If the current in an electric bulb increases by 1 %, what will be the change in the power of a bulb ? [Assume that the resistance of the filament of a bulh remains constant]

Options

A. increases by 1%
B. decreases by 1 %
C. increases by 2%
D. decreases by 2%

Answer

C

Explanation

increases by 2%
Power of a bulb P = `I^(2) ` R , where R = Resistance 0 force
Now, increase or 1% of current , new current
I. = I + 1% , I = I + 0.01 I
I. = 1.01 I
New Power P. = `I.^(2) ` R
P. = `(1.01I)^(2) R = 1.02 I^(2)` R
`(P.)/(P) = (1.02 I^(2)R)/(I^(2) R)`
P. = 1.02 P
= P + 0.02 P = P + 2% P
Change in the power of bulb = 2%