# Resonance in LRC circuit

Discussion in 'Homework Help' started by Davrum, Jun 22, 2015.

1. ### Davrum Thread Starter New Member

May 14, 2015
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0
I'm studying for an exam in a few days and trying to get my head around resonance. This is a question from last year's exam for the same paper.

What's throwing me is that from examples we've done before I expected to find a Q value of 50-100. Is the value I came up with lower than a "normal" Q value? I've done the question twice (once yesterday and against just now) and get the same thing. Are my methods on point or am I missing something?

2. ### mxg2579 New Member

Jun 10, 2015
15
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Edit: Sorry I made a comment then realized I glanced over one part.

Honestly, this work looks fine to me. Maybe someone else can help find a problem. So you say you expect a Q value of 50-100, is that what the answer should be? Or is that just an estimate you make?

3. ### MrAl Distinguished Member

Jun 17, 2014
3,605
754
Hi,

You should probably show some of your work clearly so we can follow it.
I take it they want the Q as the circuit acts like a filter, with the output taken across C.
In this case, we should get a Q near 50 but i wont say what it is yet.

4. ### mxg2579 New Member

Jun 10, 2015
15
7
The only thing I can think of that makes me second guess is where you have $I=\frac{V}{R}$, should it not be $I=\frac{V}{Z}$. Since you have a capacitance and an inductor it should be impedance. Maybe I'm wrong but that's all I can think of.

5. ### panic mode Senior Member

Oct 10, 2011
1,407
344
what is Z when circuit is in resonance?

DickCappels likes this.
6. ### mxg2579 New Member

Jun 10, 2015
15
7
Ah yes, I forgot about that tid-bit. The impedances cancel. My mistake! Sorry!

7. ### MrAl Distinguished Member

Jun 17, 2014
3,605
754
Hi,

No need to be sorry. Maybe someone else was thinking the same thing and didnt realize that Z=R sometimes at resonance so by your post and replies they will have learned something too. So you actually did some people a favor by suggesting I=V/Z, and that's still true, it's just that Z=R sometimes. So i look at it as a benefit that you suggested that. This shows some of the beauty of conversation at work.

My only question was if we are really to find out the Q relative to the filtering action, by taking the output across the capacitor. I had to guess that was the case from the questions posted in the first post. The Q would then be close to 50.

Last edited: Jun 23, 2015
8. ### Davrum Thread Starter New Member

May 14, 2015
8
0
Thanks for the feedback guys.

MXG, for me the expected value was between 50-100 simply because that's what I knew from the examples we worked in class, so I'm not sure if my low Q is an error or it's simply the case that for some circuits a Q of this value is possible.

Mr Al, I'm not sure whether it's intended as a filter or not, the only info is what's posted there in the image. The same with my working - I haven't done any extra working that I haven't shown. I'm not sure how to make in any more clear, unless you mean I should lay out the calculations vertically instead of horizontally? Normally I would, but I thought horizontally would save space in this case.

9. ### MrAl Distinguished Member

Jun 17, 2014
3,605
754
Hi again,

Ok, no problem. I see now from your formula that it looks like they want the Q relative to the response across the cap, so we're good to go i guess. So you should be able to come up with a Q of near 50. Just in case this isnt really the case (which i doubt now) try getting that Q anyway and see if you can come up close to 50, and that will tell you if you are doing it right or not.
What did you get for the required capacitor value? If you didnt get the right cap value then that would mean you may have not been using the right frequency for the resonant frequency, or even something else not quite right yet, so what did you use for the resonant frequency?

10. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,660
1,522
I tend to think of Q in terms of the overall circuit ....

Q = fr / BW

BW = fr / Q

BW = f upper (-3 dB) - f lower (-3 dB)