(a) Given:
Power, P = 3 kW = 3 1000 = 3000 W
Potential difference, V = 240 V
We know that:
P = V I
Current, I = P / V
= 3000 / 240 = 12.5 A
Thus, a current of 12.5 A is taken by an electric geyser of 3 kW, working on 240 V mains.
(b) A fuse of 13 A should be used in this geyser circuit because the rating of the fuse used in an appliance should be slightly more than the normal current drawn through it.