A light bulb is rated 100W for a 220V supply. The resistance of the bulb and the peak voltage of the source respectively are

Options

A. 242`Omega`and 311 V
B. 484 `Omega` and 311 V
C. 484 `Omega` and 440 V
D. 242 `Omega` and 440 V

Answer

B

Explanation

`R = (V^(2))/(P) = ((220)^(2))/(100) = 484 Omega`
`V_("peak") = sqrt(2) V_(rms) = sqrt(2) xx 220 = 311` V