Current Resonance: Explaining I1 Existence After Switch Closure

[Bassalisk]

[PLAIN]http://pokit.etf.ba/get/6e89526dc84a619992d2d771c7cc7fb6.jpg

After switch is closed, Xl=Xc. Current resonance kicks in. Thats from the set of the problem. Now there is no current I no more right?

But I1 still exists. How come? How come you don't have current in that one branch and you have current in branches where capacitor and inductor are? Can someone please try to near current resonance effect to me?

thanks

 

[sophiecentaur]

Firstly, there will be a finite resistance in the resonant loop and energy will constantly be supplied. If you want a realistic answer to the question you have to have a finite Q.
Secondly, it takes time (related to the Q of the circuit) for the Voltage across C to reach its maximum as the energy oscillating within the tuned circuit builds up.

For an answer to what happens on either side of resonance, you need to write out the expression for the effecive impedance of the R' LC (R' is the series resistance) and look at the resulting 'potential divider' ratio. Clearly, the volts across the C will be zero at infinite f and zero at zero f. You'll get the well known bell shape resonance curve - sharper as the R' gets less.

 

[Bassalisk]

Lets assume that we have all realistic elements. And the circuit is in resonance, current resonance.

I know when system is in voltage resonance, you have that inductor and capacitor "eat" each other up, because of phase difference (not this picture, elements have to be in series).

But what is going on when the circuit is in current resonance?

 

[Averagesupernova]

The same thing that is going on in series resonance is happening in parallel resonance. Well, kinda. I've never heard it called current resonance. Current is circulating in the tank circuit (cap and inductor in parallel). With ideal components, NO current will flow through R. With a series resonance circuit, large a current will flow in the cap and inductor as well, but since everything is in series a large current would also flow in the rest of it.
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I keep it straight by thinking of it this way: Resonant circuits behave the opposite of what is intuitive. Meaning that normally when we add components in parallel the impedance of the network goes down, and when we add components in series, the impedance of the network goes up. But when dealing with resonance, a parallel resonant tank circuit is a high impedance and a series resonant tank circuit is a low impedance. Opposite of what is intuitive, at least for me.

 

[Jiggy-Ninja]

Lets assume that we have all realistic elements. And the circuit is in resonance, current resonance.

I know when system is in voltage resonance, you have that inductor and capacitor "eat" each other up, because of phase difference (not this picture, elements have to be in series).

But what is going on when the circuit is in current resonance?

The exact same thing, except with current instead of voltage.

Assuming ideal components, at any given point in time, the current through the inductor is exactly equal, but opposite in direction to (180 degrees out of phase with), the current flowing through the capacitor.

Apply Kirchoff's Current Law to the node just above the tank circuit. If the inductor has 5mA flowing through it, the capacitor must have -5mA. In order for KCL to be true, there cannot be any current from the resistor. The impedance of the tank circuit is infinite.

With practical components, the winding resistance of the inductor adds a small phase shift, so that the inductor current is no longer exactly 180 degrees out of phase with the capacitor current. This means that at resonance, there will still be some small amount of current flowing, so the impedance of practical tank circuit will be a very large, but finite number.

 

[sophiecentaur]

I am not familiar with the term "current resonance" and it's not on the first page of Google Hits, either.
The two resonance conditions that I learned are series resonance and parallel resonance, which terms describe themselves unambiguously.
There is no essential difference between them in a practical circuit. The L and C are doing exactly the same in both cases and it's just how you choose to connect to the circuit that determines whether you see a high impedance or a low impedance at resonance.
I find it hard to see why people would rather discuss this simple circuit with arm waving and 'words' when the Maths describes it all perfectly and with no flannel.

 

[Jiggy-Ninja]

I am not familiar with the term "current resonance" and it's not on the first page of Google Hits, either.
The two resonance conditions that I learned are series resonance and parallel resonance, which terms describe themselves unambiguously.
There is no essential difference between them in a practical circuit. The L and C are doing exactly the same in both cases and it's just how you choose to connect to the circuit that determines whether you see a high impedance or a low impedance at resonance.
I find it hard to see why people would rather discuss this simple circuit with arm waving and 'words' when the Maths describes it all perfectly and with no flannel.

Series resonance = Voltage resonance (where cap and inductor voltages cancel)
Parallel resonance = Current resonance (where cap and inductor currents cancel)

I've never seen those terms before either, but I figured it out immediately from context. And there's nothing wrong or ambiguous about those terms either. It's just that one set is based on the configuration and one is based on the physics.

And there's a limit to how informative math is. Math just gives you a "black box" explanation that let's you know exactly how something will behave, but it doesn't tell you why it will behave that way. It doesn't always give you an intuitive understanding of the physics involved, which can be just as important.

 

[Bassalisk]

Thanks Jiggy. In my language you say current and voltage resonance, because of conditions that happen in each case. You have been very helpful, I got it now.

Thanks all.

 

[sophiecentaur]

Series resonance = Voltage resonance (where cap and inductor voltages cancel)
Parallel resonance = Current resonance (where cap and inductor currents cancel)

I've never seen those terms before either, but I figured it out immediately from context. And there's nothing wrong or ambiguous about those terms either. It's just that one set is based on the configuration and one is based on the physics.

And there's a limit to how informative math is. Math just gives you a "black box" explanation that let's you know exactly how something will behave, but it doesn't tell you why it will behave that way. It doesn't always give you an intuitive understanding of the physics involved, which can be just as important.

It's interesting that you "immediately figured it out" because I could just have easily concluded that voltage resonance would be for a parallel case, where the volts add to produce a voltage maximum. A current resonance would be the series case where the currents add to provide a low impedance and a current maximum. As a matter of fact, which, does the OP imply, is actually which?
Unless you are going to define a new term with some precision, why introduce it?
Maths gives you far more than a 'black box' description. It tells you what will happen on resonance (with an indication of the actual values involved) and off resonance. Don't try to fool yourself that you can get a "why" explanation of how a capacitor or inductor works without recourse to calculus (implicit or explicit). If you 'have a feeling' about how a capacitor behaves in a circuit with AC signals, then it will have to have been because you have already accepted the notion of a time derivative.

 

[Bassalisk]

It's interesting that you "immediately figured it out" because I could just have easily concluded that voltage resonance would be for a parallel case, where the volts add to produce a voltage maximum. A current resonance would be the series case where the currents add to provide a low impedance and a current maximum. As a matter of fact, which, does the OP imply, is actually which?
Unless you are going to define a new term with some precision, why introduce it?
Maths gives you far more than a 'black box' description. It tells you what will happen on resonance (with an indication of the actual values involved) and off resonance. Don't try to fool yourself that you can get a "why" explanation of how a capacitor or inductor works without recourse to calculus (implicit or explicit). If you 'have a feeling' about how a capacitor behaves in a circuit with AC signals, then it will have to have been because you have already accepted the notion of a time derivative.

I believe math is mainly a TOOL to help you solve problems and understand certain things. Ofc EE cannot work without math, nobody says that. Math generalizes things. You can apply derivative to any sense. Velocity, then current dq/dt etc. I agree with Jiggy that intuition is just as important as good calculus.

I just came back from my Electric circuits exam, and there was a question addressing current resonance. I learned the intuition, and that problem was no biggie.
And yes I have a good math sir, very good imho. But math got me only so far.

Correct me if I am wrong, but aren't elements on parallel connection always on same voltage? Having said that, there is no reason to think a current resonance as series resonance.

 

[sophiecentaur]

I believe math is mainly a TOOL to help you solve problems and understand certain things. Ofc EE cannot work without math, nobody says that. Math generalizes things. You can apply derivative to any sense. Velocity, then current dq/dt etc. I agree with Jiggy that intuition is just as important as good calculus.

I just came back from my Electric circuits exam, and there was a question addressing current resonance. I learned the intuition, and that problem was no biggie.
And yes I have a good math sir, very good imho. But math got me only so far.

Correct me if I am wrong, but aren't elements on parallel connection always on same voltage? Having said that, there is no reason to think a current resonance as series resonance.

Maths is more than a Tool. It is a Language, which describes many things far more accurately and concisely than ordinary words can. You say it "generalises" things. But, by the same token, ordinary language does that, too. The first thing that any intuitive approach to Science does is to use Metaphor and Simile - generalising the idea. If you rely on intuition then you will end up making mistakes - but you know that. For a very simple situation, you could come to a correct conclusion without explicitly using Maths but I bet you wouldn't really believe your conclusion until you had actually confirmed it with the Maths. And, for a network of components involving more than a half dozen simple elements, you wouldn't stand a cat's chance just arm waving. Let's face it, even a simple potential divider (there's a clue in the name) is a pretty mathematical circuit. If you 'see' that working by just looking at it, it will be because you have already done the Maths and are bright enough for that bit of Maths to have been assimilated into your automatic Science-Reasoning.

Yes, parallel elements will always have the same voltage - at resonance, this voltage will be a Maximum, no?. Hence my instant assumption that parallel resonance was Voltage resonance. I see that the "current resonance" was a term to indicate that a lot of current was sloshing around inside the loop but, as you can't see that without disturbing the loop, how would you know? I am not surprised that the terms are not in common use and I assume that they have come from a source which may have meant well but didn't take into the account the possibility of misinterpretation - definitely not mainstream. [edit: not long-established mainstream, at least]
I have no objection to someone using a new term but there is always the risk of misunderstanding if it's not defined fully - which it wasn't.

 

[Jiggy-Ninja]

It's interesting that you "immediately figured it out" because I could just have easily concluded that voltage resonance would be for a parallel case, where the volts add to produce a voltage maximum. A current resonance would be the series case where the currents add to provide a low impedance and a current maximum. As a matter of fact, which, does the OP imply, is actually which?

The first implication comes from the first post, talking about "current resonance" with a big picture of a tank circuit. Just remember your context clues lesson from English (or whatever your native language is) class.

Second implication is here:

I know when system is in voltage resonance, you have that inductor and capacitor "eat" each other up, because of phase difference (not this picture, elements have to be in series).

Unless you are going to define a new term with some precision, why introduce it?

Apparently it's a translation issue. It might be tough to remember, but not everyone speaks English as their first language, and the terms probably aren't exactly the same.

Maths gives you far more than a 'black box' description. It tells you what will happen on resonance (with an indication of the actual values involved) and off resonance. Don't try to fool yourself that you can get a "why" explanation of how a capacitor or inductor works without recourse to calculus (implicit or explicit). If you 'have a feeling' about how a capacitor behaves in a circuit with AC signals, then it will have to have been because you have already accepted the notion of a time derivative.

My point is that just math has limited utility. The "feeling", "intuition", or whatever you call it, can help connect the abstract math to reality, and make it easier to understand the formulas, derive new ones, re-derive ones you forgot, or notice when the answer is other than it should be. (Such as doing a voltage divider calculation and ending up with an answer bigger than the input voltage)

Math is very, very, very important, but it's not th only thing you need.

We could probably argue about this back and forth for days, but I'd rather not since the possibility of misinterpretation is really high.

Yes, parallel elements will always have the same voltage - at resonance, this voltage will be a Maximum, no?. Hence my instant assumption that parallel resonance was Voltage resonance. I see that the "current resonance" was a term to indicate that a lot of current was sloshing around inside the loop but, as you can't see that without disturbing the loop, how would you know? I am not surprised that the terms are not in common use and I assume that they have come from a source which may have meant well but didn't take into the account the possibility of misinterpretation - definitely not mainstream. [edit: not long-established mainstream, at least]
I have no objection to someone using a new term but there is always the risk of misunderstanding if it's not defined fully - which it wasn't.

When hooked up to an AC constant voltage source, it will be maximum only if there is another element (like a resistor) in series with the tank to bleed off voltage. If the tank is the only thing in the circuit, the voltage will be constant, even though the current varies.

 

[sophiecentaur]

I think you have justified my preference for Maths, when you use terms like "eat each other up" and "bleed off voltage", neither of which I have ever read in a textbook or published paper. Both of those terms (and the two we were discussing) are very ready to be misinterpreted. The simple statement of the voltage ratio in terms of the the component values in a mathematical expression is universally understandable and open to no doubt. As a language, Maths may be a bit hard to learn but it really takes a lot of beating for getting ideas across. I wonder if that could be why it is so popular with Scientists and Engineers?

I wonder why one would ever want to "hook up" a parallel resonant circuit to a constant voltage source. What would it achieve?

 

[Bassalisk]

Oh my God already! Of course I don't literally take that "eat each other up" term! Its just MUCH easier to write that, and do not digress from the problem, instead of writing set of the equations!
Stop taking everything literally.

 

[sophiecentaur]

But what does it mean to someone reading it from cold? It may be easier to write it but the point is, surely, to write something that doesn't need special, personalised translation. If you want to write purple passages then chose somewhere more suitable. If you know what you're talking about then you surely realize that a "set of equations" is not needed - just the right terms.

 

[Bassalisk]

But what does it mean to someone reading it from cold? It may be easier to write it but the point is, surely, to write something that doesn't need special, personalised translation. If you want to write purple passages then chose somewhere more suitable. If you know what you're talking about then you surely realize that a "set of equations" is not needed - just the right terms.

ok ok you do have a point there. I will from now on make a problem more professionally stated.

 

[sophiecentaur]

You know, I'm not just being grumpy and pernickity.
Imagine you knew very little electronics and were trying to follow this thread. The first thing you'd do would be to try a web search on the whacky expressions and where would that get you? Confused?

 

[Jiggy-Ninja]

I think you have justified my preference for Maths, when you use terms like "eat each other up" and "bleed off voltage", neither of which I have ever read in a textbook or published paper. Both of those terms (and the two we were discussing) are very ready to be misinterpreted. The simple statement of the voltage ratio in terms of the the component values in a mathematical expression is universally understandable and open to no doubt. As a language, Maths may be a bit hard to learn but it really takes a lot of beating for getting ideas across. I wonder if that could be why it is so popular with Scientists and Engineers?

I'll admit, "bleeding off" voltage is probably horribly bad terminology. It's the figure of speech I use internally, and I know what I mean, but it's not standard. What would be better in that situation? Dissipating? Dropping?

I wonder why one would ever want to "hook up" a parallel resonant circuit to a constant voltage source. What would it achieve?

It would achieve nothing useful. However, for people that are learning, understanding comes first, usefulness and practicality second.

Oh my God already! Of course I don't literally take that "eat each other up" term! Its just MUCH easier to write that, and do not digress from the problem, instead of writing set of the equations!
Stop taking everything literally.

In his defense, clarity is important. EE stuff is difficult enough as is. Ambiguous terminology just complicates things.

 

[Bassalisk]

in my defense xD I am 19 years old and i just started learning this stuff

 

[Averagesupernova]

As someone who has actually worked in EE (not saying none of you have, although I do wonder) I will say that there is a lot of 'slang' tossed around like this. Although I've never heard of current resonance I would have assumed it meant a parallel tank circuit since it will appear as a low impedance and draw a fair amount of current from a voltage source. When a person switches jobs and moves to a different EE job there are a lot of slang terms used at the previous employer that your new coworkers will not understand. You will most likely find that your new coworkers have a few slang terms of their own that you have never heard of.
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I don't find that the phrase 'bleed off voltage' is hard to understand at all. Technically current is bled away such as by using a bleeder resistor in parallel with a large capacitor. But the result is that the voltage on the cap is reduced down to nothing.

 

[sophiecentaur]

As someone who has actually worked in EE (not saying none of you have, although I do wonder) I will say that there is a lot of 'slang' tossed around like this. Although I've never heard of current resonance I would have assumed it meant a parallel tank circuit since it will appear as a low impedance and draw a fair amount of current from a voltage source. When a person switches jobs and moves to a different EE job there are a lot of slang terms used at the previous employer that your new coworkers will not understand. You will most likely find that your new coworkers have a few slang terms of their own that you have never heard of.
-
I don't find that the phrase 'bleed off voltage' is hard to understand at all. Technically current is bled away such as by using a bleeder resistor in parallel with a large capacitor. But the result is that the voltage on the cap is reduced down to nothing.

Parallel tuned circuits are High impedance - not low impedance. The two susceptances are equal and opposite - giving zero susceptance - infinite reactance.

But I agree that slang - especially in a very general forum like this one, with many readers who have limited experience of English - doesn't help at all in dealing with difficult topics. I often find that the use of catch-all slang can cover up for lack of understanding and consequent lack of confidence with the correct terms.

 

[Bob S]

How come, with 21 posts, there are neither equations nor illustrations of the current waveform in the capacitor?

See illustration of current waveform in thumbnail. The switch is closed at 50 us. Positive current is flowing into (charging) capacitor.

Bob S

 

[Bassalisk]

How come, with 21 posts, there are neither equations nor illustrations of the current waveform in the capacitor?

See illustration of current waveform in thumbnail. The switch is closed at 50 us. Positive current is flowing into (charging) capacitor.

Bob S

name of the software please?

Give me a sec to try to read this graph.

 

[Bob S]

name of the software please?

It is LTspiceIV free download from www.linear.com.

 

[sophiecentaur]

How come, with 21 posts, there are neither equations nor illustrations of the current waveform in the capacitor?

See illustration of current waveform in thumbnail. The switch is closed at 50 us. Positive current is flowing into (charging) capacitor.

Bob S

But you are showing the step response and the OP has an AC source. A different question, surely. The waveform actoss the capacitor will be a continuous sinusoid. The amplitude will be the only thing different for different frequencies.

 

[Averagesupernova]

Parallel tuned circuits are High impedance - not low impedance. The two susceptances are equal and opposite - giving zero susceptance - infinite reactance.

Jeez I did it again. You are correct.

 

[Bob S]

But you are showing the step response and the OP has an AC source. A different question, surely. The waveform actoss the capacitor will be a continuous sinusoid. The amplitude will be the only thing different for different frequencies.

The OP showed a + sign by the voltage source, and a ~ inside the voltage source, but did not state in the text that it was an ac voltage source, and what its frequency was, compared to the LC resonance frequency. Furthermore, the OP did not state at what phase of the ac the switch was closed.

I attach a special case of an ac circuit, where the voltage source is the same frequency as the LC resonant frequency. See thumbnail. The red trace is the current in R1, and the green trace is the current in C1. The switch is closed at 300 us. before 300 us, the current in R1 and C1 is the same, so the red trace hides the green trace. After 300 us, the current in C1 builds up, and the current in R1 is reduced.

The voltage across C1 and L1 is the same, and the current in L1 and C1 is nearly the same.

Parallel tuned circuits are High impedance - not low impedance. The two susceptances are equal and opposite - giving zero susceptance - infinite reactance.

If the LC circuit is infinite reactance and it is driven at the LC resonant frequency (25 kHz), why is it still drawing current through R1 after the switch is closed?

Bob S

 

[Jiggy-Ninja]

The OP showed a + sign by the voltage source, and a ~ inside the voltage source, but did not state in the text that it was an ac voltage source, and what its frequency was, compared to the LC resonance frequency. Furthermore, the OP did not state at what phase of the ac the switch was closed.

OP said that X_L and X_C were equal.

I attach a special case of an ac circuit, where the voltage source is the same frequency as the LC resonant frequency. See thumbnail. The red trace is the current in R1, and the green trace is the current in C1. The switch is closed at 300 us. before 300 us, the current in R1 and C1 is the same, so the red trace hides the green trace. After 300 us, the current in C1 builds up, and the current in R1 is reduced.

The voltage across C1 and L1 is the same, and the current in L1 and C1 is nearly the same.
If the LC circuit is infinite reactance and it is driven at the LC resonant frequency (25 kHz), why is it still drawing current through R1 after the switch is closed?

Bob S

It's not exactly th resonant frequency. By my calculation, at 25 kHz there is 37.2 nS of inductive susceptance (26.9 M-ohms reactance) in the tank circuit.

Admittedly, that seems way too small. It would only let a current on the order of nA through, which doesn't look like enough to cause the deflection in your graph, with looks to be about 1mA peak.

Are the switch and inductor in the simulation ideal? If there is a small amount of resistance, it will throw off the phase of the inductor's current, so that even at resonance it won't completely cancel the capacitor's current, leaving a small, nonzero admittance.

If my calculations are correct (and they might very well not be), it would only take a switch + winding resistance of about 9 ohms to let 1mA of current through at resonance, which seems to be a reasonable amount.

 

[sophiecentaur]

Jeez I did it again. You are correct.

I have done it myself lots of times. Just haven't actually pressed the button. Ha

 

[sophiecentaur]

The voltage across C1 and L1 is the same, and the current in L1 and C1 is nearly the same.
If the LC circuit is infinite reactance and it is driven at the LC resonant frequency (25 kHz), why is it still drawing current through R1 after the switch is closed?

Bob S

Yet another PF contributor is questioning the basic theory on evidence from a simulator. Tut Tut, when will you boys learn?

This would not be a problem if you were actually to build the circuit in analogue components because you would get a 'very high and very sharp' peak as you swept the frequency of your analogue generator and you'd 'believe' the theory.

There is no question that, at resonance, there is infinite impedance. BUT your simulation is dealing with idealised components. You didn't include any series resistance in your resonant loop so the resonance peak is very very sharp (to the limit of the accuracy of your digital calculations). It cannot be surprising when a calculation which is, in effect, subtracting a massive number from another massive number, fails to give a non-zero answer. Your simulation (or snare and delusion, as I refer to them) has, in fact, given you the slightly off-resonance result. btw, it would be interesting to know the phase of the current - which would tell yo which side of resonance the simulator thought we were.

Anyone would think I was a technophobe!