Can someone correct my view of resonance.

Thread Starter

Xenon02

Joined Feb 24, 2021
355
Hello !

I've got a question about resonance. I know that the circuit is in resonance when XL = XC. So they are reduced to 0.

1660994991880.png
For example here :
If I have here ZL = ZC two times. Like in series resonance, that it is a short-circuit.
Does it mean that I reduce every ZL and ZC to zero that the only Z here is R ? or just part of it. I mean I know that when ZL = ZC then I have short-circuit. So for example does it mean that C1,L1 and L2 are now as short circuit ? Or is it that all components C1,L1,L2,L3,C2,L4,L5 are short-circuit 2 times and are open circuit 2 times as a whole ?

Also I know that in resonance current is in phase with voltage in voltage source/current source. So I'm not sure.

EDIT:

I'm also not sure here :

1660995425077.pngBecause when there is resonance then this C and L are open circuit ? like in parallel resonance ?

And here :

1660995531584.png
I know that Im{Y} = 0 and Y is not = 0 so I can as well calculate Im{Z} = 0?

If there wasn't a resistor the Im{Y} and Y = 0 so Im{0} cannot be equal 0 right ?
If I had here resistor not in parallel but in series with voltage source then als Im{Y} = 0 but Y = 0 so Im{Z} is not equal 0.
 
Last edited:

Papabravo

Joined Feb 24, 2006
19,300
XL and XC are not zero at resonance, they are equal. Being equal is different than being zero. It is true that series and parallel combinations of inductors and capacitors can have a combined impedance of zero. This cannot happen with resistors or with a combination of one reactive element and one or more resistors. It can only happen with reactive components that have the opposite sign for their imaginary part.
 

BobTPH

Joined Jun 5, 2013
5,774
You are trying to apply the analysis of a simple series or parallel LC circuit to circuits that are more complicated than that. You cannot do that, you have to analyze the new circuit in total.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
XL and XC are not zero at resonance, they are equal. Being equal is different than being zero. It is true that series and parallel combinations of inductors and capacitors can have a combined impedance of zero. This cannot happen with resistors or with a combination of one reactive element and one or more resistors. It can only happen with reactive components that have the opposite sign for their imaginary part.
Ah yes sorry my bad you are right here.
I meant here combinations that are zero :D Of course for Imaginary parts Im.


You are trying to apply the analysis of a simple series or parallel LC circuit to circuits that are more complicated than that. You cannot do that, you have to analyze the new circuit in total.
I know is that when there is resonance then Imaginary part is reduced to 0 in total Z or Y.
Also I know that when there is resonance then the total of Z or Y is Re and doesn't consist Im. So most likely it has only resistance.

Also I know that when Z - max or Y - min then this is parallel resonance so open circuit and when Z - min or Y - max then it is series resonance.
So I cannot use it the same here in more complicated ? Isn't one of those branches a short circuit ? Like in simple RLC where in series it is a short circuit.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
Show me which L and C in series you talking about. Circle them on your schematic.
1661000685880.png

Something like this ?

I mean maybe I'm wrong but if this is short circuit then what about the rest in red color ?
I know that resonance is when all reaktance XL and XC must be reduced to 0 Im{Y} or Im{Z} = 0 And as I know the one in red circle are not reduced to 0.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
Am I wrong here ? Because I don't really understand here. All I know is that Imaginary is reduced to 0. And then the circuit acts like it has only resistance.
 

Jony130

Joined Feb 17, 2009
5,419
From the simulation results, we can see that at the parallel resonance we have an open circuit. Because Vout = 60dB = 1000V (1A * 1kΩ).
And at the series resonance, we have Vout = -7dB ≈ 0.447V. So, the resistance at series resonance is equal to Rs = 0.447V/1A = 0.447Ω.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
From the simulation results, we can see that at the parallel resonance we have an open circuit. Because Vout = 60dB = 1000V (1A * 1kΩ).
And at the series resonance, we have Vout = -7dB ≈ 0.447V. So, the resistance at series resonance is equal to Rs = 0.447V/1A = 0.447Ω.
But what here is in series resonance ? One of those branches ? Or is it the whole (red +orange circle). And this whole can be 2 times series resonance ?
Because all I know is that resonance is when every Imaginary substitute is reduced to 0. And there is only Re. So even if one of those branches is a series resonance then what about the second branch ? There imaginary isn't reduced.
 

