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In detector, output circuit consists of $R=10 \mathrm{k} \Omega$ and $C=100 \mathrm{pF}$. The frequency of carrier signal which it can detect is
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Verified Answer
The correct answer is:
$>>1 \mathrm{MHz}$
Given, $R=10 \mathrm{k} \Omega=10^{4} \Omega$
$$
C=100 \mathrm{pF}=100 \times 10^{-12} \mathrm{~F}
$$
Time constant of $R C$ circuit,
$$
\tau=R C=10^{4} \times 100 \times 10^{-12}=10^{-6} \mathrm{~s}
$$
For demodulation $\frac{1}{f_{C}} \ll R C \Rightarrow f_{C} \gg \frac{1}{R C}$
or $\quad f_{C} \gg \frac{1}{10^{-6}}=10^{6} \mathrm{~Hz}$
$\therefore \quad f_{\mathrm{C}} \gg 1 \mathrm{MHz}$
$$
C=100 \mathrm{pF}=100 \times 10^{-12} \mathrm{~F}
$$
Time constant of $R C$ circuit,
$$
\tau=R C=10^{4} \times 100 \times 10^{-12}=10^{-6} \mathrm{~s}
$$
For demodulation $\frac{1}{f_{C}} \ll R C \Rightarrow f_{C} \gg \frac{1}{R C}$
or $\quad f_{C} \gg \frac{1}{10^{-6}}=10^{6} \mathrm{~Hz}$
$\therefore \quad f_{\mathrm{C}} \gg 1 \mathrm{MHz}$
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