(d) an electric iron is rated 2 kw at 220 v. Calculate the capacity of the fuse that should be used for the electric iron.
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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
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