Đáp án:

 C

Giải thích các bước giải:

* Khi mắc R // V1: 

Ta có: 

\(\begin{array}{l}
\dfrac{{{U_1}}}{{{R_1}}} + \dfrac{{{U_1}}}{R} = \dfrac{{U - {U_1}}}{{{R_2}}}\\
 \Rightarrow \dfrac{8}{{{R_1}}} + \dfrac{8}{R} = \dfrac{{U - 8}}{{{R_2}}}\\
 \Rightarrow 8\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{R}} \right) = \dfrac{U}{{{R_2}}}
\end{array}\)

* Khi mắc R // V2:

Ta có: 

\(\begin{array}{l}
\dfrac{{{U_2}}}{{{R_2}}} + \dfrac{{{U_2}}}{R} = \dfrac{{U - {U_2}}}{{{R_1}}}\\
 \Rightarrow \dfrac{{10}}{{{R_2}}} + \dfrac{{10}}{R} = \dfrac{{U - 10}}{{{R_1}}}\\
 \Rightarrow 10\left( {\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{R}} \right) = \dfrac{U}{{{R_1}}}\\
 \Rightarrow \dfrac{{\dfrac{U}{{{R_2}}}}}{{\dfrac{U}{{{R_1}}}}} = \dfrac{8}{{10}} \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{4}{5} \Rightarrow {R_2} = \dfrac{5}{4}{R_1}
\end{array}\)

* Khi không mắc R:

Ta có: 

\(\begin{array}{l}
{R_2} = \dfrac{5}{4}{R_1} \Rightarrow {U_2}' = \dfrac{5}{4}{U_1}' = \dfrac{5}{4}.12 = 15V\\
 \Rightarrow U = {U_1}' + {U_2}' = 12 + 15 = 27\Omega 
\end{array}\)