InterviewSolution
Saved Bookmarks
| 1. |
If voltage across a bulb rated 220 volt-100 watt drops by `2.5 %` of its value, the percentage of the rated value by which the power would decrease isA. 0.2B. 0.025C. 0.05D. 0.1 |
|
Answer» Correct Answer - C `"Power", R=(V^(2))/(R)` As the resistance of the bulb is constant `therefore (DeltaP)/(P)=(2DeltaV)/(V)` `"% decrease in power"=(DeltaP)/(P)xx100=(2DeltaV)/(V)xx100` `2xx2.5%=5%` |
|