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LC natural response intuition 2

We can predict the shape of voltage and current in an LC circuit by tracking the motion of charge as it flows back and forth. Created by Willy McAllister.

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  • male robot donald style avatar for user usmanmuhammad9422
    You said in this video at time that the current will continue to flow in the same direction until it reaches 0. What I don't understand is that how come the current continues to flow even though there is no voltage around the capacitor.
    (5 votes)
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    • purple pi purple style avatar for user APDahlen
      Hello Usman,

      There is energy stored in the inductor. This energy is defined by the flow of current. Its as if the inductor has momentum. Once current is flowing the inductor will do everything is can to keep that flow constant..

      You are correct, the capacitor is "empty" is has no energy whatsoever. But at this moment the inductor has all of the energy and it is doing everything it can to keep the current constant. Consequently, the capacitor is driven negative.

      Regards,

      APD
      (9 votes)
  • mr pink red style avatar for user manavpandya31
    The current and voltage versus time graph will keep going on like we plotted it. Does that mean in ideal circuit, current will always keep flowing in this circuit ? What happens in real circuits ?
    (2 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      The ideal LC circuit rings forever. There is no mechanism for energy to leave the system, so it just keeps going.

      In real circuits there is always some small resistance in the circuit. That changes it from an LC to an rLC. Every time charge passes through the little resistor a little bit of energy is lost (the resistor warms up a little bit). Eventually current and voltage die out to zero. The RLC circuit is covered in the next set of articles and videos.

      A good analogy to an electrical LC circuit is a bell. If you strike a bell it rings with a pure-sounding sine wave tone. It will ring for a very long time until the vibrations give out due to mechanical resistances in the system.

      I suggest you take a look at these revised and improved articles on LC and RLC natural response. https://spinningnumbers.org/t/topic-natural-and-forced-response.html#lc
      (5 votes)
  • starky seed style avatar for user Hazem Hesham
    Excuse me at , , I thought the main reason there's a current is the charge's higher concentration on one plate of the capacitor , once some charges flow to the other plate and the have equal charges , the flow should cease .
    But How exactly does the inductor do to keep a steady current flowing ?? this is so strange to hear do you have further detailed explanation ?
    (3 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      Once some charge flows to the other plate and the plates have equal charges, the voltage becomes zero (as shown in the sketched response vs. time), but the current does not go instantly to zero. Think about the pendulum analogy for the LC circuit. When the bob (the pendulum's weighted end) swings back to bottom dead center, it's position is back to 0, but the velocity is not 0 (the pendulum does not suddenly stop when it gets to the bottom of the swing, it keeps going.) In the pendulum analogy, the position corresponds to voltage and the velocity corresponds to current.
      (2 votes)
  • orange juice squid orange style avatar for user ConnorBecz
    Is this circuit kind of like an AC power source?
    (2 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      The LC circuit is similar to an AC power source in that the both create sine waves. They differ in that the AC power source actually delivers power to whatever it is connected to, whereas the LC circuit can't deliver power in a sustained way. If you connect the LC to something the stored charge immediately drains off and the sine wave stops.

      Think of the LC as like a pendulum swinging back and forth. If you leave it alone it keeps going, but if you touch it the pendulum stops.

      The AC power source is more like a bicycle being pedaled by a strong athlete. The back wheel goes round and round. If you try to stop it with your hand it will keep going due to the strength of the rider.
      (4 votes)
  • aqualine ultimate style avatar for user Rob Cyr
    So here we start with a capacitor that is already energized somehow, and an inductor that has zero stored energy, and what follows in the video makes sense.

    In my head I've wondered about what would happen if the inductor already had some energy stored in it as well. In particular, (especially since there's no resistor in the circuit) what would happen if we had energy stored in the inductor pushing current clockwise, and the energy stored in the capacitor is trying to push that same current counterclockwise?
    At that point are we allowed to use superposition to figure out which one would "win"?
    (2 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      If the inductor starts with some stored energy, that means there is a current flowing in it. At in the video is where I sketch the starting current. In the video, i(t<0) = 0. If the current is something other than 0, this is where you would sketch in the non-zero value. The rest of the intuitive process goes on from there. You will get the same sine wave shape for current. It will just start at a different point on the sine curve.
      (2 votes)
  • piceratops ultimate style avatar for user xiaohui Liu
    At , I thought the voltage of the capacitor decreases instead of the voltage of the inductor therefore di/dt should not be negative?
    (2 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      After the switch closes, there's only one voltage (v) in the circuit, across both the capacitor and the inductor. When the capacitor voltage goes down, so does the inductor voltage, because they are connected to each other.

      At the point in the video you are thinking about, , the capacitor voltage has dropped a little, to a smaller positive value. Looking over at the inductor, we see that if the voltage is smaller (but still positive), the slope of the current, di/dt, has a smaller (but still positive) value. That's why the current sketched in the plot leans over a little to a less positive slope.
      (2 votes)
  • leaf green style avatar for user Tae Il Oh
    someone convince me!
    not sure why v started from v not since the inductor will prevent a sudden leap voltage.
    So I thought that the voltage would gradually become v not.
    and I know I'm wrong.. somebody convince me somehow
    (2 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      An inductor prevents sudden leaps of current (not voltage). It is the capacitor that prevents leaps of voltage. The current in the inductor is 0 before the switch closes. After the switch closes, the voltage from the capacitor instantly appears across the inductor, and the inductor current gradually rises from 0 to some value.
      (2 votes)
  • duskpin tree style avatar for user SnappyRiffs
    Can you also talk about RL(resistor inductor) circuits?
    (1 vote)
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  • leafers ultimate style avatar for user Ngoc Dat Pham Nguyen
    How about B and B zero ? What does it affect in the LC circuit?
    (1 vote)
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Video transcript

- [Voiceover] We've been working on an intuitive description of the natural response of an LC circuit, and in the last video, we got everything set up, and now, we're ready to close the switch. Let's close our switch. And now our switch is closed again, and what happens? Well, we had V naught. We had some voltage on our capacitor, and that means now we have some voltage on our inductor. That's what the switch closer did. So all of the sudden, our inductor finds itself with some voltage across it, so let's look at this equation here, the inductor equation. And it says that V naught, that's the starting voltage from the capacitor that we borrowed, is L times the slope of the current. Alright, so now we have a finite voltage, and then we have a finite value on L, and that means that DI, DT is some number, and if we put positive charge up here, we know that V naught is positive. So let's sketch that. Let's start a sketch of what's going on here. We're gonna plot together on the same plot the current in blue, and we'll put the voltage in orange, and this is the time axis here. So, let's go back and check. What was our voltage when we started? Our voltage was V naught when the switch closed. So I'll mark that out, and what was the current when we started? The current was zero, so here's a zero current down here. And now we're gonna sketch in what we think happens right after the switch closes. So we decided that there was gonna be some kind of a DI, DT. DI, DT equals V naught over L. I'm just gonna sketch this in. There's gonna be some rising slope. There's gonna be some positive slope on the current. Alright, so we took our first step. Now, the other thing that's happening here is that Q, what is Q doing? Let's track our charge. Q is actually gonna start flowing out of this guy. This is the current that we're talking about. DI, DT, this is a finite current, and that current is made of the charge that's in the capacitor, so that current starts flowing around and running around to the bottom side of the capacitor, right? That means the amount of charge on the top is going down, which means that the voltage is going down. So let's sketch in some voltage. The voltage is gonna start going down because the charge is leaving the capacitor. Alright, so far, so good. So let's keep going back and forth between the inductor and capacitor, and see how the voltage and current changes, and the way we do that is we just look at the equation we had. Now, the voltage on the inductor is not V naught anymore. It's changed to DI, DT equals whatever V is over L, and we notice that V went down a little bit. So that means what? That means the slope of the current went down a little bit. So let me make a little shallower version of the slope here. Alright, now what happened next to the voltage? A little more charge has left the capacitor, headed over to the opposite plate, so the voltage is gonna continue to go down. So now we're just gonna edge forward moment to moment and see what happens. Every time the voltage goes down, the slope of the current gets a little shallower, right? And then the voltage continues to drop because the current's leaving, and eventually, that voltage is gonna reach zero, and at the point the voltage reached zero, the slope of the current is zero. The slope of the current is zero. So right when this happens, the slope of the current is zero. Now, the value of the current is some number. Alright, so now we have some value of current. We have a zero voltage. Now, when we have zero voltage, let's go back up here. When this value of voltage is zero here, when this guy is zero, that means that Q is zero, and what that means is that the charge on either side of the capacitor is equalized. The same charge exists on the top and the bottom of the capacitor. That's what this means here. So in our next moment, what's gonna happen next? Does the voltage go flat? Does it go flat sideways? Does it go up? Does it go down? What does it do? What does the current do? Let's figure it out. So I see we have a finite current. Okay, what that means is the charge is continuing to flow in what direction? In this direction here, charge is continuing to flow like this, and what's happening now is we're getting excess plus charges flowing on this side of the capacitor. So we're actually building up positive charge on the bottom plate of the capacitor, and what that means is, the charge is reversed, so the voltage is reversed, and that tells us that we're gonna have a negative voltage. It's gonna start to grow here. I have a negative voltage. I go back to my inductor equation. This value of V right here, this is now negative, so that means the DI, DT has a negative slope, and negative slope looks like it's sloping down like this. It starts to roll over and slope down, and a little less current means a little less slope on the voltage. Now this positive current's gonna continue to keep flowing. It's gonna continue to deliver charge to the bottom edge, to the bottom plate of the capacitor. At some point, that current will actually go through zero again, and if we look at this situation, this is where we started from. This is almost the situation we started in. There's zero current, and there's a-- This time it's a negative voltage. We started at a high voltage, and we went to a negative voltage. This voltage here happens to be minus V naught, and the reason is is that all the charge that was on the top now has made a trip all the way around to the bottom. So the situation is perfectly reversed from when we started. Now, because of that symmetry, we're not gonna make this video any longer. I'm just gonna say that this situation will repeat itself. Basically, the voltage will curl back down like that, and the current will do this. Woops, let's just keep sketching in. So we've repeated the same story that we did at the beginning part of the curve over here, just with the voltage sign reversed. And now we reach another state. Now we have zero volts and we have a negative current here, and what we look at is we duplicated this situation here, just with the opposite sign of the current. There's positive current, and here's zero voltage. Now we have zero voltage and a negative current, so the whole story I told over here is gonna be in reverse. This charge now is gonna reverse direction. Let's take out all our arrows, and all the charge we had is gonna start flowing back around this way like that. This positive charge that we had is gonna flow around and meet up with the negative charge that we accumulated up here, and that's what's going on in this part of the cycle, and we can continue on. Basically, this is gonna go the same exact story like that, and this, the current, will do the same, and we began. We've come back to the same exact point as the starting point. Now we've replicated our starting point. So what's gonna happen next? Well, what happens next is the same thing that happened in the beginning, and this curve actually repeats itself as long as we let this circuit sit here like this. So I don't have to continue with the story. It's basically a repetition of everything that happened here, and what we're gonna get is a rocking motion on the voltage, back and forth, and we're gonna get a rocking motion on the current, back and forth, as these energy storage elements change stored charge for energy in the magnetic field, and back to stored charge, and then it rocks back and forth like that. So this is our intuitive explanation of what happens in an LC circuit when we let it do its natural response. There'll be this rocking motion back and forth, and now in the next video, we're gonna go through and see how this comes out mathematically.