In a marriage hall, there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. If the voltage of the electric main is 220 V, then the minimum fuse capacity (in A ) of the building should be

Question

In a marriage hall, there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. If the voltage of the electric main is 220 V, then the minimum fuse capacity (in A ) of the building should be (A) 20 A (B) 15A (C) 10A (D) 25A

Tardigrade Admin · Accepted Answer

Total power is (.15× 45.)+(.15× 100.)+(.15× 10.)+(.2× 1000.)=4325 W So current is =(4325/220)=19.66 A Ans is 20 A

Q. In a marriage hall, there are $15$ bulbs of $45 W,$ $15$ bulbs of $100 W,$ $15$ small fans of $10 W$ and $2$ heaters of $1 kW .$ If the voltage of the electric main is $220 V,$ then the minimum fuse capacity (in $A$ ) of the building should be

$20 A$

$15 A$

$10 A$

$25 A$

Total power is

$(15 \times 45) + (15 \times 100) + (15 \times 10) + (2 \times 1000) = 4325 W$
So current is $= \frac{4325}{220} = 19.66 A$
Ans is $20 A$

Q. In a marriage hall, there are 15 bulbs of 45W, 15 bulbs of 100W, 15 small fans of 10W and 2 heaters of 1kW. If the voltage of the electric main is 220V, then the minimum fuse capacity (in A ) of the building should be

20A

15A

10A

25A