Comparison of EMF of Two Cells with a Potentiometer

To compare the emf values of two different cells, $\varepsilon_1$ and $\varepsilon_2$, the potentiometer wire CD is connected in series with a battery Bt, a rheostat Rh, and a key K to form the primary circuit (Figure 2.28). End C of the wire is connected to the common terminal M of a DPDT (double-pole double-throw) switch, whose other common terminal N connects through a galvanometer G and a high resistance HR to a sliding jockey; the two cells to be compared are wired to the switch's two alternative throw positions, $M_1,N_1$ (for $\varepsilon_1$) and $M_2,N_2$ (for $\varepsilon_2$), with the positive terminals of Bt, $\varepsilon_1$ and $\varepsilon_2$ all connected to the same end, C.

Throwing the DPDT switch to $M_1,N_1$ brings cell $\varepsilon_1$ into the secondary circuit, and the jockey is adjusted for zero galvanometer deflection to find balancing length $l_1$. The switch is then thrown to $M_2,N_2$ to bring in cell $\varepsilon_2$, and the jockey is rebalanced to find length $l_2$ -- crucially, on the SAME wire, with the SAME current I and resistance-per-length r throughout, since neither the primary circuit nor the wire is touched between the two measurements. Using $\varepsilon=Irl$ (equation 2.58) for each cell in turn:

Dividing equation (2.59) by equation (2.60) cancels the unknown I and r completely, leaving

$\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{l_1}{l_2} \qquad (2.61)$ …