In an AM detector circuit,R = 2 text{ k}Omega | vedclass

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(B) For an $AM$ detector to work effectively,the time constant $RC$ must satisfy the condition: $\frac{1}{f_c} \ll RC \ll \frac{1}{f_m}$,where $f_c$ i

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(B) For an $AM$ detector to work effectively,the time constant $RC$ must satisfy the condition: $\frac{1}{f_c} \ll RC \ll \frac{1}{f_m}$,where $f_c$ is the carrier frequency and $f_m$ is the message frequency.Given $R = 2 	imes 10^3 \Omega$ and $C = 1 	imes 10^{-6} 	ext{ F}$.The time constant $RC = (2 	imes 10^3) 	imes (1 	imes 10^{-6}) = 2 	imes 10^{-3} 	ext{ s} = 2 	ext{ ms}$.For effective demodulation,the condition $RC \approx \frac{1}{f_c}$ is typically considered for the upper limit of the carrier frequency.Calculating $f_c = \frac{1}{RC} = \frac{1}{2 	imes 10^{-3}} = 500 	ext{ Hz} = 0.5 	ext{ kHz}$.Among the given options,$1 	ext{ kHz}$ is the closest value that satisfies the condition $RC > \frac{1}{f_c}$ (i.e.,$2 	ext{ ms} > 1 	ext{ ms}$).Thus,the circuit will demodulate a carrier frequency of $1 	ext{ kHz}$ effectively.

In an $AM$ detector circuit,$R = 2 \text{ k}\Omega$ and $C = 1 \mu\text{F}$. This circuit will demodulate a carrier frequency of ......... $\text{kHz}$ effectively.

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