Hiệu điện thế giữa hai đầu tụ điện là:

\(\begin{array}{l}{U_C} = I.{Z_C} = \dfrac{{U{Z_C}}}{Z} = \dfrac{{U{Z_C}}}{{\sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }} = \dfrac{{U{Z_C}}}{{\sqrt {{{\left( {{R^2} + {Z_L}} \right)}^2} - 2{Z_L}{Z_C} + {Z_C}^2} }}\\ \Rightarrow {U_C} = \dfrac{U}{{\sqrt {\left( {{R^2} + {Z_L}^2} \right)\dfrac{1}{{{Z_C}^2}} - 2{Z_L}\dfrac{1}{{{Z_C}}} + 1} }}\,\,\left( 1 \right)\end{array}\)

Từ (1), ta có: \(\left( {{R^2} + {Z_L}^2} \right)\dfrac{1}{{{Z_C}^2}} - 2{Z_L}\dfrac{1}{{{Z_C}}} + 1 - {\left( {\dfrac{U}{{{U_C}}}} \right)^2} = 0\)

Với giá trị của dung kháng \(\left\{ \begin{array}{l}{Z_{{C_1}}} = \dfrac{{125}}{3}\,\,\Omega \\{Z_{{C_2}}} = 125\,\,\Omega \end{array} \right.\), cho cùng 1 giá trị hiệu điện thế: \({U_{{C_1}}} = {U_{{C_2}}} = 100\,\,\left( V \right)\)

Khi \({Z_C} \to \infty \) thì \({U_C} = U = 72,11\,\,V = 20\sqrt {13} \,\,V\)

\( \Rightarrow 1 - {\left( {\dfrac{U}{{{U_C}}}} \right)^2} = 1 - {\left( {\dfrac{{20\sqrt {13} }}{{100}}} \right)^2} = 0,48\)

Theo định lí Vi – et, ta có:

\(\begin{array}{l}\left\{ \begin{array}{l}{x_1} + {x_2} = \dfrac{{ - b}}{a}\\{x_1}{x_2} = \dfrac{c}{a}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\dfrac{1}{{{Z_{{C_1}}}}} + \dfrac{1}{{{Z_{{C_2}}}}} = \dfrac{{2{Z_L}}}{{{R^2} + {Z_L}^2}}\\\dfrac{1}{{{Z_{{C_1}}}}}.\dfrac{1}{{{Z_{{C_2}}}}} = \dfrac{{0,48}}{{{R^2} + {Z_L}^2}}\end{array} \right.\\ \Rightarrow {R^2} + {Z_L}^2 = \dfrac{{0,48}}{{\dfrac{1}{{{Z_{{C_1}}}}}.\dfrac{1}{{{Z_{{C_2}}}}}}} = \dfrac{{0,48}}{{\dfrac{1}{{\dfrac{{125}}{3}}}.\dfrac{1}{{125}}}} = 2500\\ \Rightarrow \dfrac{1}{{\dfrac{{125}}{3}}} + \dfrac{1}{{125}} = \dfrac{{2{Z_L}}}{{2500}} \Rightarrow {Z_L} = 40\,\,\left( \Omega  \right)\\ \Rightarrow R = \sqrt {2500 - {{40}^2}}  = 30\,\,\left( \Omega  \right)\end{array}\)

Chọn A.