How exactly do inductors work in an LC circuit?

[Zack K]

I understand Faraday's law and about induced electric fields created by a changing magnetic fields, etc.

But what causes the current to oscillate in an LC circuit, with no battery? If you picture that there is current going into an inductor, and that current is decreasing over time, then you would see a curly electric field being produced in the direction of the current flow. (I uploaded an image for my reasoning)

Also, why would the current decrease at all? If inductors "oppose" change, wouldn't it make sense that the curly electric pushes the current to counteract the decreasing Coulombic electric fields?

 

[willem2]

The decreasing current in the inductor does indeed produce a change in the magnetic field that makes the current continue.
The current however is going into the capacitor and the voltage across the capacitor will increase, and that will make the current stop and then change sign.

 

[Dale]

But what causes the current to oscillate in an LC circuit, with no battery?

$v= L\frac{d}{dt}i$ and $i=-C\frac{d}{dt}v$
so $i=-LC\frac{d^2}{dt^2}i$
and therefore the current oscillates

 

[tech99]

I understand Faraday's law and about induced electric fields created by a changing magnetic fields, etc.

But what causes the current to oscillate in an LC circuit, with no battery? If you picture that there is current going into an inductor, and that current is decreasing over time, then you would see a curly electric field being produced in the direction of the current flow. (I uploaded an image for my reasoning)

Also, why would the current decrease at all? If inductors "oppose" change, wouldn't it make sense that the curly electric pushes the current to counteract the decreasing Coulombic electric fields?

An inductor is rather similar to a mass in mechanics; it has an inertia action. If we try to stop a moving mass, it opposes us with a force which equals the braking force we are applying (Newton). So there is an equilibrium, but it only exists when deceleration is taking place.

 

[Zack K]

The decreasing current in the inductor does indeed produce a change in the magnetic field that makes the current continue.
The current however is going into the capacitor and the voltage across the capacitor will increase, and that will make the current stop and then change sign.

Why would the capacitor the capacitor gain voltage if it was initially charged with no battery?

What I was trying to get at was what drives the current in the opposite direction? There has to be some electric or magnetic field that causes it to reverse. Near the end of one period, the current and the capacitor potential is nearly zero. If it was the electric field of the inductor , wouldn't the field have to be in the same plane as the circuit in order to drive current?

 

[anorlunda]

Why would the capacitor the capacitor gain voltage if it was initially charged with no battery?

Because the current from the inductor also goes through the capacitor. Current in a capacitor makes its voltage change.

Did you understand @Dale 's answer in #3?

 

[Zack K]

Because the current from the inductor also goes through the capacitor. Current in a capacitor makes its voltage change.

During a period, there is current going into the capacitor but that doesn't mean an increase in potential because it matters what the net charge over the capacitor plates are. So that doesn't make sense to me.

Did you understand @Dale 's answer in #3?

Yes I mean I understand but I don't know where it came from. What I want to know is what microscopic aspect of it causes the current to want to suddenly change directions.

 

[anorlunda]

During a period, there is current going into the capacitor but that doesn't mean an increase in potential because it matters what the net charge over the capacitor plates are. So that doesn't make sense to me.

That remark doesn't make sense to me. Current through a capacitor changes the charge on the capacitor plates. "Net charge" is always zero, one plate has an excess of plus the other plate has an excess of minus. When we say that the capacitor is charged, we mean that there is a differential between the plates.

What do you think charged capacitor means?

 

[Zack K]

That remark doesn't make sense to me. Current through a capacitor changes the charge on the capacitor plates. "Net charge" is always zero, one plate has an excess of plus the other plate has an excess of minus. When we say that the capacitor is charged, we mean that there is a differential between the plates.

What do you think charged capacitor means?

I mean of how a capacitor is initially fully charged, and over time the electric field of the capacitor weakens because the charge buildups cancel each other out. So then over time the potential and the current goes to zero.

 

[anorlunda]

I mean of how a capacitor is initially fully charged, and over time the electric field of the capacitor weakens because the charge buildups cancel each other out. So then over time the potential and the current goes to zero.

That sounds completely wrong, but maybe you are thinking of a different scenario than I am. With DC, the current goes to zero, but the voltage stays the same.
$I=C\frac{dV}{dt}$

Perhaps you are thinking about leakage in an imperfect capacitor.

Can you draw a circuit that demonstrates what you're talking about? Use a battery, resistor, and a capacitor. Take a picture of it with good lighting, then use the UPLOAD button to post it.

 

[Zack K]

Can you draw a circuit that demonstrates what you're talking about? Use a battery, resistor, and a capacitor. Take a picture of it with good lighting, then use the UPLOAD button to post it.

