Labo Élec
electrical lab simulator
Reminder sheet
In AC, power has three faces: active (useful), reactive (exchanged) and apparent (what the cable carries). cos φ tells which share is truly useful.
In AC we distinguish the apparent power S = U·I (VA), the active power P = U·I·cos φ (W, the only one producing work or heat) and the reactive power Q = U·I·sin φ (var, exchanged with coils and capacitors). They form a triangle: S² = P² + Q².
The power factor cos φ = P/S measures the efficiency of the installation: a low cos φ requires more current for the same useful power, hence more line losses. It is corrected by adding capacitors in parallel.
| Quantity | Symbol | Unit |
|---|---|---|
| Apparent power | S | VA |
| Active power | P | W |
| Reactive power | Q | var |
| Quantity | Symbol | Unit |
|---|---|---|
| Power factor | cos φ | - |
| Quantity | Symbol | Unit |
|---|---|---|
| Capacitance to add | C | F |
| Phase shifts before / after | φ₁, φ₂ | rad |
A single-phase motor draws 5 A at 230 V with cos φ = 0.8. Find S, P and Q. What capacitance should be placed in parallel to raise the power factor to 0.95 (50 Hz)?
S = U · I = 230 × 5 = 1 150 VA; P = S · cos φ = 1 150 × 0.8 = 920 W; Q = S · sin φ = 1 150 × 0.6 = 690 var (check: 920² + 690² = 1 150²).
tan φ₁ = 0.75 (cos φ = 0.8); tan φ₂ = 0.329 (cos φ = 0.95).
C = P · (tan φ₁ − tan φ₂) / (U² · ω) = 920 × 0.421 / (230² × 314) = 387 / 16.6 × 10⁶ = 23 µF.
Power triangle: S² = P² + Q², cos φ = P / S.