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+1 vote
39.1k views
in Physics by (36.3k points)

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:

(1) 10 A

(2) 25 A

(3) 15 A

(4) 20 A 

by (30 points)
Total power is = (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000) = 4325 W So current is = 4325/220 =19.66

2 Answers

+2 votes
by (33.0k points)
selected by
 
Best answer

Answer is (4) 20 A

220 I = P = 15 × 45 + 15 × 100 + 15 × 10 + 2 × 103

I = 4325/220 = 19.66

I ≈ 20 A

+2 votes
by (51.9k points)

Answer is (4) 20 A

Total power is = (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)

= 4325 W

So current is = 4325/220 =19.66 A

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