This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.
6.6 First Order RL Circuit.
Loading...
来自 Georgia Institute of Technology 的课程
Linear Circuits 1: DC Analysis
125 个评分
Explore Related Videos
At Coursera, you will find the best lectures in the world. Here are some of our personalized recommendations for you
从本节课中
Module 6
This module introduces the transient response behavior of RC and RL circuits after a switch in a circuit is changed.
与讲师见面
Dr. Bonnie H. FerriProfessor
Electrical and Computer Engineering
Dr. Joyelle HarrisDirector of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer
[SOUND]
[MUSIC]
Welcome back to Linear Circuits, this is Doctor Weitnauer.
This lesson is on First-Order RL Circuits.
That is, circuits that have only resistors and one, effectively, one inductor.
We will find the time response of the First-Order RL Circuit, and
we'll plot that response using the time constant.
This builds on the Norton's Equivalent circuit shown below,
and the IV relationship for the inductor V = L di/dt.
And, the standard form of the first order differential equation and
it's solution, which is in the lesson on RC first order circuits.
The first step is to find the equivalent RL circuit,
by that I mean that we will get the Norton's equivalent
of the part of the circuit that is connected to the inductor.
That's this part, and we find the Norton's equivalent.
Then you reconnect the inductor and analyze the smaller circuit.
Let's get that differential equation for the RL circuit.
Using KCL and v = L di/dt,
find the differential equation for the current i.
We're first going to use KCL, up here, and
I'm going to denote this current in this branch, is iR.
And then we can write that isc,
now this is KCL at this node, is equal to "iR" plus "i".
Now "iR" is the voltage, v,
over Rth and just keep that i.
The next thing we do is substitute v = L di/dt in for the v here.
So we'll have
iSC = I/Rth (L
di/dt) + i.
So at this point, we have a differential equation in terms of i, but
it remains to put it into standard form.
So to do that we isolate the di/dt term,
in other words, we have it coming first with the unity coefficient.
In order to get a unity coefficient, it's clear that I'm going to have to divide
every term in the equation above by l/Rth.
Or in other words, multiply every term by Rth over l.
So when I do that,
I will have di/dt +
Rth/l i = Rth/l isc.
Okay, there's our differential equation in standard form.
And this coefficient here, you recall, is 1 over tau.
That means that tau = L over Rth,
and we note that when L is in henries and
RTh is in ohms, the units of tau are going to be in seconds.
And this is our K, and we're going to need the product K tau.
You can see that when we multiply those things,
we simply get the short circuit current for isc.
Now that we have the differential equation, we write the solution.
We recall the standard form and its solution, and
then I'm repeating here the RL equation in standard form so
that we can identify the time constant tau,
which is L/Rth, and the Ktau, which is just isc.
Now, I'll emphasize that once you've found your Norton's equivalent,
you really don't have to get the differential equation.
You can just read off the Norton's equivalent, the Rth,
and put that with the L to calculate your tau.
And then you can get your K tau,
it's just the short circuit current of the Norton's equivalent.
And then the solution is this.
All right, let's do an example.
We'll find the current i that's flowing through the inductor for
t greater than or equal to 0, given the initial value of 10 amps.
Notice that we've already got our Norton's equivalent.
So recall that if we're going to get the solution for i of t,
we're going to use this expression right here which involves the isc, which is
the short circuit current, which is in rk7 amps and the initial current which is 10.
Then we're going to have our time constant, Rth over L.
Also, we know that if Rth is in Ohms and L is in Henrys,
and tau is going to be in seconds.
To be able to write down the solution, for i(t), we first need to calculate the tau,
which is L, which is 4x10 to
the -3 Henrys over Rth which, is 500 Ohms, and
that's going to be 8 microseconds.
We identify our isc, that is 7 amps.
Then we can write it down, i(t) is equal to 7,
plus the initial current is 10 minus 7
e to the minus t over tau, which is 8 microseconds.
And this is in Amps and it's for t greater than or equal to zero.
Now, if we want to plot it,
Versus time, i t in Amps versus time, what we do is we
identify the starting value, which in this case is ten.
The ending value, which is 7,
draw the horizontal asymptote for the ending value or the final value.
Then we identify the time constant point, all right.
You break this interval into thirds and come over a little ways,
drop down, and that's your time constant, which is eight microseconds.
So your curve is going to start at the ten,
it's going to go through that point where the time constant is, and
then it will finish off by approaching the asymptote, and
it's going to do this in an exponential form, like so.
And I want to point out that the final value of 7 amps makes sense if you
consider that when t equals infinity, the circuit reaches DC steady state.
When that happens, the voltage across the inductor goes to 0.
And, so it becomes a short circuit, and all of the current's going to flow through
the inductor, none of it will flow through the 500 Ohms.
And so that means that the final value of current will be the value of isc,
which is 70 Ohms.
Now, here's a quiz for you.
Can you identify which of these three is the correct time constant for
the circuit below?
To get the answer for this quiz, all we have to do is identify the time constant,
tau, which is equal to L over R.
RL is 30 times 10 to the -3 Henries over 600,
which I'm going to write as 6 times 10 squared ohms.
Remember if L is in henries, and R is in ohms, and
this is Rth, then the tau will be in seconds.
And so, simplifying this, we will find that we have 50 microseconds.
So the answer is B.
To sum up this lesson, the key concepts are that there's a general
approach to finding this solution for a first order RL circuit.
Replace the circuit connected to the inductor with it's Norton's equivalent.
Then, simply identify the values of tau, and k tau.
Tau being L over Rth, and K tau being the short circuit current, and
then plug those into the solution.
You do not need to derive the differential equation.
You already have the solution, you just need these parameters.
And then, when you plot the solution,
remember to first put your horizontal asymptote in for
the final value, mark where the time constant is, and that's going to be where
the solution is two thirds of the way from its initial value to its final value.
Thanks. [SOUND]
