Correct Answer: Option D

Explanation

The maximum current drawn by the cooker is given by;
I = \(\frac{P}{V}\)
∴ I = \(\frac{\text{3000}}{200}\)
I = 15A
The fuse rating = I \(\times\) 125% (the fuse is rated at 125% of the normal operating current)

Hence, the fuse rating = 15A \(\times\) (\(\frac{125}{100}\)) = 18.75amps

Therefore, the best fuse is 20A.

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