{F}$.
11. The correct option is (B) $184\mu \text{F}$.

Concept tested: Power factor correction with capacitor

Published 7 September 2026 · How this solution was prepared and checked