Several electric bulbs designed to be used on a 220 V line are rated 10 W . How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Given,

Potential Difference, V = 220 V

Power rating of each bulb, P = 10 W

Total current, I = 5 A

Let the number of lamps in parallel be n.

First, we will find the resistance of each bulb.

We know that,

$P=V\times I\phantom{\rule{0ex}{0ex}}$

$P=V\times \frac{V}{R}$ $(\because V=I\times R,I=\frac{V}
{R})\phantom{\rule{0ex}{0ex}}$

$P=\frac{{V}^{2}}{R}\phantom{\rule{0ex}{0ex}}$

$R=\frac{{V}^{2}}{P}\phantom{\rule{0ex}{0ex}}$

$R=\frac{(220{)}^{2}}{10}\phantom{\rule{0ex}{0ex}}$

$R=\frac{48400}{10}\phantom{\rule{0ex}{0ex}}$

$R=4840\Omega$

Resistance of 1 bulb = $4840\Omega$

For a flow of 5A current, we need to find the equivalent resistance of the circuit.

By Ohm's Law

$V=I\times {R}_{E}\phantom{\rule{0ex}{0ex}}$

${R}_{E}=\frac{V}{I}\phantom{\rule{0ex}{0ex}}$

${R}_{E}=\frac{220}{5}\phantom{\rule{0ex}{0ex}}$

${R}_{E}=44\Omega$

We know that, 

In parallel combination, the equivalent resistance is given by-
$\frac{1}{{\mathrm{R}}_{\mathrm{p}}}=\frac{1}{{\mathrm{R}}_{1}}+\frac{1}{{\mathrm{R}}_{2}}+\frac{1}{{\mathrm{R}}_{3}}+.......$

$\therefore \frac{1}{{\mathrm{R}}_{\mathrm{E}}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+.......$ (n times)

$\frac{1}{{\mathrm{R}}_{\mathrm{E}}}=n\times \frac{1}{\mathrm{R}}\phantom{\rule{0ex}{0ex}}$

$\frac{1}{44}=n\times \frac{1}{4840}\phantom{\rule{0ex}{0ex}}$

$\therefore \mathrm{n}=\frac{4840}{44}\phantom{\rule{0ex}{0ex}}$

$\mathrm{n}=\frac{1210}{11}\phantom{\rule{0ex}{0ex}}$

$\mathrm{n}=110\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

So, 110 lamps can be connected in parallel.

Updated on: 10-Oct-2022

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