Question

There are $12$ bulbs of $40W,4$ bulbs of $100W,7$ fans of $60W$ and $1$ heater of $1kW$ in a large building. The voltage of the electric mains for the building is $230V.$ The minimum capacity of the main fuse of the building should be ______ $A.$

Answer 1

Answer: 10

$P_{\text{total }}=V\times I$
$\left[12\times 40\right]+\left[4\times 100\right]+\left[7\times 60\right]+\left[1\times 1000\right]=V\times I$
$480+400+420+1000=230\times I$
$\therefore I=\frac{2300}{230}$
$=10A$