Q 12-03-227JEE MainJEE Main 2020 (7 Jan, Shift 2)Easy
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 the electric main supply is $220$ V. The minimum fuse capacity (rated value) of the building will be:
Answer: (D) $20$ A
Total power with everything on:
$$P = 15\times45 + 15\times100 + 15\times10 + 2\times1000 = 675 + 1500 + 150 + 2000 = 4325\ \text{W}$$
$$I = \frac{4325}{220} \approx 19.7\ \text{A}$$
The fuse rating must be at least this, so $20$ A.
Solution by Sreeraj P, M.Sc Physics