Question

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

• A. 20 %
• B. 2.5 %
• C. 5 %
• D. 10 %

Answer 1

Answer: (C)

5 %

Answer 2

: Power, $P = \frac{V^{2}}{R}$

As the resistance of the bulb is constant

$\therefore\frac{\Delta P}{P} = \frac{2\Delta V}{V}$

% decrease in power

$= \frac{\Delta P}{P} \times 100 = \frac{2\Delta V}{V} \times 100$

$= 2 \times 2.5\% = 5\%$