- Asked 2/21
- 2065 Chaitra (old course) · 8 marks
30 kW shunt generator: emf, copper loss, efficiency (100 W)
Short answer
Load current 150 A, shunt current 4 A, armature current 154 A. Generated emf E = 200 + 154×0.05 = 207.70 V. Copper losses: armature 1185.8 W, shunt 800 W, total 1985.8 W. Efficiency = 30000/(30000+1985.8+100) × 100 = 93.50%.
A dc shunt generator gives full load output of 30 kW at a terminal voltage of 200 V. The armature and shunt field resistances are and respectively. The iron and friction losses are 100 W. Calculate (i) generated emf (ii) copper losses (iii) efficiency.
For a shunt generator and .
(i) Generated emf
(ii) Copper losses
(iii) Efficiency
Total losses copper loss iron and friction loss .
Answer: (i) ; (ii) copper loss ; (iii) .
✓ Answer checked by Claude Opus 5.5 (high effort)
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