This simulator turns a load into a power triangle. Real power (kW) sets the horizontal side, the load angle sets how much reactive power (kVAR) the load draws, and the hypotenuse is the apparent power (kVA) the source has to carry. Add a correction capacitor and the blue kVAR side shrinks, the hypotenuse shortens and the power factor climbs toward 1.0.
• A power triangle with kW (green), kVAR (blue) and kVA (orange) sides, the angle φ marked, and the power factor printed beside it. • Two scrolling waveforms whose phase gap follows the load angle and narrows as correction is added. • Sliders for real power (10-500 kW), load angle (0-60 degrees) and correction capacitor (0 up to the load's kVAR). • An information card listing real power, corrected reactive power, apparent power and power factor with a rating of excellent, acceptable or poor, plus notes, formulas and a worked example.
Reactive power drawn by the load is Q = P × tan(φ). A correction capacitor supplies reactive power locally, so the remaining reactive power is the load's kVAR minus the capacitor's kVAR. Apparent power is √(P² + Q²) and power factor is P divided by S. Because capacitor kVAR only subtracts from an inductive load's kVAR, the simulator stops the capacitor slider at the load's own reactive power so you cannot over-correct into a leading power factor.
A low power factor means the source, transformers and conductors must carry more current than the real power alone would need. Utilities often charge commercial customers for that, and large motors are a common cause because their windings are inductive. Correction capacitors near the motors cut the reactive current the utility has to supply. The readout turns green at 0.95 or better, amber between 0.85 and 0.95 and red below 0.85.
A right triangle whose horizontal side is real power in kW, vertical side is reactive power in kVAR and hypotenuse is apparent power in kVA. The angle between real and apparent power is the power-factor angle, and power factor equals the cosine of that angle.
Find the load's kVAR before and after using Q = P × tan(arccos(pf)), then subtract. For a 60 kW load, correcting from 0.75 to 0.95 lagging takes about 52.9 − 19.7 ≈ 33 kVAR, which is the worked example in the simulator.
In this model the capacitor only offsets an inductive load's reactive power. Going past that point would swing the circuit to a leading power factor, which the lab does not simulate, so the slider range is limited to the reactive power the load actually draws.
No. It is a single-load, steady-state power-triangle model. It does not include harmonic distortion, capacitor switching transients, resonance or unbalanced three-phase loads.