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JEE Main20207 Jan 2020Evening ShiftPhysicsCurrent ElectricityActual

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 k W . The voltage of electric main supply is 220 V . The minimum fuse capacity (rated value) of the building will be:

Options

  1. A5 A
  2. B25 A
  3. C15 A
  4. D20 A

Correct answer

D. 20 A

Step-by-step solution

Total power is 15 × 45 + 15 × 100 + 15 × 10 + 2 × 1000 = 4325 W So current = 4325 220 = 19 . 66   A ≈   20   A so we have to use a fuse which can tolerate atleast 20 A current.

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