Finding resistance in an LC Circuit (FM Radio)

[Renaldo]

Homework Statement

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.)

Homework Equations

I_m = V_m/[(R)²+(ωL-1/ωC)²]^(1/2)

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.

I_O = 2I_m

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

If I set this equal to 2V_m/[(R)²+(ωL-1/ωC)²]^(1/2) (or, 2I_m) 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.

 

[rude man]

Homework Statement

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.

I_O = 2I_m

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

If I set this equal to 2V_m/[(R)²+(ωL-1/ωC)²]^(1/2) (or, 2I_m) 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.

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.

 

[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 I_O

I_O = 2I₁, because 1/2 of I_O = 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

 

[rude man]

"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 I_O

I_O = 2I₁, because 1/2 of I_O = 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

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, got to hit the sack.

 

[rude man]

bump

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.