Question 263: Current Electricity

Question

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 :

Accepted Answer

Total power is
= (15 ÃÂ 45) + (15 ÃÂ 100) + (15 ÃÂ 10) + (2 ÃÂ 1000)
= 4325 W
$$ 	herefore $$ Current = $${{4325} \over {220}}$$ = 19.66 A $$ \simeq $$ 20 A

Answer

15 A

Answer

20 A

Answer

25 A

Answer

10 A

Current Electricity

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 :

Choose the correct answer: