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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 the electric mains is 220 V. The minimum capacity of the main fuse of the building will be

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12 A

14 A

8 A

10 A

answer is A.

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Detailed Solution

12 A

Voltage - 220V

40W bulbs- 15 - 600 Watt

100W bulbs- 5 - 500 Watt

80W fans - 5 - 400 Watt

1 kW heater - 1000 Watt

Total power- 600+ 500+400+1000 = 2500 Watt

Therefore, minimum capacity of the main fuse of the building will be P/V = 2500/220 = 11.36

Approximately equal to 12 A.

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