The tuned circuit of an oscillator in a simple AM transmitter employs a 250 $μ$ H 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

210 to 230

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258 to 278

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308 to 328

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118 to 128

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Solution

The correct option is C 308 to 328
$f_{C} = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π \sqrt{250 \times 10^{- 6} \times 1 \times 10^{- 9}}}$
or
$f_{C} = 318.471 k H z$
$f_{U S B} = 318.471 + 10 = 328.471 k H z$
$f_{L S B} = 318.471 - 10 = 308.471 k H z$
Hence the correct option is C.

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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

The correct option is C 308 to 328fC=12π√LC=12π√250×10−6×1×10−9orfC=318.471kHzfUSB=318.471+10=328.471kHzfLSB=318.471−10=308.471kHzHence the correct option is C.

The tuned circuit of an oscillator in a simple AM transmitter employs a 250 $μ$ H 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

The correct option is C 308 to 328
$f_{C} = \frac{1}{2 π \sqrt{L C}} = \frac{1}{2 π \sqrt{250 \times 10^{- 6} \times 1 \times 10^{- 9}}}$
or
$f_{C} = 318.471 k H z$
$f_{U S B} = 318.471 + 10 = 328.471 k H z$
$f_{L S B} = 318.471 - 10 = 308.471 k H z$
Hence the correct option is C.