Explain the method of measuring internal resistance using a potentiometer.

Explanation

The end C of the potentiometer wire is connected to the positive terminal of the battery B and the negative terminal of the battery is connected to the end D through a key `K_1`. This forms the primary circuit.
The positive terminal of the cell `xi` whose internal resistance is to be determined is also connected to the end of the wire. The negative terminal of the cell `xi` is connected to a jockey through a galvanometer and a high resistance. 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 `CJ = l_1` is measured. Since the cell is in open circuit, its emf is
`xi prop l_1" ...(1)"`
A suitable resistance (say, `10 Omega`) is included in the resistance box and key `K_2` is closed. Letr 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 = (di R)/(R+r)` When this potential difference is balancing on the potentiometer wire, let `l_2` be the balancing length.
Then `(di R)/(R+r) prop l_2`
From equations (1) and (2)
Substituting 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 with increases of external resistance connected across its terminals.