If voltage across a bulb rated 220 ~V , 100 ~W drops by 2.5 % of its rated value, then the percentage of the rated value by which the power would decrease is

Step-by-step solution

Given, P=100 ~W V=220 ~V We know that, P= V^2 R where, R is resistance of bulb. R= V^2 P = 220 220 100 =484 Now, according to question, voltage drops by 2.5 % of its rated value. New voltage, V^ =220-2.5 % of 220 =220- 2.5 100 220=214.5 ~V aligned & New power, P^ = (V^ )^2 R = 214.5 214.5 484 =95.06 ~W & aligned % decrease in power & = P-P^ P 100 & = 100-95.06 100 100 5 % aligned aligned