Q54 / 75 · JEE Main 2025 (4 Apr Shift 2) · Physics

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 :

  • A
    np
  • B
    $\frac{\mathrm{p}}{\mathrm{n}^2}$
  • Correct answer
    C
    $\frac{\mathrm{p}}{\mathrm{n}}$
  • D
    p

Solution

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

Hence, the correct answer is option: 'C'.

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