Question

A house wiring supplied with a $220\mathrm{V}$ supply line is protected by a fuse of $9\mathrm{A}$. Find the maximum number of bulbs ( $60\mathrm{W}$ each) connected in parallel that can be turned on.

Answer 1

Step 1: Given data.

Given that bulbs are connected in parallel, so the voltage across each bulb is equal.

Step 2: Calculation.
$V=220volts$, Fuse rating is 9 amp (means can work safely up to $9\mathrm{amp}$ ), and each bulb is $60\mathrm{W}$.
Power, $P=V\times I$ Here, V is voltage and I is current.
$60=220\times I$
$I=\frac{60}{220}A$ (required current by a single lamp)
For ' n ' lamps current $=n\times I=9\mathrm{A}$
$n=\frac{9\times 220}{60}$
$n=33.$
So, a total of 33 bulbs are required for safe lighting.