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JEE Main 2014 Paper

Section: Physics

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Question : 10 of 90

Marks: +1, -0

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

12 A
14 A
8 A
10 A

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\cong12

Amp.

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