Effect of low power factor on equipment ratings

At a given real power P and voltage V, current and kVA are inversely proportional to power factor:

I   = P / (√3 V cos φ)
kVA = kW / cos φ

So when pf falls, every component carrying this current must be larger, and losses rise.

1. Transformers

• The rating of a transformer is in kVA, limited by winding current (heating) and core flux.
• Lower pf → more kVA for the same kW, so a bigger transformer is needed.
• Example: 800 kW load at pf 1.0 needs 800 kVA; at 0.8 it needs 1000 kVA; at 0.6 it needs 1333 kVA.
• Copper loss ∝ I² ∝ 1/cos²φ – at pf 0.7 the copper loss is about 2 times that at unity pf.
• Voltage regulation becomes poor with lagging pf (reactive voltage drop adds).

2. Generators (alternators, DG sets)

• Rated in kVA at a stated pf (usually 0.8 lag). Below rated pf, the field current needed to hold voltage rises above design, so the alternator must be de-rated or a larger machine bought.
• The prime mover is sized for kW; at low pf the prime mover is under-used while the alternator is at full current – poor utilisation.
• Larger excitation system needed; poorer voltage regulation.

3. Cables and conductors

• Cable size is chosen on current. Low pf increases current, so a larger cross-section is needed:

Cross-section ∝ I ∝ 1/cos φ
Copper volume (for same loss) ∝ 1/cos²φ

• Higher I²R loss and voltage drop along the cable.

4. Switchgear and protective devices

• Breakers, busbars, CTs and fuses must be rated for the higher current, adding to cost.

Summary table

Numerical illustration: a 400 kW, 400 V load at 0.6 pf draws I = 400 000 / (√3 × 400 × 0.6) ≈ 962 A, but at 0.95 pf only ≈ 608 A. The transformer could be 421 kVA instead of 667 kVA.

Hence power factor correction with capacitors (target 0.95–0.98) is used to reduce equipment ratings, losses and kVA charges.