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

Let
power(P)=2 kW
=2000W
potential difference(V)=220V
Therefore, Current(I)=?
We know that
P=VI
Therefore
I=P÷V
=2000÷220
=9.09 Ampere
Therefore the must have the capacity of 10 Ampere to withstand for an electric iron.