Q.Describe how a potentiometer is used to compare the emfs of two cells by connecting the cells individually.

[!TLDR] Connect both cells' positive terminals to A and their negatives to a two-way key; bring each cell into circuit in turn and find its own balancing length; the ratio of the two lengths e …

This is potentiometer Method I of Section 9.4.2. A potentiometer circuit is set up with a driving battery of emf $\varepsilon$, a key K and a rheostat across wire AB (A at higher potential than B). The two cells to be compared, $\varepsilon_1$ and $\varepsilon_2$, have their POSITIVE terminals both connected to point A, and their negative terminals connected to the two outer terminals of a two-way key $K_1K_2$, whose central terminal goes to a galvanometer and then a jockey touching the wire. With K closed and $K_1$ closed ($K_2$ open), only $\varepsilon_1$ is in circuit; the jockey is tapped along the wire to find its null point, at length $l_1$ from A, giving $\varepsilon_1 = k\,l_1$ (k the wire's potential gradient). Then $K_1$ is opened and $K_2$ closed instead, bringing $\varepsilon_2$ into the SAME circuit; its own null point at length $l_2$ gives $\varepsilon_2 = k\,l_2$. …

Describe the two-way-key circuit that swaps one cell at a time into the same galvanometer branch, then divide the two balan …

Trying to close BOTH $K_1$ and $K_2$ at once -- the whole method depends on only ONE cell being in the galvanometer circuit at any g …