JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Current Electricity

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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) $\frac{p}{n}$ ✓ Correct
(D) $p$

Answer: (C)

Solution

A bulb rated to draw power

p

at mains voltage

V

has resistance

R = \frac{V ^{2}}{p}

. Joining

n

such bulbs in series gives a total resistance

n R

, and the combination is placed across the same mains voltage

V

. The power drawn is

\frac{V ^{2}}{n R} = \frac{V ^{2}}{n} \cdot \frac{p}{V ^{2}} = \frac{p}{n}

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option

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