RLC cir - Resonant Freq

MikeML

Joined Oct 2, 2009
5,444
...
And for the bandwidth they quote at 22 Hz, which is very close to the more accurate 22.1 Hz, but then when they go to calculate the upper and lower frequency they divide the bandwidth by 2 and then subtract and add to the resonant frequency of 32.5 and get a result that is a little more off than we might want to see...
Mr Al, If you analyze the circuit as posted on the Tutorial web page, with a zero-output impedance voltage source in parallel with R=60 Ohms, L=200mH, C=120uf there is no resonance. Ergo all discussion about current magnitude, bandwidth, resonant frequency on that Tutorial page can only apply if the circuit is reconfigured as I show in post #11.

What started this is that the OP took the circuit as drawn on the Tutorial page, put it into LTSpice and couldn't see any resonant behavior. No surprise! LTSpice nailed it! Now if the circuit had been drawn correctly in the first place, we wouldn't be having this discussion...

Oh, and btw: the reason that there can be no resonance with a pure voltage source connected across what would otherwise be a parallel tuned circuit, is that resonance depends on energy being transferred from the inductor to the capacitor and back again. If the voltage at the common node of the circuit is defined by the pure voltage source, then no energy can transfer.
 
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JoeJester

Joined Apr 26, 2005
4,390
Mike,

If your looking for agreement that the voltage is the same across every component in a parallel connection, you won't get an argument from me.

Resonance does occur in the parallel circuits ... only the current has a nadir vice a zenith, which occurs in the series resonant circuit.
 

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JoeJester

Joined Apr 26, 2005
4,390
No surprise! LTSpice nailed it!
No it didn't. That person didn't know what they were looking for. They sought a voltage, when the current was required.

The only thing LTSpice nailed was the fact that the voltage is the same in a parallel circuit.
 

MikeML

Joined Oct 2, 2009
5,444
Joe, I don't see how your latest sim has anything to do with the topic being discussed. Since you seem to be interested in the currents through the resistor, inductor, and capacitor, then look at these two cases. The first is the circuit topology exactly as shown on the Tutorial page. The second is the way I believe the author of the tutorial should have drawn the circuit to agree with what his analysis shows...


196e.jpg

So, look at the currents in the left plot pane. Do you see any "resonant" or "oscillatory" behavior??? Look at the dashed lines which are the phase of the currents. Do you see any frequency dependence?

All I see is pure I(L1)= 100/(2*Pi*F*L1) and I(C1) = 100*2*Pi*f*C1.
Compare that to the right pane...

Now convince me that LTSpice is wrong or is somehow showing behavior that is not real!



btw Joe,

you completely blew this off:
"the reason that there can be no resonance with a pure voltage source connected across what would otherwise be a parallel tuned circuit, is that resonance depends on energy being transferred from the inductor to the capacitor and back again. If the voltage at the common node of the circuit is defined by the pure voltage source, then no energy can transfer."
 
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JoeJester

Joined Apr 26, 2005
4,390
Mike,

The cited tutorial website is not the only site out there with that same circuit when discussing parallel resonance.

wiki and Georgia state University are two and I haven't looked through the other 3.6 million, including AAC's ebooks, that use a voltage source when discussing parallel resonant circuits. You're going to have to swing your hammer far and wide if your mission is to correct that error.

I've yet to find a definition of resonance that requires energy transfer. The common definition is when inductive reactance equals capacitive reactance. GSU offered three definitions concerning resonance in parallel circuits ... found at http://hyperphysics.phy-astr.gsu.edu/hbase/electric/parres.html#c1

There are some simulation software packages where you can enter the "internal" resistance of the voltage source as well as the series dc resistance of coils, so the mere absence of the series resistances doesn't mean much, although I will agree with you that they should be present in a tutorial. BTW, someone commented on your comment at the OPs cited tutorial website.

With a "pure voltage source", a theoretical device, you still have a value of current in both the capacitor and inductor, which by the mere fact that inductive reactance equals capacitive reactance and they cancel each other out at one specific frequency. I never have the expectation of oscillation (transfer of energy from inductor to capacitor and vice versa) in parallel "resonant" circuits being fed by a continuous wave. The circuit is continuously excited and has no chance of oscillation.
 

t_n_k

Joined Mar 6, 2009
5,455
A bit of "spice" for the analytically inclined.
In the attached simulation, a parallel LCR circuit with an initial condition [initial inductor current =10mA] is allowed to 'relax' naturally.
The natural voltage response is used to drive an identical parallel LCR circuit via an ideal 1:1 voltage "transducer". The driven circuit has no initial conditions. What current flows from the transducer output into the parallel LCR circuit?
The result might be surprising to some.

