Đặt điện áp xoay chiều u = U0cosωt (ω t...

C

- Hướng dẫn giải

Ta có:

\(\begin{array}{l}
\omega _o^2 = {\omega _L}{\omega _C};{\omega _L} = \sqrt {\frac{1}{{LC}} - \frac{{{R^2}}}{{2{L^2}}}} \\
 \Rightarrow \left\{ \begin{array}{l}
{\omega _1} = 2\pi {f_1} = 2\pi {n_1}p = 40\pi rad/s\\
{\omega _2} = 2\pi {f_2} = 2\pi {n_2}p = 120\pi rad/s
\end{array} \right.
\end{array}\)

Điện áp hiệu dụng hai đầu cuộn cảm:

\(\begin{array}{l}
{U_L} = \frac{{\omega \phi .\omega L}}{{\sqrt {{R^2} + {{(\omega L - \frac{1}{{\omega C}})}^2}} }} = \frac{{{\omega ^2}\phi L}}{{\sqrt {{R^2} + {{(\omega L - \frac{1}{{\omega C}})}^2}} }}\\
{U_{L1}} = {U_{L2}} \Rightarrow \frac{{\omega _1^2\phi L}}{{\sqrt {{R^2} + {{({\omega _1}L - \frac{1}{{{\omega _1}C}})}^2}} }} = \frac{{\omega _2^2\phi L}}{{\sqrt {{R^2} + {{({\omega _2}L - \frac{1}{{{\omega _2}C}})}^2}} }}\\
 \Leftrightarrow 160\left( {\frac{1}{{LC}} - \frac{{{R^2}}}{{2{L^2}}}} \right) = \left( {81\omega _1^2 - \omega _2^2} \right) + \frac{1}{{{L^2}{C^2}}}\left( {\frac{{81}}{{\omega _1^2}} - \frac{1}{{\omega _2^2}}} \right)\\
{\omega _L} = \sqrt {\frac{1}{{LC}} - \frac{{{R^2}}}{{2{L^2}}}}  = 48\pi rad/s;{\omega _1} = 40\pi rad/s;{\omega _2} = 120\pi rad/s;{\omega _o} = \frac{1}{{LC}}\\
 \Rightarrow {\omega _o} = 156,12rad/s
\end{array}\)