Đáp án:

${U_{{V_1}}} = \dfrac{{366}}{7}V$ 

Giải thích các bước giải:

Ta có:
$\begin{array}{l}
{I_3} = \dfrac{{{U_{{V_3}}}}}{{3R}} = \dfrac{6}{{3R}} = \dfrac{2}{R}\\
{I_{{V_3}}} = \dfrac{{{U_{{V_3}}}}}{{{R_V}}} = \dfrac{6}{{{R_V}}}\\
{I_2} = {I_3} + {I_{{V_3}}} = \dfrac{2}{R} + \dfrac{6}{{{R_V}}}\\
{U_{{V_2}}} = {U_2} + {U_{{V_3}}} = {I_2}.2R + {U_{{V_3}}}\\
 \Leftrightarrow {U_{{V_2}}} = 2R\left( {\dfrac{2}{R} + \dfrac{6}{{{R_V}}}} \right) + {U_{{V_3}}}\\
 \Leftrightarrow 18 = 4 + \dfrac{{12R}}{{{R_V}}} + 6\\
 \Leftrightarrow 8 = \dfrac{{12R}}{{{R_V}}} \Leftrightarrow \dfrac{R}{{{R_V}}} = \dfrac{2}{3}\\
 \Rightarrow {R_V} = \dfrac{3}{2}R
\end{array}$

Số chỉ vôn kế v1 là:
$\begin{array}{l}
{I_1} = {I_2} + {I_{{V_2}}} = \dfrac{{{U_{{V_2}}}}}{{2R + {R_V}}} + \dfrac{{{U_{{V_2}}}}}{{{R_V}}} = \dfrac{{18}}{{2R + \dfrac{3}{2}R}} + \dfrac{{18}}{{\dfrac{3}{2}R}}\\
 \Leftrightarrow {I_1} = \dfrac{{36}}{{7R}} + \dfrac{{12}}{R} = \dfrac{{120}}{{7R}}\\
{U_{{V_1}}} = {U_1} + {U_{{V_2}}} \Leftrightarrow {U_{{V_1}}} = 2R.{I_1} + {U_{{V_2}}}\\
 \Leftrightarrow {U_{{V_1}}} = 2R.\dfrac{{120}}{{7R}} + 18 = \dfrac{{366}}{7}V
\end{array}$