Power Factor Calculator

What does this calculator do?

The power factor tells you what fraction of the apparent power is turned into useful active power.

Worked example

P = 900 W and S = 1000 VA give PF = 0.9.

Practical tips

• A valid result lies between 0 and 1.
• A low value means more current for the same useful power.
• For real correction, distinguish inductive from capacitive loads.

Common mistake to avoid

A value close to 1 does not guarantee low harmonic distortion; this simplified calculation assumes sinusoidal conditions.

Complete power factor formulas

In AC circuits there are three powers, which form the power triangle: active power P (W), which does work or produces heat; reactive power Q (var), which is exchanged with magnetic or electric fields; and apparent power S (VA), which is what the network “sees”: S = V · I.

• Power factor: PF = P / S
• With sine waves: PF = cos φ, where φ is the phase angle between voltage and current
• Apparent power: S = √(P² + Q²)
• Reactive power: Q = √(S² − P²) = S · sin φ
• Angle: φ = arccos(PF)
• Current required: I = P / (V · PF) in single-phase
• Capacitor correction: Q_c = P · (tan φ₁ − tan φ₂) and C = Q_c / (2π · f · V²)

With non-linear electronic loads (switch-mode supplies without PFC, variable-speed drives), the current contains harmonics: PF = P / S is still valid, but it no longer equals cos φ.

Typical values (S = 1000 VA)

The last column shows how much more current you need to deliver the same active power as with PF = 1 (it is 1 / PF). At PF = 0.5 the conductors carry twice the current.

Worked example: correcting a motor’s power factor

A single-phase motor draws 1500 W at 230 V, 50 Hz, with PF = 0.75. The goal is 0.95.

1. Initial current: I = 1500 / (230 × 0.75) ≈ 8.70 A.
2. tan φ₁ = tan(arccos 0.75) ≈ 0.882 and tan φ₂ = tan(arccos 0.95) ≈ 0.329.
3. Reactive power to compensate: Q_c = 1500 × (0.882 − 0.329) ≈ 830 var.
4. Capacitance: C = 830 / (2π × 50 × 230²) ≈ 50 µF (use the nearest standard AC-rated capacitor with a suitable voltage rating).
5. Final current: I = 1500 / (230 × 0.95) ≈ 6.86 A, about 21 % less.

The active power does not change: the motor does the same work, but the installation carries less current and has lower losses.

Common mistakes

• Confusing power factor with efficiency: efficiency compares mechanical and electrical power; PF compares active and apparent power. A motor can be efficient and still have a low PF.
• Entering P greater than S: that is physically impossible; check your measurements or units (kW versus VA).
• Overcompensating: too much capacitance makes the load capacitive and the PF drops again, now leading.
• Correcting harmonic-rich loads with plain capacitors: they can resonate with the grid; such loads need detuned filters or active correction.