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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.
$P_{t o t a l}=P_{1}+P_{2}+P_{3}+P_{4}$ , where $P_{1},P_{2},P_{3},P_{4}$ represents power of $45W$ bulb, power of $100W$ bulb, power of small fan and power of heater respectively.
$\Rightarrow P_{t o t a l}=\left(\right.15\times 45\left.\right)+\left(\right.15\times 100\left.\right)+\left(\right.15\times 10\left.\right)+\left(\right.2\times 1000\left.\right)=4325W$
So current is $=\frac{4325}{220}=19.66A\approx20A$