Question

(a) An electric iron is rated at 230 V, 750 W. Calculate (i) the maximum current, and (ii) the number of units of electricity it would use in 30 minutes.
(b) Which of the following fuse ratings would be suitable for this electric iron?
1 A, 3 A, 5 A, 13 A

Answer 1

(a) Here,
Power, P = 750 W
Potential difference, V = 230 V

(i) We know that:
P = V $\times$ I
Current, I = P / V
I = 750 / 230 = 3.26 A
Thus, the value of maximum current is 3.26 A.

(ii) Given:
Power, P = 750 W = 0.750 kW
Time, t = 30 minutes = 30 / 60 hours = 1 / 2 h
We know that:
Electrical energy, E = P $\times$ t
= 0.750 $\times$ (1 / 2) = 0.375 kWh or 0.375 units
Thus, the number of units of electricity is 0.375.

(b) A fuse rating of 5 A would be suitable for this electric iron because the fuse used by an appliance should be slightly more than the normal current drawn through it. In this case, the normal current drawn is 3.26 A.