Quy bài toán từ \({\rm{f}} \to \infty \)

+ TH1: Xét \({U_C} = U \Leftrightarrow {Z_{C1}} = Z \Rightarrow {R^2} = 2{Z_{{\rm{L}}1}}{Z_{C1}} - Z_{L1}^2\)

+ TH2: Xét \({{\rm{U}}_{\rm{L}}} = {\rm{U}} \Rightarrow {{\rm{Z}}_{{\rm{L}}2}} = {\rm{Z}} \Rightarrow {{\rm{R}}^2} = 2{{\rm{Z}}_{{\rm{L}}2}}{{\rm{Z}}_{{\rm{C}}2}} - {\rm{Z}}_{{\rm{C}}2}^2 = 2{{\rm{Z}}_{{\rm{L}}1}}{{\rm{Z}}_{{\rm{C}}1}} - {\rm{Z}}_{{\rm{L}}1}^2\)

\(\frac{{2\;{\rm{L}}}}{{\rm{C}}} - \frac{1}{{{{({\rm{C\omega }})}^2}}} = \frac{{2\;{\rm{L}}}}{{\rm{C}}} - {({\rm{L\omega }})^2} \Rightarrow 1 = {\left( {{\rm{LC\omega }}{{\rm{\omega }}_0}} \right)^2} \Rightarrow {\rm{LC\omega }}{{\rm{\omega }}_0} = 1\)

Đồng thời \(\cos {{\rm{\varphi }}_2} = \frac{1}{{\sqrt 3 }} \Rightarrow \sin {{\rm{\varphi }}_2} = \frac{{\sqrt 6 }}{3} = \frac{{{U_{\rm{L}}} - {U_C}}}{U} = \frac{{{U_L} - {U_C}}}{} = 1 - \frac{}{} \Rightarrow \frac{}{} = 1 - \frac{{\sqrt 6 }}{3}\)

\(\frac{1}{{{\mathop{\rm LC}\nolimits} {\omega ^2}}} = 1 - \frac{{\sqrt 6 }}{3}\left( {{\rm{\omega }} = {{\rm{\omega }}_0} + 150{\rm{\pi }}} \right) \Rightarrow {\mathop{\rm LC}\nolimits} = \frac{{{{\rm{\omega }}^2}}}(2)\)

Ta có: \(1 = \left[ {(3 + \sqrt 6 )\frac{{{{\rm{\omega }}_0}}}{{\rm{\omega }}}} \right] \Rightarrow {\rm{\omega }} = (3 + \sqrt 6 ){{\rm{\omega }}_0} \Rightarrow {\rm{f}} = (3 + \sqrt 6 ){{\rm{f}}_0} \Rightarrow {\rm{f}} \approx 16,9\;{\rm{Hz}}\). Đáp án. 16,9.