100 Ω
200 Ω
300 Ω
500 Ω
Correct answer is B
Power of the lamp = 60W = (V2)/R and the voltage rate = 120v.
R = (V2)/60 = (120 X 120)/60 = 240 Ω, the fuse wire that should ensure current does not exceed this value should be rated 200 Ω
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