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Current time:0:00Total duration:10:59

- [Voiceover] Now what I want
to to do for the RC circuit is a formal derivation of exactly what these two curves look like. And then we'll have a
precise definition of the natural response. What I want to do now is
draw our circuit again. There's R, there's C. And we have some initial voltage, we said, on the capacitor, which is V naught. And what we want to find is V of T. And there's R, and there's C. So let me also label the
currents in these guys. This is the current in the resistor, and this is the current in the capacitor. So now I'm gonna write
the two voltage current relationships for these two components. And Ohm's Law tells me
that equals one over R times V. And the current in a
capacitor, this is iR, current in a capacitor
equals C times dv, dt. That's the two currents in our devices. Now, gonna use Kirchhoff's
Current Law on this node right here, and that says
that all the currents flying out of that node
have to add up to zero. So let's do that. Let's say iC plus iR equals zero. This is KCL, Kirchhoff's Current Law. C dv, dt, plus one over
R, times V, equals zero. Now I'm just gonna divide through by C just to be tidy. Dv, dt plus one over RC,
times V, equals zero. So what we have here,
this expression here is referred to as an ordinary
differential equation. It has a derivative in it,
and that's what makes it a differential equation,
and you can go look up ODE and search for this in
Khan Academy and elsewhere and in the mathematical
videos, you'll find out different ways to solve this equation. Now we're gonna do the same thing here, and we're gonna try to find
an expression for V of t. We're gonna define V of
t in a way that makes this equation true. So we plug the function
in here, whatever it is, and then we take its derivative
and plug that in here and that has to make this
whole expression true. There's a couple of different
ways to do this, and the one that we're gonna use
here is we're gonna guess at an answer. So what we need is, we need
a function whose derivative looks kind of like
itself, because this thing has to add up to zero. So that means that the function
V and the function dv, dt they have to be of the
same form so that they have a chance of adding up to zero. The function that I know
whose derivative looks like itself is the exponential. So we're gonna propose that
there's gonna be an exponential and the way we do it is say,
we'll put in some constant here that we can adjust,
and then we'll put some constant out here. So this is our guess. Our guess is gonna be that
the solution for V of t is gonna be something of
the form, some constant times e to another constant, s, times t. The way we test our guess
is to put it back into the equation and see if it works. And if it does, then we made a good guess. If it doesn't work,
then we have to go back and try something else. So one of the things we
need, we're gonna plug v of t in right here. We're gonna need derivative
of v with respect to time. So let's take a derivative here. Equals d dt of k, e to the st. And that equals K, and the S comes down. And e to the st remains. So we just took the derivative
of our proposed solution and now we're gonna plug
it into the equation and see what happens. So now I'm gonna plug in
the derivative that we had, plus the proposed solution,
and see if our equation comes out true. So dv, dt is sk, e to the st, plus one over RC stays here. And we plug in V, and V was ke to the st. And that'll all equal zero. Now I'm gonna factor out,
here's the common term for each one of these, so
we'll just factor that out. So now we get ke to the st,
times S plus one over RC equals zero. Okay, now let's take
a little peek at this. We have to make this
zero by the way we pick K and S. How can we make this zero? There's three terms in
here, there's three terms in the product. There's this, this, and that. Any three of these terms could be zero. The first one, K is
zero, that's a boring one because it says if we have
a circuit with no energy it sits there with no energy. So that's not so interesting. Now, e to the st, we
can make that go to zero but that takes a long time
because we have to let t go to infinity. S would have to be a negative
number, and t would have to go to infinity, which
is a long time to wait for something to happen,
so that's not a very interesting solution either. So here's the interesting one. The interesting one is
when S plus RC equals zero so the interesting solution
is S plus one over RC equals zero, and that says that
S equals minus one over RC. That makes this solution true. And if that's the case,
now we can say v of t, we're making good progress here, k, some k, times e to the minus t over RC. Now something really
interesting just happened here. Look up here where we
have this RC term here. See this right here? This whole exponent has to
have no units by the time we evaluate it. That means that RC has
to have units of time. RC is a measure, it's in seconds. RC is the units of R times
C, ohms times farads, is actually seconds. This whole exponent has to
have no units by the time we evaluate it. And that means that RC
has to have units of time. And it's actually ohms times farads, comes out to be seconds. Now we solve for S, and that means we only have to find K. Let's work on K now. K is a variable here, time is a variable, and voltage is a variable. If we knew V and T at
the same time, we would be able to figure out K. And there's a time when we do know that. And that is t equals zero. At t equals zero we know that
the battery was hooked up and there's a bunch of
charge on the capacitor so we know that the
voltage on the capacitor at time zero, is equal to V naught. So now we have a pair of variables here. We know v and t at the same time, so let's plug that into our
proposed solution and see how k pops up. So we're gonna put these two
into our proposed solution and see what happens to k. V, at time equals zero,
is V naught, and that's equal to k times e to
the minus t, minus zero at time zero, over RC. Now this whole expression
is e to the zero, or one. So it drops out of the expression and all we get is k equals V naught. K is the starting
voltage on the capacitor. And that gives us to our
final answer, which is v of t equals V naught, starting voltage, e to the minus time over RC. We can go one small step
further and find out what is i of t. Well, i of t is equal to v of t over R, that's just Ohm's Law, and so i of t equals V naught over R,
e to the minus t over RC. That is the natural
response of an RC circuit.