In a large building,there are 15 bulbs of 40 | vedclass

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(B) First,calculate the total power consumption $(P_{	ext{total}})$ of all appliances:$P_{	ext{total}} = (15 	imes 40\ W) + (5 	imes 100\ W) + (5

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$12$

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(B) First,calculate the total power consumption $(P_{	ext{total}})$ of all appliances:$P_{	ext{total}} = (15 	imes 40\ W) + (5 	imes 100\ W) + (5 	imes 80\ W) + (1 	imes 1000\ W)$$P_{	ext{total}} = 600\ W + 500\ W + 400\ W + 1000\ W = 2500\ W$Using the formula for power $P = V 	imes I$,where $V = 220\ V$:$I = \frac{P_{	ext{total}}}{V} = \frac{2500}{220} \approx 11.36\ A$Since the fuse must handle the total current,the minimum capacity of the main fuse should be the next higher integer value,which is $12\ A$.

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 voltage of the electric mains is $220\ V$. The minimum capacity of the main fuse of the building will be ................ $A$.

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