Đáp án:

\( - 4 + 4\sqrt {10} \left( V \right)\)

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
{I_{{V_3}}} = \dfrac{{{U_{{V_3}}}}}{{{R_V}}} = \dfrac{8}{{{R_V}}}\\
{U_{{R_3}}} = {I_{{V_3}}}.R = 8.\dfrac{R}{{{R_V}}}\\
{U_{{V_2}}} = 8 + 8.\dfrac{R}{{{R_V}}}\\
{I_{{V_2}}} = \dfrac{{{U_{{V_2}}}}}{{{R_V}}} = \dfrac{8}{{{R_V}}} + 8.\dfrac{R}{{R_V^2}}\\
{I_{{R_2}}} = {I_{{V_2}}} + {I_{{V_3}}} = \dfrac{8}{{{R_V}}} + 8.\dfrac{R}{{R_V^2}} + \dfrac{8}{{{R_V}}} = 8.\dfrac{R}{{R_V^2}} + \dfrac{{16}}{{{R_V}}}\\
{U_{{V_1}}} = {U_{{V_2}}} + {U_{{R_2}}}\\
 \Rightarrow 10 = 8 + 8.\dfrac{R}{{{R_V}}} + R\left( {8.\dfrac{R}{{R_V^2}} + \dfrac{{16}}{{{R_V}}}} \right)\\
 \Rightarrow 2 = 8{\left( {\dfrac{R}{{{R_V}}}} \right)^2} + 24.\dfrac{R}{{{R_V}}}\\
 \Rightarrow \dfrac{R}{{{R_V}}} = \dfrac{{ - 3 + \sqrt {10} }}{2}\\
 \Rightarrow {U_{{V_2}}} = 8 + 8.\dfrac{R}{{{R_V}}} =  - 4 + 4\sqrt {10} \left( V \right)
\end{array}\)