Natural response driven parallel LCR.jpg
 

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MrAl

Joined Jun 17, 2014
11,389
Mr Al, If you analyze the circuit as posted on the Tutorial web page, with a zero-output impedance voltage source in parallel with R=60 Ohms, L=200mH, C=120uf there is no resonance. Ergo all discussion about current magnitude, bandwidth, resonant frequency on that Tutorial page can only apply if the circuit is reconfigured as I show in post #11.

What started this is that the OP took the circuit as drawn on the Tutorial page, put it into LTSpice and couldn't see any resonant behavior. No surprise! LTSpice nailed it! Now if the circuit had been drawn correctly in the first place, we wouldn't be having this discussion...

Oh, and btw: the reason that there can be no resonance with a pure voltage source connected across what would otherwise be a parallel tuned circuit, is that resonance depends on energy being transferred from the inductor to the capacitor and back again. If the voltage at the common node of the circuit is defined by the pure voltage source, then no energy can transfer.
Hi Mike,

My first thought is to disagree as i have in the past (current can pass through a short circuit easily), but since you are arguing so adamantly i'll give this a more in depth look. Sometimes we over look things the first time around so i'll take more time to look at it.

In the mean time i offer you two simple experiments to perform in the circuit simulator of your choice.

To start, you'll have to run the original circuit with the 100v source set at 32.4874 Hz and look at the current through the R, L, and C.

Experiment #1:
Using the circuit you have in post #12, adjust the voltage source until the voltage across the cap and inductor equals exactly 100 volts at 32.4874 Hz. Compare the response of the L and C to that of the original circuit with everything in parallel.

Experiment #2:
Put everything back in parallel except for the voltage source. Put 1 ohm in series with the voltage source before reconnecting it. The circuit will now be the same as the original except with a 1 ohm resistor in series with the voltage source when before there was only one resistor the 60 ohm resistor.
Next adjust the voltage source until the voltage across the R, L, and C (all in parallel again) reaches exactly 100 volts at 32.4874 Hz.
Compare the response of the R, L and C to the response of the R, L and C in the original circuit with no extra series resistor of 1 ohm.

When comparing circuits, be sure to look at both the voltage across the L and C (and R in #2) and the current in L and C (and R in #2).
Also, you may wish to vary the frequency of the source to slightly below and slightly above the frequency given above, but when doing this the voltage across the L and C (and R in #2) must be readjusted to maintain exactly 100v. Record the currents in L and C and R for each frequency.

What i'll do is the analysis again myself without a circuit simulator using general circuit analysis and post results.
 
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MrAl

Joined Jun 17, 2014
11,389
Hello again,

Ok what i am seeing here may or may not be technically called "resonance", depending on your definition.

First, we see a dip in the current response so we can call it resonance because of that alone.

What this circuit lacks however, is the energy exchange between the cap and inductor during resonance. We dont see that. It's simply a driven circuit in this respect, in that the inductor and capacitor do exactly what the voltage source tells them to do, and they have no control over each others energy as we see in the more typical resonance.
In this circuit if we disconnect the capacitor we still see the same current in the inductor, and if we disconnect the inductor we still see the same current in the capacitor. This is not typical in a resonant circuit, which normally depends on both cap and inductor to set the resonant frequency and each others current levels.

However, with both elements in place, their reactances are equal and opposite so we do see the effect of resonance. If we change either L or C value we have to change the frequency to match to get this effect again.

To sum up, i would tend to call this, "electrical resonance" as opposed to "physical resonance" which exhibits both types of resonance in the same circuit. The electrical effects are there while the physical (energy exchange) effects are not there. To be sure this circuit would have been better illustrated with a current source rather than voltage source.

Doubts about resonance like this come up from time to time, and there are different ways of looking at it. Some argue that the only true resonance is physical resonance, but there have been professional papers published that suggest that not only are other forms of resonance acceptable, they are mathematically closely related to physical resonance.

