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Current time:0:00Total duration:6:39

- [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.