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  4. In a marriage hall, there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. If the voltage of the electric main is 220 V, then the minimum fuse capacity (in A ) of the building should be

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Q. In a marriage hall, there are $15$ bulbs of $45 \, W,$ $15$ bulbs of $100 \, W,$ $15$ small fans of $10 \, W$ and $2$ heaters of $1 \, kW.$ If the voltage of the electric main is $220 \, V,$ then the minimum fuse capacity (in $A$ ) of the building should be

NTA AbhyasNTA Abhyas 2022

Solution:

Total power is given by, sum of power of all the electrical components,
i.e. $P_{\text {total }}=P_1+P_2+P_3+P_4$, where $P_1, P_2, P_3, P_4$ represents power of $45\, W$ bulb, power of $100\, W$ bulb, power of small fan and power of heater respectively.
$\Rightarrow P_{\text {total }}=(15 \times 45)+(15 \times 100)+(15 \times 10)+(2 \times 1000)=4325\, W$
So current is $=\frac{4325}{220}=19.66 \,A \approx 20 \,A$