For the purpose of education maybe this circuit was a good example, or a bad example, depending on how you look at it. If you want to look at physical resonance then it's a bad example, but if you would like to look at all forms of resonance then it shows how we can have the effects of resonance without physical resonance.
 
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A bit of "spice" for the analytically inclined.
In the attached simulation, a parallel LCR circuit with an initial condition [initial inductor current =10mA] is allowed to 'relax' naturally.
The natural voltage response is used to drive an identical parallel LCR circuit via an ideal 1:1 voltage "transducer". The driven circuit has no initial conditions. What current flows from the transducer output into the parallel LCR circuit?
The result might be surprising to some.

View attachment 72442
Consider the complex plane. The impedance of the circuit has a zero at the origin and two complex poles not on the jw axis, due to the resistor. Normally, to obtain the impedance vs. frequency, we would evaluate the impedance expression as we move along the jw axis from some low frequency to some higher frequency.

Since the natural time response is at the complex frequency of a pole, where the impedance is infinite, the current from the transducer output will be zero.
 
Thanks for replying.
I make the transducer output current to be 10mA DC ad infinitum.
You can probably tell that I didn't actually do the analysis--I gave an educated guess. So now the issue for me is, do I want to do the work of a actual analysis? I can't help getting sucked in, can I?
 

MikeML

Joined Oct 2, 2009
5,444
Ok, let me try this one more time:

Let us discuss this circuit:

196e.jpg

Suppose that the initial current in the inductor is zero and that the initial voltage at V(a)=100. Initially, the energy stored in the capacitor is 1/2*C*V^2 = 0.5*120e-6*100^2 = 0.6J. Some time later, the residual energy in the capacitor will be zero and all of it will have transferred into the inductor (there are no dissipative elements in this circuit). The peak energy stored in the inductor = 1/2*L*I^2, so solving for peak current in the inductor, I= sqrt(0.6/0.5*0.2) = sqrt(6) = 2.449A

The time for a complete cycle of energy transfer from capacitor to inductor and back to the capacitor will be the familiar 2*Pi*sqrt(LC) = 2*3.14159*sqrt(0.2*120e-6) = 30.78ms (Frequency =1/30.88 = 32.49Hz). This oscillatory cycle is what I consider "resonance".

Lets see what LTSpice shows us: V(a) oscillates from +100V to -100V and back to +100V with a period ~= 31ms). I(L1) starts at zero, goes to +2.4A, back to zero and then to -2.4A, etc... Things go on forever because there is no dissipation.

196f.jpg

Now, let us add in the 60Ohm resistor in parallel with the inductor and capacitor. We would expect the circuit to behave as before, with the resonance damped due to the energy lost each cycle in the resistor. If we repeat the simulation with the resistor, no one should be surprised at the results:

196g.jpg

We now see the familiar damped oscillation, same frequency as before. Note that energy is constantly being dissipated as heat losses in the resistor, but it takes several cycles before none is left. We can ask LTSpice to plot the power in the resistor = V(a)*I(R1) , and it can also show us the integral of V(a)*I(R1) over time, which is energy. Note that after 80ms, by which time things have mostly damped out, most of the initial energy the capacitor had (0.6J) has been transferred to the 60 Ohm resistor (0.59984J, which is within the resolution of integrating the waveform)...

196h.jpg

Now, let us add in the voltage source (but set to zero V) . I choose to include 1mOhm of resistance between the voltage source and node a. Let us see how long it would take to dissipate the initial energy stored in the capacitor (hasn't changed, still 0.6J). Note that the initial discharge current for the capacitor (initial voltage is 100V) is determined by the parallel combination of 1mOhm and 60Ohms = 0.99998mOhms. So the initial discharge current would be 100V/0.9998mOhm ~= 100kA.

Now ask what is the time constant? Tau = RC = 1e-3*120e-6 = 0.12e-6 = 0.12us, so we would expect all the energy in the capacitor to be dissipated in the resistors by say 5 time constants, so about 5*0.12u = 0.6us. So let us see what LTSpice shows...

196i.jpg

As predicted, V(a) starts at 100V and dies out to zero in 600ns. The capacitor current starts at -100Ka, almost all of which flows in resistor R2. Only tiny currents (so small that they have to be plotted on their own plot axis) flow in the inductor and R1 (see I(L1) on the lower plot pane). Again, I ask LTSpice to plot the power dissipated in resistor R2 = V(a,v)*I(r2) = dark Blue trace, and to integrate that to find energy. Note that in 600ns, 0.590J went into R2 (the other 0.010J went into R1 and L1).

So here is the bottom line. With R2=1mOhm, almost all of the energy that existed in the capacitor is gone in 600ns. For the "parallel resonant circuit" to exhibit any oscillatory behavior (where energy transfers from capacitor to inductor back to the capacitor), there would have to be some energy left at the end of the normal period (30ms) which is 1/frequency where the circuit "oscillates".

Another way to say this is that the circuit is so highly overdamped, the "oscillation" dies out in a tiny fraction of the first period. Note that this happened as a consequence of R2=1mOhm being added in parallel with the circuit. What do you expect would happen if we make R2 even smaller (V1 more like an ideal voltage source)?

So back to the tutorial. The equations developed there basically show that there is one frequency where the inductive reactance cancels the capacitive reactance, and if you plot the current that flows from the voltage source into the circuit, there is one frequency where the residual impedance is just the 60Ohm resistor R1. To which I say, so what? This is not resonance, it is simply an algebraic trick...

Here a definition of electrical resonance I quote from here:

"Resonance of a circuit involving capacitors and inductors occurs because the collapsing magnetic field of the inductor generates an electric current in its windings that charges the capacitor, and then the discharging capacitor provides an electric current that builds the magnetic field in the inductor. This process is repeated continually. An analogy is a mechanical pendulum."

Can you do anything useful with this circuit, like build an oscillator, or a filter, or an rf coupling network, or a pi network, etc? I do not think so. I do not consider this a useful resonant circuit where there is a constant interchange of energy from capacitor to inductor and vice versa. The elements are effectively decoupled from each other by being paralleled with a voltage source.
 

MrAl

Joined Jun 17, 2014
11,389
Ok, let me try this one more time:

Let us discuss this circuit:

View attachment 72462

Suppose that the initial current in the inductor is zero and that the initial voltage at V(a)=100. Initially, the energy stored in the capacitor is 1/2*C*V^2 = 0.5*120e-6*100^2 = 0.6J. Some time later, the residual energy in the capacitor will be zero and all of it will have transferred into the inductor (there are no dissipative elements in this circuit). The peak energy stored in the inductor = 1/2*L*I^2, so solving for peak current in the inductor, I= sqrt(0.6/0.5*0.2) = sqrt(6) = 2.449A

The time for a complete cycle of energy transfer from capacitor to inductor and back to the capacitor will be the familiar 2*Pi*sqrt(LC) = 2*3.14159*sqrt(0.2*120e-6) = 30.78ms (Frequency =1/30.88 = 32.49Hz). This oscillatory cycle is what I consider "resonance".

Lets see what LTSpice shows us: V(a) oscillates from +100V to -100V and back to +100V with a period ~= 31ms). I(L1) starts at zero, goes to +2.4A, back to zero and then to -2.4A, etc... Things go on forever because there is no dissipation.

View attachment 72463

Now, let us add in the 60Ohm resistor in parallel with the inductor and capacitor. We would expect the circuit to behave as before, with the resonance damped due to the energy lost each cycle in the resistor. If we repeat the simulation with the resistor, no one should be surprised at the results:

View attachment 72464

We now see the familiar damped oscillation, same frequency as before. Note that energy is constantly being dissipated as heat losses in the resistor, but it takes several cycles before none is left. We can ask LTSpice to plot the power in the resistor = V(a)*I(R1) , and it can also show us the integral of V(a)*I(R1) over time, which is energy. Note that after 80ms, by which time things have mostly damped out, most of the initial energy the capacitor had (0.6J) has been transferred to the 60 Ohm resistor (0.59984J, which is within the resolution of integrating the waveform)...

View attachment 72465

Now, let us add in the voltage source (but set to zero V) . I choose to include 1mOhm of resistance between the voltage source and node a. Let us see how long it would take to dissipate the initial energy stored in the capacitor (hasn't changed, still 0.6J). Note that the initial discharge current for the capacitor (initial voltage is 100V) is determined by the parallel combination of 1mOhm and 60Ohms = 0.99998mOhms. So the initial discharge current would be 100V/0.9998mOhm ~= 100kA.

