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. 0.2
B. 0.025
C. 0.05
D. 0.1

Answer

(c )

Explanation

Here, `V = 220 V, P = 100 W, Delta V// V = 2.5%`
As `P = (V^2)/(R ) :. (DeltaP)/(P) = (2 DeltaV)/(V) + (DeltaR)/(R )`
(As resistance of bulb is constant, `:. DeltaR = 0)`
Thus ` = (DeltaP)/(P) = (2 DeltaV)/(V) =2xx2.5 = 5%`