A. 100 Ω
B. 200 Ω
C. 300 Ω
D. 500 Ω
Correct Answer:
Option B – 200 Ω
Explanation
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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