In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of electric mains is 220 V. The minimum capacity of the main fuse of the building will be: (a) 8 A (b) 10 A (c) 12 A (d) 14 A

Answer: (c). Total power = 15(40) + 5(100) + 5(80) + 1000 = 600 + 500 + 400 + 1000 = 2500 W. I = (2500)/(220) ≈ 11.4 A. Minimum fuse rating above this is 12 A. So the answer is (c).

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Electrical Power & Heating Effect

In a large building, there are $15$ bulbs of $40\text{ W},\ 5$ bulbs of $100\text{ W},\ 5$ fans of $80\text{ W}$ and $1$ heater of $1\text{ kW}.$ The voltage of electric mains is $220\text{ V}.$ The minimum capacity of the main fuse of the building will be:

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Answer: (c)

Total power $= 15(40) + 5(100) + 5(80) + 1000 = 600 + 500 + 400 + 1000 = 2500$ W.
$I = \frac{2500}{220} \approx 11.4$ A.
Minimum fuse rating above this is 12 A. So the answer is (c).

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