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

Solution :
Given, $$e = 2\;V$$  and $$l = 4\,m$$
Potential drop per unit length
$$\phi = \frac{e}{l} = \frac{2}{4} = 0.5\;V/m$$
I
$$e' = \phi {l_1}\,.......\left( {\text{i}} \right)\,\,\,\left( {e' = {\text{emf of the given cell}}} \right)$$
II
$$V = \phi {l_2}\,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii),
$$\eqalign{ & \frac{{e'}}{V} = \frac{{{l_1}}}{{{l_2}}} \cr & \therefore e' = I\left( {r + R} \right) \cr} $$
and $$V = IR$$   for the second case
$$\eqalign{ & \frac{{I\left( {r + R} \right)}}{{IR}} = \frac{{{l_1}}}{{{l_2}}} \cr & {\text{So,}}\,\,r = R\left( {\frac{{{l_1}}}{{{l_2}}} - 1} \right) \cr & = 9.5\left( {\frac{3}{{2.85}} - 1} \right) \cr & = 9.5\left( {1.05 - 1} \right) \cr & = 9.5 \times 0.05 = 0.475 \simeq 0.5\,\Omega \cr} $$