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

Answer 1

The minimum fuse capacity for the building is D. 20 A.

Step-by-step explanation:

To determine the minimum fuse capacity for the building, we need to add up the power ratings of all the devices plugged into the circuit. First, let's convert the power ratings into watts:

• 15 bulbs of 45 W each: 15 * 45 = 675 W
• 15 bulbs of 100 W each: 15 * 100 = 1500 W
• 15 small fans of 10 W each: 15 * 10 = 150 W
• 2 heaters of 1 kW each: 2 * 1000 = 2000 W

Now, let's add all the power ratings together: 675 + 1500 + 150 + 2000 = 4325 W. Since the voltage of the electric main is 220 V, we can divide the total power by the voltage to find the current:

Current (I) = Power (P) / Voltage (V)

Current (I) = 4325 W / 220 V = 19.66 A

The minimum fuse capacity (rated value) needed for the building will be greater than or equal to 19.66 A. Therefore, the correct answer is option D. 20 A.