Đáp án:

\(R = \dfrac{{{e_1}{r_2} - {e_2}{r_1}}}{{2\left( {{e_2} - {e_1}} \right)}}\)

Giải thích các bước giải:

a) Áp dụng định luật Ôm cho từng đoạn mạch: 

\(\begin{array}{l}
\left\{ \begin{array}{l}
{U_{AB}} = {e_1} - {I_1}{r_1}\\
{U_{AB}} = {e_2} - {I_2}{r_2}\\
{U_{AB}} = {I_3}R
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{I_1} = \dfrac{{{e_1} - {U_{AB}}}}{{{r_1}}}\\
{I_2} = \dfrac{{{e_2} - {U_{AB}}}}{{{r_2}}}\\
I = \dfrac{{{U_{AB}}}}{R}
\end{array} \right.\\
I = {I_1} + {I_2} \Rightarrow \dfrac{{{U_{AB}}}}{R} = \dfrac{{{e_1} - {U_{AB}}}}{{{r_1}}} + \dfrac{{{e_2} - {U_{AB}}}}{{{r_2}}}\\
 \Rightarrow {U_{AB}} = \dfrac{{\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}}}\\
 \Rightarrow \left\{ \begin{array}{l}
{I_1} = \dfrac{{{e_1} - \dfrac{{\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}}}}}{{{r_1}}}\\
{I_2} = \dfrac{{{e_2} - \dfrac{{\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}}}}}{{{r_2}}}\\
I = \dfrac{1}{R}.\dfrac{{\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}}}
\end{array} \right.
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
{I_1} = {I_2} \Rightarrow \dfrac{{{e_1} - \dfrac{{\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}}}}}{{{r_1}}} = \dfrac{{{e_2} - \dfrac{{\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}}}}}{{{r_2}}}\\
 \Rightarrow {e_1}{r_2} - \dfrac{{\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}}}.{r_2} = {e_2}{r_1} - {r_1}.\dfrac{{\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}}}\\
 \Rightarrow {e_1}{r_2}\left( {\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}} \right) - \left( {\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}} \right).{r_2} = {e_2}{r_1}\left( {\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{R}} \right) - {r_1}\left( {\dfrac{{{e_1}}}{{{r_1}}} + \dfrac{{{e_2}}}{{{r_2}}}} \right)\\
 \Rightarrow {e_1} + \dfrac{1}{R}.{e_1}{r_2} - {e_2} = {e_2} + \dfrac{1}{R}.{e_2}{r_1} - {e_1}\\
 \Rightarrow \dfrac{1}{R}\left( {{e_1}{r_2} - {e_2}{r_1}} \right) = 2\left( {{e_2} - {e_1}} \right)\\
 \Rightarrow R = \dfrac{{{e_1}{r_2} - {e_2}{r_1}}}{{2\left( {{e_2} - {e_1}} \right)}}
\end{array}\)