Question

The capacity of an electric motor is 1.5kw what happens if it is used in 5a domestic electric

Answer 1

Answer:

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.

Explanation:

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