Two cells of same $emf$ $E$ but internal resistances $r_1$ and $r_2$ are connected in series to an external resistor $R$ (as shown in the figure). What should be the value of $R$ so that the potential difference across the terminals of the first cell becomes zero?

(R = R_1 - R_2) The equivalent internal resistance of the series combination is $r = r_1 + r_2$.
The equivalent $emf$ of the series combination is $E' = E + E = 2E$.
Therefore,the current $I$ flowing in the circuit is given by:
$I = \frac{E'}{R + r} = \frac{2E}{R + r_1 + r_2}$
The potential difference $V_1$ across the terminals of the first cell is given by:
$V_1 = E - I r_1$
Given that the potential difference across the first cell is zero,we set $V_1 = 0$:
$0 = E - I r_1$
$E = I r_1$
Substituting the expression for $I$:
$E = \left( \frac{2E}{R + r_1 + r_2} \right) r_1$
$1 = \frac{2 r_1}{R + r_1 + r_2}$
$R + r_1 + r_2 = 2 r_1$
$R = r_1 - r_2$
Thus,the required value of the external resistance is $R = r_1 - r_2$.