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The carrier freqeuncy of a station is 40...

The carrier freqeuncy of a station is 40 MHz. A resistor of 10k `Omega` and capacitor of 12 pF are available in the detector circuit. The possible value of C will be

A

12

B

8.2

C

5.6

D

All of these

Text Solution

Verified by Experts

The correct Answer is:
D
Here, `R=10kOmega=10xx10^(3)Omega`
`C=12pF=12xx10^(-12)F`
`v_(c)=40MHz=40xx10^(6)Hz=4xx10^(7)Hz`.
Time period of carrier frequency
`T_(c)=(1)/(v_(c))-(1)/(4xx10^(7))=2.5xx10^(8)s`
time constant of C-R circuit,
As `T_(c)lt tau`, therefore, circuit is good enough for detection All the values of C given in the options are possible.
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