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:

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: (A) 25 A (B) 15 A (C) 10 A (D) 20 A

Tardigrade Admin · Accepted Answer

220 I = P = 15 Ã 45 + 15 Ã 100 + 15 Ã 10 + 2 Ã 103 I = (4325/220) = 19.66 I ≃ 20 A

Q. 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:

$25 A$

$15 A$

$10 A$

$20 A$

$220 I = P = 15 \times 45 + 15 \times 100 + 15 \times 10 + 2 \times 1 0^{3}$
$I = \frac{4325}{220} = 19.66$
$I ≃ 20 A$

Q. In a building there are 15 bulbs of 45W,15 bulbs of 100W, 15 small fans of 10W and 2 heaters of 1kW. The voltage of electric main is 220V. The minimum fuse capacity (rated value) of the building will be:

25A

15A

10A

20A