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* Hướng dẫn giải

Khi:

\(\begin{array}{l} {\lambda _1} = 10m \Rightarrow {\lambda _1} = 2\pi \sqrt {L{C_1}} \\ \Rightarrow {C_1} = \frac{{\lambda _1^2}}{{4{\pi ^2}{c^2}L}}{\mkern 1mu} = \frac{{{{10}^2}}}{{4{\pi ^2}{{.3}^2}{{.10}^{16}}{{.8,8.10}^{ - 6}}}} = {3,2.10^{ - 13}}F \end{array}\)

Do C₁0⇒ ghép C_x nối tiếp C₀:

\(\begin{array}{l} \frac{1}{{{C_1}}} = \frac{1}{{{C_x}}} + \frac{1}{{{C_0}}}{\mkern 1mu} \Rightarrow {C_x} = 0,32pF\\ \begin{array}{*{20}{l}} {Khi{\mkern 1mu} {\lambda _2} = 50m}\\ { \Rightarrow {C_2} = \frac{{\lambda _2^2}}{{4{\pi ^2}{c^2}L}}}\\ {\:\:\:\:\:\:\:\:\: = \frac{{{{50}^2}}}{{4{\pi ^2}{{.3}^2}{{.10}^{16}}{{.8,8.10}^{ - 6}}}}}\\ {\:\:\:\:\:\:\:\:\: = {{1,6.10}^{ - 12}}F = 1,6pF} \end{array} \end{array}\)

Do C₂0 nên Cx nối tiếp C₀ suy ra C_x=1,6pF.