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.