Explanation:
Here, \({V=E-I r_{1} \Rightarrow 0=E-I r_{1}(\because V=0)}\)
\(\therefore \;\;\;I = \frac{E}{{{r_1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\)
Current in circuit, \({I=\dfrac{E+E}{r_{1}+r_{2}+R}}\)
\(I = \frac{{2E}}{{{r_1} + {r_2} + R}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
From eqns. (1) and (2) we get\(\begin{aligned}& \dfrac{E}{r_{1}}=\dfrac{2 E}{r_{1}+r_{2}+R} \Rightarrow 2 r_{1}=r_{1}+r_{2}+R \\& \therefore R=r_{1}-r_{2}\end{aligned}\).
So correct option is (2)