Now ask what is the time constant? Tau = RC = 1e-3*120e-6 = 0.12e-6 = 0.12us, so we would expect all the energy in the capacitor to be dissipated in the resistors by say 5 time constants, so about 5*0.12u = 0.6us. So let us see what LTSpice shows...

View attachment 72466

As predicted, V(a) starts at 100V and dies out to zero in 600ns. The capacitor current starts at -100Ka, almost all of which flows in resistor R2. Only tiny currents (so small that they have to be plotted on their own plot axis) flow in the inductor and R1 (see I(L1) on the lower plot pane). Again, I ask LTSpice to plot the power dissipated in resistor R2 = V(a,v)*I(r2) = dark Blue trace, and to integrate that to find energy. Note that in 600ns, 0.590J went into R2 (the other 0.010J went into R1 and L1).

So here is the bottom line. With R2=1mOhm, almost all of the energy that existed in the capacitor is gone in 600ns. For the "parallel resonant circuit" to exhibit any oscillatory behavior (where energy transfers from capacitor to inductor back to the capacitor), there would have to be some energy left at the end of the normal period (30ms) which is 1/frequency where the circuit "oscillates".

Another way to say this is that the circuit is so highly overdamped, the "oscillation" dies out in a tiny fraction of the first period. Note that this happened as a consequence of R2=1mOhm being added in parallel with the circuit. What do you expect would happen if we make R2 even smaller (V1 more like an ideal voltage source)?

So back to the tutorial. The equations developed there basically show that there is one frequency where the inductive reactance cancels the capacitive reactance, and if you plot the current that flows from the voltage source into the circuit, there is one frequency where the residual impedance is just the 60Ohm resistor R1. To which I say, so what? This is not resonance, it is simply an algebraic trick...

Here a definition of electrical resonance I quote from here:

"Resonance of a circuit involving capacitors and inductors occurs because the collapsing magnetic field of the inductor generates an electric current in its windings that charges the capacitor, and then the discharging capacitor provides an electric current that builds the magnetic field in the inductor. This process is repeated continually. An analogy is a mechanical pendulum."

Can you do anything useful with this circuit, like build an oscillator, or a filter, or an rf coupling network, or a pi network, etc? I do not think so. I do not consider this a useful resonant circuit where there is a constant interchange of energy from capacitor to inductor and vice versa. The elements are effectively decoupled from each other by being paralleled with a voltage source.
Hello,

The energy exchange mechanism was already discussed in post #28, perhaps you did not see that post yet?
It was mentioned that what this circuit lacks is an exchange of energy between cap and inductor. But there is still behavior worth noting.

Back in post #21 you said that there is no point in discussing bandwidth because it's not a resonant circuit, but there is still bandwidth even if you dont want to accept the electrical resonance.

Is this circuit useful? Well, in it's present form im not sure, but if we put a series resonant circuit just ahead of it it makes a double bandpass filter.
 

JoeJester

Joined Apr 26, 2005
4,390
Can you do anything useful with this circuit, like build an oscillator, or a filter, or an rf coupling network, or a pi network, etc? I do not think so.
The "theoretical" voltage source doesn't exist, so if you needed a bandpass filter with the band of this one, then yes it's useful.

You nor I get to decide what is useful for someone else. However, I would not try to market it for profit because it could be an extremely limited market for that specific product.
 

MikeML

Joined Oct 2, 2009
5,444
...so if you needed a bandpass filter with the band of this one, then yes it's useful.
No, it is not. If you really want a bandpass filter, you would build it like this:


196p.jpg

You nor I get to decide what is useful for someone else. ...
Ok, but if you are going to put up a Web Page purporting to teach newbies how "resonance" works, don't you think that the circuit you post as an example should in fact exhibit resonance, and that it should be a basis of practical applications, which the circuit posted here is not.
 
Ok, but if you are going to put up a Web Page purporting to teach newbies how "resonance" works, don't you think that the circuit you post as an example should in fact exhibit resonance...
You're using imprecise language again. The circuit on the web page does exhibit resonance--"current resonance", not "voltage resonance". You keep saying that it doesn't exhibit "resonance", and nobody disagrees with you that it doesn't exhibit "voltage resonance".

The current from the voltage source, if plotted versus frequency, shows a pronounced minimum at a frequency given by fr = 1/(2*pi*SQRT(L*C))--just like you would expect from a circuit whose impedance vs. frequency shows a resonant peak. But you say:

So back to the tutorial. The equations developed there basically show that there is one frequency where the inductive reactance cancels the capacitive reactance, and if you plot the current that flows from the voltage source into the circuit, there is one frequency where the residual impedance is just the 60Ohm resistor R1. To which I say, so what? This is not resonance, it is simply an algebraic trick...
Are you saying that the current from the voltage source does not behave as the algebra indicates? Again you use the unqualified word "resonance". True enough, there is no "voltage resonance" but there is "current resonance". According to the algebra, the current from the voltage source shows a minimum at a particular frequency, just the sort of behavior associated with resonance.

...that it should be a basis of practical applications, which the circuit posted here is not.
It is the basis of a very important practical application. Consider the circuit under discussion without the capacitor:

ParCir6.png

There will be some reactive current that the voltage source will have to supply. Suppose we wished to minimize that reactive current. We could add a capacitor in parallel with the rest of the circuit. If the value of the capacitor is such as to have the same reactance as the inductor, the reactive current supplied by the source will be minimized--it will be zero in the ideal case. The value of the capacitor will be given by solving this equation for C: f = 1/(2*pi*SQRT(L*C)), where f is the frequency of the AC source. That's the same capacitor that would resonate with L at the frequency f.

The electric grid that supplies our homes has a very low impedance. If the L and R above represent an highly inductive industrial load, placing a capacitor in parallel with the grid at the load reduces the reactive current the grid has to supply. The capacitor, in effect, is tuning the inductive part of the load to the frequency of the grid--a low Q parallel resonance. Even if the grid impedance were zero, the low Q "current resonance" would perform the same function of power factor correction, which is just another name for tuning the load to "current resonance".
 

JoeJester

Joined Apr 26, 2005
4,390
Ok, but if you are going to put up a Web Page purporting to teach newbies how "resonance" works, don't you think that the circuit you post as an example should in fact exhibit resonance, and that it should be a basis of practical applications, which the circuit posted here is not.
I had already agreed with you once on that topic. There are many websites out there that exhibited that circuit or one like it with a theoretical voltage source. If your going to swing that hammer, you'll be very busy contacting a lot of websites to correct their mistake. In fact, if you go on that journey, I probably don't have enough minutes left on this side of mother earth to congratulate you on your accomplishment.

As far as the practical circuit, the one the OP posted didn't fit that criteria. It's kinda funny that all the text kept mentioned the proper way to analyze a parallel circuit, yet a lot of posts have been dedicated to chastising them for an incorrect diagram. I wish we'd go after the authors of books with such vigor.

Would you have been happier if the circuit exhibited a higher Q? I look at that circuit in the same fashion as why schools retain using the uA741 Op Amp for analysis . For training purposes only.

That circuit didn't exhibit resonance because there wasn't proper excitation. That doesn't mean it couldn't be resonant. The whole question was about the inductor current at the frequency of resonance. Can I assume you read the whole page and the diagram is the ONLY problem you have with it? You did illustrate that it could resonate if it wasn't so heavily damped with that 60 ohm resistor.

On edit:
I did an image search at google on parallel resonance. Attached is a 13 page printout that shows numerous "incorrect" diagrams. If your going to start swinging that hammer, I recommend you start with identifying the circuits from a google search. Some may be duplicates.
 

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MikeML

Joined Oct 2, 2009
5,444
...
I did an image search at google on parallel resonance. Attached is a 13 page printout that shows numerous "incorrect" diagrams. If your going to start swinging that hammer, I recommend you start with identifying the circuits from a google search. Some may be duplicates.
Actually, you just made my day. A surprising number of those parallel circuits shown in your document do exhibit "resonance" because energy can move from the inductor to the capacitor and back again... These are the ones that are driven by a voltage source in-series with a resistor, a current-source driving the parallel RLC, or those where there are resistor(s) in series with either the inductor, the capacitor, or both. All of these circuits have more than two circuit nodes...

The ones that have only two circuit nodes (ideal voltage source in parallel with RLC) are the non-resonant ones...
 

MikeML

Joined Oct 2, 2009
5,444
... the low Q "current resonance" would perform the same function of power factor correction, which is just another name for tuning the load to "current resonance".
Well, you call that current resonance; I simply call it what it is: power factor correction, accomplished by cancelling inductive reactance by paralleling it with capacitive reactance. This is a great example of superposition, not resonance.
 
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