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 k W …

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 supply is 220V . The minimum fuse capacity (rated value) of the building will be:
  1. 5A
  2. 25A
  3. 15A
  4. 20A

Solution

Total power is 15×45+15×100+15×10+2×1000
=4325W
So current =4325220=19.66 A≈ 20 A

so we have to use a fuse which can tolerate atleast 20 A current.

Asked in: JEE Main 2020 (07 Jan Shift 2)

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