Question

An electric oven of 2kW power rating is operated in a domestic electric circuit (220V) that has a current rating of 5A. What result do you expect? Explain.

Answer 1

SOLUTION:-

We know that,

$P = \frac{ {V}^{2} }{R} = > 2 \times 10 { }^{3} = \frac{220 \times 220}{R} \\ \\ = > R= \frac{220 \times 220}{2 \times {10}^{3} } \: \: or \: \: i = \frac{V}{R} \\ \\ = > i = \frac{ \frac{220}{220 \times 220} }{2 \times {10}^{3} } \\ \\ = > i = \frac{220 \times 2 \times {10}^{3} }{220 \times 220} \\ \\ = > i = \frac{2000}{220} = \frac{100}{11} \\ \\ = > i = 9.09 \: ampere$

As this stage, due to very high value of electric current, the oven will not function and would be damaged.

Hope it helps ☺️

Answer 2

Given

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

Potential Difference (V) = 220 V

Solution

We have P = VI

⇒I=P/V

    = 2000/220

    = 9.09A

Explanation

• 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.

hope it helps...✌️