## In This post you will get answer of Q.

In a detector, output circuit consists of $R = 10, k Omega $ and $C = 100 mu F.$ The frequency of carrier signal it can detect is

Solution:

## Here $hspace10mm R = 10, k Omega = 10^4 Omega$

$hspace20mm C = 100pF = 10^{-10}F $

Time constant of the circuit

$hspace20mm tau = RC = 10^4 times 10^{-10} = 10^{-6}s$

For demodulation.$frac{1}{f_c}

or $hspace10mm f_c >> frac{1}{RC}, or, f_c >>frac{1}{10^{-6}} Hz$

Therefore,frequency of carrier signal must be much greater than 1 MHz.