A single‑phase AC load consumes $P=7\text{kW}$ at a rms voltage $V=415\text{V}$ and frequency $f=50\text{Hz}$. The load has a lagging power factor $\cos \varphi =0.78$. A capacitor is connected in parallel with the load to improve the overall power factor to unity. Determine the required capacitance (in $\mu \text{F}$) and state the rms current drawn from the source after correction.

Step-by-step solution

1. Calculate the apparent power before correction: $S=\frac{P}{\cos \varphi }=\frac{7000}{0.78}\approx 8.97\text{kVA}$.
2. Find the initial reactive power: $Q=\sqrt{S^2-P^2}=\sqrt{(8.97{)}^2-(7{)}^2}\text{kVAr}\approx 5.62\text{kVAr}$ (lagging).
3. To achieve unity power factor, the capacitor must supply an equal magnitude of reactive power: $Q_C=5.62\text{kVAr}$ (capacitive).
4. Use the formula $Q_C=V^2\omega C$ with $\omega =2\pi f=314.16\text{rad/s}$:
5. $C=\frac{Q_C}{V^2\omega }=\frac{5.62\times 10^3}{(415{)}^2\times 314.16}\approx 1.04\times 10^{-4}\text{F}$.
6. Convert to microfarads: $1.04\times 10^{-4}\text{F}=104\mu \text{F}$.
7. After power‑factor correction, the current is purely real: $I_{\text{rms}}=\frac{P}{V}=\frac{7000}{415}\approx 16.87\text{A}$.

Concept tested: Power‑factor correction using a capacitor

Published 20 September 2026 · How this solution was prepared and checked