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

V = 220 Volt, P = 100 watt, `R = V^(2)/P = 484 Omega`
`V. = 220-220 xx 2.5/100 = 214.5` V
`P. =V^(2)/R = (214.5)^(2)/484 = 95.0625` Watt
% decrease in power `=(100 - 95.0625)/100 xx 100 %`
`= 4.93 = 5%`