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An electric iron consumes 1 kW electric power when operated at 220V. What should be the current rating of the fuse?

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Here, P = 1 kW = 1000 W 

V = 220 V 

∴Current, I = \(\frac{P}{V}\)\(\frac{1000}{220}\) = 4.5 A

Hence, to tolerate a current of 4.5 A, minimum of 5 A current rating fuse should be used.

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