LC natural response intuition 2

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. Now we're ready to close the switch. Let's close our switch, and now our switch is closed again.

What happens? Well, we had ( V_{0} ), we had some voltage on our capacitor, and that means now we have some voltage on our inductor. That's what the switch closing did. So, all of a sudden, our inductor finds itself with some voltage across it.

Let's look at this equation here, the inductor equation. It says that ( V ), that's the starting voltage from the capacitor that we borrowed, is ( L ) times the slope of the current. All right, so now we have a finite voltage, and we have a finite value on ( L ), and that means that ( DT ) is some number.

If we put positive charge up here, we know that ( V_{KN} ) is positive, so let's sketch that. Let's start a sketch of what's going on here. We're going to plot together on the same plot the current, current in blue, and we'll put the voltage in orange. This is the time axis here.

So, let's go back and check what was our voltage when we started. Our voltage was ( V_{KN} ) when the switch closed, so I'll mark that out. What was the current when we started? The current was zero, so here's a zero current down here.

Now we're going to sketch in what we think happens right after the switch closes. We decided that there was going to be some kind of a ( \frac{di}{dt} = \frac{V_{KN}}{L} ) so there's going to be some—I'm just going to sketch this in—there's going to be some rising slope; there's going to be some positive slope on the current.

All right, 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 ); it's actually going to start flowing out of this guy. This is the current that we're talking about, ( \frac{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 going to start going down because the charge is leaving the capacitor.

All right, so far, so good? So, let's keep going back and forth between the inductor and capacitor and see how the voltage and current change. We do that by just looking at our equation. Now the voltage on the inductor is not ( V_{KN} ) anymore; it's changed to ( \frac{di}{dt} = \frac{V}{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. Let me make a little shallower version of the slope here. All right, now what happened next? A little more charge has left the capacitor, headed over to the opposite plate, so the voltage is going to continue to go down.

So now we're just going to 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? Then the voltage continues to drop because the current's leaving, and eventually, that voltage is going to reach zero.

At the point the voltage reaches 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. All right, 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, when this guy is zero, that means that ( Q ) is zero. 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 going to 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.

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. What that means is the charge is reversed. So, the voltage is reversed, and that tells us that we're going to have a negative voltage that is going to 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 ( \frac{di}{dt} ) has a negative slope, and a negative slope looks like it's sloping down like this. It starts to roll over and slope down. A little less current means a little less slope on the voltage.

Now this positive current is going to continue to keep flowing. It's going to 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. 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 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 ( -V_{KN} ), and the reason 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 going to make this video any longer. I'm just going to say that this situation will repeat itself. Basically, the voltage will curl back down like that, and the current will do this. Oops, 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. Now we reach another state; now we have 0 volts and we have a negative current here. What we look at is we've 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 going to be in reverse. This charge now is going to reverse direction.

Let's take out all our arrows, and all the charge we had is going to start flowing back around this way, like that. This positive charge that we had is going to flow around and meet up with the negative charge that we accumulated up here. And that's what happens; that's what's going on in this part of the cycle.

We can continue on. Basically, this is going to go the same exact story like that, and the current will do the same. We've again come back to the same exact point as the starting point. Now we've replicated our starting point.

So what's going to 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 we don't have to continue with the story; it's basically a repetition of everything that happened here.

What we're going to get is a rocking motion on the voltage, back and forth, and we're going to get a rocking motion on the current, back and forth, as these energy storage elements change storage charge for energy in the magnetic field and back to storage charge. 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 will be this rocking motion back and forth, and now in the next video, we're going to go through and see how this comes out mathematically.