Welcome back to Linear Circuits, last 
time we talked about RC circuits, so this 
time we get to do RL circuits. 
And the aim is basically the same as last 
time, except now we're using these 
differential equations to show the 
behavior of RL circuits instead of RC 
circuits. 
And as I mentioned before, the previous 
lesson covered the RC ciircuits With 
resistors and capacitors. 
Now we're doing the exact same thing but 
using inductors. 
The lesson objectives also are the same. 
We're generating the differential 
equation, identifying initial and final 
conditions, solving the differential 
equations, and graphing the result. 
So let's look at the behavior of RL 
circuits. 
Again we have a source. 
This time it's current source. 
It's going to go through here, and as 
this current source comes through, 
there's going to be a current divider. 
This current here, iL, is going to be 
dependent upon what the existing magnetic 
field happens to be. 
Through this inductor here. 
Since this inductor doesn't want to, it's 
current to change very much. 
So, if I make a sudden change to this 
source, or for example, I have a switch 
and I close a switch suddenly. 
It's going to try and send current 
through it a different rate. 
And so, in order to deal with that We can 
kind of start looking at this. 
This is closed so what's going to happen? 
Well initially, most of the current is 
going to go through this resistor 
because, the inductor doesn't want to 
change its current rate. 
But over time, the voltage that's 
applied, because using Kircchoff's, using 
Ohm's Law we see a current going through 
this resistor is going to say, V is going 
to be equal to whatever this resistor 
current is times the resistance. 
This voltage is going to drive The 
current and start to change the magnetic 
field in the inductor. 
But as that happens, this voltage is 
going to decrease. 
And notice the L and v R, because they 
are parallel, these two values have to be 
equal. 
So as this voltage decreases, so does 
this one. 
And the current decreases, and this 
current increases. 
Until such time that it reaches a 
steady-state. 
And that steady-state is when all of the 
current is going through the inductor and 
none of the current is going through the 
resistor. 
But this current is going to start by 
changing somewhat quickly and then over 
time will start to get slower and slower. 
And so again were going to see that 
exponential behavior In this case, we're 
going to i sub s as our maximal value, 
and we'll assume that the starting 
current is zero because right now time, 
before time equals zero, this is actually 
a gap. 
Now that we have some intuition in what's 
going on, let's look at some examples. 
Again, we'll start by looking at initial 
and final conditions. 
So, before time t equals 0, we just have 
an inductor and a resistor put together. 
So, any current going through this 
inductor is eventually going to be used 
up. 
As it loses energy, or loses power in 
this resistor. 
So, we can say this has been. 
It's initial position for a long time. 
The initial voltage is equal to zero. 
As is the current. 
After we close this switch and we let it 
sit for a long time. 
Again, remember that this is going to 
start to look like a wire, which that 
means that all of the current is going to 
go this direction. 
So, I look at my current after a long 
time, this is going to be equal to i sub 
s. 
We also know that the voltage after a 
long time Is also going to be equal to 0. 
Now, this problem's a little bit 
trickier. 
Because what we're actually going to 
solve for is vL. 
We're not going to solve for iL. 
And so, the way that we're going to do 
that is we're going to find iL. 
And then we will use the fact that V 
equals L d i l, d t to find the voltage. 
It can be an interesting exercise to go 
through this derivation again and solving 
using the Vs directly instead of solving 
for the is. 
And this will probably become a little 
bit more clear as we do the example. 
So, in working the example, we need to 
generate our differential equation. 
So, we start with from what we know. 
First of all, that vL equals L diL dt, 
and then iR is equal to v over R, using 
Ohm's law. 
And then Kirchhoff's current law tells us 
that iS has to equal iR plus iL. 
Putting all of that together, we see that 
is equals L over R, diL dt plus iL, and 
we're going to then rearrange this a 
little bit, so it's in the same format as 
we were using before. 
So now, R over L is A, R over Lis is our 
K, and that tal is going to be L over R. 
So if we take L over R, we're going to 
take 10 million, divided by 5 k, it's 
going to give us 2 time 10 to the 
negative 6, which are microseconds. 
So, that one time constant is going to be 
2 microseconds. 
Putting it all together, we get this for 
our current, we get this for our voltage. 
So, the voltage is i sub s times R times 
e to the negative t divided by tau. 
This is a little bit trickier than 
before. 
Because we noticed that initially, our 
voltage across our inductor is zero, 
because all of the current's been 
consumed by the resistor over time, so 
it's going to be zero. 
But we also know that this equation is 
going to start us off at five volts. 
