Two cells each of emf e but internal resistances `r_1` and `r_2` are connected in series through an external resistance R. IF the potential difference across the first cell is zero while current flows,the relation of R in terms of `r_1` and `r_2` is

Explanation

When the two cells are connected in series the emf of the said circuit=E+E=2E
Total resistance of the circuit `=r_1+r_2+R`.
So current through the circuit, `I=(2E)/(r_1+r_2+R)`
Therefore, potential difference between the two ends of the first cell, `V_1=E-Ir_1`.
According to the question, `V_1=0`
i.e., `E=Ir_1=(2E)/(r_1+r_2+R).r_1`
or,`r_1+r_2+R=r_1or,R=r_1-r_2`