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.