Question

You've decided to build a radio to listen to your favourite FM radio station, which broadcasts at 102.9 MHz. For the tuner, you'll be using an RLC circuit, but the only inductor you happen to have on hand has a value of 0.870 μH.

Unfortunately, there's another radio station — which you don't want to listen to — at a nearby frequency of 103.3 MHz. To prevent any interference from this 103.3 MHz radio station, you want the peak current from it to be no more than 0.160% of the peak current from your favourite 102.9 MHz station. What value of resistor should you use in your RLC circuit? (Assume the two stations have the same strength.)

Hint:You'll have to first calculate the value of the capacitor so that the radio tunes to 102.9 MHz.

Answer 1

Step 1: find the resonance frequency to listen your favourite station:

f0 = 1/(2*pi*sqrt (LC))

C = 1/(4*pi^2*f0^2*L)

Given that f0 = 102.9 MHz = 102.9*10^6 Hz

L = inductance = 0.870*10^-6 H

So,

C = 1/(4*pi^2*(102.9*10^6)^2*0.870*10^-6)

C = 2.75*10^-12 F = 2.75 pF

Step 2: Now suppose resistor has a value of R, max current in I0 and max voltage is V0, then for 102.9 MHz, max current in your favourite station will be:

I0 = V0/Z

for favourite station, Z = R, since XL = Xc

I0 = V0/R

for another station of 103.3 MHz frequency, max current will be

I1 = V1/Z, where

V1 = V0, since same source

Z = sqrt (R^2 + (XL - Xc)^2)

XL = 2*pi*f*L = 2*pi*103.3*10^6*0.870*10^-6 = 564.68 Ohm

Xc = 1/(2*pi*f*C) = 1/(2*pi*103.3*10^6*2.75*10^-12) = 560.26 Ohm

Z = sqrt (R^2 + (564.68 - 560.26)^2) = sqrt (R^2 + 19.54)

So,

I1 = V0/sqrt (R^2 + 19.54)

Now given that

I1 = 0.160% of I0

I1 = 0.00160*I0

V0/sqrt (R^2 + 19.54) = 0.00160*V0/R

1/sqrt (R^2 + 19.54) = 0.00160/R

R/0.0016 = sqrt (R^2 + 19.54)

R^2/0.0016^2 = R^2 + 19.54

R^2 - R^2*0.00160^2 = 19.54*0.00160^2

R^2 = 19.54*0.00160^2/(1 - 0.00160^2)

R^2 = 0.0000500225

R = sqrt 0.0000500225 = 0.00707 ohm

R = 7.07*10^-3 Ohm = Value of resistor