Question

You've decided to build a radio to listen to your favourite FM radio station, which broadcasts at 97.30 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 0.210 μH. Unfortunately, there's another radio station — which you don't want to listen to — at a nearby frequency of 96.90 MHz. To prevent any interference from this 96.90 MHz radio station, you want the peak current from it to be no more than 0.150% of the peak current from your favourite 97.30 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 97.30 MHz.

Answer 1

Given,

Inductor, L = 0.210 $\mu$H

Resonant frequency, f = 97.30 MHz

Let capacitor be C

Now,

We know,

f = 1/(2$\pi$)

VLC

=>
VLC
= 1/(2$\pi$*f) = 1 / (2*3.14*97.30*10⁶)

=> = 1.64 * 10^-9

VLC

=> LC = (1.64 * 10^-9 )²

=> LC = 2.69 * 10^-18

=> 0.210 * 10^-6 * C = 2.69 * 10^-18

=> C = (2.69 * 10^-18) / (0.210 * 10^-6)

          = 12.81 * 10^-12 F

Now,

For resonant frequency, Z = R

=> Current, I = V/R

Now,

For frequency, f = 96.90 MHz

Impedence, Z' = $\sqrt{R^{2} + (X_{C}-X_{L})^{2}}$

Now,

X_L - X_C = 2$\pi$fL - 1/(2$\pi$fC)

              = (2*3.14*96.90*10⁶*0.210*10^-6) - 1 / (2*3.14*96.90*10⁶*12.81*10^-12)

              = 127.8 - 128.3 = -0.5

Now,

Z' = $\sqrt{R^{2} + (-0.5)^{2}}$

    = $\sqrt{R^{2} + 0.25}$

So, current, I' = V/Z'

                      = V / $\sqrt{R^{2} + 0.25}$

Now,

Current for 96.90MHz should be 0.15% of that for 97.30 MHz

Thus,

=> (0.15/100)*V/R = V/$\sqrt{R^{2} + 0.25}$

=> 0.0015 / R = 1/$\sqrt{R^{2} + 0.25}$

=> R / 0.0015   = $\sqrt{R^{2} + 0.25}$

=> (R / 0.0015)² = R² + 0.25

=> R² * 0.44*10⁶ = R² + 0.25

=> R² * 0.44*10⁶ = 0.25

=> R² = 0.25 / (0.44*10⁶) = 0.57 * 10^-6

=> R = $\sqrt{0.57 *10^{-6}}$

          = 0.75 * 10^-3 ohms