--- title: "Power factor and why keep it high" subject: "Basic Electrical Engineering" chapter: 7 key: "semester_1/basic_electrical_engineering:power-factor-importance-high" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/power-factor-importance-high site: BCT Engineering --- # Power factor and why keep it high Basic Electrical Engineering · Chapter 7: Power in AC Circuits - Asked in: 2073 Shrawan, 8 marks - Video: [Power Factor Explained - The basics what is power factor pf](https://www.youtube.com/watch?v=Tv_7XWf96gg) — The Engineering Mindset - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/power-factor-importance-high - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-7 ## Question Define power factor and explain why in general it should be kept on high as possible in power supply system. ## Short answer Power factor cos φ = P/S. Keeping it high reduces current drawn, minimizing I²R losses, voltage drops, and equipment size/costs, while allowing full capacity utilization and avoiding utility penalties. Low pf demands larger kVA rating for same kW. ## Answer **Power factor** (cos φ) is the cosine of the angle between voltage and current. It equals the ratio of true power to apparent power, P/S = kW/kVA. It shows what fraction of the supplied kVA does useful work. #### Why it should be as high as possible For a given load power P at voltage V, the current is I = P/(V cos φ). At a low power factor the supply must also carry an extra "wattless" (reactive) current that only flows back and forth and does no useful work. - **Large current for the same power.** I = P/(V cos φ) for single phase, or P/(√3 V_L cos φ) for three phase. A low cos φ means a larger current. - **Higher copper loss.** Loss = I²R rises as 1/cos²φ in the cables, transformers and generator windings. - **Poor voltage regulation.** The large current causes a bigger voltage drop (IR and IX) in lines and transformers, so the load voltage falls. - **Larger equipment size and cost.** Generators, transformers, switchgear and cables must be rated in kVA, not kW, so they are bigger and costlier for the same useful power. - **Reduced capacity.** The same kVA rating delivers less kW (kW = kVA × cos φ), so a plant cannot supply as many loads. - **Lower efficiency and heavier tariff.** Utilities usually charge a penalty on low power factor (below about 0.8 to 0.9). #### Example A 10 kW single-phase load at 400 V needs: - at cos φ = 1: I = 10000/400 = 25 A - at cos φ = 0.5: I = 10000/(400 × 0.5) = 50 A, so twice the current and four times the line loss. Hence power companies require industrial consumers to keep the power factor high (typically above about 0.85 to 0.9), usually by fitting capacitor banks near the load. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which statement correctly defines power factor? A. The ratio of apparent power to true power, kVA/kW B. The ratio of true power to apparent power, kW/kVA = cos φ C. The ratio of reactive power to true power D. The sine of the angle between voltage and current Answer: B. Power factor is cos φ = P/S = kW/kVA, the fraction of supplied kVA that does useful work. (Key point: Definition of power factor) 2. For a fixed load power P and voltage V, what happens to the line current when the power factor falls? A. It decreases because I = P/(V cos φ) has cos φ in the numerator B. It stays the same because P is fixed C. It increases because I = P/(V cos φ) D. It increases only if the voltage also increases Answer: C. I = P/(V cos φ), so a smaller cos φ needs a larger current for the same power. (Key point: Current at low pf) 3. How does the copper loss vary with power factor for a fixed load power? A. It is proportional to cos φ B. It rises as 1/cos²φ C. It is independent of cos φ D. It rises as 1/cos φ Answer: B. Loss = I²R and I is proportional to 1/cos φ, so loss goes as 1/cos²φ. (Key point: Copper loss) 4. Why must generators and transformers be bigger for a low-pf load of the same kW? A. They are rated in kVA, and kVA = kW/cos φ is larger B. They are rated in kW, and kW rises at low pf C. Low pf raises the supply frequency D. Low pf makes the voltage higher Answer: A. Equipment is rated in kVA; for the same useful kW a low pf means a higher kVA demand, so larger and costlier equipment. 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