# Maximazation of current in LC circuit

1. Oct 2, 2011

### miko1977

Hello guys

This is my first post and i am glad to be a member of this forum. There is something bugging me for alot of time. I have a problem understanding why the current of an LC oscilator circuit is maximized when the capacitor is fully discharged. Of course when it is explained from the energy perspective it makes sence since the energy of th circuit is transformed totally to magnetic energy. But what about the dynamical approach of the problem. Why the electrons drift velocity maximizes (and so the current) when the capacitor is discharged???
I know that when the switch is closed there are two electric fields in the conductors, the electric field due to the charged capacitor and the opposing induced electric field from the coil. What is the relationship between these fields? Which is bigger? I suppose the electric field of the capacitor, that is why the current increases. I am not sure. Please help!
Minas

2. Oct 2, 2011

### Philip Wood

Are you able to set up the equation:
$-L\frac{dI}{dt}= \frac{Q}{C}$ ?
Here, I is the current, which is the rate of increase of charge ±Q on the capacitor plates (that is the rate of transfer of charge through the inductor from the +plate of the capacitor to the - plate). So $I=\frac{dQ}{dt}$
So $-L\frac{d^{2}Q}{dt^2}= \frac{Q}{C}$
The general solution to this equation is $Q=Q_0 sin(ωt + ε)$
in which $ω^2=\frac{1}{LC}$
So $I =Q_{0}ω cos(ωt + ε)$

3. Oct 3, 2011

### miko1977

I am aware of the approach, you are essentially using the first Kirchoff rule which is based on thw conservation of the charge. I need an explanation of what happens dynamically. Why the electrons reach their maximum drift velocity when the capacitor is discharged?

4. Oct 3, 2011

### Philip Wood

Hallo Miko. I posted in error - the post wasn't complete. I was going to point out that we have a perfect electrical analogue to a mass-spring system doing SHM. The form of the equations is identical.

The displacement x (from equilibrium) in the mechanical system is analogous to the charge Q on one capacitor plate. When x is zero, the mass is in the middle of its swing, and dx/dt is at its greatest. When Q is zero (i.e the capacitor is discharged), dQ/dt (i.e. current) is at its greatest.

You can go a step further and regard the inductor as like the mass in the mechanical system, and the capacitor as like the spring. When Q = 0, the spring is neither stretched nor compressed, but the inductance (inertia) keeps the current flowing, so the capacitor starts charging in the other direction. Dropping the analogy, the inductor produces a back-emf which opposes change in current, so the current doesn't suddenly cease when the capacitor is empty.

Some of this is a bit hand-waving. Hope it helps.

5. Oct 3, 2011

### miko1977

Hey Philip

The analogy is always helpful and i am aware of it. But it doesn't give a realistic explanation of why the drift velocity of the electrons maximizes when the capcitor is discharged. I am still in dark.

Thanks anyway
\m/

6. Oct 3, 2011

### f95toli

The drift velocity of the electrons has nothing (well, very little) to do with the amount of current flowing in a circuit. It is better to forget about electrons altogether when trying to understand what happens at the circuit level.

7. Oct 3, 2011

### Philip Wood

I'm sorry to have to say that the only response I can give (other than the mechanical analogy) is to refer you back to the equations I gave in my first post on this thread. Do you understand where they come from, i.e. the laws of electromagnetic induction, the concepts of emf, pd, inductance and capacitance? If you do understand, then you should see where the equations come from and how they lead inescapably to the current being a maximum when the charge on the capacitor is zero. If you don't understand, do say...

The drift velocity is, of course, proportional to the current. But if you're looking for an explanation in terms of free electrons, we can regard the emf induced in the inductor as work done per unit charge on the free electrons in the wire, which implies the existence of forces on them, due to an electric field arising – Faraday's Law - from the changing flux through the coil.

8. Oct 3, 2011

### miko1977

hi f9toli
I don't get it when you say that the current has nothing to do with the electrons drift velocity. From what i know it is proportional to the current. I know that viewing a circuit at a microscopic level (electrons) is not the most efficient way but then again it is something i need to clarify for personal reasons.
Minas

9. Oct 3, 2011

### miko1977

Yes Philip
I do understand your explanations very well and i know that the results are unquestionable. But i am taking this different microscopic approach on explaining what happens. It is something that i always try to do when dealing with a phenomenon. This bugs me for alot of time!! Maybe what i am searching is a "mechanical" explanation of what happens to the electrons in the conductors similar to what happens to a body attached on a spring.

10. Oct 3, 2011

### Philip Wood

What did you make of my second paragraph, where I talked about forces on free electrons in the conductor? Isn't that style of explanation along the lines you wanted? There are also Coulomb-type attraction and repulsion forces due to the charges on the capacitor plates. Consider both these forces, correlate them with the equations, and surely you've got what you wanted! If not, I'm baffled.

I suppose you might want a mechanical picture of what exerts these forces. In that case, you're back in the wonderful nineteenth century world of 'ethers' made of spinning vortices, and so on. Superseded for good reason, I'm afraid.

11. Oct 3, 2011

### miko1977

what you explained in your second paragrapgh (faradays law, induced electric field, work done on free electrons e.t.c) i have mentioned in my first post also. Which means that i am already aware of all that. You suggest me to consider both forces and correlate them. Well that is my question from the begining (if you read carefully my first post). An analysis based on these forces that proves that the electrons are accelerating until the capacitor is fully discharged.

12. Oct 3, 2011

### f95toli

This question comes up quite frequently and there is (unfortunately) no easy answer to what happens at the microscopic level, even the simplest explanation of what happens in a real material require a fair amount of quantum mechanics (and that does not include scattering etc).
Anyway, note that the average drift velocity in a given direction of electrons in a typical metal is only a few cm/s, whereas the speed of an electrical signal is equal to the speed of light in that metal (for e.g. copper about 0.8c). Hence, for a typical circuit (excluding things like ballistic transport, single electron pumps etc) the two are not directly related. Note also that electrical transport even for free electrons is a collective phenomenon, meaning you can't really think of current as (number of electrons)/s either (unless you are talking about something like a CRT).

13. Oct 3, 2011

### Philip Wood

Miko. Well that's certainly put me in my place.

I leave you with (1/C)vL$\ddot{v}$ = 0.

in which v is the drift velocity.

And this: Q/C and -LdI/dt are the line integrals along the wire of the Coulomb and Faraday E fields. Assume these fields act parallel to the wire and are the same in magnitude all the way along it. Hence you have the relative magnitudes of the two fields.

Last edited: Oct 3, 2011
14. Oct 3, 2011

### miko1977

Now we are getting somewhere. Thanks allot man!