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Question

The tuned circuit of an oscillator in a simple AM transmitter employs a 250 μH coil and 1nF condenser. If the oscillator output is modulated by audio frequency upto 10KHz. The frequency range occupied by the side bands in KHz is
  1. 258 to 278
  2. 118 to 128
  3. 308 to 328
  4. 210 to 230

A
210 to 230
B
308 to 328
C
258 to 278
D
118 to 128
Solution
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fC=12πLC=12π250×106×1×109
or
fC=318.471kHz
fUSB=318.471+10=328.471kHz
fLSB=318.47110=308.471kHz
Hence the correct option is C.

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