Verified Solution • 2025–26 Session moderate

QJEEM20Ph9c72: JEE Main 2020 Physics - Class 12 JEEM20Ph9c72 Solution: In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 …

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

Multiple Choice Options

  • A 15 A
  • B 20 A
  • C 25 A
  • D 10:00 AM

Step-by-Step Verified Solution

Total power is

= (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)

= 4325 W

∴\therefore Current = 4325220{{4325} \over {220}} = 19.66 A ≃\simeq 20 A

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