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

Options

A. `20%`
B. `2.5%`
C. `5%`
D. `10%`

Answer

C

Explanation

power : `P=(V^(2))/(R )`
For small variation in the value of power after differentiation:
`(DeltaP)/(P)xx 100=(2xx DeltaV)/(V ) xx 100%`
`=2xx2.5=5%`
`:.` Power decreased by 5%