Question

How can e.m.f. of two cells be compared using potentiometer?

Answer 1

The standard cell (S) of known emf and the cell (X) of unknown emf are connected in parallel ( through a double switch, galvanometer and movable jocky) to a conducting wire AB of uniform cross section. By using double switch, first the standard cell S is placed in circuit. Jocky is then moved along wire AB and its position is adjusted so that no current flows through galvanometer.
Let AD be the length of the conducting wire for this position of jocky.
$E_S\propto l\left(AD\right)$
Now, the cell (X) of unknown emf is placed in circuit. Jocky is then moved along wire AB and its position is adjusted so that no current flows through galvanometer. Let AD' be the length of the conducting wire for this position of jocky.
$E_X\propto l\left(A{D}^{\prime }\right)$
$\frac{E_S}{E_X}=\frac{l\left(AD\right)}{l\left(A{D}^{\prime }\right)}$
From the above equation, the emf of cell (X) can be calculated.