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Compute the LC product of a tuned amplifier circuit required to generate a carrier wave of $1 \mathrm{MHz}$ for amplitude modulation.
Solution:
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Verified Answer
As given that, the frequency of carrier wave is $1, \mathrm{f}=1 \mathrm{MHz}$ Formula for the frequency of tuned amplifier,
$$
\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}
$$
So equating both of them So,
$$
\begin{aligned}
&\frac{1}{2 \pi \sqrt{\mathrm{LC}}}=1 \mathrm{MHz} \\
&\sqrt{\mathrm{LC}}=\frac{1}{2 \pi \times 10^6}
\end{aligned}
$$
Squaring both side
$$
\mathrm{LC}=\frac{1}{\left(2 \pi \times 10^6\right)^2}=2.54 \times 10^{-14} \mathrm{sec}^2
$$
So, the product of LC is $2.54 \times 10^{-14} \mathrm{sec}^2$.
$$
\mathrm{f}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}}
$$
So equating both of them So,
$$
\begin{aligned}
&\frac{1}{2 \pi \sqrt{\mathrm{LC}}}=1 \mathrm{MHz} \\
&\sqrt{\mathrm{LC}}=\frac{1}{2 \pi \times 10^6}
\end{aligned}
$$
Squaring both side
$$
\mathrm{LC}=\frac{1}{\left(2 \pi \times 10^6\right)^2}=2.54 \times 10^{-14} \mathrm{sec}^2
$$
So, the product of LC is $2.54 \times 10^{-14} \mathrm{sec}^2$.
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