JEE Main 2025 · Physics · Current Electricity · easy

Twenty-four identical cells, each of emf 1.5 V and internal resistance 1 Ω, are arranged in m parallel rows of n cells in series each, with mn = 24…

Single correct option

Twenty-four identical cells, each of emf

1.5\ \mathrm{V}

and internal resistance

1\ \Omega

, are arranged in

m

parallel rows of

n

cells in series each, with

mn = 24

, and connected to an external resistance of

6\ \Omega

. The largest current that can be obtained through the external resistance is

Show answer and solution

Answer: (B) $1.5\ \mathrm{A}$

The current is

I = \dfrac{nE}{R + nr/m}

. It is a maximum when the external resistance equals the equivalent internal resistance:

R = nr/m

, i.e.

n/m = 6

. With

mn=24

this gives

n=12

m=2

. Then

I = \dfrac{12\times1.5}{6+6} = 1.5\ \mathrm{A}

. Other groupings such as

n=24,m=1

n=6,m=4

give only

1.2\ \mathrm{A}

, confirming the maximum.

Original question written in the JEE Main 2025 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake