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LC natural response derivation 1

We begin the derivation of the natural response of the LC circuit, by modeling it with a 2nd-order differential equation. Created by Willy McAllister.

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  • blobby green style avatar for user Emmen Bramasta
    Is it correct to say that in this LC circuit, the capacitor is also the voltage source? Would it be different if we have a separate voltage source instead?
    (2 votes)
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    • purple pi purple style avatar for user APDahlen
      Hello Emmen,

      Both the inductor and the capacitor will appear to be the voltage source. However, they never do so at the same time. Think of this as an energy storage problem. The energy constantly sloshes back and forth between the two devices.

      In radio terms we call this a tank circuit as in water sloshing back and forth in a tank. The rate of energy exchange is known as the resonant frequency. The LC circuit is often found in radio frequency system as part of the station tuning mechanism.

      Things get really interesting when you add separate voltage sources to this LC circuit. In general the LC circuit loves voltage sources that operate at the natural frequency. It will try to short out anything else.


      (4 votes)
  • starky seed style avatar for user Hazem Hesham
    Hello ,
    I have a problem understanding why the voltage on the capacitor and the inductor is the same , at least it shouldn't be the same just as you close the switch .
    Any help ?
    (2 votes)
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    • leaf red style avatar for user Swoozienoah101
      Hello Hazem,

      Before time zero the capacitor was charged to some voltage and the inductor is "empty" which is to say, no current is flowing in the inductor. At time zero the switch is closed. Instantaneously the voltage on both devices is equal. The voltage then oscillates as Willy describes in the video.

      Recall that the voltage on an inductor is defined as:

      V = Ldi/dt

      This allows an instantaneous voltage jump when the switch is closed as the change in current is instantaneous (square wave).

      Or stated another way, the inductor likes to keep current constant. Before the switch was closed the current was zero. Immediately after the switch is closed the inductor current is still zero. The only way this can happen is if the inductor's voltage instantaneously rises to the capacitor voltage.

      Please leave a comment below if you have any questions.


      (2 votes)
  • blobby green style avatar for user tobbenda
    At Why is the current direction "reversed"? What is the reason for the minus sign in the equation?
    (2 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      The negative sign is a small accident of how i is defined in the main LC circuit diagram on the left.

      At I draw a capacitor over on the far right. The i-v equation for a capacitor is defined assuming the current, i_c, is coming down into the positive voltage terminal of the cap. With this current direction the capacitor equation is i_c = C dv/dt.

      Now look at the direction of i_withnosubscript in the LC circuit on the left side of the screen. Notice it points in the opposite direction of i_c. It is flowing up out of the positive terminal of the cap. That means -i = i_c.

      In the main drawing on the left there is no variable called i_c, there is just i. To write the capacitor i-v relationship correctly using i as the current variable you substitute in -i in place of i_c. This is how the negative sign gets introduced.
      (3 votes)
  • duskpin tree style avatar for user halahleis
    At , shouldn't the expression for voltage on the capacitor include a constant? (shouldn't we add the initial voltage on the capacitor?)
    (1 vote)
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    • old spice man green style avatar for user Willy McAllister
      You can include an initial voltage on the capacitor if you want (and an initial current in the inductor. For this video I assumed both were 0, just for simplicity's sake.

      Once you have gone through this solution it is interesting to go back and include those initial conditions on L and C. You will find that you get the same sinusoidal response but with a phase shift (the sine wave will start a little later or earlier than the 0 initial condition LC.
      (2 votes)
  • blobby green style avatar for user Rajab Ali
    At , 2nd derivative for the numerator is d^2i but in the denominator it is dt^2? Why is that?
    (1 vote)
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  • blobby green style avatar for user aandras
    In the previous video (intuition #2), the current was flowing into the same direction, and yet the equation for the capacitor did not need the minus sign. What is the reason for that?
    (1 vote)
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Video transcript

- [Voiceover] In this video, we're gonna begin the derivation of the LC Natural Response, the response of an inductor capacitor circuit. This is a difficult derivation, but it really pays off in the end. There's a really fun surprise at the end, and that is, this is where sine waves are born. We're gonna end up with sine waves at the end of this. And that's a really nice result because these are everywhere in electronics, and in the natural world. We're gonna start out by saying, we're gonna put some charge on this capacitor here. And we're gonna have the current in this circuit, i, we're gonna have that start at zero. So i zero is zero, but this is the current through the circuit, and that means that current exists over here as well. Let's put a switch in this circuit. And we're gonna close that switch at t equals zero. So we have one variable as i, and the other variable we're gonna use is v, and v is this voltage here, after the switch closes. So we wanna find i and v. And for this we're gonna focus on finding i. Once we find i it's straightforward to find e. So this is gonna be our independent variable, is the current. So let's begin the analysis. At the moment, t equals zero, the charge that exists here is gonna start flowing in the circuit, and both voltage and current are gonna start to change. So let's write some expressions, let's start out by writing some things that we know are true about our two components. And we'll start with the capacitor. So we'll start by writing an expression for our capacitor, and there's a little sign flip that happens here, so we gotta be a little bit careful. So if I have v on a capacitor, and i in a capacitor, I would say that i capacitor equals c, dv, dt. Now if we look here, vc is the same as v here, positive sign at the top, positive sign at the top, negative sign at the bottom, so vc and v are the same. But i, we have to be careful, i is in the opposite direction of the current that I picked for independent variable, so that's upside down. So i, there's a negative sign we get to flip here, equals negative c, dv, dt. So that's the IV equation for our particular circuit. We have that little extra negative sign. And now I wanna write the integral form of the capacitor IV equation, which is v equals one over c times the integral of idt. And remember our minus sign. This is the important minus sign to bring along. Okay so let's do the voltage across the inductor, the inductor voltage is same variable v, is L, di, dt. So there's no extra minus sign here. This is assigned according to the sign convention for passive components. So now we have an expression for v on the capacitor, and we have an expression for v on the inductor, and we know that that's the same voltage. So let's set these two things together, so we can say that L, di, dt, equals minus one over c, times the integral of idt. All we did was set the two identical voltages equal to each other. And now I'm gonna manipulate this a bit, it says that L, di, dt, plus one over c, the integral of idt equals zero. We just gathered terms on one side, and we said the other side became zero. Now since I'm not used to having integrals inside equations, I'd rather have derivatives, since I have some experience with differential equations, let's, if we take a derivative of this whole equation, so here's what we're gonna do. We're gonna take a derivative with respect to time, of this whole thing here, and that gives us, we get the second derivative of the first term, so we get L, second derivative of i with respect to time, plus one over c, and then, the derivative of the integral of idt, turns into just i, and the derivative of zero, on this side, is zero. So this is now the differential equation for the LC circuit. And it's called, it has a name, it's called a second order homogeneous ordinary differential equation. It's a differential equation because it has derivatives in it, it's homogeneous because it only has derivatives of i with respect to t and nothing else. The indicator is that this side is equal to zero, there's no forcing term over here on this side. So when you can write the equation this way, we say it's homogeneous. And it's called a second order equation, because it has this second derivative right here. So now we've set up our second order homogeneous ODE, and in the next video, we'll go about solving this. We're gonna go through it step by step.