Derive relation between e.m.f. and terminal potential difference and thus find expression for internal resistance.
Consider a cell of e.m.f. $\varepsilon$ and internal resistance $r$ connected to an external resistance $R$ through a key (K).
Case 1 : When key (K) is closed , current is drawn from the cell by the circuit, which is given by
$$I = \frac{\varepsilon}{R+r}$$
R and r are in series, so (R + r) is the equivalent resistance of the circuit.
$\varepsilon = IR + Ir$ ...(i)
According to Ohm's law :
That is, $V = IR$ ...(ii)
Hence equation (i) becomes $\varepsilon = V + Ir$
$V = \varepsilon - Ir$...(iii)
This shows that the terminal potential difference of the cell is less than the e.m.f. of the cell.
Now the voltmeter connected across the cell will read the value as $V$ which is less than the value of e.m.f. ($\varepsilon$).
Case 2 : When key (K) is open, $I = 0$
Hence eqn. (iii) becomes $V = \varepsilon$
Thus, terminal potential difference between the electrodes of the cell is equal to the e.m.f. of the cell in an open circuit.
Determination of internal resistance :
We know $I = \frac{\varepsilon}{R+r}$
Using this value in eqn. (iii), we get
$V = \frac{\varepsilon R}{(R+r)}$
$r= \left(\frac{\varepsilon}{V} - 1\right)R$ ...(iv)
Thus, knowing the values of $\varepsilon$, $V$ and $R$,
we can determine the value of the internal resistance ($r$) of the cell.