Đáp án:

\(1,{62.10^{ - 8}}F \ge C \ge 1,{01.10^{ - 9}}F\)

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
\omega  = \dfrac{1}{{\sqrt {LC} }} \Rightarrow 2\pi f = \dfrac{1}{{\sqrt {LC} }}\\
 \Rightarrow f = \dfrac{1}{{2\pi \sqrt {LC} }}\\
12500 \le f \le 50000\\
 \Rightarrow 12500 \le \dfrac{1}{{2\pi \sqrt {LC} }} \le 50000\\
 \Rightarrow 12500 \le \dfrac{1}{{2\pi \sqrt {0,01.C} }} \le 50000\\
 \Rightarrow {8.10^{ - 5}} \ge 2\pi \sqrt {0,01.C}  \ge {2.10^{ - 5}}\\
 \Rightarrow 1,{62.10^{ - 8}}F \ge C \ge 1,{01.10^{ - 9}}F
\end{array}\)