JEE Main 2014PhysicsCurrent ElectricityMediumMCQ

JEE Main 2014 — Current Electricity Question with Solution

JEE Main 2014 (06 Apr)

Question

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:

Choose an option

Show full solutionCorrect option: C
Correct answer
C12 A

Step-by-step explanation

The power generated by each electrical device is calculated below

Item Number Power Consumed
40 W bulb 15 40×15=600 Watt
100 W bulb 5 100×5=500 Watt
80 W fan 5 80×5=400 Watt
1000 W heater 1 1000 Watt

The total power consumed=2500 Watt

So the minimum current capacity

i=PV=2500220=12511=11.36≅12 A

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About this question

This is a previous-year question from JEE Main 2014, covering the Current Electricity chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.