--- title: "Number of lamps an alternator can add" subject: "Electric Machines II" chapter: 1 key: "semester_5/electric_machines_ii:lamps-added-850kva-alternator" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/lamps-added-850kva-alternator site: BCT Engineering --- # Number of lamps an alternator can add Electric Machines II · Chapter 1: Three Phase Synchronous Generator - Asked in: 2069 Chaitra, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/lamps-added-850kva-alternator - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1 ## Question A 850 kVA, 380 V, 50 Hz 3-phase alternator delivers 400 kW to an 3-phase induction motor at a power factor of 0.8 lagging. Calculate the number of 100 W lamps which can be added to the alternator so that the alternator does not overload beyond its capacity. ## Short answer Present load: P=400 kW, Q=300 kVAR, S=500 kVA. Maximum active power at 850 kVA with same Q: P_max = √(850² - 300²) = 795.3 kW. Additional lamp load = 795.3 - 400 = 395.3 kW. Number of 100 W lamps = 3953. ## Answer Lamps are resistive (unity pf), so they add only active power; the alternator limit is its rating, 850 kVA. #### Present load (induction motor) $$ \begin{aligned} P_1 &= 400\ \text{kW} \\ Q_1 &= P_1\tan\phi = 400\times\frac{0.6}{0.8} = 300\ \text{kVAR} \\ S_1 &= \frac{400}{0.8} = 500\ \text{kVA} \end{aligned} $$ #### Maximum active power at 850 kVA with the same kVAR $$ \begin{aligned} P_{max} &= \sqrt{S^2 - Q_1^2} = \sqrt{850^2 - 300^2} = 795.30\ \text{kW} \\ \text{Lamp load} &= 795.30 - 400 = 395.30\ \text{kW} \\ \text{Number of lamps} &= \frac{395.30\times10^3}{100} = 3953 \end{aligned} $$ ```text S = 850 kVA .'| .' | Q = 300 kVAR .' | O-------+--------+ 400 kW 395.3 kW lamps ``` New pf of the alternator $= 795.3/850 = 0.936$ lagging. **Answer: about 3953 lamps of 100 W can be added (395.3 kW).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why do the lamps add only active power? A. They are resistive loads at unity pf B. They are inductive loads C. They are capacitive loads D. They have no power rating Answer: A. Lamps are resistive, so they add kW but no kVAR; the motor's reactive load stays 300 kVAR. (Key point: Lamps add only P) 2. What is the reactive power of the 400 kW motor at 0.8 pf? A. 400 kVAR B. 500 kVAR C. 240 kVAR D. 300 kVAR Answer: D. Q = P tan(phi) = 400 x 0.6/0.8 = 300 kVAR. (Key point: Reactive power of the motor) 3. What is the maximum total active power within 850 kVA? A. 850 kW B. 795.3 kW C. 550 kW D. 500 kW Answer: B. P = sqrt(850^2 - 300^2) = 795.3 kW since Q is fixed. (Key point: Maximum active power) 4. How many 100 W lamps can be added? A. About 3953 B. About 5500 C. About 4000 minus 400 D. About 7953 Answer: A. Extra load = 795.3 - 400 = 395.3 kW = 395300 W / 100 W = 3953 lamps. 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