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LC natural response derivation 4

In this final step of the derivation, we find two initial conditions and use them to come up with a sinusoidal solution for the LC natural response. Created by Willy McAllister.

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Video transcript

- [Voiceover] So now we're gonna use the initial conditions to figure out our values, our two constant values A1 and A2. That is in our proposed solution for current for the LC circuit. So one thing we need to do, because this is a second order equation. We need to have two initial conditions for the variable that we're studying here. So we're studying I. Right now we have one initial condition for I, and because we have a second order equation. That means we need two ic's for I. So we have one initial condition right here, and what we'd like to know is what is di dt at time equal zero? So the other piece of information we have is this v. V knot at time equal zero. So let's use that, and we'll just plug that straight into the inductor equation. so the inductor equation at t equals zero. The voltage across the inductor is v knot, and that equals L times di dt. Alright and that means the di dt equals v knot over L. So now I have two initial conditions in terms of I. There's one, and there's one there. And we can use these now to go after A1 and A2. First off let's plug in I for time equals zero, and then see if we can work our something over here. So that means that time equals zero. The current is zero and that equals A1 cosign omega knot times, t is zero, times zero plus A2 times sine of omega knot times zero, and what does this evaluate to? Okay this is sine of zero and sine of zero is zero, and cosine of zero is one. So that comes up with zero equals A1. Okay, and A1 equals zero means that this entire term of our solution just dropped out. Alright let me write what we end up with. I equals A2 sine omega knot t. This whole term here just dropped out of the solution. So here's our proposed solution down here. Now we need to go after A2, let's do that. As you might suspect, we're gonna use our second initial condition to do that. So to use our initial condition we need di dt. So let's take di dt of this. So we're gonna take d dt of this whole equation, and on the left side we'll get di dt, and on the other side we'll get d dt of A2 sine omega knot t. Okay so far so good. Let's roll it down again. So let's take that derivative. We get di dt equals, A2 comes out of the derivative, and the derivative of sine omega knot t with respect to t is omega knot times cosine omega knot t, and we apply our initial condition. Let's go to t equals zero, and we know the di dt was v knot over L equals A2 omega knot cosine of omega knot times zero, and cosine of zero goes to one, and so we can solve for A2. A2 equals V knot over L, and omega knot goes down here. So now we solve for our second adjustable parameter and we can write I. I was A2 sine omega knot t. So let's fill it in for A2. I equals A2 is v knot over L omega knot times sine omega knot t, and I wanna go back now. I wanna write this a little bit different. I wanna go back and plug in our value for omega knot. So if we remember, we said omega knot equals one over LC and the square root of that whole thing. So now L omega knot equals square root of one over LC times L, and that equals square root of L squared over LC, and that equals square root of L over C. Lastly I'll write one over L omega knot equals square root of C over L, just the reciprocal, and now we can write I equals square root of C over L times V knot sine of omega knot t, and that is the solution for the natural response of an LC circuit. It's in the form of a sine wave and the frequency is determined by omega knot. Which is the two component values, and the amplitude is determined by the energy we started with. Which is represented here by V knot, and the ratio of the two components again. So this is why I said at the beginning that this is where sine waves are born.