Welcome back to Linear Circuits. Last time from Doctor Ferry, you learned
a little about first ordered differential equations.
So today we're actually going to use them for something real and apply them to
something. In this case it'll be RC circuits or
circuits with resistors and capacitors in them.
So the aim of this class, is to allow you to use differential equations to show the
behavior of an RC circuit, as the system changes.
So the previous lesson you learned some basics about solving simple first order
linear differential equations. And let's start apply them to something
real, RC circuits, and then it turns out that RL circuits or circuits that have
resistors and inductors in them, behave very similarly.
And so you might notice a lot of similarity between this lesson, and the
next lesson. The objectives for this lesson are to
allow you to generate a differential equation from the circuit, identify
initial and final conditions, solve the differential equation, and then graph the
result. So we'll start by looking at the behavior
of circuits with resistors and capacitors in them.
So we'll take a look at this very simple example.
Here we have a resistance and a capacitance.
And we have labeled here a current. Now I know that this current, due to
Ohm's law. It's going to be equal to Vs minus Vc,
all divided by R. So Vs is this voltage, i, Vc is this
voltage, so if I want to know this current, I can calculate it that way.
Which, this current happens to also be equal to ic.
So lets think of what the system will do. We've already talked about if we were to
analyze this circuit in a steady state. That this capacitor is going to look like
an open circuit. And this voltage is going to then be
equal to V sub s. Since there is no where for the current
to flow, no voltage drop cross this resistor.
Because of that, we know what our final solutions going to be but, how does it
get there? That's what this lesson is going to be
about. And since we know something about this
current. Initially, let's say that this initial
which is zero to begin with, but times zero.
There's a big difference in voltage between the two sides of the resistor.
So this is a fairly strong current. And as this current is flowing, this
capacitor is getting charged. But as this capacitor charges Vc gets
bigger, so the difference between Vs and Vc get smaller, so this current
decreases. And as this current decreases, this
capacitor still continues to charge up, but at a slower rate.
And so it turns out that if we look at the behavior of this system, our
capacitor, voltage is going to start at zero.
And it's going to start by increasing very rapidly, but as it starts to
increase, the current decreases, and so it starts to tapers off.
So it has kind of an exponential curve look to it.
And I know that it has kind of this limit, this upper limit of V sub s.
Cause I know the final value, if my system has been in this state for a very
long time. Let's actually do the math to see where
this comes from. So our first example is this one.
We have a few different resistances, a few different capacitance's.
But the first thing we're going to start with, is what we do already know?
The first thing we're going to need to look at, then, is what are, where are we
starting, and where are we getting to? So here, here you see that we have little
switch that we're opening at time t equal 0.
So let's consider what's going on before the switch is opened.
We're going to assume this capacitor has been there for a very long time.
But this is a wire, since the switch is closed.
That means that any charge that's built up on this capacitor, is going to go
away. So, my Vc, at 0, before the switch is
opened is getting, is going to be equal to 0.
Because any charge is just going to be able to go from one plate to the other.
So we'll have no initial charge. After a long time of this switch being
open however, it goes back to the, the previous example that we were looking at.
And so I know that after a very long time, and then this voltage is going to
be equal to 5 volts, because that's what the source is.
So, we know where, where we're starting and where we're heading.
Let's now derive a differential equation for this circuit.
So let's start with what we know. First of all, we know that the current i,
is equal to C times dv dt. Well, dv c dt.
Right, the voltage across the capacitor. We also know that the current here, is
going to be equal to V sub s minus V sub c over R.
Just like we were looking at in the, the first problem that we showed.
Putting these two things together, we see that Vs minus Vc over R is C dvc dt.
We'll do a little bit of manipulation to get it into this format.
And this is similar to the format that Doctor Ferry presented, when presenting
differential equations. So, here we have a differential.
The constanse times our function is equal to some constant.
So here, 1 over RC is a, and Vs over RC is K.
What we're going to do, is we're going to say let tau be equal to RC.
And you already talked about tau in the previous lesson.
And here it means the same thing that it did before, the time constant.
Doing that, and putting this into the equation we showed here, we see that Vc
is equal to Vs times the quantity 1 minus e to the minus t over tau.
