Hiệu điện thế giữa hai đầu tụ điện là:

\({U_C} = I.{Z_C} = \frac{{U{Z_C}}}{Z} = \frac{{U{Z_C}}}{{\sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }} = \frac{{U{Z_C}}}{{\sqrt {{{\left( {{R^2} + {Z_L}} \right)}^2} - 2{Z_L}{Z_C} + {Z_C}^2} }}\)

\( \Rightarrow {U_C} = \frac{U}{{\sqrt {\left( {{R^2} + {Z_L}^2} \right)\frac{1}{{{Z_C}^2}} - 2{Z_L}\frac{1}{} + 1} }}{\mkern 1mu} {\mkern 1mu} \left( 1 \right)\)

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

Với giá trị của dung kháng \(\left\{ {\begin{array}{*{20}{l}}{{Z_} = \frac{3}{\mkern 1mu} {\mkern 1mu} \Omega }\\{{Z_} = 125{\mkern 1mu} {\mkern 1mu} \Omega }\end{array}} \right.\), cho cùng 1 giá trị hiệu điện thế: \({U_} = {U_} = 100{\mkern 1mu} {\mkern 1mu} \left( V \right)\)

Khi \({Z_C} \to \infty \) thì \({U_C} = U = 72,11{\mkern 1mu} {\mkern 1mu} V = 20\sqrt {13} {\mkern 1mu} {\mkern 1mu} V\)\( \Rightarrow 1 - {\left( {\frac{U}{}} \right)^2} = 1 - {\left( {\frac{{20\sqrt {13} }}} \right)^2} = 0,48\)

\( \Rightarrow \left( {{R^2} + {Z_L}^2} \right)\frac{1}{{{Z_C}^2}} - 2{Z_L}\frac{1}{} + 0,48 = 0\)

Theo định lí Vi – et, đặt \[x = \frac{1}{}\] ta có: \(\left\{ {\begin{array}{*{20}{l}}{{x_1} + {x_2} = \frac{{ - b}}{a}}\\{{x_1}{x_2} = \frac{c}{a}}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{1}{} + \frac{1}{} = \frac{{2{Z_L}}}{{{R^2} + {Z_L}^2}}}\\{\frac{1}{}.\frac{1}{} = \frac{{0,48}}{{{R^2} + {Z_L}^2}}}\end{array}} \right.\)

\( \Rightarrow {R^2} + {Z_L}^2 = \frac{{0,48}}{{\frac{1}{}.\frac{1}{}}} = \frac{{0,48}}{{\frac{1}{{\frac{3}}}.\frac{1}}} = 2500\)\( \Rightarrow \frac{1}{{\frac{3}}} + \frac{1} = \frac{{2{Z_L}}} \Rightarrow {Z_L} = 40{\mkern 1mu} {\mkern 1mu} \left( \Omega \right)\)

\( \Rightarrow R = \sqrt {2500 - {{40}^2}} = 30{\mkern 1mu} {\mkern 1mu} \left( \Omega \right)\).

Đáp án: \(30\Omega \)