Question

An electric iron is rated 2kw at 220V. calculate the capacity of the fuse that should be used for the electric iron...?

Answer 1

P=VI
I=P/V
I=2000/220
I=9.09A
in this case a fuse of 10A should be used

Answer 2

The capacity of the fuse is 9.09 A.

Given:

Power (P) = 2 kW  

Since 1 kW = 1000 W

Therefore P = 2 kW  =  2000 W

Potential difference (V) = 220 V

Current = Need to calculate  

Since:

$P=V \times I$

Therefore $I=\frac{P}{V}$

Plug the given values:  

$I=\frac{2000 \mathrm{W}}{220 \mathrm{V}}=9.09 \mathrm{A}$

Hence the capacity of the fuse that should be used for the electric iron rated 2kw at 220V will be withstand with 10 A capacity.