And we know that tau, because it's equal to R times C, Is equal to 2K times 500
times 10 to the negative 6. So this ends up becoming 1000.
So the time constant just being 1. The time constant, it turns out, is going
to be in seconds. Because it's a time, time constant.
And if you actually go through and look at the units of resistance, and the units
of capacitance, and you multiply them together, you do get seconds.
I'm not going to through the derivation of that, however.
Let's see what this looks like on a graph.
So, we already decided that we, sort of knew where we were headed.
That we were going to start at 0. Then we're going to end up here at 5
volts. And because of this exponential behavior,
we know that in one time constant, we're going to make it around 2 3rds of the
way. So we'll kind of guess at 2 3rds, maybe
say it's about there. At 2 time constants, we're going to go
another 2 3rds of the way. So another 2 3rds might be approximately
here, and then 3 time constants again another 2 3rds.
It turns out that one way to help you sketch this, is that if you look at the
time constant where you cross this line, because of the way the exponential is
it's own derivative, it's going to start off, kind of following along that curve.
So, to kind of sketch it out, I believe that my curve is going to look something
like this. So let's fill it in, and see how close we
were. Not too bad.
At 1 time constant we're going to be 63.2% of the way toward the final value,
after 2 time constants 86.5%, and 3 time constants and so on, is going to follow
the same way. Because of this 5 minus 5 over e, and 5
minus 5 over e squared behavior. And so that's what we're going to expect,
and we can see that our initial and our final conditions do match what we what we
calculated initially. Now, we're going to move through another
example, and we're going to go through this next example a little more quickly.
In this case, we again have a switch. Now it's closing at time t equal 0, and
we have two different resistors, but let's look at our initial and final
conditions. Before this resist, before this switch is
closed, we have this current source that is going to flow through here.
This is going to look like an open circuit, so Vc is just going to be this 1
ml amp and times 12 Kohms. So Vc, before we close the switch, is
going to be 12 Volts. After the switch is closed, now, these
two resistors could be seen as being in parallel.
And their equivalent resistance would be 3 kilohms.
And again, after a long time the capacitor will look like an open circuit.
And vC after a very long time, is going to be equal to 3 volts.
Let's see how we get our differential equations for this one.
Capacitance here is 1 picofarad, 12 kilohms, 4 kilohms, and so on.
We, again, are going to use some things we know, in this case, Kirchhoff's
current law. I know that this source current, has to
equal the sum of the currents going through each of these different devices.
And I know the equations for each of these devices, for the resistors, it's
Ohm's law. And then for this capacitors, it's C
dvdt. So we manipulate these equations, and I
show the derivations here. We're going to use R eq to represent
this. Which is the same thing as when we were
calculating equivalent resistances, in parallel.
And finally we're going to get this equation.
So again, we have our K, we have our a, and our tau is just going to be R eq
times C. So it corresponds to the tau where it was
presented before, giving us this final solution.
And is times Req is going to be 3, and the is times R1 is going to be 12.
Our time constant, tau, Is equal to 1 over, or sorry, it's just R eq times C,
so 3K ohms times c, which is one picofarad.
So it's going to give us 1 microsecond for 1 tao.
Again, we're going to graph it. So each of these tau's now, instead of
corresponding to 1 second, so this is 1 microsecond, and this is the equation
we're looking to, to show. Now, what it turns out to be, is that
initially, we're going to start at 12 volts.
We're going to go down to 3 volts. By looking at this equation, you can kind
of identify why. Again, 2 3rd's down, and the first time
constant. Another 2 3rd's, and 2 time constants.
Another 2 3rd's and 3. So, if we kind of trace a little curve
along this line like this. We're a little, we're pretty close.
That was a little sloppy there, especially near the end.
But again, in the first time constant, we've gotten 63.2% of the way to our
final value, and in 2 time constants, 86.5, and so on.
So, we've now been able to get some intuition about how RC circuits behave.
And we identified their initial and final conditions.
And then we found differential equations, to find out what's going on in the
interim. We were able to then solve these
differential equations, and then graph the results.
In the next lesson were applying the same methods and techniques, to circuits with
resistors and inductors in them. And we'll see that the behavior is very
similar, but we'll see that there's some differences as well.
So until then, take care.