Q. If voltage across a bulb rated 220 volt-100 watt drops by $2.5\%$ of its rated value, the percentage of the rated value by which the power would decrease is

Solution:

Power, $P = \frac{V^2}{R}$
As the resistance of the bulb is constant
$\therefore \, \, \frac{\Delta V^2}{P}=\frac{2\Delta V}{V}$
$\%$ decrease in power $ =\frac{\Delta P}{P} \times 100=\frac{2\Delta V}{V} \times 100$
$\hspace40mm = 2 \times \, 2.5 \% = 5\% $

You must select option to get answer and solution

Questions from AIPMT 2012

Questions from Current Electricity

Physics Most Viewed Questions

8. The hot wire ammeter measures

Rajasthan PMT 2004 Alternating Current