How Does Frequency Alter Output Voltage in an RLC Circuit?

[Alouette]

Homework Statement

A radio tuner built by connecting an L=0.58mH inductor and variable capacitor in series with an antenna. The driving voltage is provided by the radiowaves that pass through the antenna, and the total internal resistance of the circuit is R=2.6Ohms.

The variable capacitor is adjusted to tune the radio to 540kHz because of a capacitance C1 = 1.5 x 10^(-10), and the quality factor of the circuit with C1 is 756.

(a) If another station operates at 490kHz with the same amplitude at your location, what is the ratio of the output voltage due to this background signal to the output of the signal for which the circuit has been tuned? The output voltage is the voltage across the capacitor.

(b) Find this same ratio if the other station operates with the same amplitude at frequency 535kHz

Homework Equations

q = (XL) / (R) = 2*∏*(L/R)*f [this is the voltage across the capacitor, f = frequency]

The Attempt at a Solution

Well since the only variable is frequency, I thought that the ratio of output voltage is basically the ratio of the frequencies. As in this:

(2*∏*(L/R)*490kHz) / (2*∏*(L/R)*540kHz)

So for both (a) and (b), the ratios would have been:

(a) 490/540 & (b) 535/540

Yet this gives me too high of a number (both are ~ 0.9). Am I missing a factor in my equation?

 

[gneill]

Homework Statement

A radio tuner built by connecting an L=0.58mH inductor and variable capacitor in series with an antenna. The driving voltage is provided by the radiowaves that pass through the antenna, and the total internal resistance of the circuit is R=2.6Ohms.

The variable capacitor is adjusted to tune the radio to 540kHz because of a capacitance C1 = 1.5 x 10^(-10), and the quality factor of the circuit with C1 is 756.

(a) If another station operates at 490kHz with the same amplitude at your location, what is the ratio of the output voltage due to this background signal to the output of the signal for which the circuit has been tuned? The output voltage is the voltage across the capacitor.

(b) Find this same ratio if the other station operates with the same amplitude at frequency 535kHz

Homework Equations

q = (XL) / (R) = 2*∏*(L/R)*f [this is the voltage across the capacitor, f = frequency]

Hmm. That would be the Q-factor for a coil with inductance L and resistance R. How do you figure it's the voltage across the capacitor of a series RLC circuit?

The Attempt at a Solution

Well since the only variable is frequency, I thought that the ratio of output voltage is basically the ratio of the frequencies. As in this:

(2*∏*(L/R)*490kHz) / (2*∏*(L/R)*540kHz)

So for both (a) and (b), the ratios would have been:

(a) 490/540 & (b) 535/540

Yet this gives me too high of a number (both are ~ 0.9). Am I missing a factor in my equation?

I think you're missing an entire equation What's the expression for the magnitude of the voltage across the capacitor in a series RLC circuit given a unit stimulus (1V @ frequency f)?

HINT: It's convenient to work symbolically with reactances XL, XC and plug in the actual values later.