In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be :

Question

In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be : (A) 8 A (B) 10 A (C) 12 A (D) 14 A

Tardigrade Admin · Accepted Answer

15 × 40+5 × 100+5 × 80+1000=V × I 600+500+400+1000=220 I I=(2500/220)=11.36 I=12 A

Q. In a large building, there are 15 bulbs of 40{\,} W, 5 bulbs of 100 W, 5 fans of $80 W$ and 1 heater of $1 kW$ . The voltage of the electric mains is $220 V$ . The minimum capacity of the main fuse of the building will be :

8 A

10 A

12 A

14 A

$15 \times 40 + 5 \times 100 + 5 \times 80 + 1000 = V \times I$
$600 + 500 + 400 + 1000 = 220 I$
$I = \frac{2500}{220} = 11.36$
$I = 12 A$

Q. In a large building, there are 15 bulbs of 40{\,} W, 5 bulbs of 100 W, 5 fans of 80W and 1 heater of 1kW. The voltage of the electric mains is 220V. The minimum capacity of the main fuse of the building will be :

8 A

10 A

12 A

14 A

15×40+5×100+5×80+1000=V×I 600+500+400+1000=220I I=2202500​=11.36 I=12A

Q. In a large building, there are 15 bulbs of 40{\,} W, 5 bulbs of 100 W, 5 fans of $80 W$ and 1 heater of $1 kW$ . The voltage of the electric mains is $220 V$ . The minimum capacity of the main fuse of the building will be :

$15 \times 40 + 5 \times 100 + 5 \times 80 + 1000 = V \times I$
$600 + 500 + 400 + 1000 = 220 I$
$I = \frac{2500}{220} = 11.36$
$I = 12 A$