To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of 5Ω , the balancing length is 250 cm . When the cell is shunted by 20Ω , the balancing len

Question

To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of 5Ω , the balancing length is 250 cm . When the cell is shunted by 20Ω , the balancing length of potentiometer wire is 400 cm . The internal resistance of the cell is

TestHub · Accepted Answer

The internal resistance of a cell determined using a potentiometer is given by the formula:

$r = R (\frac{l _{0}}{l} - 1)$

where $l_{0}$ is the balancing length for the cell without a shunt, and $l$ is the balancing length when shunted by a resistance $R$ .

For the first case, $R_{1} = 5Ω$ and $l_{1} = 250$ cm:

$r = 5 (\frac{l _{0}}{250} - 1)$

$\frac{r}{5} + 1 = \frac{l _{0}}{250}$

$l_{0} = 50 r + 250$

For the second case, $R_{2} = 20Ω$ and $l_{2} = 400$ cm:

$r = 20 (\frac{l _{0}}{400} - 1)$

$\frac{r}{20} + 1 = \frac{l _{0}}{400}$

$l_{0} = 20 r + 400$

Equating the two expressions for $l_{0}$ :

$50 r + 250 = 20 r + 400$

$30 r = 150$

$r = 5Ω$

Answer: $5Ω$

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