Cho đoạn mạch AB mắc nối tiếp theo thứ tụ gồm điện trở thuần R

Chọn C

$U_{Cmax}\iff {\overrightarrow{U}}_{RL}\perp \overrightarrow{U}\Rightarrow \left\{\begin{array}{l}{\left(\frac{u_{RL}}{U_{RL}\sqrt{2}}\right)}^2+{\left(\frac{u}{U\sqrt{2}}\right)}^2=1\\ \frac{1}{U_{RL}^2}+\frac{1}{U^2}=\frac{1}{U_R^2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{\left(\frac{25\sqrt{6}}{U_{RL}\sqrt{2}}\right)}^2+{\left(\frac{75\sqrt{6}}{U\sqrt{2}}\right)}^2=1\\ \frac{1}{U_{RL}^2}+\frac{1}{U^2}=\frac{1}{{75}^2}\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}U=150\left(V\right)\\ U_{RL}=75\sqrt{3}\left(V\right)\end{array}\right.$