If voltage across a bulb rated 220 volt-100 watt drops by `2.5 %` of its 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

d

Explanation

As given `(Delta V)/(V) xx 100 = 2.5%`
Since `P = (V^(2))/(R )`….(i)
So by differentiating, we get, `Delta P = (2V DeltaV)/(R ) `….(ii)
Dividing (ii) by (i)
`(Delta P)/(P) = (2VDelta V)/(R) xx (R )/(V) = (2DeltaV)/(V) = 2 xx 2.5 = 5%`