Explain the determination of the internal resistance of the cell using voltmeter.

Explanation

(i) The measure the internal resistance of a cell, the circuit connections are made as shown in Figure. The end C of the potentiometer wire is connected to the positive terminal of the battery B and the positive terminal of the battery is connected to the end D through a key `K_(1)`. This forms the primary circuit.
(ii) The positive terminal of the cell `xi` whose internal resistance is to be determined is also connected to the end C of the wire. The negative terminal of the cell `xi` connected to a jockey through a galvanometer and a high resistance.

(iii) A resistance box R and key `K_(2)` are connected across the cell `xi`. With `K_(2)` open, the balancing point J is obtained and the balancing length `C_(J)=l_(1)` is measured. since the cell is in open circuite, its emf is
`xi=l_(1) " "..(1)`
(iv) A suitable resistance (say, `10 Omega`) is included in the resistance box and key `K2` is closed. Let r be the internal resistance of the cell. The current passing through the cell and the resistance R is given by
`I=(xi)/(R+r)`
The potential difference across R is
`V=(xiR)/(r=r)`
(v) When this potential difference is balanced on the potentiometer wire, let `l_(2)` be the balancing length.
Then `(xiR)/(R+r)" ".....(2)`
From equations (1) and (2)
`(R+r)/(R)=(l_(1))/(l_(2))`
`1+(r)/(R)=(l_(1))/(l_(2))`
`r=R[(l_(1)-l_(2))/(l_(2))]`
`:. r=R[(l_(1)-l_(2))/(l_(2))]`
(iv) Subsituting the values of the `R, l_(1) and l_(2)` the internal resistance of the cell is determined. The experiment can be repeated for different values of R. It is found that the internal resistance of the cell is not constant but increases with increases of external resistance connected across its terminals.