If voltage across a bulb rated 22 volt 100 watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is
A. 0.05
B. 0.1
C. 0.2
D. 0.025

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1 Answers

Correct Answer - 0.01
`PpropV^(2)` So `(DeltaP)/(P)=2(DeltaV)/(V)=5%`

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