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.

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

Answer 1

Answer: (A)

$12$A

Item	No.	Power
$40$W bulb	$15$	$600$ Watt
$100$W bulb	$5$	$500$ Watt
$80$W fan	$5$	$400$ Watt
$1000$W heater	$1$	$1000$ Watt

Total Wattage $=2500$ Watt
So current capacity $i=\frac{P}{V}=\frac{2500}{220}=\frac{125}{11}=11.36\cong 12$ Amp.