In the first case,
\(2=\frac{\varepsilon}{2+r}\) ....\((i)\)
In the second case,
\(0.5=\frac{\varepsilon}{9+r}\) ....\((ii)\)
Divide \((i)\) by \((ii),\) we get
\({\frac{2}{0.5}=\frac{9+r}{2+r} \Rightarrow 4+2 r=4.5+0.5 r} \)
\({1.5 r=0.5 \Rightarrow r=\frac{0.5}{1.5}=\frac{1}{3} \,\Omega}\)