Question

Describe the experiment to find internal resistance of a cell by Potentiometer under the following points.
(i) Labelled electric circuit diagram.
(ii) Derivation of formula
(iii) Observation table
(iv) Two precautions.

Answer 1

(i) Labelled electric circuit diagram: Potentiometer wire, Rh$\to$ variable resistance, C- accumulator, D-experimental cell, G-galvanometer, R.B- resistance box, $K_1,K_2$-keys, J-jockey.
Note: Direction of current is shown by arrows.
(ii) Derivation of formula: Let the e.m.f. of the experimental cell is E and its internal resistance is r. If the resistance R from the resistance box is connected with this cell then the potential difference at the ends of the cell becomes V.
$E=I(R+r)$ and $V=IR$
then $r=\left(\frac{E}{V}-1\right)R$ ..........(1)
Let the potential gradient of the potentiometer be v.
When R is not connected, then let the balance point is found at distance $l_1$ from A, then
$E=v{l}_1$ .........(2)
Now when R is connected with the cell, then let the balance point is obtained at a distance $l_2$ from A. The potential difference at the ends of the cell
$V=v{l}_2$ ..........(3)
Substituting eqns. (2) and (3) in (1),
$r=\left(\frac{v{l}_1}{v{l}_2}-1\right)R=\left(\frac{l_1}{l_2}-1\right)R$
(iii) Observation table:

S.No.	When cell is in open circuit balancing length $l_1$(cm)	When cell is in closed circuit balancing length $l_2$(cm)	Resistance used from resistance box R(ohm)	Internal resistance $r=R\left(\frac{l_1}{l_2}-1\right)$ohm

(iv) Two precautions: (1) the diameter of potentiometer wire should be unifrom throughout, otherwise potential gradient will not remain same everywhere.
(2) Shunt should be connected with the galvanometer before null position but near the null point the shunt should be removed.