An electric bulb is rated 60W, 220V. Calculate the resistance of its filament when it is operating normally
-
A.
296.7(Omega) -
B.
400.0(Omega) -
C.
512.2(Omega) -
D.
806.7(Omega)
Correct Answer: Option D
Explanation
P = (frac{v^2}{P} = frac{220^2}{100} = 806.7Omega)