A cell, shunted by a 8 $Ω$ resistance, is balanced across a wire of potentiometer length 3m. The balancing length is 2 m when the cell is shunted by 4 $Ω$ resistance. The value of internal resistance of the cell will be _______ $Ω$ .

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answer is 8.

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Detailed Solution

$\frac{V_{1}}{V_{2}} = \frac{K \times 3}{K \times 2} = \frac{3}{2}$

Also, $\frac{V_{1}}{V_{2}} = \frac{E - i_{1} r_{1}}{E - i_{2} r_{2}} = \frac{E - \frac{E}{8 + r} \times r}{E - \frac{E}{4 + r} \times r}$
$\frac{V_{1}}{V_{2}} = \frac{8 (4 + r)}{4 (8 + r)}$
From (i) & (ii), we get …(i)
$\begin{array}{l} \frac{3}{2} = \frac{8}{4} (\frac{4 + r}{8 + r}) \Rightarrow \frac{3}{4} = \frac{4 + r}{8 + r} \\ \Rightarrow 24 + 3 r = 16 + 4 r \Rightarrow 8 = r \Rightarrow r = 8 Ω \end{array}$