In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the building will be:
(1) 10 A
(2) 25 A
(3) 15 A
(4) 20 A
Answer is (4) 20 A
220 I = P = 15 × 45 + 15 × 100 + 15 × 10 + 2 × 103
I = 4325/220 = 19.66
I ≈ 20 A
