Đặt điện áp xoay chiều \(u = U\sqrt 2 \cos 2ft\,\,\,\left( V \right)\)

Ta có: \(\left\{ \begin{array}{l}{U_R} = 120V\\{U_L} = 180V\\{U_C} = 20V\end{array} \right.\)

\( \Rightarrow U = \sqrt {U_R^2 + {{\left( {{U_L} - {U_C}} \right)}^2}}  \\= \sqrt {{{120}^2} + {{\left( {180 - 20} \right)}^2}}  = 200V\)

\( \Rightarrow \dfrac{{{U_R}}}{{{U_C}}} = \dfrac{R}{{{Z_C}}} = \dfrac{{120}}{{20}} = 6\)

\( \Rightarrow \dfrac{{{U_C}}}{{{U_L}}} = \dfrac{{{Z_C}}}{{{Z_L}}} = \dfrac{{20}}{{180}} = \dfrac{1}{9}\)

+ Khi giảm tần số của nguồn 2 lần thì \(\left\{ \begin{array}{l}{Z_C}' = 2{Z_C}\\{Z_L}' = \dfrac{{{Z_L}}}{2}\end{array} \right.\)

 \(\begin{array}{l}\dfrac{{{U_R}'}}{{{U_C}'}} = \dfrac{R}{{{Z_C}'}} = \dfrac{R}{{2{Z_C}}} = \dfrac{6}{2} = 3\\ \Rightarrow {U_R}' = 3{U_C}'\end{array}\)

\(\begin{array}{l} \Rightarrow \dfrac{{{U_C}'}}{{{U_L}'}} = \dfrac{{{Z_C}'}}{{{Z_L}'}} = \dfrac{{2{Z_C}}}{{\dfrac{{{Z_L}}}{2}}} = \dfrac{{4{Z_C}}}{{{Z_L}}} = \dfrac{4}{9}\\ \Rightarrow {U_L}' = \dfrac{9}{4}{U_C}'\end{array}\)

\(\begin{array}{l}U = \sqrt {{U_R}{'^2} + {{\left( {{U_L}' - {U_C}'} \right)}^2}} \\ = \sqrt {9{U_C}{'^2} + {{\left( {\dfrac{9}{4}{U_C}' - {U_C}'} \right)}^2}}  = \dfrac{{13}}{4}{U_C}'\\ \Rightarrow {U_C}' = \dfrac{4}{{13}}U = \dfrac{4}{{13}}.200 = 61,5V\end{array}\)