And after one time constant go down to 2 
3rds. 
After two time constant another 2 3rds. 
After three cons, time constant another 2 
3rds and so on. 
So, we're going to expect it to look 
something like that. 
A little soft at the beginning, but 
[INAUDIBLE]. 
So it's interesting to notice that here 
the ohm doesn't jump. 
Where it was at zero and suddenly it's 
times equal zero and you close the switch 
and it jumps up to five volts. 
But remember voltage, voltages jumping in 
conductors is just fine. 
We just can't have the currents 
instantaneously change. 
Let's see how we did. 
Pretty close. 
A little bit sloppy on the curve this 
time, it's a little hard to free hand 
draw sometimes but we see That, we see 
Two thirds drop, So after one time 
constants, Which is to microsecond, we've 
dropped 62.3% of the way and then 86.5 
after two time constants, and so on. 
Now we're going to look at something a 
little bit new. 
Instead of having just a switch that we 
close or open, now we're going to have a 
switch that we toggle. 
In this example, we're only going to 
assume that the switch was in its first 
position, here, for a very long time 
before we switch it into our second 
position here. 
It can be an interesting thing, if you do 
this analysis, to see what would happen 
if you toggle this switch back and forth, 
so it might be something good to discuss 
on the forums. 
Again, we're going to start by looking at 
our initial and our final conditions. 
The initial condition is with the switch 
in this position. 
Voltage here going through here, this is 
going to effect a short circuit or a 
wire. 
So, this voltage is going to be zero or 
solving for iL. 
We're going to solve for a current this 
time and since it's in that configuration 
after awhile we have 12 volts. 
Going across 3 kg ohms, so we see that 
that is going to result in a current of 4 
mA after a long time. 
When I put in the second configuration, 
it's in the same basic situation, you 
have a voltage source, a resistance and 
so in that circumstance this current is 
going to be equal to Minus one half of a 
million. 
So, we have our initial current and our 
final current. 
And remember that the current cannot 
instantaneously change in an inductor. 
So, we'll be looking to make sure that 
that holds in our solution. 
In generating the difference of 
equations, we'll start with The L di dt 
and here, we see this. 
Il in the second position is going to be 
equal to this, so we're using the V2 
minus Vl divided by R2. 
And putting all of that together, you get 
this equation, and then we're going to 
rearrange it into our canonical form. 
So this is now our a, this is now our k, 
and now tau is equal to L over R 2. 
So using the differential equation 
solution that we have, i L is equal to v 
2 over R 2 times quantity 1 minus e to 
the minus t over tau, plus V1 over R1 e 
to the minus t over tap /g. 
So this V2 over R2 is going to become 
minus one half mA and the V1 over R1 is 
going become 4 mA. 
So that's what we see here. 
So what we're going to expect to see is 
we're going to start up here at 4 
milliamps. 
Now, this, down here at minus 1 milliamp, 
is where we're headed. 
So one time constant will go 2 steps away 
from the 4 milliamps to the minus one 
half milliamp. 
So, approximately here. 
2 time constants about there, 3 time 
constants. 
Somewhere around there. 
And then we can make our exponential 
curve, but we'll just draw it out. 
We're pretty close. 
Again, the same quantities, 63%, 86% 
after two time clusters, and you see the 
same behavior, the same exact behavior. 
So once you find your time constant, it's 
pretty easy the kind of guess what the 
curve is going to look like, but we've 
also seen that you can derive it 
mathematically directly. 
And this matches all of the things were 
expecting to see here. 
It can also be interesting to see what 
would happen, if instead of just using a 
constant source here, we changed into a 
function, and if you know something of 
differential equations You can probably 
see how you can go about doing that and 
solving that. 
But, that kind of goes beyond the scope 
of what we'll be covering in this class. 
If you do have questions about it though, 
I encourage you to go to the forums and 
ask them there. 
As we can discuss some things that, 
perhaps, go beyond the material that was 
covered in the lectures. 
So, to summarize. 
We got some intuition about how our L 
circuits behave. 
We identified their initial and final 
conditions and used differential 
equations to see what happens while it 
transitions. 
Then we also graphed the results. 
In the next lesson, we'll start talking 
about second order differential 
equations. 
And this is what's going to result when 
we have RLC circuits. 
Circuits that have resistors, inductors, 
and capacitors all together. 
And it's a little bit more interesting 
than just the first order. 
And these concepts with then be carried 
on into module three, where we'll start 
talking a little bit about frequency 
response and filters. 
Until next time.