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I am having some trouble understanding the behaviour of electrons inside an RC circuit. Let me explain:

  1. One capacitor plate accumulates a lot of electrons, which pushes electrons away into the negative/ground from the other plate.
  2. I disconnect the charged capacitor.
  3. I insert it into the LC circuit.
  4. Because of the attraction of the electrons on the one plate to the electron-deprived protons on the other plate, they start to "travel" in that direction through the wire.
  5. The electrons hit the inductor, which creates a magnetic field due to the change in current, which pushes back against the electrons, but lets some pass through.
  6. Now the discharge of the capacitor has completed and the magnetic field in the inductor collapses and causes electrons from the wire to be pushed towards the (previously) negative plate.
  7. The plates have now switched and the same amount of electrons that were on the one plate before are now on the other.

Questions:

  • Regarding 4) Since there is a lack of electrons on the one plate, won't it simply attract free electrons from the wire (that is connected between the end of the inductor and the "negative" plate) to fill the "holes"?
  • Regarding 5) As I understand it, there is always some electrons passing through, it's just the number that changes?
  • Regarding 6) The plate that is being charged by the inductor (the initially "negtative" plate) should already have a slightly positive charge from the current that passed through the inductor while it was building the magnetic field. But this would only be true if my assumption about point 4) is true (the first question).

Could somebody try to clarify this, and tell me if my assumptions are correct?

EDIT: My main question is how the overcharge from the one plate can create an overcharge on the other. Since the amount of positive charge on the one plate of the capacitor should match the amount of negative charge on the other, shouldn't they just cancel each other out once the discharge is complete? How does the inductor in between influence the current so that it ends up in a net positive charge? Are the electrons that were displaced by the charge still there to refill the space?

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First, there are so many free electrons in metal compared to the number of electrons involved in a given cycle of current that individual electron behavior has a minuscule effect. It's better to think of "charge" as a continuous amount of stuff, and "current" as a flow. There are places where this isn't true (dim-light photodiodes come to mind), but this isn't one of them.

Regarding 4, yes, the electrons get pulled into the capacitor's positive plate. But the lack of electrons pulls in more electrons from further down the wire, etc., and current flows. The "signal" that starts the current flowing propagates down the wire at roughly the speed of light, so unless your circuit is smaller than about 1/10th of a wavelength at your frequency of interest, you can just assume that the current starts flowing simultaneously, everywhere.

Regarding 5, yes, but you're confusing yourself by trying to count electrons.

Regarding 6, I'm not sure where you're coming from. In the circuit as you describe it, the current in the coil will reach its maximum value when the voltage on the cap is zero. This happens when the charges on the plates are balanced. Because the current is moving in the coil it has built up a magnetic field, and that energy gets returned to the capacitor in the form of a voltage.

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  • \$\begingroup\$ Thanks, that cleared it up a bit. The main thing I still don't understand is how the overcharge from the one plate can create an overcharge on the other. Since the amount of positive charge on the one plate of the capacitor should match the amount of negative charge on the other, shouldn't they just cancel each other out once the discharge is complete? How does the inductor in between influence the current so that the charge can be positive on one plate instead of just equal to the other plate? \$\endgroup\$ – pewlowz Jul 29 at 20:47
  • \$\begingroup\$ Because the magnetic field which the inductor creates, is polarized. Supplying the current to create that field, creates the field in one of two possible orientations. When that current is exhausted, the magnetic field begins to collapse. A collapsing magnetic field creates an opposite electrical field, which forces current into the capacitor in reverse. Thus the opposite plate of the capacitor gets "charged" and the cycle repeats. \$\endgroup\$ – rdtsc Jul 29 at 21:48
  • \$\begingroup\$ @rdtsc thanks for your answer. My trouble is with understanding how there can be a surplus at all in the end. Let's say I have plate A at +5 and plate B is at -5. Doesn't the charge from Plate A simply fill the 5 holes in Plate B and then we land at 0? How can there be more than zero when the inductor has finished discharging? \$\endgroup\$ – pewlowz Jul 29 at 21:58
  • \$\begingroup\$ Think of the current (not the electrons) as rushing through the coil with some momentum. It takes them a while to stop. So when the capacitor voltage is at zero, the current through the coil is high -- and it doesn't want to stop. So it starts jamming charge into the capacitor in the reverse polarity from what it was before. \$\endgroup\$ – TimWescott Jul 29 at 22:01
  • \$\begingroup\$ @TimWescott I guess I'm having trouble understanding the role of the coil. Does the coil just delay the displacement of charges and thus enable the oscillation in the first place? Wouldn't that also mean that just a capacitor discharging into itself should cause the same behaviour just at a much higher frequency? \$\endgroup\$ – pewlowz Jul 29 at 22:13
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The LC circuit is analogous to a swinging pendulum. When you charge the capacitor, it is equivalent to pulling the pendulum to one side.

When you connect the capacitor to the coil, it's equivalent to letting go of the pendulum. It swings toward the resting position, but when it gets there, it has nonzero velocity, and this momentum carries it to the same height on the other side.

The kinetic energy of the pendulum bob is equivalent to the energy stored in the magnetic field coil by the flow of current. This flow continues until the charge on the capacitor is high enough to stop it — and this charge results in an equal but opposite voltage from what you started with.

As soon as the current stops flowing in the first direction, the new charge on the capacitor starts pushing it in the other direction, just like the pendulum bob immediately starts swinging back toward you.

In the LC circuit, you are continually trading energy stored in the electric field in the capacitor for energy stored in the magnetic field of the inductor. With the pendulum, you are trading the potential energy of lifting the bob in a gravity field for the kinetic energy of its motion at the lowest point of its swing.

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  • \$\begingroup\$ What would be the equivalent of the "nonzero velocity" in the circuit? \$\endgroup\$ – pewlowz Jul 29 at 22:16
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    \$\begingroup\$ @pewlowz nonzero velocity is nonzero amps. The higher the current, the faster the flow velocity. The number of electrons in the wires doesn't vary, only their speed is changing. When current hits zero, the electrons halt in place. The wires are always full of electrons. Electrons can move along, but in wires their number doesn't change. Circuits are like leather belts, where amperes is the belt-speed. AC is when the entire belt goes back and forth, as a unit. DC is when all the electrons rotate along as a unit, like a flywheel or drive-belt. \$\endgroup\$ – wbeaty Jul 30 at 1:29
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My main question is how the overcharge from the one plate can create an overcharge on the other.

I think I understand your question. The initial conditions are with the capacitor fully charged (all energy in the system is present as potential energy on the capacitor and not as potential energy in the inductor) prior to connection into an ideal capacitor/inductor "tank." I believe you want to know why the capacitor ever manages to get an oppositely arranged charge.

It's not terribly complex (if I skip over some details, anyway.) This all takes place in four phases. Let's put up a schematic to simplify the discussion a little:

schematic

simulate this circuit – Schematic created using CircuitLab

Note that I've "grounded" one end of things and just labeled the opposite side \$V\$. There is also a current in the inductor that I've labeled \$I\$.

Phase 1: At the start of phase 1, let's say the capacitor's voltage is at its maximum magnitude, called \$+V_\text{MAX}\$, and the inductor's current is \$I=0\:\text{A}\$. As you probably already know, the capacitor attempts to discharge itself through the inductor but the inductor limits the rate of change in current. So the current starts at zero and gradually builds up over time, allowing the capacitor to discharge -- but slowly at first and then increasing as time proceeds. Meanwhile, as the capacitor discharges, the remaining voltage decreases and this gradually lowers the rate of increase in the inductor current. That current still increasing in phase 1, but as the capacitor discharges the rate of increase is at an ever-decreasing magnitude. Eventually, the capacitor's voltage is zero and the rate of change in the current of the inductor reaches its peak value, \$I=+I_\text{MAX}\$.

Phase 2: The inductor now has reached \$I=+I_\text{MAX}\$ right at the point where the capacitor's voltage is \$V=0:\text{V}\$. At this point, all of the potential energy stored in the capacitor at the start of phase 1 is now present as potential energy stored in the inductor. Now, the inductor's current can't just stop. So it continues and, as it does so, it starts charging the capacitor, again. But you have to remember that since the current is still positive in the inductor, the change in the voltage on the capacitor continues in the same direction as before -- it continues to go in the negative direction. So now the capacitor's voltage at \$V\$ is moves into negative territory. (It's just continuing its prior change, which was "declining.") As the capacitor charges towards it's reversed maximum voltage, now, there is a increasingly negative voltage at \$V\$ and this reverses the sign of the current change in the inductor, so the inductor's current declines now and moves towards zero. Eventually, the capacitor's reversed voltage reduces the inductor's current completely so that \$I=0\:\text{A}\$, again. But now, at this point, the capacitor's voltage is \$V=-V_\text{MAX}\$.

Phase 3: This is just like phase one, except the capacitor discharges oppositely through the inductor, changing the direction of the growing inductor current, \$I\$.

Phase 4: This is just like phase two, except again that the signs are reversed.

At the end of phase 4, you are right back at the start of phase 1.

That's the qualitative view. The quantitative one would have some mathematics involved. But the basic idea doesn't change.

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  • \$\begingroup\$ I guess my issue is more conceptual. The simplest way to express it for me: Let's say we have 10V across the capacitor in the beginning. So +5 on one plate, -5 on the other. Shouldn't that amount to 0, like when you discharge a capacitor into itself without an inductor? Why don't the charges cancel each other out? \$\endgroup\$ – pewlowz Jul 29 at 22:53
  • \$\begingroup\$ @pewlowz You are creating in your mind some kind of invisible "zero" that exists. Perhaps in the middle of the capacitor??? It's a lot easier if you stop trying to create artificial complexities where none are needed. (Where exactly in the schematic that I wrote out for you would you find this "zero point" located?) Another way to say this, though, is that you can just call the "-5" as if it were 0. This would make the +5 instead a +10, right? It's all relative. Nature doesn't understand absolutes, just differences. I think you need to break yourself of this artifice that you are stuck on. \$\endgroup\$ – jonk Jul 29 at 23:05
  • \$\begingroup\$ I would say 0 means an equal number of electrons and protons on both plates. Suppose we have 100 electrons and 100 protons on each plate when not charged. We charge it, and one plate has 180 electrons and 100 protons and one plate has 20 electrons and 100 protons. When connected 80 electrons should be rushing over to equalize the charge back to 100 electrons and 100 protons on both plates. No voltage exists and nothing should happen. Now the coil seems to prevent that somehow. And i dont understand how it does that. \$\endgroup\$ – pewlowz Jul 29 at 23:37
  • \$\begingroup\$ @pewlowz So your concern is about why the electrons can't just "rush over" but instead must "take their time getting there" because of some mysterious (to you) property of the inductor? Is the only question you have about why an inductor can impede an instantaneous neutralization of both plates on the capacitor? \$\endgroup\$ – jonk Jul 29 at 23:50
  • \$\begingroup\$ Yes, that is what I'm trying to figure out. Because this is what causes the oscillation, no? If the capacitor were to neutralize as it does without the coil, nothing would happen. \$\endgroup\$ – pewlowz Jul 29 at 23:54

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