There are ' $n$ ' number of identical electric bulbs, each is designed to draw a power $p$ independently from the mains supply. They are now joined in series across the mains supply. The total power drawn by the combination is :
Solution
<p>$$\begin{aligned}
& \mathrm{R}_{\mathrm{s}}=\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3+\ldots \ldots \ldots+\mathrm{R}_{\mathrm{n}} \\
& \frac{\mathrm{~V}^2}{\mathrm{P}_{\mathrm{s}}}=\frac{\mathrm{V}^2}{\mathrm{P}}+\frac{\mathrm{V}^2}{\mathrm{P}}+\ldots \ldots . .+\frac{\mathrm{V}^2}{\mathrm{P}_{\mathrm{n}}} \\
& \mathrm{P}_{\mathrm{s}}=\frac{\mathrm{P}}{\mathrm{n}}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Electrical Power
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