[SOLVED] Series LC with DC supply

In a series LC circuit (R≈0), how is possible to oscillate even with DC supply applied ? How can the capacitor to discharge even with the DC source applied ? The capacitor should block DC and act as open circuit...
For 1 V input, capacitor voltage goes to 2 V and then to 0 and so on.

I know that solving it's differential equation get that analytical result, but do not understand how that physically could be possible.

When you suddenly apply a DC voltage to the circuit, the capacitor starts to charge through the inductor.
This stores energy in the inductor equal to 1/2 LI^2.

When the capacitor voltage equals the supply voltage, the inductive reactance keeps the current flowing and charging the capacitor due to the energy stored in the inductor.
This continues until all the inductive energy has been transferred to the capacitor, equal to 1/2 CV^2.

It turns out that this ends up charging the capacitor to twice the supply voltage.

Now, with the inductor current stopped, the capacitor starts to discharge back through the inductor, storing the capacitor energy back into the inductor as the inductor current increases in the opposite direction.

When the capacitor voltage equals the supply voltage the inductor keeps the current moving until all the inductive energy is transferred back to the supply and the capacitor is at 0V (and we're back at square one).

Now the power supply starts things over again from the beginning.

In a lossless system this oscillation will continue indefinitely in a sinusoidal fashion with the capacitor and inductor transferring the energy back and forth.

Note that when the inductor has maximum energy (current), the capacitor energy (voltage) is zero, and when the capacitor has maximum energy (voltage), the inductor energy (current) is zero.

Make sense?

It will not work continuously but only for a short while when the voltage is applied. Ideally, once you have charged the capacitor you can remove the battery and short the wire. The LC will continue to oscillate.

In reality, you will lose energy by radiation (every tank circuit is a radiator) and oscillations will die a natural death.

How they keep on oscillating USED to puzzle me too! I try to explain the other way.

At the moment the power is applied: consider the potentials at various points. The current in the inductor is increasing and it is inducing a opposing voltage V that makes voltage at the inductor battery junction close to zero. One end of the capacitor gets -V and it induces +V on the other side.

When the current has increased for some time the capacitor is getting charged and soon after the voltage across the capacitor is high enough to drive the inductor the other way. The voltage across the inductor has now decreased to a smaller value.

Now you know what happens.

If you connect any point of this circuit to ground it will not work. That is the reason you need to remove the battery for a proper oscillation.

That capacitor voltage goes upto 2V is an illusion. You need to measure the voltage across the capacitor (what you are actually doing is to measure the voltage across the LC). You cannot use an electrolytic capacitor for this experiment (capacitor voltage swings both ways)

The inductor and wiring are resistors that waste some of the stored energy by making heat so the LC rings and stops ringing soon.

Another way to see it is to understand how an inductor beahave.
An analog behaviour would be the inertia force of a mass object. When you start to move your car it's difficult at the beginning because of the weight of the car. So the speed will increase little by little. Then when you want to stop the car,the speed will also decrease little by little.
The same thing happens in an inductor. When a voltage is applied to an inductor, it stores magnetic energy little by little, but when the voltage is turned off, the inductor will return it's stored energy little by little.
So in your LC circuit, energy will be transferred each time from inductor to capacitor and then back to the inductor.
https://en.wikipedia.org/wiki/LC_circuit#/media/File:Tuned_circuit_animation_3_300ms.gif

In reality LC circuit doesn't exist, they are RLC because of the résistance of the wire. If you add à series résistor in your circuit, a small amount of energy will be dissipated throught the inductor at each tranfer of energy.
Another thing: The résonnance in your circuit is not ignitiated by the DC voltage but by the rising pulse when your turn on the DC voltage. An LC tank reacts to a voltage variation and not to a DC source. Why? Because inductors stores energy only from varying current, this is why u=Ldi/dt.
You can see the LC tank as an energy storage. Idealy, when a varying voltage is applied to it, it stores energy endlessly. But in reality the energy is dissipated in the series résistor.

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Another thing: replace your DC source with an AC source of 1V and at frequency of 503Hz.
What happens? you excitate your LC tank at résonnance. So each variation of your source accumulate energy in your LC tank.
How is it possible? An analog way to see it is a baby on a swing. If you push the baby at the right frequency, the swing will go higher and higher each time you push. Same thing with the LC tank, you "push" it at the right moment, so it it stores more and more energy.

Audioguru said:

The inductor and wiring are resistors that waste some of the stored energy by making heat so the LC rings and stops ringing soon.

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In an ideal experiment it is permissible to consider a circuit without resistance and under such conditions the LC should oscillate indefinitely. No!

The tank is also a radiator and it loses energy by radiation and there is no theoretical escape from that. You can make the resistance go to zero by using superconductors but you will not be able to stop radiation from the tank circuit.

I though that radiation appeared only in resonnance and that out of resonnance there was only heat losses (DC résistance+Proximity effect+skin effect+eddy currents from environment).

