# Finding resistance in an LC Circuit (FM Radio)

1. Apr 7, 2013

### Renaldo

1. The problem statement, all variables and given/known data

The figure shows a simple FM antenna circuit in which L = 8.22 µH and C is variable (the capacitor can be tuned to receive a specific station). The radio signal from your favorite FM station produces a sinusoidal time-varying emf with an amplitude of 12.2 µV and a frequency of 88.3 MHz in the antenna.

(a) To what value, C0 ,should you tune the capacitor in order to best receive this station?

(b) Another radio station's signal produces a sinusoidal time-varying emf with the same amplitude, 12.2 µV, but with a frequency of 88.1 MHz in the antenna. With the circuit tuned to optimize reception at 88.3 MHz, what should the value, R0, of the resistance be in order to reduce by a factor of 2 (compared to the current if the circuit were optimized for 88.1 MHz) the current produced by the signal from this station? (When entering units, use ohm for Ω.)

2. Relevant equations

Im = Vm$/$[(R)2+(ωL-1/ωC)2](1/2)

3. The attempt at a solution

Part (a) was no problem for me. The answer is 3.95E-13.

Part (b) has me stumped. I don't know what "circuit tuned to optimize reception at 88.3 MHz" means. I assumed that, in an optimized circuit, (ωL-1/ωC) would = 0.

It has to be twice the current in the non-optimized circuit.

IO = 2Im

IO = Vm/R (Because I assumed that ωL-1/ωC = 0, the equation simplifies to Ohm's Law).

If I set this equal to 2Vm$/$[(R)2+(ωL-1/ωC)2](1/2) (or, 2Im) things cancel and I lose my Resistance variable. This leads me to believe that my initial assumption as to the meaning of optimized circuit is wrong. I don't know where to proceed from here.

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2. Apr 7, 2013

### rude man

That equation applies to 88.3 MHz but not to 88.1 MHz.

"Optimized at 88.3" means wL = 1/wC at 88.3. It's not optimized at 88.1.

Recalculate the impedance of the RLC circuit at 88.1 MHz, then adjust R until you get 1/2 the current you got at 88.3.

3. Apr 7, 2013

### Renaldo

"Recalculate the impedance of the RLC circuit at 88.1 MHz, then adjust R until you get 1/2 the current you got at 88.3."

I guess I could do that pretty easily if I could determine the current in the optimized circuit. I can determine an equation for it, but it comes with two variables, I and R.

The equation is I = V/R, Ohm's Law.

This is the current in the optimized circuit. I rename it IO

IO = 2I1, because 1/2 of IO = I1.

I1 is the current in the non optimized circuit. It equals the equation I used earlier. It's a pain to type up.

The V's on both sides cancel.

I re-arrange the equations and solve for R.

My final equation is:

[ωL - (1/ωC)]/[3^(1/2)]= R

This produces a wrong answer.

ω = 83100000(2∏)
C = 3.97E-13

4. Apr 8, 2013

### rude man

EDIT: I just realized they didn't give you a value for R at 88.3 MHz (part a). That means the answer to part b will be a function of R and can't be a number.

Looks like you tried to come up with a new value of C. That is not the idea. In part b, L and C are the same as in part a but R has to change to make the current half of what it was in part a.

More tomorrow, gotta hit the sack.

Last edited: Apr 8, 2013
5. Apr 8, 2013

### rude man

OK, you found i = V/R when f = 88.3 MHz. You did that by setting wL = 1/wC.

Now write the equation for impedance at 88.1 MHZ LEAVING L AND C ALONE, but introducing a different R = R'. Then what is the impedance Z(jw) for the new network with the new R' at w = 2 pi 88.1 MHz? That will get you the ratio Z(88.1)/Z(88.3). Z(88.3) = R as we already know. What is Z(88.1)?

Finally, what must Z(88.1)/Z(88.3) be in order for the current at 88.1 to be half that at 88.3? And from that, what is R'? There will be R in the answer.

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