Ina detector, output circuit consists of `R=10kOmega` and `C=100 pF`, the frequency (in MHz) of a carrier signal it can detect must be greater than:

Answer

1

Explanation

Here, `R=10kOmega = 10^4 Omega ,C = 100 pF = 10^(-10)` F
Time constant of the circuit,
`tau = RC = 10^4xx 10^(-10) = 10^(-6)s`
For demodulation,`1/v_clt lt RC` or `v_c gt gt 1//RC`
or `v_c gt gt 1/(10^(-6))=10^6 Hz =1MHz`
`:. ` Frequency of carrier signal must be greater than 1 MHz.