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Q.

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by $5 Ω$ .When a resistance of $15 Ω$ is used for shunting , null point moves to 300cm.The internal resistance of the cell is ______ $Ω$ .

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answer is 5.

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Detailed Solution

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$as r = R (\frac{l_{0}}{l} - 1)$
When $R = 5 Ω, l = 200 cm$
$So r = 5 (\frac{l_{0}}{200} - 1)$ ......(1)
When $R = 15 Ω, l = 300 cm$
So $r = 15 (\frac{l_{0}}{300} - 1) \dots$ ....(2)
From equation (1) and (2)
$\begin{array}{l} 5 (\frac{l_{0}}{200} - 1) = 15 (\frac{l_{0}}{300} - 1) \\ \Rightarrow \frac{l_{0}}{200} - 1 = \frac{3 l_{0}}{300} - 3 \\ \Rightarrow \frac{l_{0}}{100} - \frac{l_{0}}{200} = 2 \Rightarrow \frac{l_{0}}{200} = 2 \end{array}$
So from Eqn (i)
$r = 5 (\frac{l_{0}}{200} - 1) = 5 (2 - 1) = 5 Ω$

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