Mạch điện xoay chiều gồm cuộn dây có (L,R0) và hai tụ điện C1, C2

Đặt: \(C_1 + C_2 = x,C_1.C_2 = y\)
Khi mắc song song: \(\Rightarrow \omega ^2 = \frac{1}{L.C}\Leftrightarrow (48 \pi)^2 = \frac{1}{L.(C_1 + C_2)} = \frac{1}{Lx}\)
Khi mắc nối tiếp: \(\omega ^2 = \frac{1}{L.C}\Leftrightarrow (100 \pi)^2 = \frac{1}{L.(\frac{C_1C_2}{C_1 + C_2})} = \frac{1}{L.\frac{y}{x}}\)

Ta có: 
\(\Rightarrow (\frac{48 \pi}{100 \pi})^2 = \frac{\frac{y}{x}}{x} = \frac{y}{x^2} \Leftrightarrow 12 x = 25 \sqrt{y} \Leftrightarrow 12C_1 + 12C_2 = 25 \sqrt{C_1.C_2}\)

\(\Rightarrow \bigg \lbrack \begin{matrix} C_1 = \frac{16}{9}C_2\\ C_1 = \frac{9}{16}C_2\end{matrix}\)
\(* C_1 = \frac{16}{9}C_2\)
\(\Rightarrow (\frac{48 \pi}{\omega })^2 \frac{\frac{y}{x}}{x} = \frac{y}{x^2}\Leftrightarrow 12 x = 25\sqrt{y}\Leftrightarrow 12C_1 + 12 C_2 = 25\sqrt{C_1.C_2}\)

\(\Rightarrow \bigg \lbrack\begin{matrix} C_1 = \frac{16}{9}C_2\\ C_1 = \frac{9}{16}C_2\end{matrix}\)
\(* C_1 = \frac{16}{9}C_2\)
\(\Rightarrow (\frac{48 \pi}{\omega _3})^2 = \frac{\frac{1}{L.(C_1 + C_2)}}{\frac{1}{C_1}} = \frac{16}{25}\Leftrightarrow \omega _3 = 60 \pi\)
\(* C_1 = \frac{9}{16}C_2\)
\(\Rightarrow (\frac{48 \pi}{\omega _3})^2 = \frac{C_1.L}{(C_1 + C_2).L} = 80 \pi (loai)\)