In a building,there are 15 bulbs of 45 ; W,15 | vedclass

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(D) The total power $P$ consumed in the building is the sum of the power of all appliances.$P = (15 	imes 45) + (15 	imes 100) + (15 	imes 10) + (2

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$20$

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(D) The total power $P$ consumed in the building is the sum of the power of all appliances.$P = (15 	imes 45) + (15 	imes 100) + (15 	imes 10) + (2 	imes 1000)$$P = 675 + 1500 + 150 + 2000 = 4325 \; W$Using the formula $P = V 	imes I$,where $V = 220 \; V$ is the voltage:$I = \frac{P}{V} = \frac{4325}{220} \approx 19.66 \; A$Since the fuse must handle the total current,the minimum fuse capacity should be the next standard integer value,which is $20 \; A$.

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

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