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A tuned circuit of a transistor oscillator unit has an inductance of \(5 \mathrm{mH}\) and a capacitance of \(5 \mathrm{pF}\). The natural frequency of the oscillator is
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Verified Answer
The correct answer is:
\(1 \mathrm{MHz}\)
For tuned circuit of a transistor oscillator unit,
\(\begin{aligned}
& L=5 \mathrm{mH}=5 \times 10^{-3} \mathrm{H} \\
& C=5 \mathrm{pF}=5 \times 10^{-12} \mathrm{~F}
\end{aligned}\)
\(\therefore\) Natural frequency of tuned oscillator,
\(\begin{aligned}
f & =\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{5 \times 10^{-3} \times 5 \times 10^{-12}}} \\
& =\frac{1}{2 \pi \sqrt{2.5 \times 10^{-14}}} \\
& =\frac{10^7}{2 \pi \sqrt{2.5}}=\frac{10}{9.93} \times 10^6 \mathrm{~Hz} \\
& =1.006 \times 10^6 \mathrm{~Hz} \\
& \approx 1 \times 10^6 \mathrm{~Hz}=1 \mathrm{MHz}
\end{aligned}\)
\(\begin{aligned}
& L=5 \mathrm{mH}=5 \times 10^{-3} \mathrm{H} \\
& C=5 \mathrm{pF}=5 \times 10^{-12} \mathrm{~F}
\end{aligned}\)
\(\therefore\) Natural frequency of tuned oscillator,
\(\begin{aligned}
f & =\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{5 \times 10^{-3} \times 5 \times 10^{-12}}} \\
& =\frac{1}{2 \pi \sqrt{2.5 \times 10^{-14}}} \\
& =\frac{10^7}{2 \pi \sqrt{2.5}}=\frac{10}{9.93} \times 10^6 \mathrm{~Hz} \\
& =1.006 \times 10^6 \mathrm{~Hz} \\
& \approx 1 \times 10^6 \mathrm{~Hz}=1 \mathrm{MHz}
\end{aligned}\)
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