Two cells of internal resistance \( r_{1} \) and \( r_{2} \) and of same emf are connected in series across a resistor of resistance \( R \). If the terminal potential difference across the cells of internal resistance \( r_{1} \) is zero, then the value of \( \mathrm{R} \) is

Since the two cells are connected in series, therefore, total emf is \( E=E_{1}+E_{2} \)
Since the emf is same, therefore, \( E=2 E\left[\because E_{1}=E_{2}=E(\right. \) say \( \left.)\right] \)
Total resistance \( =R+r_{1}+r_{2} \)
Thus, current in circuit, \( I=\frac{2 E}{R+r_{1}+r_{2}} \)

Now, terminal potential difference across cell of internal resistance \( r_{1} \) is \( E-I r_{1} \)
Given, this terminal potential difference is zero, therefore,
\(E-I r_{1}=0 \)
\(E-\frac{2 E}{R+r_{1}+r_{2}} \times r_{1}=0 \)
\(\Rightarrow E=\frac{2 E r_{1}}{R+r_{1}+r_{2}} \)
\(\Rightarrow R+r_{1}+r_{2}=2 r \)
\(\Rightarrow R=2 r_{1}-r_{1}-r_{2} \)
\(\Rightarrow R=r_{1}-r_{2}\)