Ta có bước sóng
\(\lambda = c.T = 2\pi c\sqrt {LC} \)
Vậy \({\lambda _{\max }} = c.T = 2\pi c\sqrt {{L_{\max }}{C_{\max }}}\)
\( \Rightarrow {L_{\max }} = \dfrac{{\lambda _{\max }^2}}{{4{\pi ^2}{c^2}{C_{\max }}}}\\ = \dfrac{{{{1000}^2}}}{{4{\pi ^2}{{({{3.10}^8})}^2}{{.860.10}^{ - 12}}}} = 0,{33.10^{ - 3}}H\)
\({\lambda _{\min }} = c.T = 2\pi c\sqrt {{L_{\min }}{C_{\min }}} \)
\(\Rightarrow {L_{\min }} = \dfrac{{\lambda _{\min }^2}}{{4{\pi ^2}{c^2}{C_{\min }}}}\\ = \dfrac{{{{10}^2}}}{{4{\pi ^2}{{({{3.10}^8})}^2}{{.15.10}^{ - 12}}}} \\= 1,{87.10^{ - 6}}H\)
