The tuned circuit of an oscillator in a simple AM transmitter employs a 250 micro henry Coil and 1 nf condenser. If the oscillator output is modulated by audio frequency upto 10 KHz, the frequency range occupied by the side bands in KHz is
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a
210 to 230
b
258 to 278
c
308 to 328
d
118 to 128
answer is C.
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Detailed Solution
Given : L=250×10−6H , C=1×10−9F and fm=10KHz=10×103Hz Carrier Wave frequency, fc=12πLC=12π250×10−6×10−9 = =106×722=0.318×106 ⇒fc=318×103Hz = 318 KHz Hence, Lower side band =fc−fm =318-10=308 KHz Upper side band =fc+fm =318+10=328 KHz