Question

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

Answer 1

Ans: In this problem we have

Power of the oven (P) = 2 kW = 2000 W

Potential Difference (V) = 220 V

We have P = VI

⇒I=P/V

    = 2000/220

    = 9.09A


Here the current drawn by the electric oven is 9.09 A, which exceeds the safe limit of the circuit that is 5A. Therefore electric fuse will melt and break the circuit.

Answer 2

Answer:

The current I = 9.0909 A, flowing in the conductor exceeds the safe limit of 5A, thus the fuse will blow and the circuit will open.

Explanation:

Given,

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

Because the current flowing in the conductor exceeds the safe limit of 5A, the fuse will blow and the circuit will open.