The tuning circuit in an FM radio receiver is a series RLC circuit with a 0.200

Question

The tuning circuit in an FM radio receiver is a series RLC circuit with a 0.200 H inductor

The receiver is tuned to a station at 103.3 MHz . What is the value of the capacitor in the tuning circuit?

FM radio stations are assigned frequencies every 0.2 MHz, but two nearby stations cannot use adjacent frequencies. What is the maximum resistance the tuning circuit can have if the peak current at a frequency of 103.7 MHz , the closest frequency that can be used by a nearby station, is to be no more than 0.200 % of the peak current at 103.3 MHz ? The radio is still tuned to 103.3 MHz , and you can assume the two stations have equal strength.

Explanation / Answer

part 1:

tuning frequency is the resonant frequency of the RLC circuit

which is givne by 1/(2*pi*sqrt(L*C))

hence using the values as provided,

1/(2*pi*sqrt(0.2*10^(-6)*C))=103.3*10^6

==>C=1.1868*10^(-11) F

part 2:

let voltage in the RLC circuit is V.

then peak current at 103.3 MHz is given by V/R A

at 103.7 MHz, capacitive reactance=-j/(2*pi*103.7*10^(6)*1.1868*10^(-11))=-j 129.32 ohms

inductive reactance=j (2*pi*103.7*10^6*0.2*10^(-6))=j 130.3132 ohms

let maximum resistance be R.

then maximum impedance=sqrt(R^2+(130.3132-129.32)^2)=sqrt(R^2+0.98644)

peak current=V/sqrt(R^2+0.98644)

peak current at 103.3 MHz station will be V/R (as it is in resonance)

it is given that V/sqrt(R^2+0.98644)=0.2*0.01*V/R

==>sqrt(R^2+0.98644)=500*R

==>R^2+0.98644=250000*R^2

==>R=sqrt(0.98644/249999)=1.9854*10^(-3) ohms

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