# Width of Resonance in RLC-circuit

• Marcus95
In summary: This is a rather ugly equation, which yields:This is the equation you are looking for. The solution is ##\omega = \pm \sqrt{2} \omega_0##. Something must be wrong, because ##\omega## cannot be negative, and this doesn't produce the required result either.
Marcus95

## 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! :)

#### Attachments

• 15608783_1325861530786701_1524560008_o.jpg
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Marcus95 said:
By using the fact that the current through the capacitor must be the same as the total current supplied by the voltage source
How do you prove this 'fact' ? At resonance, the current through the resistor is ##V_0\over R## I seem to remember.

BvU said:
How do you prove this 'fact' ? At resonance, the current through the resistor is ##V_0\over R## I seem to remember.

Well, all components are in series. Thus, by Kirchhoff's current law (esentially conservation of fundamental charge) the same current must run through all components. This is the "total current supplied by the voltage source", which at resonance when the reactive parts of the impedance cancle is indeed ##V_0/R##.

If you make the substitution ω←##\frac 1{\sqrt {LC}}## in your expression for ## |V_c|## what do you get for Vc at resonance?

BvU
Marcus95 said:
Well, all components are in series. Thus, by Kirchhoff's current law (esentially conservation of fundamental charge) the same current must run through all components. This is the "total current supplied by the voltage source", which at resonance when the reactive parts of the impedance cancle is indeed ##V_0/R##.
Yes, my bad. Focused on the ##V_c## but O2 already pointed at that.

Marcus95
NascentOxygen said:
If you make the substitution ω←##\frac 1{\sqrt {LC}}## in your expression for ## |V_c|## what do you get for Vc at resonance?

Oh, there's a misstake, isn't it? I thought it would be ##\frac{V_0L}{R\omega_0^2} ## but it should be ##\frac{V_0L\omega_0}{R} ##, shouldn't it?
But does that change anything at all? Isn't the basic equation still ## \sqrt{(\frac{R\omega_0}{R})^2 + \omega^2 - \omega_0^2} = \sqrt{2} \cdot \frac{R\omega_0}{R} ##

Last edited:
Umm, is that an R in the numerator and an R in the denominator??

What I intended earlier, exactly:

If you make the substitutions ω←##\frac 1{\sqrt {LC}}## and ωo←##\frac 1{\sqrt {LC}}## in your expression for ## |V_c|## what do you get for Vc at resonance?
it should be ##\frac{V_0L\omega_0}{R} ##, shouldn't it?
Right, that's VC at resonance. And this can be written ##\frac {V_O}Q## where Q is the Q of the circuit.

NascentOxygen said:
Umm, is that an R in the numerator and an R in the denominator??
No sorry! That was a typo. It should be L in the denominator of course!

Right, that's VC at resonance. And this can be written ##\frac {V_O}Q## where Q is the Q of the circuit.

What is Q in this context and how does it help us?
Isn't the basic equation still ## \sqrt{(\frac{R\omega_0}{L})^2 + \omega^2 - \omega_0^2} = \sqrt{2} \cdot \frac{R\omega_0}{L} ## with soultion ##\omega = \sqrt{2} \omega_0 ## which cannot be right?

Earlier in the maths, have you said that ##{(\omega\ – \omega_0)}^2\ =\ \omega^2 – \omega_0^2## ?

NascentOxygen said:
Earlier in the maths, have you said that ##{(\omega\ – \omega_0)}^2\ =\ \omega^2 – \omega_0^2## ?
No I haven't, but I seem to have completely missed the square! And a factor 2 in one of the lines...

Okay, So I have tried reworking it. If I am not wrong, the equation I have to solve to find at which ##\omega## the amplitude is ##1/\sqrt{2}## of the maximum is:
## \sqrt{(\frac{R\omega_0}{L})^2 + (\omega^2 - \omega_0^2)^2} = \sqrt{2} \cdot \frac{R\omega_0}{L} ## . This is a rather ugly equation, which yields:
## \omega^2 = \omega_0^2 - \frac{R^2}{2L^2} \pm \sqrt{2\omega_0^4 - \omega^2 + \frac{R^4}{4L^4}} ##. This seems rather far away from proving ##\Delta \omega = 2L/R ##, especially because taking the square-root of that expression will give an algebraic jumble... Have I done something wrong?

We'll go with your first equation in post #11.

