1.

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 .

Answer»

<P>`5A`
20 A
25 A
15 A

Solution :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


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