Question

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

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

Answer 1

The correct option is D $20A$

Let the power consumed by the building be $P$.

power consumed by $15$ bulbs each of $45W=(45W\times 15)$

power consumed by $15$ bulbs each of $100W=(100W\times 15)$

power consumed by $15$ fans of each $10W=(10W\times 15)$

power consumed by $2$ heaters of each $1000W=(1000W\times 2)$

Then total power consumed $P=(45W\times 15)+(100W\times 15)+(10W\times 15)+(1000W\times 2)$
$P=4325W$

power $P=VI$
where supply voltage $V=220V$ ,

$I=$ current in the main supply wire of building
Therefore, $P=220I$
$\Rightarrow 220I=4325W$
$\Rightarrow I=19.66A\simeq 20A$
the minimum rated capacity of fuse must be equal to this current so that it can allow this current without breaking the circuit when all equipments are working together.

Hence, option (d) is correct.