potentiometer

Question

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of

(i) infinity (ii) 9.5 $\Omega $ the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is

Moderate

Solution

The internal resistance of the cell is

$r=\left(\frac{{l}_{1}}{{l}_{2}}-1\right)R$

Here, ${l}_{1}=3\mathrm{m},{l}_{2}=2.85\mathrm{m},R=9.5\Omega $

$\therefore r=\left(\frac{3}{2.85}-1\right)(9.5\Omega )=\frac{0.15}{2.85}\times 9.5\Omega =0.5\Omega $

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