--- title: "Induction motor improving power factor" subject: "Industrial Power Distribution and Illumination" chapter: 10 key: "semester_6/industrial_power_distribution_and_illumination:induction-motor-pf-improvement" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/induction-motor-pf-improvement site: BCT Engineering --- # Induction motor improving power factor Industrial Power Distribution and Illumination · Chapter 10: Electrical Energy Audit in Industry - Asked in: 2074 Bhadra, 8 marks - Video: [Power Factor Improvement in Power System](https://www.youtube.com/watch?v=D8s9vQZhDN0) — Smart Educare - 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/induction-motor-pf-improvement - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-10 ## Question An induction motor improves the power factor of a load of 500 KW from 0.707 lagging to 0.95 lagging. Simultaneously the motor carries a load of 100 KW. Find: a) The leading KVAR supplied by motor b) KVA rating of the motor and c) Power factor at which the motor operates ## Short answer Motor supplies 302.9 kVAR leading. Its kVA rating is 319 kVA, operating at 0.313 leading power factor. ## Answer Note: an induction motor always draws lagging kVAR, so it cannot improve the power factor. The machine that does this while also driving a mechanical load is an **over-excited synchronous motor**; the solution below treats the motor as such. **Given:** load P1 = 500 kW at 0.707 lag; motor load Pm = 100 kW; final pf = 0.95 lag. **Reactive power of the original load:** ```text tan(cos⁻¹0.707) = 1.0003 Q1 = 500 × 1.0003 = 500.15 kVAR (lagging) ``` **Combined load after adding motor:** ```text P = 500 + 100 = 600 kW tan(cos⁻¹0.95) = 0.3287 Q = 600 × 0.3287 = 197.21 kVAR (lagging) ``` #### a) Leading kVAR supplied by motor ```text Qm = Q1 − Q = 500.15 − 197.21 = 302.94 kVAR (leading) ``` #### b) kVA rating of the motor ```text Sm = √(Pm² + Qm²) = √(100² + 302.94²) = √(10 000 + 91 773) = 319.02 kVA ``` #### c) Power factor of the motor ```text cos φm = Pm / Sm = 100 / 319.02 = 0.313 (leading) ``` **Phasor (power triangle) picture:** ```text Q (kVAR) load ↑ 500.15 lag | motor ↓ 302.94 lead ----------------------- total 197.21 lag → pf 0.95 at 600 kW ``` **Answer: (a) 302.9 kVAR leading, (b) about 319 kVA, (c) about 0.31 leading.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the solution treat the 'motor' as an over-excited synchronous motor rather than an induction motor? A. An induction motor always draws lagging kVAR, so it cannot supply leading kVAR B. An induction motor cannot drive a 100 kW mechanical load C. A synchronous motor has a higher efficiency D. An induction motor draws only real power Answer: A. An induction motor needs magnetising current and so always absorbs lagging kVAR. Only an over-excited synchronous motor can supply leading kVAR while also driving a mechanical load. (Key point: Which machine improves pf) 2. How is the leading kVAR supplied by the motor found? A. Qm = Q1 + Q (sum of initial and final lagging kVAR) B. Qm = Q1 - Q (initial lagging kVAR minus final lagging kVAR) C. Qm = Pm x tan(cos^-1 0.95) D. Qm = P x 0.95 Answer: B. The motor must cancel the part of the original lagging kVAR that is not left in the final state, so Qm = 500.15 - 197.21 = 302.94 kVAR leading. (Key point: Leading kVAR of motor) 3. What real power is used to find the final lagging kVAR at pf 0.95? A. 500 kW only B. 100 kW only C. 600 kW (load plus motor load) D. 400 kW Answer: C. The motor's 100 kW mechanical load adds to the 500 kW load, so total P = 600 kW and Q = 600 x 0.3287 = 197.21 kVAR. (Key point: Final total power) 4. With Pm = 100 kW and Qm = 302.94 kVAR leading, what are the motor kVA rating and pf? A. About 319 kVA and 0.95 leading B. About 403 kVA and 0.25 leading C. About 319 kVA and 0.31 leading D. About 203 kVA and 0.49 leading Answer: C. S = sqrt(100^2 + 302.94^2) = 319 kVA and pf = P/S = 100/319 = 0.313 leading. 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