Question

In the given detector circuit, the suitable value of carrier frequency is :-

(A) $\ll {10^9}Hz$
(B) $\ll {10^5}Hz$
(C) $\gg {10^9}Hz$
(D) None of these.

Answer 1

In this detector circuit, using the values for the resistance and the capacitance, we can find their product. Now using the condition for the carrier frequency in detector circuit that is, $\dfrac{1}{{{f_c}}} \ll RC$, we can find the suitable value of the carrier frequency.

Formula Used In this solution we will be using the following formula,
$\dfrac{1}{{{f_c}}} \ll RC$
where ${f_c}$ is the carrier frequency,
$R$ is the resistance and
$C$ is the capacitance

Complete Step by Step Solution
For a detector circuit, the carrier frequency, the resistance and the capacitance are all related by the formula,
$\dfrac{1}{{{f_c}}} \ll RC$
Now we can take reciprocal on both the sides. So we will get,
${f_c} \gg \dfrac{1}{{RC}}$
The inequality sign also gets reversed on taking the reciprocal.
Now we are given in the diagram in the question that the resistance has a value,
$R = 1000\Omega$
And the capacitance is given as,
$C = 1 \times {10^{ - 12}}F$
Therefore, the RHS of the previous equation is the product of the resistance and capacitance. So n doing the product we get,
$R \times C = 1000 \times 1 \times {10^{ - 12}}$
So this gives us a value of,
$R \times C = {10^{ - 9}}$
Therefore, we can substitute this value in the equation and get,
${f_c} \gg \dfrac{1}{{{{10}^{ - 9}}}}$
Hence this becomes,
${f_c} \gg {10^9}Hz$
So the carrier frequency should have a suitable value of $\gg {10^9}Hz$

So the correct answer is option C.

Note The carrier frequency is the frequency of the carrier wave that is modulated to transmit signals.
A carrier signal is used for efficient transmission in order to reduce the wavelength. The carrier signal is usually a simple single frequency sinusoid. A detector circuit is used to extract information from a modulated radio frequency current or voltage.