Trong mạch chọn sóng vô tuyến, khi mắc tụ điện có...

A

- Hướng dẫn giải

Phương pháp giải:

Bước sóng : \(\lambda  = cT = c.2\pi \sqrt {LC} \)

Giải chi tiết:

Ta có : \(\left\{ \begin{array}{l}{\lambda _1} = c.2\pi \sqrt {L{C_1}} \\{\lambda _2} = c.2\pi \sqrt {L{C_2}} \\\lambda  = c.2\pi \sqrt {LC} \end{array} \right. \Rightarrow \left\{ \begin{array}{l}\frac{1}{{\lambda _1^2}} = \frac{1}{{4{\pi ^2}{c^2}.L{C_1}}}\\\frac{1}{{\lambda _2^2}} = \frac{1}{{4{\pi ^2}{c^2}.L{C_2}}}\\\frac{1}{{\lambda _{}^2}} = \frac{1}{{4{\pi ^2}{c^2}.L{C_{}}}}\end{array} \right.\)

\({C = \frac{{{C_1}.{C_2}}}{{{C_1} + {C_2}}} \Rightarrow \frac{1}{{{\lambda ^2}}} = \frac{1}{{4{\pi ^2}{c^2}.L\frac{{{C_1}.{C_2}}}{{{C_1} + {C_2}}}}}}\)

\({ \Rightarrow \frac{1}{{{\lambda ^2}}} = \frac{{{C_1} + {C_2}}}{{4{\pi ^2}{c^2}.L{C_1}.{C_2}}} = \frac{1}{{4{\pi ^2}{c^2}.L{C_2}}} + \frac{1}{{4{\pi ^2}{c^2}.L{C_1}}}}\)

\({ \Rightarrow \frac{1}{{{\lambda ^2}}} = \frac{1}{{\lambda _1^2}} + \frac{1}{{\lambda _2^2}}}\)

\({ \Leftrightarrow \frac{1}{{{\lambda ^2}}} = \frac{1}{{{{30}^2}}} + \frac{1}{{{{40}^2}}} \Rightarrow \lambda  = 24cm}\)

Chọn A.