Correct Answer: Option D

Explanation

1 KW = 1000 watts, power P = IV, thus for a mains of 250 V, the maximum permissible Current I = \(\frac{\text{P}}{\text{V}}\) = \(\frac{1000}{250}\) = 4 A
Thus, the fuse rating in the plug should be 4 A

Putting safety into consideration, the appropriate rating of the fuse should be 5A

That is = 4 x 1.25 = 5A ( 1.25 = safety factor)

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