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 the electric mains is 220 V. The minimum capacity of the main fuse of the building will be:

Question

Aicharya · Accepted Answer

The power generated by each electrical device is calculated below

Item	Number	Power Consumed
$40 W$ bulb	$15$	$40 \times 15 = 600 Watt$
$100 W$ bulb	$5$	$100 \times 5 = 500 Watt$
$80 W$ fan	$5$	$80 \times 5 = 400 Watt$
$1000 W$ heater	$1$	$1000 Watt$

The total power consumed $= 2500 Watt$

So the minimum current capacity

$i = \frac{P}{V} = \frac{2500}{220} = \frac{125}{11} = 11.36 ≅ 12 A$

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…

Solution