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JEE MAIN Physics QUESTION #5587
Question 1

A large building has the following electrical loads all running simultaneously at 220 V mains:

DeviceQuantityPower Each
Bulbs1540 W
Bulbs5100 W
Fans580 W
Heater11000 W

What is the minimum capacity of the main fuse required?

  • 12:00 AM✔️
  • 14 A
  • 8:00 AM
  • 10:00 AM
Correct Answer Explanation

Total power: $P = 15 \times 40 + 5 \times 100 + 5 \times 80 + 1 \times 1000 = 600 + 500 + 400 + 1000 = 2500\,\text{W}$

Total current: $I = \frac{P}{V} = \frac{2500}{220} \approx 11.36\,\text{A}$

The fuse must safely carry this current. The smallest standard value that is greater than or equal to 11.36 A from the options is 12 A. Answer: A.