Q.

In a building there are 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10W and 2 heater of 1 kW. The voltage of electric main is 220V. the minimum fuse capacity (rated value) of the building will be:

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a

10 A

b

20 A

c

25 A

d

15 A

answer is B.

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Detailed Solution

All the components are connected in parallel with the voltage source. Voltage across each each component is 220VTotal power consumed =  15×45+15×100+15×10+2×1000  =4325W So current is  i=PV=4325220=19.66A∴ A fuse of capacity 20A should be used.
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In a building there are 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10W and 2 heater of 1 kW. The voltage of electric main is 220V. the minimum fuse capacity (rated value) of the building will be: