An electric bulb is rated 60W, 220V. Calculate the resistance of its filament when it is operating normally
- 296.7\(\Omega\)
- 400.0\(\Omega\)
- 512.2\(\Omega\)
-
806.7\(\Omega\)
The correct answer is: D
Explanation
R = \(\frac{v^2}{P} = \frac{220^2}{100} = 806.7\Omega\)