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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2025
  • Easy
  • MCQ

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 main supply. The total power drawn by the combination is :

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Correct Answer

(C) \(\frac{\mathrm{p}}{\mathrm{n}}\)

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Detailed explanation

\(\begin{aligned} & R_s=R_1+R_2+R_3+\ldots \ldots..+R_n \\ & \frac{V^2}{P_s}=\frac{V^2}{P}+\frac{V^2}{P}+\ldots \ldots.+\frac{V^2}{P_n} \\ & P_s=\frac{P}{n}\end{aligned}\)
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