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RC natural response - intuition

We put some charge on a capacitor in an RC circuit and observe what happens to the voltage and current. This is called the natural response. Created by Willy McAllister.

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  • leaf green style avatar for user sabarishsankar97
    how the voltage across capacitor is vo? Some amount has to drop across resistor isn't it?
    (10 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      With the switch in the initial position (connected to the battery) for a long time, a long time ago a current flowed that charged up the capacitor to Vo. Now consider the time in the recent past, just before we flip the switch. The voltage on both ends of the resistor is the same, Vo. The voltage across the resistor is Vo - Vo = 0 volts. Ohm's Law tells us the current through the resistor is i = V/r = 0/R = 0 amps. So just before we flip the switch, the initial conditions are Vc = Vo, Vr = 0, and i = 0.
      (7 votes)
  • blobby green style avatar for user nurulainsimbolon.rizal
    does frequency response and natural response the same thing?
    (5 votes)
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    • purple pi purple style avatar for user APDahlen
      Hello Nural(a),

      Not exactly.

      Let's use the analogy of a wine glass. Ref: https://www.youtube.com/watch?v=BE827gwnnk4

      The "natural response" term is easy. This is the frequency that the wine glass naturally "rings" at. In the video this is the frequency that ultimately destroys the glass. Know that the wine glass analogy is applicable to electronics circuits containing resistors, inductors, and capacitors. The difference is that the electronic circuits don't blow up... Usually...

      "Frequency response" generally implies a chart that describes how the circuit responds to a broad range of frequencies. In the video this is like changing the frequency and noting how the glass responds to each frequency. For example, the glass does not respond to low frequencies but responds greatly to the natural frequency. One way to present this information is to use Bode plots. Ref

      https://en.wikipedia.org/wiki/Bode_plot

      Please leave a comment below if you have any questions.

      Regards,

      APD
      (8 votes)
  • eggleston blue style avatar for user dena escot
    during discharge , the voltage across the capacitor will not be reversed?

    and why does the charge on the upper plate will go to neutralize with the negative charge in the lower plate?
    (3 votes)
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    • old spice man green style avatar for user Willy McAllister
      With the RC circuit the voltage on the capacitor does not reverse (does not change sign). At the outset, the battery places a positive charge on the top capacitor plate. That means there is a charge difference between the top and bottom plate. That charge difference, q, causes a voltage to appear on the capacitor, v = q/C.

      When you throw the switch the battery is removed from the circuit and there's a new path created that allows charge to flow from the top plate, through the resistor R, and onto the bottom plate.

      Over time, there's less and less excess charge on the top plate, and more and more charge on the bottom plate. Eventually there is an identical amount of charge on both plates, which means there is no "excess" charge anywhere. When that happens the voltage on the capacitor falls to v = 0/C = 0.

      At that point there's no more charge on the top plate that could possibly flow to the bottom plate and make it take on a negative voltage. The capacitor voltage never reverses.
      (3 votes)
  • boggle blue style avatar for user Aidan
    So do you have to review this to yourself or do you remember it all? If you do, that is impressive because I got like 30 something pages of notes..I can't remember everything but I am getting the basic concepts.
    (2 votes)
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  • blobby green style avatar for user Josiah Jagelman
    Why is i(t) = 0 before the switch is thrown? Shouldn't it be a negative V/R?
    (3 votes)
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Video transcript

- [Voiceover] Now we're going to cover a really important circuit in electronics, it's the resistor capacitor circuit, or RC circuit. And in particular, in this video, we're going to talk about the natural response of an RC circuit. The natural response is what happens when you put some initial energy into the circuit. And in the case of an RC circuit, it represents charge that's stored on the capacitor at the beginning of the analysis. We're going to go through an intuitive description of what happens in this circuit when this switch is changed from this position, where we're connecting a battery to the RC combination, and then we're going to basically move the switch over to here, and watch what happens as that capacitor discharges back through the resistor. And that discharge pattern is called the natural response. So let's look at the circuit as it's sitting here right now. Let's say that the switch has been in this position for a really long time. Now sometime in the past, there was some sort of a current that flowed out of here. That current flowed through the resistor, and on to this capacitor, right here, and that left us with some charge. There's some charge here. What that looks like, let's do a close up picture of the capacitor. As charge flowed in here, it piled up on this side of the capacitor, and there's a corresponding negative charge that collected on the other side. This amount of charge matches this amount of charge. So that's what we mean when we say that q, or charge, is collecting on a capacitor. And for a capacitor structure, we know that the amount of charge here, is equal to the capacitance value times the voltage. So as more and more charge accumulates here, the voltage goes up. C is fixed, it's a property of the capacitor. If charge goes up, then v goes up. So let's go back over here and see what happens as this current flows onto the capacitor, charge accumulates, and eventually, this voltage here will rise up to be plus and minus v naught. And that that point, when this point is at v naught, and this point is at v naught, the current through the resistor stops, so the current here will be zero, and the voltage at this point will be v naught. So I want to plot that right now. We'll begin our sketch of what this response looks like. So this will be our time axis. This will be v of t. Which is this voltage here. And this will be i of t. And we're going to label i of t to be this current here. And you'll see why I picked that direction in a minute. So, at the beginning, before I throw the switch. We're going to throw the switch in this direction right here in a minute. But before we do that, we have v naught on the capacitor, and we have i of t equals to zero, so let me fill those in. Before time equals zero. This is equal to v naught. And this current down here is zero. So now we're ready to throw the switch. Let me do that. We'll erase this. And then put the switch in this position here. And now let me clean this up. I'm going to erase the battery here. The battery's done it's job for us, which was to initially put some charge on our capacitor, so we don't need this anymore. So here's our simplified circuit. And what we have here, there's a bunch of charge stored up on C, and what's going to happen? It's going to start rushing out of the capacitor, and going back through the resistor, to basically come over here and neutralize the charge on the bottom plate. What's going to happen is over here, the charge is going to leave, it's going to go over through the resistor, it's going to come back in here, and these pluses and minuses will then be neutralized. And that means if there's no net charge, the voltage is going to go to zero. So we can draw that. What we're going to say is after all this charge rushes out of here, goes through the resistor, and ends up on the back side of the capacitor, neutralizing the charge in the circuit, that means the voltage in the circuit is going to be zero. So we can draw that. We draw the long time from now out here. It's going to sketch into something like that. It's going to be low. Likewise, the current is going to end up at zero, when we're finished here. And that's what happens with every natural response. It basically, the energy in the circuit is allowed to die to zero. So let's see if we can fill in what's going to happen in between. While this voltage here started at v naught, and it's going to end up at zero. The amount of current flowing out of here is going to be proportional to the voltage across this resistor, right? The amount of current flowing out is one over r times v. So if v is high, then i is going to be high. And as v goes down, as we use up our charge, the current is going to go down. So I can sort of guess what's going to happen here, it's going to basically come down, and end up somehow at zero, after a while. Now we don't know how long it's going to take yet, but it's going to have some shape like this. So let's look at what the current is going to do. The current right now, when the switch was thrown, is at zero. As soon as that switch is thrown, as soon as we connect it right here, all of a sudden, we have v naught on this side of the resistor, and zero on this side of the resistor. And so there's going to be a sharp current increase going through this resistor here. So we expect this to go up like that. To jump up to some value. What we think the current is going to look like, the current is actually over here. It's directly proportional to the voltage, so as the voltage goes down, the current is going to go down. And we can take a guess, it'll have the same shape as the voltage, so it's going to go down something like that. Until it dies out to zero. So this is our forecast for what the natural response of an RC circuit looks like, and in the next video, we will actually derive really precise expressions for what those two curves look like.