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

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(D) The total power $P$ consumed in the building is the sum of the power of all appliances.$P = (15 \times 45) + (15 \times 100) + (15 \times 10) + (2 \times 1000)$$P = 675 + 1500 + 150 + 2000 = 4325 \; W$Using the formula $P = V \times I$,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 $20 \; A$.

Class 12 Physics Current Electricity Electrical Energy and Power

Medium

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: .......... $A$

JEE MAIN 2020 All JEE MAIN

A
$10$
B
$25$
C
$15$
D
$20$

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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: .......... $A$

Similar Questions

The heat produced by a $100\, W$ heater in $2\, minutes$ will be equal to:

The emf of a battery is $2\, V$ and its internal resistance is $0.5 \,\Omega$. The maximum power which it can deliver to any external circuit will be ............. $watt$.

In India,electricity is supplied for domestic use at $220\,V$. It is supplied at $110\,V$ in the $USA$. If the resistance of a $60\,W$ bulb for use in India is $R$,then the resistance of a $60\,W$ bulb for use in the $USA$ will be:

Two electric bulbs have tungsten filaments of the same length. If one of them is rated $60\ W$ and the other $100\ W$,then:

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