In a large building, there are 15 bulbs of 40W, 5 bulbs of 100W, 5 fans of 80W and 1 heater of 1kW. 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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10A

12A

14A

answer is C.

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Detailed Solution

Power of 15 bulbs $= P_{1} = 15 \times 40 = 600 W$

power of 5 bulbs $= P_{2} = 5 \times 100 = 500 W$

power of 5 fans $= P_{3} = 5 \times 80 = 400 W$

power of 1 heater $= P_{4} = 1 \times 1000 = 1000 W$

total power= $P = P_{1} + P_{2} + P_{3} + P_{4} = 2500 W$

$P = V \times i$

$i = \frac{P}{V} = \frac{2500}{220} = 11.3 ≃ 12 A$

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