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. 0.1
B. 0.2
C. 0.025
D. 0.05

Answer

D

Explanation

We have the following relation
`P = (V^(2))/(R )` …(i)
And for small changes we can differentiate this relation as follows:
`(dP)/(dV) = (2V)/(R ) rArr dP = (2V)/(R ) dV rArr Delta P = (2V)/(R ) Delta V` …(ii)
Dividing equation (ii) from equation (i) and then multiplying both sides by 100, we get the following:
`(Delta P)/(P) xx 100 = 2 (Delta V)/(V)xx 100` ...(iii)
Note that we calculate errors also using similar result. From equation (iii) we can see that change in power will be double to that in case of voltage. Hence option (d) is correct