Your equation has (ω - ωo) so keep that together and try again to isolate that term on one side of the equation. Show your working here.

BTW, I do obtain their final answer.

https://www.physicsforums.com/attachments/holly-1756-gif.110502/

Marcus95
NascentOxygen said:
We'll go with your first equation in post #11.
Your equation has (ω - ωo) so keep that together and try again to isolate that term on one side of the equation. Show your working here.
Okay! So we have:
## \sqrt{(\frac{R\omega}{L})^2 + (\omega^2 - \omega_0^2)^2} = \sqrt{2} \cdot \frac{R\omega_0}{L} ## .
(Note: My equation in #11 had a misstake, one of the ##\omega_0## is actually supposed to be ##\omega##... (as seen above) Is this a problem or did you calculate with it?)

Thus:

##(\frac{R\omega}{L})^2 + (\omega^2 - \omega_0^2)^2 = 2 \cdot (\frac{R\omega_0}{L})^2 ##
##(\omega^2 - \omega_0^2)^2 + \frac{R}{L})^2 (\omega^2 - \omega_0^2) - \frac{R}{L})^2 \cdot \omega_0^2 = 0 ##
here we can use the quadratic formula:
##\omega^2 - \omega_0^2 = \frac{(\frac{R}{L})^2}{-2} \pm \sqrt{(\frac{(\frac{R}{L})^2}{-2})^2 + (\frac{R}{L})^2 \cdot \omega_0} ##

now I would like to express this as a single term, to then be able to express ##\omega## in some nice way, but I am stuck...
BTW, I do obtain their final answer.

That's good to hear! That means I only have to figure out how to get there!

I think since we are interested in small range ##\Delta\omega=\omega-\omega_0<<\omega_0## of values around ##\omega_0##, the following approximations are valid ##\omega+\omega_0=2\omega_0+\Delta\omega \approx 2\omega_0##. Hence ##\omega^2-\omega_0^2=2\Delta\omega \omega_0##. Use that and start from equation at post 11, it doesn't matter if it is ##\omega## or ##\omega_0## in there because ##R\omega=R\omega_0+R\Delta\omega\approx R\omega_0## since R is small and ##\Delta\omega## is not very big.

Last edited:
Marcus95
Delta² said:
I think since we are interested in small range ##\Delta\omega=\omega-\omega_0<<\omega_0## of values around ##\omega_0##, the following approximations are valid ##\omega+\omega_0=2\omega_0+\Delta\omega \approx 2\omega_0##. Hence ##\omega^2-\omega_0^2=2\Delta\omega \omega_0##. Use that and start from equation at post 11, it doesn't matter if it is ##\omega## or ##\omega_0## in there because ##R\omega=R\omega_0+R\Delta\omega\approx R\omega_0## since R is small and ##\Delta\omega## is not very big.
Thank you very much! I finally got it now
##(\frac{R\omega}{L})^2 + (\omega^2 - \omega_0^2)^2 = 2 \cdot (\frac{R\omega_0}{L})^2 ##
## \rightarrow (2\Delta\omega\omega_0)^2 = (\frac{R\omega_0}{L})^2##
giving ##\Delta\omega = R/2L = 1/T ##, thus the total width of the resonance is ##2/T## as was to be shown.

Delta2

## 1. What is the Width of Resonance in an RLC-circuit?

The Width of Resonance in an RLC-circuit refers to the range of frequencies at which the circuit exhibits maximum current or voltage. It is also known as the bandwidth of the circuit.

## 2. How is the Width of Resonance calculated in an RLC-circuit?

The Width of Resonance can be calculated using the formula: Δf = R / 2ΠL, where R is the resistance, L is the inductance, and Δf is the width of resonance.

## 3. What is the significance of the Width of Resonance in an RLC-circuit?

The Width of Resonance is important in determining the frequency response of the circuit. It indicates the range of frequencies at which the circuit will be most efficient in transferring energy.

## 4. How does the Width of Resonance affect the behavior of an RLC-circuit?

The behavior of an RLC-circuit is affected by the Width of Resonance as it determines the frequency range at which the circuit will experience a sharp increase in current or voltage, resulting in amplification or oscillation.

## 5. Can the Width of Resonance be adjusted in an RLC-circuit?

Yes, the Width of Resonance can be adjusted by changing the values of the resistance, inductance, or capacitance in the circuit. By altering these parameters, the frequency response of the circuit can be modified, resulting in a different Width of Resonance.

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