SK245230 said:

You can see the LC tank as an energy storage. Idealy, when a varying voltage is applied to it, it stores energy endlessly. But in reality the energy is dissipated in the series résistor.

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Consider an ideal pendulum with a period of 1sec. As it is without any friction (ideal approximation), it will continue to oscillate indefinitely.

Now put an unit charge on the bob of the pendulum: as the pendulum moves, the charge also moves and it produces a current and the current produces a magnetic field and the bob becomes a source of electromagnetic radiation.

The electromagnetic radiation causes the bob to lose its energy and slowly it comes to a halt (inspite of having zero mechanical friction). The electromagnetic radiation will have a frequency of 1Hz.

When you say, that it comes to a halt and that the radiation will have a frequency of 1Hz. Do you mean that it will stop oscillating or it will continue to oscillate at 1Hz?

SK245230 said:

When you say, that it comes to a halt and that the radiation will have a frequency of 1Hz. Do you mean that it will stop oscillating or it will continue to oscillate at 1Hz?

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This is an example of an electric generator; the bob (and the pendulum) will come to rest soon (well, you may have to wait for a long time). The amplitude will keep on decreasing and the intensity of the radiation will also decrease (but not the frequency) and the amplitude will asymptotically approach zero or the rest point. And all this will happen without any mechanical friction and just because of the charge placed on the bob of the pendulum.

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SK245230 said:

I though that radiation appeared only in resonnance and that out of resonnance there was only heat losses (DC résistance+Proximity effect+skin effect+eddy currents from environment).

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There are two terms in radiation: spontaneous emission and stimulated emission. Both are non-zero but can vary widely. We are however discussing only theoretically...

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SK245230 said:

An analog way to see it is a baby on a swing. If you push the baby at the right frequency, the swing will go higher and higher each time you push. Same thing with the LC tank, you "push" it at the right moment, so it it stores more and more energy.

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What will happen if you push the swing slightly off the "right" frequency? Assume that the right frequency is f0 and you push the swing at f and sweep f (just for fun) between f0-delta and f0+delta. You will get the Q curve (absorption) and the dispersion (phase) and they will be certainly interesting. Just keep the coupling low (push only very little every time) and you can see the second order effects too.

But then I don't get it, what force will force attenuate the pendulum. Because if I make the sum with the Laplacian force it won't attenuate it. (Maybe my schematic is false I haven't done this awhile)

Of course in real life there is small amount of resistance and my circuit is RLC, but was referring to ideal circuit. Even though, never thought about radiaton.

And of course the ideal way to make a LC oscillate is in AC at right frequency... but that is obvious.

An other question comes to mind is this one:
If provide AC square wave supply, the LC will be tuned at some frequency, so will take that frequency from Fourier series of the sqare wave. But what happends with the other frequencies of the saquare wave written in Fourier series ? Dissipate in heat ?

A series LC passes frequencies near its resonant frequency and the capacitor blocks lower frequencies and the inductor blocks higher frequencies. Very little heat is created by the blocking.

SK245230 said:

But then I don't get it, what force will force attenuate the pendulum. Because if I make the sum with the Laplacian force it won't attenuate it

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I know, I know, it looks like black magic. The charge moving in its own magnetic field, caused by its own motion, experiences a drag (that opposed the cause- Lenz's rule) that is mainly responsible.

I try my best to avoid drawing (I am not Picasso) whenever possible but your drawing looks ok (but the current should be in the direction of motion- in case of a pendulum it should be perpendicular the long black line) and B is perpendicular to the plane of the paper (correct) and the force (p is also wrongly drawn, IMHO).

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CataM said:

If provide AC square wave supply, the LC will be tuned at some frequency, so will take that frequency from Fourier series of the sqare wave. But what happends with the other frequencies of the saquare wave written in Fourier series ? Dissipate in heat ?

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The swing example is a good example to explain. I shall try my best.

When you excite the swing, you push only a bit when it is near the end (v is close to zero and it can take up energy easily). That means you apply a narrow pulse with the same frequency as the natural frequency of the swing. This is called a pulse train. So that you need not disturb the natural frequency of the swing, you must make the width of the pulse extremely small but must transfer finite energy per pulse. Such a pulse train is called 'Dirac Comb'- see https://en.wikipedia.org/wiki/Dirac_comb. The beauty is that the Fourier transformation of the Dirac comb is also a Dirac comb. In other words, the same pulse train can excite higher harmonics too.

Other frequencies of the square wave are simply not absorbed. Hence the swing acts like a resonance filter- it will pass only the frequency it is in resonance with.

Suppose the excitation frequency is slightly different from the natural frequency of the swing. In such a case you will miss the hit and you will be forced to increase the pulse width of the excitation pulse train so that some energy will be transferred by contact with the swing. This is an example of the forced vibration close to resonance. This gives rise to absorption and dispersion phenomena.

I hope I have been clear.