In a large building , there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80W and 1 heater of 1 kW. The voltage of the electric main is 220 V (rms). The minimum capacity of the main fuse (rms value) of the building will be nearly

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6 A

8 A

12 A

14 A

answer is C.

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Detailed Solution

Item No. Power
40 W bulb 15 600 Watt
100 W bulb 5 500 Watt
80 W fan 5 400 Watt
1000 W heater 1 1000 Watt
Total Wattage = 2500 Watt
So current capacity $i = \frac{P}{V} = \frac{2500}{220} = \frac{125}{11} = 11.36 ≅ 12 A m p$

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