The series combination of two batteries both of the same emf 10 V ,  but different internal resistance of $20\Omega$ and $5\Omega$  is connected to the parallel combination of two resistors $25\Omega$ and $\text{R\hspace{0.17em}}\Omega$ .The voltage difference across the battery of internal resistance $\text{20\hspace{0.17em}}\Omega$ is zero . The value of $\text{R}\left(\text{in\hspace{0.17em}}\Omega \right)$ is,  

 

$$\begin{gathered} For 10 V, 20\Omega battery : \\ \therefore {\text{\hspace{0.17em}\hspace{0.17em}V}}_{\text{1}}={\text{\hspace{0.17em}ε}}_{\text{1}}-{\text{ir}}_{\text{1}} \\ \text{⇒}0=10-\text{i}\times 20 \\ \text{i}=0.5\text{A} \\ For 10 V, 5\Omega battery : \\ \therefore {\text{\hspace{0.17em}V}}_{\text{2}}={\text{\hspace{0.17em}ε}}_{\text{2}}-{\text{i\hspace{0.17em}r}}_{\text{2}} =10-0.5\times 5=7.5\text{V} \\ The potential difference across all three branches is equal. \\ V = V_1 + V_2 = 0 + 7.5 = 7.5 V \\ Now, i = i_1 + i_2 \\ \Rightarrow 0.5 = \frac{7.5}{25} +\frac{7.5}{R} \\ \Rightarrow 0.5=0.3+\frac{7.5}{\text{R}} \\ \Rightarrow 0.2=\frac{7.5}{\text{R}} \\ \Rightarrow \text{R\hspace{0.17em}=\hspace{0.17em}}\frac{7.5}{0.2}=37.5\Omega \end{gathered}$$