--- title: "Causes, disadvantages and improvement of low pf" subject: "Utilization of Electrical Energy" chapter: 6 key: "semester_7/utilization_of_electrical_energy:low-power-factor-causes-methods" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/low-power-factor-causes-methods site: BCT Engineering --- # Causes, disadvantages and improvement of low pf Utilization of Electrical Energy · Chapter 6: Demand Side Management - Asked in: 2079 Baisakh, 8 marks; 2069 Chaitra (asked 2 times) - Video: [Disadvantages & Cause of low power factor | Improvement of power factor | HSBTE(NSQF level-3)](https://www.youtube.com/watch?v=Z2Sx2V-PDrQ) — Seema Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/low-power-factor-causes-methods - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-6 ## Question What are the causes and disadvantages of low power factor? Mention the methods of power factor enhancement. ## Short answer Low lagging power factor is primarily caused by lightly loaded induction motors, transformers, arc furnaces and discharge lamps. It leads to larger equipment kVA ratings, increased losses and poor voltage regulation. Improvement methods include static capacitors, synchronous condensers, phase advancers and avoiding light-load operation. ## Answer **Power factor** is the cosine of the angle between voltage and current, $\cos\phi = P/S$. A low (lagging) power factor means the load draws large reactive current for the same useful power. #### Causes of low power factor 1. **Induction motors**, especially when lightly loaded (pf about 0.2–0.3 at no load, 0.8–0.9 at full load); they are the largest industrial load. 2. **Transformers** working at light load, drawing magnetising current. 3. **Arc and induction furnaces** and welding sets, which have high reactance. 4. **Discharge lamps** (fluorescent, mercury, sodium) with chokes. 5. **Supply voltage above rated**, which increases magnetising current. 6. **Power electronic loads** (rectifiers, drives) with phase-controlled firing. #### Disadvantages of low power factor For the same power, current $I = P/(\sqrt3 V \cos\phi)$ increases, so: 1. Larger **kVA rating** of generators, transformers and switchgear is needed (higher capital cost). 2. Larger **conductor size** is needed for the same loss. 3. Higher **copper losses** ($I^2R$), so lower efficiency. 4. Higher **voltage drop**, poor voltage regulation. 5. Reduced handling capacity of the system. 6. Consumers pay more under **kVA-demand tariffs** or pf penalties. #### Methods of power factor improvement 1. **Static capacitors:** connected in parallel (star or delta) with the load, supply leading kVAR. Low loss, little maintenance, no moving parts; used individually on motors or as automatic capacitor banks. 2. **Synchronous condenser:** an over-excited synchronous motor on no load acts as a variable capacitor; gives stepless control, used in large substations. 3. **Phase advancer:** an ac exciter on the rotor of an induction motor supplies magnetising ampere-turns at slip frequency, so the motor can run at unity or leading pf. 4. **Static VAR compensator / STATCOM:** thyristor or IGBT controlled devices for fast, continuous control. 5. **Good practice:** avoid running motors and transformers on light load, use correct motor size, use synchronous motors where possible. ```text Supply ---+---------+-----> to load | | === C [Load] | (lagging) --------- +---------+ capacitor supplies leading kVAR ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which load causes especially low pf when lightly loaded? A. Induction motor, pf about 0.2-0.3 at no load B. Incandescent lamp C. Resistance heater D. Synchronous motor over-excited Answer: A. Lightly loaded induction motors draw mostly magnetising current. (Key point: Causes) 2. Why does low pf need larger kVA ratings and cause higher losses? A. Voltage rises, so insulation is stressed B. For the same power I = P/(sqrt3 V cos phi) rises, so I^2R loss and voltage drop rise C. Power P itself increases D. Frequency falls Answer: B. Lower cos phi means higher current for the same kW. (Key point: Disadvantage logic) 3. Which describes static capacitors for pf improvement? A. Connected in series to supply lagging kVAR B. They give stepless control like synchronous condensers C. Connected in parallel with the load, they supply leading kVAR with low loss and little maintenance D. They are used only at no load Answer: C. Capacitors in parallel supply leading kVAR; stepless control is a feature of synchronous condensers. (Key point: Static capacitor) 4. What does a phase advancer do? A. It is a dc generator on the stator B. It reduces slip to zero C. It is a thyristor controlled capacitor D. An ac exciter on the rotor supplies magnetising ampere-turns at slip frequency, letting the induction motor run at unity or leading pf Answer: D. The phase advancer relieves the stator of magnetising current. 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