In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of electric mains is 220 V. The minimum capacity fo the main fuse of the building will be :

Options

A. 14A
B. 8A
C. 10A
D. 12A

Answer

D

Explanation

All components are connected in parallel. Maximum current is drawn from the mains when everything is switched on simultaneously. We can use formula I= P/V to find the current required when particular component is running. Total current drawn:
`I = 15 xx (40)/(220) +5 xx (100)/(220) + 5 xx (80)/(220) +1 xx (1,000)/(220)`
`rArr I = (600 + 500 + 400 + 1,000)/(220) = (2,500)/(220) = 11.4A`
We can see that option (d) is the closest. When the fuse of 12A rating is there then it will not blow even if by chance every switch in the building is turned on.