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

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:, 8 A, 10 A, 12 A, 14 A

MARKS App · Accepted Answer

The power generated by each electrical device is calculated below Item Number Power Consumed 40 W bulb 15 40 × 15 = 600 Watt 100 W bulb 5 100 × 5 = 500 Watt 80 W fan 5 80 × 5 = 400 Watt 1000 W heater 1 1000 Watt The total power consumed = 2500 Watt So the minimum current capacity i = P V = 2500 220 = 125 11 = 11 . 36 ≅ 12 A

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	$1$	$1000 Watt$

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