Which of the following apparatus will require the smallest fuse rating for its protection?
60W, 240V
60W, 40V
40W, 12V
40W, 5V
Correct answer is A
I = \(\frac{P}{V}\) = \(\frac{60}{240}\) = 0.25A
I = \(\frac{P}{V}\) = \(\frac{60}{40}\) = 1.5A
I = \(\frac{P}{V}\) = \(\frac{40}{12}\) = 3.3A
I = \(\frac{P}{V}\) = \(\frac{40}{5}\) = 8A