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SarkariResultOutBMS · 2024

Previous Year Question

A separately excited DC generator achieves maximum efficiency when terminal voltage is 220 V and induced EMF is 240 V. If the armature resistance is 0.25Ω, then what will be stray losses in the generator?
  1. 1000 W
  2. 1600 W
  3. 3200 W
  4. 2000 W

Correct answer: B

Explanation

At the stated operating point, armature current is Ia=(E−V)/Ra=(240−220)/0.25=80 A. Maximum efficiency occurs when variable copper loss equals constant/stray loss, so stray loss=Ia²Ra=80²×0.25=1600 W. This matches option B.

Formula

Ia=(240−220)/0.25=80A; stray loss=Ia²Ra=1.6kW

Step-by-step solution

Stray Loss at Maximum Efficiency explanatory diagram

Correct Answer

B — 1600 W

Concept and Reasoning

At the stated operating point, armature current is Ia=(E−V)/Ra=(240−220)/0.25=80 A. Maximum efficiency occurs when variable copper loss equals constant/stray loss, so stray loss=Ia²Ra=80²×0.25=1600 W. This matches option B.

Calculation / Key Relation

Ia=(240−220)/0.25=80A

stray loss=Ia²Ra=1.6kW

Exam Takeaway

Maximum efficiency condition: variable copper loss = constant loss.

Common Mistake

Use induced emf minus terminal voltage to obtain generator armature current.

Common mistake

Use induced emf minus terminal voltage to obtain generator armature current.

Exam tip

Maximum efficiency condition: variable copper loss = constant loss.

Quick method

Maximum efficiency condition: variable copper loss = constant loss.