# Q factor of an AM radio RLC circuit

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1. Dec 7, 2016

### RyanP

1. The problem statement, all variables and given/known data
Suppose you want to use a series RLC circuit to tune in your favorite AM radio station which broadcasts at a frequency of 1.20 MHz. You would like to avoid the obnoxious easy listening station which broadcasts at 1.10 MHz, right next to the one you like. In order to achieve this, for a given EMF from your antenna, you need the current flowing in your circuit to be 10-2 times less at 1.10 MHz than at 1.20 MHz. What is the minimum Q for this circuit? Now, note that you cannot avoid having a resistance of R = 0.1 ohm, and practical considerations also dictate that you use the minimum L possible. What values of L and C must you use?

2. Relevant equations
Q = wL/R = w/(delta w) = w/2a where a = R/2L
f = w/2pi
3. The attempt at a solution

.01 = e^(-a(x-w))
a = ln100 / (2pi * (1.1*10^6 - 1.2*10^6))

I don't know if this is right or where to go from here.

2. Dec 7, 2016

### tech99

Terman* gives universal resonance curves for engineering purposes, so it is obviously a complicated calculation to do exactly.
I have slightly adapted his formula for a case such as ours where we are more than 3Fo/Q (MHz) from resonance:-
Vfo/Vf = Q((f/fo)^2 - 1)
100 = Q (1.1/1.2)^2 - 1)
100 = Q (0.16)
Q = 100/0.16 = 625

3. Dec 7, 2016

### RyanP

I also found this formula:

I = Io / [1 + Q^2 (w^2 - wo^2) / (w * wo)] where wo is the resonant frequency. With this formula I got Q = 23.8. I'm very confused now...

4. Dec 8, 2016

### tech99

Using your formula I obtain Q = 582. I think you have a mathematical error somewhere.

5. Dec 11, 2016

### Staff: Mentor

It may be expected that you solve by analysing the 3-element series circuit, rather than quoting a formula from somewhere.

The LC product is set so as to give resonance at 1.2 MHz
At resonance, the series impedance is 0.1 Ω

Now set the impedance at 1.2 MHz to be 10 Ω, and solve for L and/or C

My answer for Q is just under 600