JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 17)
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 :
np
p
$\frac{\mathrm{p}}{\mathrm{n}^2}$
$\frac{p}{n}$
Explanation
$$\begin{aligned}
& \mathrm{R}_{\mathrm{s}}=\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3+\ldots \ldots \ldots+\mathrm{R}_{\mathrm{n}} \\
& \frac{\mathrm{~V}^2}{\mathrm{P}_{\mathrm{s}}}=\frac{\mathrm{V}^2}{\mathrm{P}}+\frac{\mathrm{V}^2}{\mathrm{P}}+\ldots \ldots . .+\frac{\mathrm{V}^2}{\mathrm{P}_{\mathrm{n}}} \\
& \mathrm{P}_{\mathrm{s}}=\frac{\mathrm{P}}{\mathrm{n}}
\end{aligned}$$
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