Question

State the working principle of Potentiometer. Explain with the help of circuit diagram, how the potentiometer is used to determine the internal resistance of the given Primary cell.
In a potentiometer arrangement, a cell of emf $1.25V$ gives a balance point at $35.0cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $63.0cm$, what is the emf of the second cell?

Answer 1

Working principle : Potential difference across length of potentiometer wire is proportional to the length of the wire.
$E=kL$ where, $E$ is the emf and $k$ is the potential gradient.
In the figure:
Cell emf $e$ whose internal resistance $r$ is to be determined is connected via resistance box R.B. through key $K_2$. This key when open, balance is obtained at $l_1\left(A{N}_1\right)$ .
So, $e=k{l}_1$
If $v$ is terminal potential difference, balance is obtained at length $l_2\left(A{N}_2\right)$.
So, $v=k{l}_2$
Now, $e=I(R+r)$ where $I$ is the current.
and $v=IR$

Dividing the equations.
$\frac{e}{v}=\frac{l_1}{l_2}=\frac{R+r}{r}$

Therefore, $r=R(\frac{l_1}{l_2}-1)$
Emf of the cell, $E_1$ $=1.25V$
Balance point of the potentiometer, $l_1$ $=35cm$
Cell is replaced by another cell of emf, $E_2$
Balance point of the potentiometer, $l_2$ $=63cm$
$\frac{E_1}{E_2}=\frac{l_1}{l_2}$

$E_2=1.25\times \frac{63}{35}=2.25V$