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 )`
As the resistance of the bulb is constant
`therefore (DeltaP)/(P)=(2DeltaV)/(V)`
% decrease in power `=(DeltaP)/(P)xx100=(2DeltaV)/(V)xx100=2xx2.5%=5%`