Đáp án:

\(\dfrac{{2\pi  + 1}}{{2\pi \left( {\pi  + 1} \right)}}\)

Giải thích các bước giải:

Đặt:

\(\begin{array}{l}
OA = OB = r \Rightarrow \left\{ \begin{array}{l}
{C_1} = 2\pi r\\
{C_2} = \pi r
\end{array} \right.\\
{R_{OA}} = {R_{OB}} = R \Rightarrow \left\{ \begin{array}{l}
{R_2} = {R_3} = R\\
{R_1} = 2\pi R\\
{R_4} = \pi R
\end{array} \right.\\
{R_{34}} = \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}} = \dfrac{{R.\pi R}}{{R + \pi R}} = \dfrac{\pi }{{\pi  + 1}}.R\\
{R_{234}} = {R_2} + {R_{34}} = R + \dfrac{\pi }{{\pi  + 1}}.R = \dfrac{{2\pi  + 1}}{{\pi  + 1}}.R\\
{I_1} = \dfrac{U}{{{R_1}}} = \dfrac{U}{{2\pi R}}\\
{I_2} = \dfrac{U}{{{R_{234}}}} = \dfrac{U}{{\dfrac{{2\pi  + 1}}{{\pi  + 1}}.R}} = \dfrac{{U\left( {\pi  + 1} \right)}}{{\left( {2\pi  + 1} \right)R}}\\
 \Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{\dfrac{U}{{2\pi R}}}}{{\dfrac{{U\left( {\pi  + 1} \right)}}{{\left( {2\pi  + 1} \right)R}}}} = \dfrac{{2\pi  + 1}}{{2\pi \left( {\pi  + 1} \right)}}
\end{array}\)