+ Ta có:

\({U_C} = \max  = \frac{{U.\frac{L}{C}}}{{R\sqrt {\frac{L}{C} - \frac{{{R^2}}}{4}} }} = 90 \Leftrightarrow \frac{{U.{Z_L}.{Z_C}}}{{R\sqrt {{Z_L}{Z_C} - \frac{{{R^2}}}{4}} }} = 90\)           (1)

+ Khi đó: \({U_{RL}} = \frac{{U\sqrt {{R^2} + Z_L^2} }}{{\sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }} = 30\sqrt 5 \)                                     (2)

+ Lấy (1) chia (2), ta có: \(\frac{{{Z_L}{Z_C}\sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }}{{R\sqrt {{Z_L}{Z_C} - \frac{{{R^2}}}{4}\sqrt {{R^2} + Z_L^2} } }} = \frac{3}{{\sqrt 5 }}\)                 (3)

+ Lại có, khi U_C = max thì:

\({Z_L} = \sqrt {\frac{L}{C} - \frac{{{R^2}}}{2}}  \Leftrightarrow Z_L^2 = {Z_L}{Z_C} - \frac{{{R^2}}}{2} \Rightarrow {R^2} = 2{Z_L}{Z_C} - 2Z_L^2\)                   (4)

+ Thay (4) vào (3), ta có:  

 \(\begin{array}{l}
\frac{{{Z_L}{Z_C}\sqrt {2{Z_L}{Z_C} - 2Z_L^2 + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }}{{\sqrt {2{Z_L}{Z_C} - 2Z_L^2} \sqrt {{Z_L}{Z_C} - \frac{{2{Z_L}{Z_C} - 2Z_L^2}}{4}\sqrt {2{Z_L}{Z_C} - 2Z_L^2 + Z_L^2} } }} = \frac{3}{{\sqrt 5 }}\\
 \Leftrightarrow \frac{{{Z_L}{Z_C}\sqrt {Z_C^2 - Z_L^2} }}{{\sqrt {{Z_L}{Z_C} - Z_L^2}  + \sqrt {{Z_L}{Z_C} + Z_L^2\sqrt {2{Z_L}{Z_C} - Z_L^2} } }} = \frac{3}{{\sqrt 5 }} \Leftrightarrow \frac{{{Z_C}}}{{\sqrt {2{Z_L}{Z_C} - Z_L^2} }} = \frac{3}{{\sqrt 5 }}\\
 \Leftrightarrow 5Z_C^2 = 18{Z_L}{Z_C} - 9Z_L^2
\end{array}\)

Chọn Z_L = 1 Þ Z_C = 0,6 hoặc Z_C = 3

+ Với Z_L = 1 Þ Z_C = 0,6. Thay vào (4) suy ra R < 0 loại

+ Với Z_L = 1 Þ Z_C = 3. Thay vào (4) suy ra R = 2

+ Thay Z_L = 1, Z_C = 3 và R = 2 vào (1) ta có: U = 84,85 V