--- title: "Importance and improvement of power factor" subject: "Basic Electrical Engineering" chapter: 7 key: "semester_1/basic_electrical_engineering:power-factor-importance-improvement" source: https://www.bctengineer.com/q/semester_1/basic_electrical_engineering/power-factor-importance-improvement site: BCT Engineering --- # Importance and improvement of power factor Basic Electrical Engineering · Chapter 7: Power in AC Circuits - Asked in: 2072 Kartik, 4 marks - Video: [Power Factor - Definition, Methods to Improve Power Factor |BEE|](https://www.youtube.com/watch?v=KQgQjmztz8s) — Trouble- Free - 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-improvement - Whole chapter: https://www.bctengineer.com/md/semester_1/basic_electrical_engineering/chapter-7 ## Question Explain the importance of power factor in an ac circuit, with suitable example. How power factor can be improved? ## Short answer Power factor (cos φ) indicates how effectively current is converted to useful power. Low PF demands larger current, increasing losses and equipment size. It can be improved by static capacitors in parallel, synchronous condensers, phase advancers, or properly sizing motors; for example, raising PF from 0.5 to 0.95 nearly halves the current. ## Answer **Importance:** Power factor (cos φ = P/S) shows how effectively the supplied current is converted into useful work. For a given power, a low power factor needs a larger current, so cables, transformers and generators become larger, losses (I²R) rise and the voltage drop increases. A high power factor means smaller kVA demand, lower loss and better regulation. #### Example A 5 kW single-phase motor at 230 V: - At cos φ = 0.5: I = 5000/(230 × 0.5) = 43.5 A, S = 10 kVA. - At cos φ = 0.95: I = 5000/(230 × 0.95) = 22.9 A, S = 5.26 kVA. The current is nearly halved and the line loss (∝ I²) falls to about one quarter, so the same line and transformer can carry more load. #### How power factor can be improved 1. **Static capacitors** connected in parallel with the inductive load. They supply the leading reactive power needed by the load. This is the most common method (capacitor banks near the load or at the substation). 2. **Synchronous condenser:** an over-excited synchronous motor running at no load, which takes a leading current. 3. **Phase advancers:** used with large induction motors. A phase advancer is an AC exciter on the motor shaft that supplies the exciting ampere-turns to the rotor circuit at slip frequency, so the motor draws less magnetising current from the supply. 4. Avoiding lightly loaded motors and transformers, and using correctly sized equipment. Capacitor needed to improve the power factor from cos φ₁ to cos φ₂ for load P: C = P (tan φ₁ − tan φ₂) / (ω V²) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a given power, what does a low power factor require? A. A smaller current B. A larger current, so bigger cables and more losses C. A higher frequency D. A lower voltage drop Answer: B. A low pf needs a larger current for the same power, increasing size, I²R loss and voltage drop. (Key point: Why high pf matters) 2. In the 5 kW, 230 V motor example, pf improves from 0.5 to 0.95. What happens to the line loss? A. It roughly doubles B. It stays the same C. It is nearly halved D. It falls to about one quarter, since loss is proportional to I² Answer: D. Current falls from 43.5 A to 22.9 A (about half), so I²R loss falls to about one quarter. (Key point: Example result) 3. What is the commonest way to improve the power factor of an inductive load? A. Connect static capacitors in parallel with the load B. Connect capacitors in series with the load C. Connect an inductor in parallel D. Run motors lightly loaded Answer: A. Parallel capacitors supply the leading reactive power needed by the inductive load. (Key point: Improvement method) 4. What is a synchronous condenser? A. An under-excited synchronous motor on full load B. An over-excited synchronous motor running at no load, taking leading current C. A bank of inductors at the substation D. A transformer with a tap changer Answer: B. An over-excited synchronous motor at no load takes a leading current and acts like a capacitor. 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