- #1

Marcus95

- 50

- 2

## Homework Statement

In an undriven RLC-circuit, the characteristic time of the capacitor, ie the time taken for the amplitude of the capacitor voltage ## V_c ## to drop by a factor e, is ## T = 2L/R ##.

We now have a RLC-circuit which is driven by a AC voltage of variable frequency ## \omega ## with amplitude ##V_0##. Show that for small R. the width of the resonance is approximately ##2/T = R/L##.The width of the frequency is defined as the angular frequency range for which the amplitude of ##V_c## is greater than ##1/\sqrt{2}## of its resonance value.

## Homework Equations

Ohm's law: ##V = I Z##

Impedance: ##X_c = 1/i\omega C ##, ## X_L = i\omega L ##, ##X_R = R##

Resonance in LC-circuit: ## \omega_0^2 = 1/LC ##

## The Attempt at a Solution

I have attached a photo of my attempt at a solution. The basic method is the following:

1. By using the fact that the current through the capacitor must be the same as the total current supplied by the voltage source, I find that the amplitude of the voltage on the capacitor can be expressed as:

## |V_c| = \frac{V_0}{\omega C} \cdot \frac{1}{\sqrt{(\frac{R\omega}{L})^2 + \omega^2 - \omega_0^2}}##

2. I find the resonance value to be: ##V_c(\omega_0) = \frac{V_0L}{R\omega_0^2} ##

(this uses the assumption that R is small)

3. I want to find the two values of ##\omega## which yield ##V_c(\omega) = \frac{V_0L}{R\omega_0^2} \cdot \frac{1}{\sqrt{2}} ##. The difference between these values should be the width we are looking for.

This gives an equation with the solution ## \omega = \pm \sqrt{2} \omega_0##. Something must be wrong, because ##\omega## cannot be negative, and this doesn't produce the required result either.

What have I done wrong? Many thanks for all help! :)