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
258 to 278
118 to 128
308 to 328
210 to 230
A
210 to 230
B
308 to 328
C
258 to 278
D
118 to 128
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Solution
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fC=12π√LC=12π√250×10−6×1×10−9 or fC=318.471kHz fUSB=318.471+10=328.471kHz fLSB=318.471−10=308.471kHz Hence the correct option is C.
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