An electric kettle is rated 240V, 1000W.
Explain the above statement.
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If a certain amount of power is dissipated for a given time, then energy is dissipated.
Energy (Power×time) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.
Energy dissipated = PtorVI×torV
2
RtorevenI
2
RtJoules.
We will exclude the time factor in this case. The power is 1 KW, that is, 1000 W and the voltage. The above formula is rewritten as I=P/V.
Therefore, the current required for the kettle I=1000/220 = 4.5 A.
Hence, to tolerate a current of 4.5 A, minimum of 5 A fuse should be used.
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