Một đoạn mạch RLC nối tiếp được mắc vào hai cực củ...

B

- Hướng dẫn giải

Từ hình, ta có:

\(\begin{array}{l}
1,5{T_1} = {2.10^{ - 2}} \Rightarrow {T_1} = \frac{4}{3}{.10^{ - 2}}\left( {\rm{s}} \right) \Rightarrow {\omega _1} = 150\pi  \Rightarrow {n_1} = 75(vong/s)\\
{T_2} = {2.10^{ - 2}} \Rightarrow {\omega _2} = 100\pi  \Rightarrow {n_2} = 50(vong/s)
\end{array}\)

+ Ta có:  

\(\begin{array}{l}
I = \frac{E}{{\sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }} = \frac{{NBS.\omega }}{{\sqrt 2 \sqrt {{R^2} + {{\left( {\omega L} \right)}^2} - 2\frac{L}{C} + {{\left( {\frac{1}{{\omega C}}} \right)}^2}} }}\\
 \Rightarrow I = \frac{{NBS}}{{\sqrt 2 \sqrt {\frac{{{R^2}}}{{{\omega ^2}}} + {L^2} - 2\frac{L}{C}\frac{1}{{{\omega ^2}}} + \frac{1}{{{\omega ^4}{C^2}}}} }} = \frac{{NBS}}{{\sqrt 2 \sqrt {\left( {\frac{1}{{{C^2}}}} \right)\frac{1}{{{\omega ^4}}} - \left( {2\frac{L}{C} - {R^2}} \right)\frac{1}{{{\omega ^2}}} + {L^2}} }}\\
 \Rightarrow I = \frac{{NBS}}{{\sqrt 2 \sqrt {\left( {\frac{1}{{{C^2}}}} \right)\frac{1}{{{\omega ^4}}} - \left( {2\frac{L}{C} - {R^2}} \right)\frac{1}{{{\omega ^2}}} + {L^2}} }} \Rightarrow \left( {\frac{1}{{{C^2}}}} \right)\frac{1}{{{\omega ^4}}} - 2\left( {\frac{L}{C} - \frac{{{R^2}}}{2}} \right)\frac{1}{{{\omega ^2}}} + A = 0
\end{array}\)                                                                                                  (*)

+ Từ phương trình (*), ta có:  

\(\begin{array}{l}
\left\{ \begin{array}{l}
\frac{1}{{\omega _1^2}} + \frac{1}{{\omega _2^2}} =  - \frac{b}{a}\\
\frac{1}{{\omega _0^2}} =  - \frac{b}{{2a}}
\end{array} \right.\\
 \Rightarrow \frac{1}{{\omega _1^2}} + \frac{1}{{\omega _2^2}} = \frac{2}{{\omega _0^2}} \Leftrightarrow \frac{1}{{n_1^2}} + \frac{1}{{n_2^2}} = \frac{2}{{n_0^2}} \Rightarrow {n_0} = 58,83(vong/s)
\end{array}\)