Cho mạch điện xoay chiều R, L, C nối tiếp.

+ Vì f1 và f2 cho cùng cosφ => cho cùng Z => cho cùng I

 \(\begin{array}{l}
{\omega _1}.{\omega _2} = \frac{1}{{LC}} \Rightarrow {\omega _2}L = \frac{1}{{{\omega _1}C}} \Rightarrow {Z_{L2}} = {Z_{C1}}(1)\\
{\cos ^2}\varphi  = \frac{{{R^2}}}{{{R^2} + {{\left( {{Z_{L1}} - {Z_{C1}}} \right)}^2}}}\\
 \Rightarrow {R^2} = {\left( {{Z_{L1}} - {Z_{C1}}} \right)^2}{R^2} = {\left( {{Z_{L1}} - {Z_{L2}}} \right)^2} = {L^2}{\left( {{\omega _1} - {\omega _2}} \right)^2}(2)\\
{U_{L - max}}:{Z_C} = \sqrt {\frac{L}{C} - \frac{{{R^2}}}{2}}  \Leftrightarrow {\left( {\frac{1}{{\omega C}}} \right)^2} = \frac{L}{C} - \frac{{{R^2}}}{2}\\
 \Rightarrow \frac{1}{{{\omega ^2}}} = LC - \frac{{{R^2}{C^2}}}{2}(3)
\end{array}\)

+ Thay (2) vào (3), ta có:  

 \(\begin{array}{l}
\frac{1}{{\omega {c^2}}} = LC - \frac{{{C^2}}}{2}\left[ {{L^2}{{\left( {{\omega _1} - {\omega _2}} \right)}^2}} \right] = LC - \frac{{{L^2}{C^2}}}{2}{\left( {{\omega _1} - {\omega _2}} \right)^2}\\
 \Rightarrow \frac{1}{{{\omega ^2}}} = \frac{1}{{{\omega _1}.{\omega _2}}} - \frac{1}{{2{{\left( {{\omega _1}.{\omega _2}} \right)}^2}}}{\left( {{\omega _1} - {\omega _2}} \right)^2}\frac{1}{{{\omega ^2}}} = \frac{1}{{9\omega _1^2}}\\
 \Rightarrow \omega  = 3{\omega _1} \Rightarrow f = 3{f_1}
\end{array}\)