What I'm thinking about is a capacitor and a resistor for example. With no battery. Initially at $t=0$ the capacitor is fully charged and has some potential across it's plates. Over time, the current falls of as $I=\frac{emf}{R}e^{\frac{-t}{RC}}$ therefore the potential of the capacitor drops as well. We can see an example here: http://www.falstad.com/circuit/e-cap.html (just imagine that there is no battery and the capacitor has an initial potential $\Delta V$

Coming back to my original question. In this RC circuit, now if we place an inductor in it, this happens.

My question is why does this happen? What force/field makes the current go the other way?

 

[Dale]

I don't know where it came from

The definition of a capacitor is $i= C \frac{d}{dt} v$. In other words, a capacitor is a device where the current through the device is proportional to the change in the voltage across it.

The definition of an inductor is $v=L\frac{d}{dt} i$. In other words an inductor is device where the voltage across the device is proportional to the change in the current through it.

In a LC circuit the current through each device is the same and the voltage across them is opposite, so when you write the expression for the circuit you get $i = -LC \frac{d^2}{dt^2} i$, which we immediately recognize as the equation of an oscillator.

What I want to know is what microscopic aspect of it causes the current to want to suddenly change directions.

All of the changes are gradual. There are no sudden changes.

 

[Zack K]

The definition of a capacitor is $i= C \frac{d}{dt} v$. In other words, a capacitor is a device where the current through the device is proportional to the change in the voltage across it.

The definition of an inductor is $v=L\frac{d}{dt} i$. In other words an inductor is device where the voltage across the device is proportional to the change in the current through it.

In a LC circuit the current through each device is the same and the voltage across them is opposite, so when you write the expression for the circuit you get $i = -LC \frac{d^2}{dt^2} i$, which we immediately recognize as the equation of an oscillator.

All of the changes are gradual. There are no sudden changes.

What field causes the current to change direction?

 

[sophiecentaur]

What field causes the current to change direction?

The Charge on the Capacitor builds up so eventually the volts increase and become greater than the emf induced in the inductor and the current reverses. The process repeats in the other direction etc. etc.
You also say "During a period, there is current going into the capacitor but that doesn't mean an increase in potential" and that is wrong. Current flowing into a capacitor will change the PD.

 

[Dale]

What field causes the current to change direction?

Both the capacitor and the inductor are necessary. It is like asking if the spring or the mass causes the velocity to reverse in a mass-spring oscillator. Without both parts you don’t get oscillation.

That said, at the moment when the current switches from positive to negative (or vice versa) the magnetic field is zero and the electric field is maximum.

 

[Mister T]

Also, why would the current decrease at all?

Something unmentioned is causing the current to decrease. Perhaps a hand is turning a knob, or a battery is discharging.

 

[Mister T]

I understand Faraday's law and about induced electric fields created by a changing magnetic fields, etc.

But what causes the current to oscillate in an LC circuit, with no battery?

Changing electric fields also induce magnetic fields.

 

[Zack K]

You also say "During a period, there is current going into the capacitor but that doesn't mean an increase in potential" and that is wrong. Current flowing into a capacitor will change the PD.

So why does the voltage of an RC circuit with no battery drop to zero over time?

The Charge on the Capacitor builds up so eventually the volts increase and become greater than the emf induced in the inductor and the current reverses

I'm talking about an LC circuit with no battery. To my understanding, a fully charged capacitor discharges and over time the fringing fields of the capacitor decrease because the original charge imbalance on the capacitor balances over time as electrons move from the negative plate, to the positive one until they cancel each other out. But the inductor reverses that?

 

[Zack K]

Something unmentioned is causing the current to decrease. Perhaps a hand is turning a knob, or a battery is discharging.

I know that's not possible obviously, but you took that out of context. Wouldn't it be reasonable to say that this inductor creates an electric field to oppose the decrease in current, and in doing so the electric field never decreases. (I uploaded an image of my reasoning).

 

[Zack K]

Both the capacitor and the inductor are necessary. It is like asking if the spring or the mass causes the velocity to reverse in a mass-spring oscillator. Without both parts you don’t get oscillation.

That said, at the moment when the current switches from positive to negative (or vice versa) the magnetic field is zero and the electric field is maximum.

And you can fundamentally explain that can't you? So the springs are made of some material who's molecules like to be positioned the way they are, and when displaced will exert a force the opposite way to go back to their original position, probably due to how they bind geometrically (I'm not sure). So what fundamental aspect of an inductor makes it want to reverse current?

 

[Dale]

And you can fundamentally explain that can't you?

I already did! Starting from the definition of a capacitor and an inductor and deriving it explicitly for you step by step. What part of the derivation did you not follow? You have not asked a single specific question about it.

Any time you get a function of the form $\frac{d^2}{dt^2}x = -k x$ you have an oscillator. Do you understand that? And if so did you understand my derivation?

So what fundamental aspect of an inductor makes it want to reverse current?

It is not just the inductor, it is the combination of both the inductor and the capacitor, as I showed above.

 

[Zack K]

I already did! Starting from the definition of a capacitor and an inductor and deriving it explicitly for you step by step. What part of the derivation did you not follow? You have not asked a single specific question about it.

I understand your derivation and I can see that it shows oscillation. I'm more wanting to know what's going on microscopically.

$\frac{d^2}{dt^2}x=-kx$. But why? Why does this oscillation occur? Why is there a restoring force? (those are rhetorical questions) All that can be explained to an extent on a somewhat fundamental level. Maybe you have already explained it but I haven't picked up on it.

You can say it in the most layman terms possible.

So an RC circuit eventually loses all it's current. An inductor now in that circuit opposes the change and tries to increase the current (to my understanding). But how exactly does it oppose this change?

 

[Dale]

I'm more wanting to know what's going on microscopically.

I have no idea what you mean by this. The size of the capacitor and inductor is not relevant.

You can say it in the most layman terms possible

Then why did you mark the thread as intermediate? If you want basic answers you should mark your thread as basic. This is highly frustrating

So an RC circuit eventually loses all it's current. An inductor now in that circuit opposes the change and tries to increase the current (to my understanding). But how exactly does it oppose this change?

Current is not a conserved quantity so it makes little sense to say that a circuit loses its current. What it loses is its energy. The energy is dissipated in the resistor as heat. For the inductor the energy is not dissipated but is temporarily stored in the inductor in its magnetic field. So the difference is that a RC circuit loses energy and a LC circuit does not.

 

[jartsa]

So an RC circuit eventually loses all it's current. An inductor now in that circuit opposes the change and tries to increase the current (to my understanding). But how exactly does it oppose this change?

It's the capacitors task to make the current change.Inductor :

1: drains the capacitor smoothly - not short circuiting
2: charges the capacitor to reverse polarity
3: drains the capacitor smoothly - current is now opposite to 1

 

[Zack K]

I have no idea what you mean by this. The size of the capacitor and inductor is not relevant.

Then why did you mark the thread as intermediate? If you want basic answers you should mark your thread as basic. This is highly frustrating

Current is not a conserved quantity so it makes little sense to say that a circuit loses its current. What it loses is its energy. The energy is dissipated in the resistor as heat. For the inductor the energy is not dissipated but is temporarily stored in the inductor in its magnetic field. So the difference is that a RC circuit loses energy and a LC circuit does not.

Sorry, I didn't mean to frustrate you.

I think I understand, but the answer wasn't to what I expected to be about so I was confused.

 

[ronburk]

What an interesting thread -- it's so hard to translate from brain to symbols and back again. I'll take a whack at it, mostly for my own edification. I am not a physicist; barely an EE on a good day.

If you hung a mass from a spring on the ceiling, pulled it down until the spring was fully extended, you wouldn't be surprised to see that the mass not only flies up until the spring is completely compressed, but then reverses direction and continues to oscillate. My theory is that you're asking for the same physical intuition about the LC circuit and how it reverses direction, for which we are not born with any natural intuition. Not being that smart, I try to think about things in the simplest possible terms, preferably basic DC if no calculus is really required.

If I hook a battery up to a capacitor, it initially looks like a short circuit. All the electrons go a runnin' towards one plate, but they can't cross the barrier, so a voltage (measured across the capacitor) starts to zoom upwards and the electrons start going slower and slower as the capacitor voltage gets closer and closer to being equal to that of the battery itself. Finally, the capacitor looks like an open circuit to the battery, or like a battery that has the same voltage as the battery that charged it. If I then short the terminals of the capacitor, it will send a current running in the reverse direction, first a whole lot of current, then less and less as the capacitor's voltage decreases because it is using up its electrons.

If I hook a battery up to a coil, it initially looks like a open circuit, 'cause the moving electrons are trying to cause a magnetic field to build up. As the field gets closer to its maximum value (associated with whatever current the battery is capable of supplying), less work is being done to increase it and the current goes faster, until eventually the coil looks like a short circuit, current is flowing as fast as the battery can supply it. Of course, if I then try to unhook the battery, the current that was supporting that magnetic field stops and the field starts collapsing, inducing a current in the reverse direction (and a big ol' spark that made many a child of the 50's think that electronics might be cool to learn).

So both the capacitor and the coil are capable of storing energy via electrons running in one direction, then releasing energy by causing electrons to run in the opposite direction. We've got the basic ingredients for oscillation.

You're proposing to charge a capacitor up to some voltage, then hook a coil to it. So at the beginning, the capacitor looks like a battery, the coil looks like an open circuit. Electrons start to flow slowly, the mag field of the coil starts to build, the voltage across the capacitor starts to drop slowly as its plate loses electrons. Electrons flow faster, because the coil is starting to look more like a short circuit, the mag field is building, and the voltage across the capacitor is dropping faster. Eventually, things reach their peak. The rising current from the capacitor can rise no more, the mag field can build no more, and when the current starts to decrease (as it must because the capacitor voltage keeps dropping), that means the mag field starts to collapse. That's where the first reversal comes in. The collapsing field supplies a reversal, and soon the electrons are flowing in the opposite direction, charging the capacitor with the opposite polarity that you originally charged it with. One half of the oscillation is complete and things are back roughly the starting point except the capacitor has an opposite charge, so it can supply its next reversal. Of course, the process symmetrically reverses itself, supplying the other half of the oscillation, putting things roughly back where it all started.

Just as the spring and mass oscillate by sloshing energy back and forth between two different storage mechanisms (the energy stored by elevating a mass, and the energy stored by extending a spring), the LC circuit also oscillates by sloshing energy back and forth between two different storage mechanisms: the voltage potential created by charging a capacitor, and the current potential created in a magnetic field by running a current through a coil.

Of course, real inductors and capacitors leak and contain resistances, the Second Law pertains, and things peter out over time rather than becoming a perpetual motion machine, but I imagine you were already clear about that.

 

[NH_EE]

Right after I posted this I finally read Ronburk's post. Except for details like battery vs capacitor and a mass vs a flywheel, we're pretty much saying the same thing.

The original question was about how the inductor works in an LC circuit. Consider what happens over the course of a complete cycle of the oscillation. Consider only an ideal capacitor and an ideal inductor, with no losses. Both the capacitor and the inductor are energy storage elements. The energy stored in the capacitor is E=(C*V^2)/2. Notice that it doesn't depend on the instantaneous current "through" (charging or discharging) the capacitor, only on the integral of the current (dv/dt = i/C). The energy stored in the inductor is E=(L*I^2)/2. Notice that it does not depend upon the instantaneous voltage across the inductor. only on the integral of the voltage (di/dt - v/L). The energy stored in the capacitor is in the electric field. The energy stored in the inductor is in the magnetic field. Electrical resonance involves continual cyclic transfer of energy back and forth between the inductor and the capacitor.

Let's say you have the inductor (I'll call it L) and capacitor (I'll call it C) in parallel, and initially you establish a DC current in L by connecting a current source across the L/C pair. To avoid complicating matters by storing charge in C, you start at zero current and slowly turn up the current so the rate of change of current in L (and thus the voltage across it) are negligible.

Now, instantaneously turn off the current source. L, being an ideal inductor, would maintain that constant current indefinitely if the voltage across it were held at zero. The current path is through C, however, so voltage is increasing across it and thus across L. The current direction in L still the same as it was with the current source operating, which means L is now drawing current from C, so the voltage is changing in the negative direction. This voltage across L causes an increasing negative rate of change in the current.

The moment the current crosses zero coincides with voltage being at negative peak, and thus the rate of change of current being at maximum. All the energy from L has been transferred into C. We have peak voltage and zero current. The oscillation has now completed its first quarter-cycle since you turned off the current source.

The negative voltage continues producing a negative rate of change in current through L (and C), di/dt = v/L. As the current (now negative in L) increases, the voltage (negative) is changing towards zero at an increasing rate, dv/dt=i/C. The moment the voltage crosses zero coincides with the current being at negative peak. All the energy that was stored in C has been transferred back into L. The second quarter-cycle has now completed.

The third and fourth quarter-cycles are just like the first two except for voltage and current polarity and direction.

If you observe the flywheel and hair-spring in a mechanical watch, the strain in the hair-spring is analogous to the current, which drives the acceleration of the flywheel. The velocity of the flywheel is analogous to voltage, which is at maximum when acceleration is passing through zero. Position of the flywheel, the integral of velocity, corresponds to strain in the hair-spring which again is analogous to current.

 

[sophiecentaur]

What I want to know is what microscopic aspect of it

The microscopic aspect of this is not relevant; discussion at that level will definitely not help you at all and I think you are wasting your time looking in that direction. (At least until you have appreciated the classical behaviour of circuit elements).
Maths describes very well what happens and it's been quoted higher up the thread. Nothing (and I mean Nothing) about Electric Circuit theory can be explained without using rigorous Maths. There is no substitute for it. Circuits that are operating at frequencies where the wavelength is many times the dimensions of the circuit can be characterised particularly well using the components that are included in the basic functional diagram - plus the occasional 'parasitic' component that is there because of the physical layout (for instance the Capacitance between the conduction wires or across the junction in a diode, or the Inductance of the spiral wrapping of an off the shelf Capacitor).