Total power is (15×45)+(15×100)+(15×10)+(2×1000)
=4325W
So current =2204325=19.66A≈20A
so we have to use a fuse which can tolerate atleast 20 A current.
JEE Main 2020 — Physics Electromagnetism
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:
Held on 7 Jan 2020 · Verified 6 Jul 2026.
5A
25A
15A
20A
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