Changing values of R in a RLC series circuit

[Twoism]

Homework Statement

A television channel is assigned the frequency range from 54 MHz to 60 MHz. A series RLC tuning circuit in a TV receiver resonates in the middle of this frequency range. The circuit uses a 16 pF capacitor. Inductor is .487 micro Henries.
In order to function properly, the current throughout the frequency range must be at least 50% of the current at the resonance frequency. What is the minimum possible value of the circuit's resistance?

Homework Equations

I=V/Z
Z=sqrt(R^2+(omegaL-(omegaC)^-1)^2)
omega=2pi(frequency)

The Attempt at a Solution

I'm kinda lost on this one.
I tried setting .5(V/Z1)=(V/Z2) and solving for R.
Z1 is using the 57MHz and Z2 is using 54 or 60 MHz, neither gave me a correct answer.

I meant to put "finding a value" instead of "changing values of," sorry.

 

[creillyucla]

It seems like your set up is correct. What is the expression you get for $$\frac{| Z_1 |}{| Z_2 |}$$?

Though it doesn't seem like you need the info I wrote up below, it might be worth checking out:

Your expression for the magnitude of the impedance of the circuit is correct, but is more instructive written in this way:

$$| Z(w) | = \sqrt{ R^2 + ( \frac{ ( \frac{\omega}{\omega_o} )^2 - 1 }{ \omega C })^2 }$$

Where $$\omega_o = \frac{1}{\sqrt{LC}}$$ , or the angular frequency at which $$| Z |$$ is a minimum. Notice is only depends on L and C, not R. This is correspondingly the frequency of maximum current.

When you plug in the values for L and C, you find that the resonant frequency ($$\frac{\omega_o}{2 \pi}$$ ) is very close to 57 MHz, right in the middle of the range given in the problem.

Your task is to find the value R such that the ratios of the currents at the edge of the frequency range are at least 50% of the current at the resonant frequency. Hope this helps.