--- title: "Effect of low power factor on equipment ratings" subject: "Industrial Power Distribution and Illumination" chapter: 10 key: "semester_6/industrial_power_distribution_and_illumination:low-pf-effect-ratings" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/low-pf-effect-ratings site: BCT Engineering --- # Effect of low power factor on equipment ratings Industrial Power Distribution and Illumination · Chapter 10: Electrical Energy Audit in Industry - Asked in: 2068 Jestha (old course), 8 marks - Video: [Effect of Low Power Factor](https://www.youtube.com/watch?v=BFEnbrCPjjw) — Actech - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/low-pf-effect-ratings - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10 ## Question How does the low power factor affect the rating of the transformers, generators and cables in distribution system of an industry? ## Short answer Low power factor increases current for the same real power. Transformers need higher kVA rating and suffer poor voltage regulation. Generators must be derated. Cables require larger cross-section, leading to higher I²R losses and voltage drop. ## Answer At a given real power P and voltage V, current and kVA are inversely proportional to power factor: ```text 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: ```text 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 | Equipment | Effect of low pf | |---|---| | Transformer | Higher kVA rating, higher Cu loss, poor regulation | | Generator | De-rating, larger exciter, under-used prime mover | | Cable | Larger size, more loss and voltage drop | | Switchgear | Higher current rating | | Consumer | Higher kVA demand charge / pf penalty | **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At a fixed kW and voltage, what happens to current and kVA when pf falls? A. Both fall B. Current rises, kVA is unchanged C. Both rise, as both are inversely proportional to pf D. Current is unchanged, kVA rises Answer: C. I = P/(sqrt3 V cos phi) and kVA = kW/cos phi, so both increase as pf decreases. (Key point: Current and kVA vs pf) 2. What kVA transformer is needed for an 800 kW load at pf 0.8? A. 800 kVA B. 640 kVA C. 1333 kVA D. 1000 kVA Answer: D. kVA = 800/0.8 = 1000 kVA, compared with 800 kVA at unity pf. (Key point: Transformer rating) 3. Why must a generator be de-rated or a bigger one bought at low pf? A. The prime mover is overloaded in kW B. The field current needed to hold voltage rises above design while the alternator is at full current, leaving the prime mover under-used C. Frequency rises D. Speed rises Answer: B. Alternators are rated in kVA at a stated pf; at lower pf the excitation and current limits are reached before the prime mover is fully loaded. (Key point: Generator de-rating) 4. How do cable cross-section and copper loss vary with pf? A. Cross-section is proportional to 1/cos phi and loss to 1/cos^2 phi B. Cross-section is proportional to cos phi C. Loss is independent of pf D. Cross-section is proportional to 1/cos^2 phi and loss to 1/cos phi Answer: A. Cable size follows current (1/cos phi) and I^2R loss follows 1/cos^2 phi, so low pf needs larger cables and gives more loss and drop. 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