Question

How will you compare the e.m.f of two cells using a potentiometer?

Answer 1

Comparison of EMFs of two given cells using potentiometer : The potentiometer wire $AB$ is connected in series with a battery $\left(Bt\right)$, Key $\left(K\right)$, rheostat $\left(Rh\right)$ as shown in figure. This forms the primary circuit. The end $A$ of potentiometer is connected to the terminal $C$ of a $DPDT$ switch (six-way key-double pole double throw). The terminal $D$ is connected to the jockey $\left(J\right)$ through a galvanometer $\left(G\right)$ and high resistance $\left(HR\right)$. The cell of emf $E_1$ is connected between terminals $C_1$ and $D_1$ and the cell of emf $E_2$ is connected between $C_2$ and $D_2$ of the $DPDT$ switch.
Let $I$ be the current flowing through the primary circuit and $r$ be the resistance of the potentiometer wire per metre length.
The $DPDT$ switch is pressed towards $C_1,D_1$ so that cell $E_1$ is included in the secondary circuit. The jockey is moved on the wire and adjusted for zero deflection in galvanometer. The balancing length is $l_1$. The potential difference across the balancing length $l_1=Ir{l}_1$. Then, by the principle of potentiometer,
$E_1=Ir{l}_1....\left(1\right)$
The $DPDT$ switch is pressed towards $E_2$. The balancing length $l_2$ for zero deflection in galvanometer is determined. The potential difference across the balancing length is $l_2=Ir{l}_2$, then
$E_2=Ir{l}_2....\left(2\right)$
Dividing $\left(1\right)$ and $\left(2\right)$ we get,
$\frac{E_1}{E_2}=\frac{l_1}{l_2}$
If emf of one cell $\left(E_1\right)$ is known, the emf of the other cell $\left(E_2\right)$ can be calculated using the relation,
$E_2=E_1\frac{l_1}{l_2}$