If voltage across a bulb rated 220 ~V , 50 ~W drops by 5 % of its rated value, the percentage of the rated value by which the power would decrease is

Step-by-step solution

The power P consumed by a bulb is related to the voltage V by the formula P = V^2 R , where R is the constant resistance of the bulb. Taking the logarithmic derivative of both sides, we get dP P = 2 dV V . Given that the voltage drops by 5 % , we have dV V = -0.05 . Substituting this into the expression for the fractional change in power: dP P = 2 (-0.05) = -0.10 . The negative sign indicates a decrease in power. Thus, the power decreases by 10 % of its rated value. Answer: 10 %