Derive a relation between internal resisance and emf of a cell.

Explanation

(i) The emf of cell `xi` is measured by connecting a high resistance voltemeter across it without connecting the external resistance R.

(ii) Since the volmeter draws very little current for deflection, the circuit may be considered as open. Hence the voltmeter reading gives the emf of the cell.
(iii) Then, external resistance R is included in te circuit. and current I is established in the circuit. the potential difference across R is equal to the potential differene across the cell (V).
(iv) The potential drop across the resistor R is
`V=IR " "....(1)`
(v) Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell `xi`. It is because, certain amount of voltage (Ir) has dropped across the internal resistance r.
Then `V= xi-Ir`
`Ir= xi-V " "...(2)`
(vi) Dividing equation (2) by equation (1), we get
`(Ir)/(IR)=(xi-V)/(V)`
`r=|(xi-V)/(V)|R`
Since `xi` V and R are known, internal resistance r cane be determined.