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An $L-C$ circuit contains $196 \mathrm{pF}$ capacitor and a $441 \mu \mathrm{H}$ inductor. The frequency of electromagnetic radiation emitted by antenna coupled to the $L-C$ circuit is
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Verified Answer
The correct answer is:
$5.41 \times 10^5 \mathrm{~Hz}$
Given, capacitance, $C=196 \mathrm{pF}=196 \times 10^{-12} \mathrm{~F}$
Inductance, $L=441 \mu \mathrm{H}$
$$
=441 \times 10^{-6} \mathrm{H}
$$
Since, frequency, $f=\frac{1}{2 \pi \sqrt{L C}}$
$$
\begin{aligned}
\therefore \quad f & =\frac{1}{2 \pi \sqrt{441 \times 10^{-6} \times 196 \times 10^{-12}}} \\
& =\frac{1}{2 \pi \times 14 \times 21 \times 10^{-9}} \\
& =5.41 \times 10^5 \mathrm{~Hz}
\end{aligned}
$$
Inductance, $L=441 \mu \mathrm{H}$
$$
=441 \times 10^{-6} \mathrm{H}
$$
Since, frequency, $f=\frac{1}{2 \pi \sqrt{L C}}$
$$
\begin{aligned}
\therefore \quad f & =\frac{1}{2 \pi \sqrt{441 \times 10^{-6} \times 196 \times 10^{-12}}} \\
& =\frac{1}{2 \pi \times 14 \times 21 \times 10^{-9}} \\
& =5.41 \times 10^5 \mathrm{~Hz}
\end{aligned}
$$
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