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In the determination of the internal resistance of a cell using a potentiometer, when the cell is shunted by a resistance “R” and connected in the secondary circuit, the balance length is found to be L₁. On doubling the shunt resistance, the balance length is found to increase to L₂. The value of the internal resistance is

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$\frac{2 R (L_{2} - L_{1})}{(L_{1} - 2 L_{2})}$

$\frac{2 R (L_{2} - L_{1})}{(2 L_{1} - L_{2})}$

$\frac{R (L_{2} - L_{1})}{(L_{1} - 2 L_{2})}$

$\frac{R (L_{2} - L_{1})}{(2 L_{1} - L_{2})}$

answer is B.

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Detailed Solution

$(\frac{E}{R + r}) R = K . I$

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