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

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20 A

15 A

10 A

25 A

answer is A.

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Detailed Solution

Total power is(15×45)+(15×100)+(15×10) + (2 x 1000) = 4325 WSo current is =4325220=19.66AAns is 20 Amp

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