In a potentiometer experiment, the balancing length obtained with a cell is 240 cm . On shunting the cell with a resistance of 2 Ω , the balancing length becomes 120 cm . The internal resistance of th

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In a potentiometer experiment, the balancing length obtained with a cell is 240 cm . On shunting the cell with a resistance of 2 Ω , the balancing length becomes 120 cm . The internal resistance of the cell is

TestHub · Accepted Answer

In the first case, the emf of the battery is

$E = {Kl}_{1}$ , where $K$ is the potential gradient and $l_{1}$ is the first balancing length.

In the second case the potential difference across the battery is

$\frac{ER}{r + R} = {Kl}_{2}$ , where $R$ is the shunt resistance used.

Dividing the above two equations, we get

$\frac{r}{R} + 1 = \frac{l_{1}}{l_{2}}$

$⇒r = 2 [\frac{240 - 120}{120}] = 2 Ω$

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