How to calculate the resonance frequency of this circuit?

In summary, LTSpice predicts that there are two resonances in this circuit, one at 30 ohms and the other at 20 ohms. The higher resonance is caused by the 30 ohm resistor and can be ignored for the purpose of finding the frequencies.
  • #1
Eeyore79
2
0
Hi,

Need help with this circuit.

I am task to find out the resonance frequency of this circuit.

R=30 ohms
L=20mH
C1=2uF
C2=3uF

If it is just R,L and C1 and I easily calculate the resonance frequency by applying the formula
f=1/2âˆâˆšLC

Now I have another capacitance C2 parallel to this original RLC1, how do I calculate the resonance?

How does this C2 affect the resonance?

Any help is appreciated. Thanks.
 

Attachments

  • RLC circuit.JPG
    RLC circuit.JPG
    4.8 KB · Views: 750
Physics news on Phys.org
  • #2
The two capacitors are in series so you can find the equivalent capacitance by [itex] \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2} [/itex]. Now you have only one capacitor!
 
  • #3
C2 is parallel to the whole of the R,L,C1.

I can just add up the Capacitance component with this formula?
1/C = 1/C1 +1/C2
 
  • #4
The way I see it,the two capacitors are in series...but if you have problem with this,I can suggest another approach.You can consider all elements as impedance(check here) and because impedances,like resistors,are simply added when in series,you can add the corresponding impedances of each of the elements.Then you can differentiate with respect to frequency and set it equal to zero.

EDIT:I should note that all elements are in series in this circuit.Those circles coming out of the two ends of the second capacitor are probably connected to a voltmeter which is ideally an infinite-resistance element and so no current goes through that branch and so all the current just goes through the second capacitor and so it becomes series with other elements!
 
Last edited:
  • #5
Eeyore79 said:
C2 is parallel to the whole of the R,L,C1.

I can just add up the Capacitance component with this formula?
1/C = 1/C1 +1/C2

No: series. You can draw any loop to look series or parallel but the topology is what counts. C1 and C2 can be treated as a singe C, as people are saying. It's not just an arbitrary circuit exercise; the 'purpose' of this sort of arrangement (two Cs ) is often to improve the Q of the resonance when an external circuit is connected (a Matching network).
 
  • #6
Eeyore79 said:
C2 is parallel to the whole of the R,L,C1.
That is "correct", but it's the hard way to it, because you have to use the impedances of the two parallel branches. If you decided one branch was R,L,C1, you have to finds the impedance Z of those components in series (which is a complex number, of course), them put the impedance of C2 in parallel with it, and then find when the impedance of the whole circuit is a minimum or maximum.

It's much easier to combine C1 and C2 in series first.
 
  • #7
People are obsessed with classifying a circuit as either parallel or series. In this case (as usual), it makes no difference what you call it. The sums give you the same answer for resonant frequency wherever you connect your test probe. Look at any simulation. They don't give you a special button to press to select Series or Parallel.
 
  • #8
This circuit has two resonances which occur at slightly different frequencies.

The resistor has an effect on the shape of the resonances but can be ignored for the purpose of finding the frequencies.

Redrawing the diagram may help explain this:
https://dl.dropboxusercontent.com/u/4222062/series%20parallel%20cct.PNG

In the top diagram the inductor is in parallel with the two capacitors in series and the feedpoint is at the junction of the two capacitors. At the resonant frequency, the input will see a high impedance.In the second diagram, the resonance is formed by the 2 uF capacitor in series with the inductor. The feedpoint will see a low impedance.

You can apply the usual resonance formula to obtain these resonant frequencies.

Note that the series resonant frequency is lower than the parallel resonant one because the capacitors are in series in the parallel one.

LTSpice gives both resonances.
 
Last edited by a moderator:
  • #9
I don't think there is enough information in the very first circuit. The impedance of the source / detector is relevant. If a voltage source is connected to the input terminals then C2 becomes irrelevant and you will have an impedance minimum (current max) at the L and C1 resonance. If you drive the input terminals with a current source then there will be a resonance with L and (C1 series C2).
@vk6kro. What does LTSpice assume in your simulation?
 
