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 is 220 V . The minimum fuse capacity (rated value) of the building will be

Option 1 -

10 A

Option 2 -

20 A

Option 3 -

25 A

Option 4 -

15 A

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago
    Correct Option - 4


    Detailed Solution:

    Total power is ( 15 * 45 ) + ( 15 * 100 ) + ( 15 * 10 ) + ( 2 * 1000 ) = 4325 W .

    So, current is 4325 220 = 19.66 A

    Answer is 20 A m p .

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