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Q. In a large building, there are $12$ bulbs of $40 \,W$, $4$ bulbs of $100\, W , 7$ fans of $60 \,W$ and $1$ heater of $1\, kW$. The voltage of the electric mains is $230 \,V$. The minimum capacity of the main fuse of the building will be _____ $A$.

Current Electricity

Solution:

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