Q.When a resistor of 5 $\Omega$ is connected across the cell, its terminal potential difference is balanced by 150 cm of potentiometer wire and when a resistance of 10 $\Omega$ is connected across the cell, the terminal potential difference is balanced by 175 cm of the same potentiometer wire. Find the balancing length when the cell is in open circuit and the internal resistance of the cell.

Comparing the balancing lengths for a cell on open circuit and on a known resistance box gives the internal resistance of the cell.

In a potentiometer experiment to find the internal resistance $r$ of a cell of emf $E$: with the key in the resistance-box circuit open, the balancing length $l_1$ gives $E \propto l_1$. With a known resistance $R$ connected across the cell (so the terminal potential difference is $V$), the balancing length $l_2$ gives $V \propto l_2$. Since $l_1/l_2 = E/V$, and the current in the circ …

In a potentiometer, balancing length is proportional to the potential difference being measured; e.m.f. gives length $l_1$, terminal p.d. (when shunted by $R$) gives length $l_2$.

In a potentiometer experiment, the balancing length is directly proportional to the potential difference (or e.m.f.) being measured (since the potential gradient along the wire is constant):

$\text{p.d.}\propto l$

When the cell (e.m.f. $\varepsilon$, internal resistance $r$) is balanced directly (open circuit), the balancing length $l_1$ corresponds to its full e.m.f.:

$\varepsilon\propto l_1$

When the cell is shunted by an external resistance $R$, it drives a current $i=\dfrac{\varepsilon}{R+r}$ through the circuit, and its terminal potential difference becomes …