Current Electricity

JEE Main level, single correct, medium

262 of 380

Question 262 of 380

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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Answer: (C) 12 A

The power generated by each electrical device is calculated below Item Number Power Consumed 40W40W bulb 1515 40×15=600Watt40\times 15=600Watt 100W100W bulb 55 100×5=500Watt100\times 5=500Watt 80W80W fan 55 80×5=400Watt80\times 5=400Watt 1000W1000W heater 11 1000Watt1000Watt The total power consumed=2500Watt=2500Watt
So the minimum current capacity
i=PV=2500220=12511=11.36≅12Ai=\dfrac{P}{V}=\dfrac{2500}{220}=\dfrac{125}{11}=11.36\cong 12A

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