A cell, shunted by a $8 Ω$ resistance, is balanced across a potentiometer wire of length $3 m$ . 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

$\frac{V_{1}}{V_{2}} = \frac{3}{2} = \frac{E - i_{1} r}{E - i_{2} r} = \frac{E - \frac{E}{8 + r} \times r}{E - \frac{E}{4 + r} \times r} \Rightarrow \frac{3}{2} = \frac{8 (4 + r)}{4 (8 + r)} \Rightarrow 24 + 3 r = 16 + 4 r \Rightarrow r = 8 Ω$

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