Lời giải:Theo bài ra ta có
\(\begin{array}{l}
C = {C_0} \Rightarrow {U_L} = {U_R} = {U_{{C_0}}} \Rightarrow R = {Z_L} = {Z_{{C_0}}}\\
\Rightarrow U = 40V
\end{array}\)
Ta có
\(\begin{array}{l}
{U_C} + {U_L} = 60V \Rightarrow {U_{{R_2}}} = {U_L}\\
\Rightarrow \left\{ \begin{array}{l}
{U_C} + {U_{{R_2}}} = 60V \Rightarrow {U_C} = 60 - {U_{{R_2}}}\\
U_{{R_2}}^2 + {\left( {{U_L} - {U_C}} \right)^2} = {40^2}
\end{array} \right.\\
\Rightarrow U_{{R_2}}^2 + U_L^2 + U_C^2 - 2{U_L}{U_C} = {40^2}\\
\Rightarrow U_{{R_2}}^2 + U_{{R_2}}^2 + {\left( {60 - U_{{R_2}}^{}} \right)^2} - 2{\left( {60 - U_{{R_2}}^{}} \right)^2}U_{{R_2}}^{} = {40^2}\\
\Rightarrow U_{{R_2}}^{} = 10.73 \approx 11V
\end{array}\)
