JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Current Electricity

A cell, shunted by a

8Ω

resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by

4Ω

resistance. The value of internal resistance of the cell will be ______

Ω

Answer: 8

Solution

In a potentiometer, the balancing length is proportional to the terminal voltage across the cell. When shunted by resistance

R_{s}

, the terminal voltage is

V = E - i r

where

i = \frac{E}{R _{s} + r}

.

Given:

\frac{V _{1}}{V _{2}} = \frac{l _{1}}{l _{2}} = \frac{3}{2}

For the two cases:

\frac{E - \frac{E r}{8 + r}}{E - \frac{E r}{4 + r}} = \frac{3}{2}

Simplifying:

\frac{\frac{E \cdot 8}{8 + r}}{\frac{E \cdot 4}{4 + r}} = \frac{3}{2}

\frac{8 ( 4 + r )}{4 ( 8 + r )} = \frac{3}{2}

16 (4 + r) = 12 (8 + r)

64 + 16 r = 96 + 12 r

4 r = 32 ⟹ r = 8 Ω

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer