Discussion in 'General Electronics Chat' started by The Electrician, Sep 1, 2014.

1. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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It seems that whenever a circuit is presented for study, the ones that elicit the most disagreement are those that contain ideal circuit elements.

For example, consider this one:

If the voltage sources are ideal (zero internal impedance), the wiring is ideal (superconductor perhaps?), and the question is "what is the current in the 1 ohm resistor", beginning students are quite flummoxed. Eventually, after much discussion, it is agreed that the question can't be answered.

Note that if anything is less than ideal, if the batteries have any internal resistance (even one of them), or if the wiring has some resistance, however small, the question has a definite answer.

The student learns that you can't connect ideal voltage sources in parallel, and you can't connect ideal current sources in series

Another similar situation is the case where two ideal capacitors, one of which is charged to some voltage, are then connected in parallel with wiring having no resistance. Calculations show that energy is lost after the connection is made. But where did the energy go? If there is any resistance in the capacitors or wiring, there is no paradox. I am aware that in the ideal case where the method of making the connection is not ideal, energy will be lost in the spark, or in radiation from the loop of wire making the connection, but that's not the point here.

The circuit from the tutorial as shown:

raises the question "Is there any interaction between L1 and C1?" Is the current Iz sloshing back and forth between L1 and C1? It would seem that if E1 is an ideal voltage source that the answer is no. Thus, it would seem that L1 and C1 are decoupled from each other, and there can be no resonance.

Yet, if we calculate and plot the impedance of the total LCR circuit, we see an definite peak in the impedance. If we analyze (or simulate) the current from the voltage source and plot it vs. frequency, it exhibits a pronounced dip in value at a frequency given by 1/(2*pi*(L1*C1)). This is the frequency we would normally associate with resonance, so apparently L1 and C1 are not decoupled from one another.

We have a paradox: L1 and C1 are decoupled, yet they are not decoupled.

The addition of the slightest non-ideality removes the paradox. Here's the ideal circuit with it's paradoxical bahavior, redrawn:

But with just a tiny non-ideality, the paradox is resolved:

Now L1 and C1 are no longer decoupled. But, even now, the "voltage resonance" is extremely low q; probably not very useful.

I haven't seen the terms "voltage resonance" and "current resonance" before, but in discussing circuits where the presence of ideal components gives rise to a paradox, they can be useful. We see that such a circuit can have one kind of resonance but not the other, and apparently for one kind of excitation components can be decoupled, but for another kind of excitation, they are not decoupled.

Any resonant circuit with non-ideal components, voltage sources with non-zero internal impedance, and current sources with less than infinite internal impedance, will always have both "voltage resonance" and "current resonance"; there will be no paradox.

http://www.electronics-tutorials.ws/accircuits/parallel-resonance.html

The tutorial could be enhanced by pointing out this paradox, and pointing out that in practical applications, a parallel resonant circuit should probably be current driven, that is, driven from a high impedance.

Last edited: Sep 3, 2014
2. ### MrAl Distinguished Member

Jun 17, 2014
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Hello there Electrician,

Thanks for starting a new thread, good idea.

What i'd like to state first is there is a misunderstanding between you and i on what we are calling what.

First, i did not miss any point because i knew you were talking about power factor correction and that the TOTAL current changes when we include the capacitor. Anybody reading this knows that, that's common knowledge. So when you say you are doing that and the total current is affected so you look at it as a another form of resonance, i have no problem with that.
The real problem is, that's not what i was talking about. I was talking about the difference between:
1. Physical Resonance, and
2. Electrical Resonance.

Physical Resonance is something we can observe by seeing an exchange of energy.
Electrical Resonance is something we can observe only by measuring an electrical quantity (such as current or voltage).

Note that in #2 we dont always see an exchange of energy, while in #1 we always see an exchange of energy in addition to seeing an electrical quantity change.

By exchange of energy i mean that the capacitor supplies the inductor with energy some of the time, and the inductor supplies the capacitor with energy some of the time. So in a way they 'share' the SAME energy, even if it is not long lasting.

In the power correction circuit, the TOTAL current changes when we add the cap, but the INDUCTOR current does not change at all. If the circuit was really exhibiting physical resonance, then the inductor current would change because the cap would sometimes take energy and sometimes supply energy, so it could not be the same.
In this kind of circuit the inductor and capacitor share energy with the line only.

As you mentioned, we see a zero phase shift with the addition of the cap, and of course that means the whole circuit acts like a purely resistive load, which is also a sign of resonance. And all along we've seen the dip in the current response as we approach the center frequency. These are really just electrical specifications though.

So the point is (and what Mike was trying to make) is that not every circuit that exhibits a max or min is a true resonant circuit, but we might still call it "electrical resonance" because it exhibits electrical behavior typical of a resonant circuit.

This fact was noted a long time ago so i cant take any credit for somehow discovering it, but i did notice this a long time ago too.

Another better example is perhaps a passive RC two stage band pass filter. There can not be any true physical resonance, but there is resonance like behavior.

Last edited: Sep 1, 2014
3. ### ErnieM AAC Fanatic!

Apr 24, 2011
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First, the "case where two ideal capacitors, one of which is charged to some voltage, are then connected in parallel" does not loose energy when the connection is made: it transfers rather quickly (instantly for the ideal case) with zero loss.

For the LRC if you do the calculations correctly there can be no "sloshing" for the ideal case. The R has a fixed non zero current , the C current is also fixed but that is fixed at zero, and the inductors' current is rising linearly towards positive infinity.

4. ### crutschow Expert

Mar 14, 2008
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But the two capacitors in parallel do lose energy. You can see this if you calculate the total capacitor energy before and after they are connected. Suppose you have two equal value capacitors (C), one discharged and one charged to V. The total energy (on the one capacitor) is 1/2 CV^2. When the capacitors are connected together (with a perfect connection) the resulting voltage is (1/2)V on both capacitors. The total energy is thus 1/2 * 2 C *((1/2)V)^2 = 1/4 CV^2.

So where did 1/2 the energy go? My understanding is that it is radiated by the infinite frequency of the step current pulse when the charge is transferred.

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5. ### MikeML AAC Fanatic!

Oct 2, 2009
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I'm staying out of this one....

6. ### MrAl Distinguished Member

Jun 17, 2014
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Hi,

We dont have to resort to radiated energy in order to explain the two capacitor transfer of energy. All we need is a tiny tiny resistance which exists between the two. Even with the smallest resistance (some short wires connecting the two) we still loose 1/2 the energy to HEAT loss in the resistance. There will always be some resistance and if you have enough wire to form an antenna then you already have enough to form a resistor.
Does anything get radiated? Im sure a little does as in any circuit even with a small current, but not half because not only is a zero resistance in and between capacitors impossible, there is no perfect antenna either.
Maybe you can theorize that half the energy would get radiated if there were no other losses, but im not sure what good that would do you because we cant even perform the experiment without wires connecting the two and perfectly conducting capacitor plates

7. ### crutschow Expert

Mar 14, 2008
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We were discussing ideal components. If you want to introduce non-ideal components into the mix, even tiny ones, that changes the discussion. The question is, theoretically how would you get rid of the excess energy between the two ideal capacitors connected by an ideal (superconducting?) wire.

8. ### nsaspook AAC Fanatic!

Aug 27, 2009
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A theoretical PEC 'perfect conductor' would violate thermodynamics and mathematics by the generation of a frozen field if initial conditions change. If the initial field is zero it will never change from zero. A real PEC 'superconductor' is an different animal because of the quantum properties of the Meissner effect.

In circuits you don't calculate the effects of a PEC, It's just a short cut to eliminate complications by assuming all common connections are at exactly the same point in space and time when changes in fields between those points are minute.

Last edited: Sep 1, 2014
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9. ### studiot AAC Fanatic!

Nov 9, 2007
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There is simply a situation where there are more unknowns than there are equations (conditions).

A couple of centuries ago when faced with this situation structural engineers didn't cry into their beer, they went out and found more equations.

If you trawl back into the dim and distant you may remember doing some applied maths or physics, and a situation in mechanics where you solved problems that were

"statically determinate" by resolving vectors , which is equivalent to applying the principle of conservation of energy.

You may also have met situations that were "statically indeterminate" and could not be solved by the equations of equilibrium.

So the engineers of the time turned to the material properties of the elements to generate additional equations to solve the system.

We should not be suprised if it is possible to construct similar situations in electrical networks, and a similar solution is in order.

Turn to the material properties.

Yes the connecting wires will have resistance, but so will the capacitors, inductors etc.
Further real capacitors and inductors will have edge losses.

That is where to look for the dissipation of the energy in the case of the two capacitors for instance.

Last edited: Sep 2, 2014
10. ### MrAl Distinguished Member

Jun 17, 2014
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This thread was not really created to discuss this topic, maybe you should start another thread for this. This thread was for discussing a resonant circuit that may or may not look like it is really a resonant circuit.

A tiny variable resistance does not really change the discussion, it allows us to transition between one set of conditions to another set of conditions in a smooth way rather than abrupt where it is hard to tell what is happening.

The limiting value of the function is still 1/2, but that is from an electrical standpoint that does not include radiation effects. That is the result when we allow the resistance to decrease to zero in a circuit that already has a resistance but has no mechanism to show radiation. Stated another way, as we allow R to gets closer to zero, the radiation would get larger because the electron acceleration would increase, but without an equation that shows the radiation we would not be able to tell if our purely electrical analog would still hold, and surely it would not because if it works correctly with resistance alone when that resistance dissipates all of the necessary power to result in a value of 1/2, as the radiation gets higher and higher that would necessarily mean more power lost from the circuit, which in turn would not allow us to get a result of 1/2.
In other words, if the energy we loose in the resistance is ER and the energy we "loose" (loose in quotes because we might want this to happen) by radiation is Er, then the total energy lost is no longer just ET=ER, it is ET=ER+Er.
This makes it possible to think that maybe ALL of the energy gets radiated with zero resistance.

This is also probably not a good topic for discussion as is just how much does a short circuit radiate. If we consider the wire to be a lumped circit element with zero resistance then it does not radiate at all. If we consider it to be a transmission line, then it radiates

Back on topic, i had talked about two different basic forms of resonance, Physical and Electrical. The Physical form is often talked about in physics books while the Electrical form is more talked about in books on electronics and circuit analysis. There is a professional paper out there somewhere that unifies the two forms, but good luck finding it.

Last edited: Sep 2, 2014
11. ### nsaspook AAC Fanatic!

Aug 27, 2009
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AT&T Archives: Similiarities of Wave Behavior

Last edited: Sep 2, 2014
12. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Check your math. The resulting voltage on two equal capacitors is sqr(2)V on both (where sqr(2) is the square root of two).

As the ideal capacitor is a lossless component you should have tossed your answer away as obviously incorrect, or as my post began before typo corrections "check your meth."

13. ### crutschow Expert

Mar 14, 2008
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Au contraire, my math (and dosage) is fine. The amount of charge on a capacitor is Q = V/C. Thus if you double the capacitance by connecting two equal value capacitors in parallel the voltage drops by 1/2 since the amount of change remains the same (can't create or destroy charge).

You calculation is only correct for 1/2 the energy on each capacitor but that, obviously, cannot be correct in this case. That's why the conversation on the missing energy.

14. ### MrAl Distinguished Member

Jun 17, 2014
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Hi,

To help clear things up, with an initial voltage of 1v across cap 1 and 0v across cap 2, the transfer equation in the time domain is simply:
v2=(1/2)*(1-e^(-t/tau))
where
v2 is the voltage that appears across C2 when the two caps are jointed by a resistor with value R, and
tau=R*C, and
C is the capacitance of either capacitor assuming they are both equal.
Note this is identical to a single cap charging through a resistor except for the multiplying (source) voltage which is 1/2 for equal value caps. and for a good reason: the initial voltage in C1 looks like a voltage source in series with it and both caps then with zero energy, and then the whole circuit looks like two caps in series with a resistor and driven by a voltage source of 1v. The two caps, being both equal and both with zero volts across them, then charge up to 1/2 the total voltage each, which comes out to 0.5 volts each.

As the time is allowed to reach a high value relative to R*C, we see the limiting voltage across C2 is exactly 0.5 volts.
If the capacitor values are different, then the final voltage across C2 is C1/(C1+C2). So if C2=3 and C1=1 then the final voltage is 0.25 volts.

Last edited: Sep 2, 2014
15. ### studiot AAC Fanatic!

Nov 9, 2007
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Thank you Spook for a fascinating video and link to ATT.

Meanwhile let us consider some elctric circuit theory about the two capacitors and show firstly that Crutschow is correct in his calculations and further that the missing energy dissipated is independent of the connection resistance.

Firstly when one of two identical capacitors is charged to voltage V the energy stored is equal to the work done in charging it.

$W = \frac{{QV}}{2} = \frac{{{Q^2}}}{{2V}} = \frac{{C{V^2}}}{2}$

If we now remove the source of charging and connect the second capacitor across the first the charges will equalise over both capacitors or the total charge, Q, is the same before and after connection of the second capacitor, by the law of conservation of charge.

But charge = Capacitance x Voltage

charge before = charge after
CVbefore = (C+C)Vafter

Vbefore = 2Vafter

The voltage across both capacitors after addition is half that of the initial charge voltage V.

Now consider a connection resistance R between the two capacitors

Kirchoff's voltage equation becomes

$RI + \frac{2}{C}\int {Idt} = 0$

differentiating

$\frac{{dI}}{{dt}} + \frac{{2I}}{{RC}} = 0$

and applying the boundary condition at t=0, I = V/R
The differential equation has the solution

$I = \frac{V}{R}{e^{ - \frac{{2t}}{{RC}}}}$

The final voltage of the second capacitor is

$\frac{{C{V^2}}}{4}$

The total energy is the resistive energy + capacitive energy

$TotalEnergy = \int\limits_0^\infty {R{I^2}} dt + \frac{C}{2}{\left( {\frac{V}{2}} \right)^2} + \frac{C}{2}{\left( {\frac{V}{2}} \right)^2}$

If we substitute the forgoing expression for I into the integral from the above solution to our differential equation this results in

total energy is

$\frac{{C{V^2}}}{4} + \frac{{C{V^2}}}{4}$

Where the first term is the resisitve energy and the second the collected energy on both capacitors.

This equals the original energy, satisfying conservation of energy and shows that half the enrgy is dissipated in the resistor.

But since R does not appear in the final value for the resistive energy it is independent of the connecting resistance.

16. ### nsaspook AAC Fanatic!

Aug 27, 2009
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The 'the missing energy' independent of resistive losses in wiring is in the kinetic energy of the charge carriers as they are accelerated. The electrons in this case will hit the plates of the capacitor and be dissipate as heat in the plates due to inelastic collisions and/or transformed into mechanical motion of the matter in the plates. Unless the potential is so great that we see accelerations close to a large fraction of light speed the losses from EM radiation will be small.

Last edited: Sep 2, 2014
17. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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What you describe is ohmic losses, which were postulated to be zero. Current thinking is that radiation accounts for the loss of energy; here are the calculations:

http://www.physics.princeton.edu/~mcdonald/examples/twocaps.pdf

Last edited: Sep 3, 2014
18. ### The Electrician Thread Starter AAC Fanatic!

Oct 9, 2007
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Small typo here. It's actually Q = C*V. But your result is correct, of course.

19. ### t_n_k AAC Fanatic!

Mar 6, 2009
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With respect to the parallel resonance discussion ......

The idea that was conveyed to me was based around driving point impedances of networks and the network poles & zeros.

If a network can be driven into resonance by a pure voltage source, the driving point impedance zeros are directly related to the time domain resonant frequencies. The driving source frequency and the impedance zeros frequencies must be one and the same for resonance to occur. In the case of the voltage driven network at resonance, the natural response of the circuit can be derived from the characteristic transient response equation resulting from the network having a short placed across the driving point terminals. In the case of the parallel RLC circuit under consideration, the resulting natural response for the network thus configured would reduce to the original network in parallel with a short circuit - which is inconsistent with expectation.

The alternative case of a network driven by an ideal current source is different. In that case the driving point impedance poles determine the resonant condition. The removal of the driving source in this case to define the circuit's natural behaviour simply requires one to consider the driving point terminals left as an open circuit. This option is entirely consistent with the parallel RLC network.

Perhaps this analytical approach could be applied to the case of a non-ideal driving voltage source - i.e one having non-zero internal resistance.

20. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Also it is possible by a bit of mental trickery to conceive of a zero internal resistance driving voltage source which can excite the parallel RLC network into resonant behaviour in which there appears to be a coupled exchange of energy between the capacitor and the inductor. I suggested this in the other thread which preceded and prompted this one.