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Question: Answered & Verified by Expert

In a marriage hall, there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. If the voltage of the electric main is 220 V, then the minimum fuse capacity (in A) of the building should be

PhysicsCurrent ElectricityNEET
Options:
  • A 20 A
  • B 15 A
  • C 10 A
  • D 25 A
Solution:
1446 Upvotes Verified Answer
The correct answer is: 20 A

Total power is


(15×45)+(15×100)+(15×10)+(2×1000)=4325 W
So current is =4325220=19.66 A

Ans is 20 A

 

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