JEE Main 2025 Apr 4 Shift 2, Physics Q2: Energy, Power And Heating Effect

JEE Main2025Apr 4, Shift 2Physics

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
$n p$
B
$\frac{p}{n ^{2}}$
C
$\frac{p}{n}$
D
$p$

Solution

Each bulb has resistance $R = \frac{V ^{2}}{p}$ , where $V$ is the mains voltage.

In series, the total resistance is $R_{s} = n R = \frac{n V ^{2}}{p}$ .

The total power drawn at the same mains voltage $V$ is

$P_{s} = \frac{V ^{2}}{R _{s}} = \frac{V ^{2}}{n V ^{2} / p} = \frac{p}{n}$ .

$∴$ Total power = $\frac{p}{n}$ .

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JEE Main 2025 Apr 4 Shift 2, Physics Q2: Energy, Power And Heating Effect

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