A) \[>>1\,\,MHz\]
B) \[0.1\text{ }kHz\]
C) \[>>1\text{ }GHz\]
D) \[{{10}^{3}}\,Hz\]
Correct Answer: A
Solution :
Here \[R=10\,k\,\Omega ={{10}^{4}}\,\Omega \] \[\operatorname{C}=100pF=1{{0}^{-10}}F\] Time constant of the circuit \[\tau \,\,=\,\,RC=1{{0}^{4}}\times {{10}^{-10}}\,\,=\,\,{{10}^{-}}^{6}\,s\] For demodulation, \[\frac{1}{{{f}_{c}}}<<RC\] Or \[{{f}_{c}}>>\frac{1}{RC}\] \[{{f}_{c}}>>\frac{1}{{{10}^{-6}}}\,\,=\,\,{{10}^{6}}\,Hz\] Therefore, frequency of carrier signal must be much greater than 1 MHz.
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