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Current Electricity

JEE Main level, single correct, medium

253 of 380

Question 253 of 380

In a building there are $15$ bulbs of $45W$ , $15$ bulbs of $100W,15$ small fans of $10W$ and $2$ heaters of $1kW$ . The voltage of electric main supply is $220V$ . The minimum fuse capacity (rated value) of the building will be:

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Answer: (D) $20A$

Total power is

\left(15\times 45\right)+\left(15\times 100\right)+\left(15\times 10\right)+\left(2\times 1000\right)

=4325W

So current

=\dfrac{4325}{220}=19.66A\approx 20A

so we have to use a fuse which can tolerate atleast 20 A current.

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Current Electricity (Electric Power and Heating Effect of Current), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

In a building there are 151515 bulbs of 45W45W45W , 151515 bulbs of 100W,15100W,15100W,15 small fans of 10W10W10W and 222 heaters of 1kW1kW1kW . The voltage of electric main supply is 220V220V220V . The minimum fuse capacity (rated value) of the building will be:

JEE Main question paper PDF

More Physics practice

In a building there are $15$ bulbs of $45W$ , $15$ bulbs of $100W,15$ small fans of $10W$ and $2$ heaters of $1kW$ . The voltage of electric main supply is $220V$ . The minimum fuse capacity (rated value) of the building will be: