Power Factor Correction

Add a shunt capacitor to supply an inductive load’s reactive current locally and reduce the source current.

V1 supplies 10 V peak at 50 Hz to R1 = 100 Ω in series with L1 = 318.31 mH. At this frequency the inductive reactance equals the resistance.

$$X_L=2\pi fL\approx100\,\Omega$$ $$\phi=\tan^{-1}\frac{X_L}{R}=45^\circ$$ $$\mathrm{PF}=\cos\phi=\frac{P}{S}\approx0.707$$

The RMS source voltage is 7.071 V and the uncorrected RMS current is 50 mA. The load consumes 0.25 W of real power and exchanges 0.25 var of inductive reactive power; its apparent power is 0.3536 VA.

$$C=\frac{Q_L}{\omega V_{\mathrm{rms}}^2}\approx15.915\,\mathrm{\mu F}$$

C1 is connected in parallel with the whole RL load. It supplies leading reactive current that cancels the load’s lagging reactive current at 50 Hz. Source RMS current drops to about 35.36 mA and the source power factor approaches 1, while load current and real power remain the same.

Read the chart. Vin is green. Isource is blue and Iload purple. The saved chart shows the final two cycles, after startup. Compare their phases: source current is in phase with voltage, while load current still lags.

Try it. Change C1’s Capacitance to 7.95774715 µF. This supplies half the reactive compensation; source power factor becomes about 0.894 lagging. Restore 15.9154943 µF. Correction is frequency-dependent.

Model limits. These calculations assume sinusoidal voltage, a linear ideal inductor and a pure resistor. With harmonics, total power factor also includes waveform distortion and is not simply the cosine of one phase angle.