--- title: "Leading kVAR supplied by synchronous motor" subject: "Electric Machines II" chapter: 2 key: "semester_5/electric_machines_ii:pf-correction-leading-kvar" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/pf-correction-leading-kvar site: BCT Engineering --- # Leading kVAR supplied by synchronous motor Electric Machines II · Chapter 2: Three Phase Synchronous Motor - Asked in: 2081 Bhadra, 8 marks - Video: [Synchronous Motor | Reactive power & power factor | Synchronous Machines | Lec-41](https://www.youtube.com/watch?v=bZU5F01n02M) — Education 4u - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/pf-correction-leading-kvar - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-2 ## Question A synchronous motor absorbing 60 kW is connected in parallel with a factory load of 240 kW having a lagging p.f. of 0.8. If the connected combined load has a p.f. of 0.9, what is the value of the leading kVAR supplied by the motor and at what p.f. is it working? ## Short answer Factory: 240 kW at 0.8 lag => 180 kVAR lag. Combined 300 kW at 0.9 lag => 145.3 kVAR lag. Motor supplies 34.7 kVAR leading. Motor kVA = √(60²+34.7²)=69.31 kVA, pf = 0.866 leading. ## Answer The synchronous motor is over-excited so it supplies leading kVAR, which cancels part of the lagging kVAR of the factory load. #### Factory load $$ \begin{aligned} P_1 &= 240\ \text{kW},\quad \cos\phi_1 = 0.8\ \text{lag},\quad \tan\phi_1 = 0.75 \\ Q_1 &= 240 \times 0.75 = 180\ \text{kVAR (lagging)} \end{aligned} $$ #### Combined load $$ \begin{aligned} P &= 240 + 60 = 300\ \text{kW} \\ \cos\phi &= 0.9 \ \Rightarrow\ \tan\phi = \tan(25.84^\circ) = 0.4843 \\ Q &= 300 \times 0.4843 = 145.30\ \text{kVAR (lagging)} \end{aligned} $$ #### kVAR from the motor $$ \begin{aligned} Q_m &= Q_1 - Q = 180 - 145.30 = 34.70\ \text{kVAR (leading)} \\ S_m &= \sqrt{60^2 + 34.70^2} = 69.31\ \text{kVA} \\ \cos\phi_m &= \frac{60}{69.31} = 0.866\ \text{leading} \end{aligned} $$ ```text kVAR (lag +) | 180 + factory load | 145.3 + combined (pf 0.9) | motor gives -34.7 kVAR +--------------------------- kW 60 240 300 ``` **Answer:** The motor supplies **34.70 kVAR leading** and works at a power factor of **0.866 leading** (input 69.31 kVA). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What lagging kVAR does the 240 kW, 0.8 lagging factory load take? A. 180 kVAR B. 192 kVAR C. 300 kVAR D. 144 kVAR Answer: A. Q1 = P tan(phi) = 240 x 0.75 = 180 kVAR. (Key point: Factory kVAR) 2. For the combined 300 kW load at pf 0.9 lagging, what is the net kVAR? A. 145.3 kVAR B. 180 kVAR C. 270 kVAR D. 34.7 kVAR Answer: A. tan(25.84 degrees) = 0.4843, so Q = 300 x 0.4843 = 145.3 kVAR. (Key point: Combined load kVAR) 3. How is the motor's leading kVAR found? A. Add 180 and 145.3 B. Subtract: 180 - 145.3 = 34.7 kVAR C. Multiply 60 by tan(0.9) D. Take 0.1 x 300 Answer: B. The motor must cancel the difference between the factory's lagging kVAR and the permitted net kVAR. (Key point: Motor kVAR) 4. What is the motor's operating pf? A. 0.866 leading B. 0.9 leading C. 0.8 lagging D. 0.866 lagging Answer: A. S = sqrt(60^2 + 34.7^2) = 69.31 kVA, so pf = 60/69.31 = 0.866 leading. 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