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 :

Solution

Physics sir · 2 steps

Add every watt in the building, divide by 220 V, and pick the smallest fuse above that current.

1. Parallel loads add. Every appliance in a building is connected in parallel across the same 220 V main. So the total power drawn is the sum of all the individual powers.

   Appliance	Count and rating	Power
   Bulbs	15 × 45 W	675 W
   Bulbs	15 × 100 W	1500 W
   Fans	15 × 10 W	150 W
   Heaters	2 × 1000 W	2000 W
   Total	675 + 1500 + 150 + 2000	4325 W

   I = P/V = (4325 W)/(220 V) ≈ 19.7 A

2. So. The fuse must carry at least 19.7 A without blowing. The smallest rating that does this is 20 A, option (b).

The idea to keep. For a fuse rating, find the total current I = Pₜₒₜₐₗ/V with everything switched on. Then choose the smallest rating that is just above it.

Answer from the NTA final answer key. Solution by PrepWiser.