Measurement of Internal Resistance of a Cell Using a Potentiometer

To measure the internal resistance $r$ of a cell using a potentiometer, the cell (of EMF $\varepsilon$) is first connected into the secondary circuit ALONE, with its own circuit left open (no current drawn from the cell beyond the negligible balance current, which is zero exactly at balance), and its balance length $l_1$ is found in the usual way:

Next, a known resistance $R$ (from a resistance box) is connected directly across the terminals of the same cell, via a separate key, so that when this key is closed, the cell now drives a current through $R$ and, in doing so, drops part of its EMF across its own internal resistance -- exactly the situation analysed in Section 3.9, where the terminal potential difference $V = \varepsilon - Ir$ is now less than the EMF. With this external resistance $R$ now connected and drawing current, the NEW balance point (for this reduced terminal PD $V$, not the full EMF) is found at length $l_2$:

Since the potential gradient $k$ is unchanged throughout (the primary circuit is not touched), dividing the two balance equations gives

But from Section 3.9, for a cell of EMF $\varepsilon$, internal resistance $r$, driving current $I=V/R$ through an external resistance $R$: $\varepsilon = V + Ir = V(1 + r/R)$, i.e. $\varepsilon/V = 1+r/R$. Combining this with $\varepsilon/V = l_1/l_2$ from above and solving for $r$:

$r = R\left(\frac{l_1}{l_2}-1\right) = R\,\frac{l_1-l_2}{l_2}$ …