Đáp án:
Bài 9:B
Ta có:
\[\begin{gathered}
{U_o} = \frac{{{Q_o}}}{C} = \frac{{{I_o}}}{{\omega C}} = \sqrt {\frac{L}{C}} {I_o} \hfill \\
{\left( {\frac{{{u_C}}}{{{U_o}}}} \right)^2} + {\left( {\frac{i}{{{I_o}}}} \right)^2} = 1 \hfill \\
\Leftrightarrow {\left( {\frac{{{u_C}}}{{{I_o}\sqrt {\frac{L}{C}} }}} \right)^2} + {\left( {\frac{i}{{{I_o}}}} \right)^2} = 1 \hfill \\
\Leftrightarrow {I_o} = 0,05A = 50mA \hfill \\
\end{gathered} \]
Bài 10: A
Ta có:
\[f = \frac{\omega }{{2\pi }} = \frac{1}{{2\pi }}\sqrt {\frac{1}{{LC}}} \Rightarrow C = \frac{1}{{4{\pi ^2}L{f^2}}}\]
\[f = \frac{1}{{2\pi }}\sqrt {\frac{1}{{L{C_{td}}}}} = \frac{1}{{2\pi }}\sqrt {\frac{1}{{L\left( {{C_1} + {C_2}} \right)}}} = \frac{1}{{2\pi }}\sqrt {\frac{1}{{L(\frac{1}{{4{\pi ^2}L{f_1}^2}} + \frac{1}{{4{\pi ^2}L{f_2}^2}})}}} \Rightarrow f = \sqrt {\frac{1}{{\frac{1}{{{f_1}^2}} + \frac{1}{{{f_2}^2}}}}} = 4,8kHz\]
