An electric oven of 2kW power rating is operated in a domestic electric circuit that has a current rating of 5A. If the supply voltage is 220V, what result do you expect? Explain.


Power (P) = 2kw =2000W

Current (I) = 5A

Voltage (V) = 220V

Power = current × voltage

P = I × V

Current = Power/voltage

I = P/V

I = 2000/220

I = 9.0909 A

As the amount of current flowing in the conductor is more than 5 A which is the safe limit, then the fuse will get broken and hence the circuit will get open.

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