Jony130

Joined Feb 17, 2009
5,419
The whole circuit.
But what here is in series resonance ?
\
At arond 35Hz it will be C2 L4||L5 and at 50Hz C1 L1||L2.
Also, don't forget that if we have two resistors connected in parallel the smallest one will "win" and dominate the whole circuit.
And this is why we have Rs = 0.447Ω ar series resonance.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
The whole circuit.
\
At arond 35Hz it will be C2 L4||L5 and at 50Hz C1 L1||L2.
Also, don't forget that if we have two resistors connected in parallel the smallest one will "win" and dominate the whole circuit.
And this is why we have Rs = 0.447Ω ar series resonance.
You said that if it is the resonance for the whole, then why you said that 35Hz is series resonance for 1 branch ? Isn't it for both branches ? Because 1 branch and another branch needs to reduce it's imaginary value to 0. So when you said that one branch is a series resonance then it means that one of those branches has 0 imaginary value but the second branch isn't in resonance so it has imaginary value.

EDIT:
It's like in the picture I've showed above. If it's only resonance for 1 branch then what with another branch ? Second branch isn't in resonance because you mentioned that 35Hz is for first branch resonance. So second branch didn't reduce it's imaginary value to 0. Because if i have equation for the total Z or Y, then those 2 branches are in the equation and if one of them is reduced to 0 then what about the second one?
 
Last edited:

Jony130

Joined Feb 17, 2009
5,419
These two branches resonate at two different frequencies. So they will have their local minimum at different frequencies.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
These two branches resonate at two different frequencies. So they will have their local minimum at different frequencies.
But in the equation for total Z or Y, both of those branches has to reduce it's imaginary value to 0.
1661013075644.png
So if only one of those branches has imaginary value equal 0 then the second branch is not reduced to 0. So there is still imaginary value, so it is not a resonance. Because resonance is when every imaginary value is reduced to 0 and there is only Re value.

So it doesn't make any sense for me. That only one branch is in resonance when you said that the whole circuit is in resonance. So the whole circuit or only 1 branch out of 2? Because both branches contain imaginary value of Z of Y.
 

Papabravo

Joined Feb 24, 2006
19,300
In the graph of post #6 there are two separate resonances, and they occur at different frequencies. If one branch has the resonance the other does not and vice versa. I'm sorry that things don't seem to make sense to you, so maybe you need to take a step back and examine the potential reasons for that. This is not easy stuff and maybe an alternate approach is called for.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
In the graph of post #6 there are two separate resonances, and they occur at different frequencies. If one branch has the resonance the other does not and vice versa. I'm sorry that things don't seem to make sense to you, so maybe you need to take a step back and examine the potential reasons for that. This is not easy stuff and maybe an alternate approach is called for.
I know. But the thing I am looking at is that resonance is when the whole circuit is in resonance/total value of Z or Y the imaginary value is 0. Like in the post #15. There is a total value impedance. So for it to work as resonance parallel then Z which is imaginary must be reduced to 0, and only R is left. But only one branch is in resonance and the other one is not so Z contains imaginary value. So how is it still resonance ?
 

Papabravo

Joined Feb 24, 2006
19,300
I know. But the thing I am looking at is that resonance is when the whole circuit is in resonance/total value of Z or Y the imaginary value is 0. Like in the post #15. There is a total value impedance. So for it to work as resonance parallel then Z which is imaginary must be reduced to 0, and only R is left. But only one branch is in resonance and the other one is not so Z contains imaginary value. So how is it still resonance ?
This is like asking how many angels can dance on the head of a pin. As you have so artfully demonstrated, a circuit can have multiple resonances at different frequencies. The better question might be: "How close can they get before they become indistinguishable?" The answer to that question might provide the insight you seek.

AFAIK, this is not something that has any intrinsic practical value. This is merely a hypothetical exercise and so the answer is mostly irrelevant.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
355
So they are not fully in resonance ?
One branch is in resonance but another is not so the whole circuit is not exactly in resonance ? Because I don't really understand what were you trying to tell me.

I mostly base my resonance knowledge from basic rule that whole imaginary value is 0 when the circuit is in resonance.

Also I made a simulation here : https://tinyurl.com/2ltbxkvu frequency different than the resonance one. And they look like in resonance because current is in phase with voltage in voltage source.

So I don't get it in mathematical way. Resonance that should reduce thhe whole value of imaginary value to 0 but only on of the branches is actualy 0 in imaginary value, but the other one is not.
 

BobTPH

Joined Jun 5, 2013
5,774
Because, like your other thread, you are taking a statement that is true for one particular case and thinking it applies to a totally different case.

Neither of the circuits you circled is a single L and C in series or parallel, so what you know about those cases does not apply, yet you insist on doing so. In your other thread, it took forever for you to understand that Ohm’s law does not apply to caoacitors. I am not interested in going through that tedious process again, hence the never mind.
 
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