RC natural response derivation (2 of 3)

Now what I want 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.

Okay, what I want to do now is real quick draw our circuit again. There's R, there's C, and we have some initial voltage we said on the capacitor which is V_KN. What we want to find is we want to find 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 going to write the two voltage current relationships for these two components. Ohm's law tells me that I = 1 / R * V, and the current in a capacitor, this is I_C, the current in a capacitor, is equal to C * dV/dt.

That's the two currents in our devices. Now I'm going to use Kirchhoff's current law on this node right here, and that says that all the currents flowing out of that node have to add up to zero. So let's do that.

Let's say I_C + I_R = 0. This is KCL, Kirchhoff's current law. C * dV/dt + 1 / R * V = 0. Now I'm just going to divide through by C just to be tidy.

dV/dt + 1 / (RC) * V = 0. 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. 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 going to do the same thing here, and what we're going to do is we're going to try to find an expression for V of T. We're going to 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 going to use here is we're going to 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 going to propose that there's going to be an exponential.

The way we do it is we 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 going to be that the solution for V of T is going to be something of the form K * e^(s * 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 going to plug V of T in right here. We're going to need the derivative of V with respect to time. So let's take a derivative here: d/dt of (K * e^(s * t)) = K * s * e^(s * t).

Now we've just taken the derivative of our proposed solution, and now we're going to plug it into the equation and see what happens.

So now I'm going to plug in the derivative that we had plus the proposed solution and see if our equation comes out true. So dV/dt is K * s * e^(s * t) + (1 / R*C) * K * e^(s * t) = 0.

Now I'm going to factor out the common term for each one of these. So we'll just factor that out. So now we get K * e^(s * t) * (s + 1 / (R*C)) = 0.

Okay, now let's take a little peek at this. We have to make this zero, by the way, we pick K and S. All right, how can we make this zero? There's three terms in here. There's this, this, and that. Any three of these terms could be zero.

Okay, the first one is K = 0; that's a boring one because it just says if we have a circuit with no energy, it sits there with no energy, so that's not so interesting.

Now e^(s * t), 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 + 1 / (RC) = 0. So the interesting solution is s + 1 / (RC) = 0, and that says that s = -1 / (R*C).

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 * e^(-t / (R*C)).

Now something really interesting just happened here. Look up here where we have this RC term. 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 * 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 = 0. At t = 0, 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_KN.

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 out. So we're going to put these two into our proposed solution and see what happens to K.

V at time equals 0 is V_KN, and that's equal to K * e^(−0 / (R*C)). Now this whole expression is e^0 or 1, so it drops out of the expression, and all we get is K = V_KN, the starting voltage on the capacitor.

And that gives us our final answer, which is V of T equals V_KN * e^(−t / (RC)). We can go one small step further and find out what I of T is. Well, I of T is equal to V of T / R; that's just Ohm's law, and so I of T = V_KN / R * e^(−t / (RC)).

That is the natural response of an RC circuit.