  • #10
sophiecentaur said:
I don't think there is enough information in the very first circuit. The impedance of the source / detector is relevant. If a voltage source is connected to the input terminals then C2 becomes irrelevant and you will have an impedance minimum (current max) at the L and C1 resonance. If you drive the input terminals with a current source then there will be a resonance with L and (C1 series C2).
@vk6kro. What does LTSpice assume in your simulation?

The simulation uses a voltage source, but the graph displays current from the source.

So a high impedance shows in the graph as a reduced current and vice versa.

The 30 ohm resistor produces considerable damping and the resonance features are quite broad.
However they are obvious.

I guess posting the simulation here might be a bit too helpful, but I can PM it if you like.

C2 does not affect the input voltage at the parallel resonant frequency, but it does affect the current, much like a power factor capacitor does.

Were you affected by the big storm last week?
 
  • #11
vk6kro said:
The simulation uses a voltage source, but the graph displays current from the source.

So a high impedance shows in the graph as a reduced current and vice versa.

The 30 ohm resistor produces considerable damping and the resonance features are quite broad.
However they are obvious.

I guess posting the simulation here might be a bit too helpful, but I can PM it if you like.

C2 does not affect the input voltage at the parallel resonant frequency, but it does affect the current, much like a power factor capacitor does.

Were you affected by the big storm last week?

So, with a voltage source, I guess you got the L C1 resonant frequency? With the Current source, the resonance would have been L C1C2 ? This is a typical situation in which the measurement will affect the result.
I believe that the results of simulations should always be used with great care. In fact, I would almost go so far as to say that you should only ever use a simulation if you could work out the answer for yourself if only you had the time and inclination. On their own, the bare results of a simulation don't necessarily 'prove' anything. I think the analytical way to approach to this very basic problem is more fruitful so, no thanks, to your offer of the simulation.

I don't think the storm affected our house at all and the family all traveled here in periods between bouts of bad weather. How about you?
 
  • #12
I find LTSpice pretty accurate and certainly quicker than me on a calculator.

In this case the resonances are spot-on the calculated frequencies.

It is especially good when opposing effects are at work.

I have used other programs where you have to know the answer to get a result from the simulation, but LTSpice is a seriously good program. Easy to use, too.

I am not in the UK but managed to watch a UK sporting team getting demolished. Doing better today.
 
  • #13
LTSpice isn an excellent package, I believe. It should give the same results as your calculator - I should hope so too! It's just that numerical solutions are only ever applicable to a particular problem and can allow you to miss the bigger picture and to get results without any serious understanding.

I don't watch much cricket (I assume that's what you refer too?) but it was somewhat galling, I have to admit. It's a bit late for them to be doing better now. We can hardly blame the Aussies for easing up a bit now they've made their point.
 
  • #14
It is frustrating to see that program pop up with an answer immediately when I have to battle through a page of simultaneous equations.

Yes that sporting team were pretty unlucky that hardly any of them got to show that they were better than they appeared. Nobody expects that to continue.
 
  • #15
A very charitable response when they were clearly all to pieces. I guess the Aussies were just more hungry.
 

1. How do I determine the values of the components in the circuit?

The values of the components in the circuit, such as the inductance and capacitance, can be determined through measurements or calculations using the circuit diagram and relevant equations.

2. What is the formula for calculating resonance frequency?

The formula for calculating resonance frequency in a series RLC circuit is fres = 1/(2π√(LC)), where L is the inductance and C is the capacitance of the circuit.

3. Can I use the same formula for parallel RLC circuits?

No, the formula for calculating resonance frequency in a parallel RLC circuit is different and is given by fres = 1/(2π√(LC)), where L is the total inductance and C is the total capacitance of the circuit.

4. How does changing the values of the components affect the resonance frequency?

Changing the values of the components, such as L and C, will change the resonance frequency of the circuit. Increasing the inductance will decrease the resonance frequency, while increasing the capacitance will increase the resonance frequency.

5. Is there a way to calculate the bandwidth of the circuit based on the resonance frequency?

Yes, the bandwidth can be calculated using the formula BW = R/L, where R is the resistance in the circuit and L is the inductance. This formula applies to both series and parallel RLC circuits.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
8
Views
2K
Replies
10
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
Replies
5
Views
783
  • Electrical Engineering
Replies
12
Views
1K
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
3K
  • Other Physics Topics
Replies
12
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
12
Views
1K
Back
Top