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

Explanation

The emf of cell `xi` is measured by connecting a high resistance voltmeter across it without connecting the external resistance R. (Fig. (a)). Since the voltmeter draws very little current for deflection, the circuit may be considered as open, So, the voltmeter reading gives the emf of the cell. Then, external resistance R is included in the circuit and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V). (Fig (b)).

The potential drop across the resistor R is
`V = IR " ...(1)"`
Due to internal resistance r of the cell, the voltmeter reads a value V, that 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)"`
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 can be determined.