In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs 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 approximately .

Options

A. `5A`
B. 20 A
C. 25 A
D. 15 A

Answer

B

Explanation

20A
Total power,
`15 xx 45 + 100 xx 15 + 10 xx 15 + 1000 xx 2 `
`P = 675 + 1500 + 150 + 2000`
`therefore P = 4325` W
Now,
`VI = 4325`
`therefore I = (4325)/(V) = (4325)/(220)`
`therefore I = 19.659 A " " therefore I = 20 `A