A. 8/3
B. 1/3
C. 3
D. 3/8
A
Ta có:
\(\begin{array}{l}
+ C = a.\varphi + b\\
\Rightarrow \frac{{{C_2} - {C_0}}}{{{C_1} - {C_0}}} = \frac{{{\varphi _2} - {\varphi _0}}}{{{\varphi _1} - {\varphi _0}}}\frac{{{C_2} - {C_0}}}{{{C_1} - {C_0}}} = \frac{{{\varphi _2}}}{{{\varphi _1}}}\\
+ f = \frac{1}{{2\pi \sqrt {LC} }} \Rightarrow C = \frac{1}{{4{\pi ^2}L{f^2}}}\\
\Rightarrow \frac{{{C_2} - {C_0}}}{{{C_1} - {C_0}}} = \frac{{\frac{1}{{f_2^2}} - \frac{1}{{f_0^2}}}}{{\frac{1}{{f_1^2}} - \frac{1}{{f_0^2}}}} = \frac{{\frac{9}{{f_0^2}} - \frac{1}{{f_0^2}}}}{{\frac{4}{{f_0^2}} - \frac{1}{{f_0^2}}}} = \frac{8}{3}\\
\Rightarrow \frac{{{\varphi _2}}}{{{\varphi _1}}} = \frac{8}{3}
\end{array}\)
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