Question

Draw a circuit diagram of a potentiometer to compare the emf of two cells. Hence, derive the formula for comparing emf of two cells.

Accepted Answer

Circuit diagram for comparing emf of two cells:

The points marked 1,2,3 form a two-way key. Consider first a position of the key where 1 and 3 are connected so that the galvanometer is connected to $E_{1}$ . The jockey is moved on the wire till at a point $N_{1}$ , at a distance $l_{1}$ from A, there is no deflection in the galvanometer. Let $ϕ$ be the potential drop per unit length of the wire. On applying Kirchhoff's second rule to this closed loop we get:

$ϕ l_{1} + 0 - E_{1} = 0$ ....(i)

Similarly, if another emf $E_{2}$ is balanced against $l_{2}$ :

$ϕ l_{2} + 0 + E_{2} = 0$ ....(ii)

Using(i) and(ii) we get:

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$ ......(iii)

(iii) is the required relation.

Draw a circuit diagram of a potentiometer to compare the emf of two cells. Hence, derive the formula for comparing emf of two cells.

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