Hi, I'm Jacob Block and I'm a ECE graduate student here at Georgia Tech.
Today, I'll be going over an extra problem on RL circuits.
Alright, in this extra problem, we're going to be looking at an RL circuit.
And the RL circuit here has a current source, three resistors, an inductor and
a switch and the switch opens at t equals 0.
So we're interested in finding i1 of t, i2 of t and iL of t.
This is a, a brief review. We need to know what an inductor is or
what happens to inductor in steady state. So an inductor has some current and
voltage across it. The voltage is related to the current by
a derivative. So the current is proportional to the
change in current. So in steady state, when you have no
change in current, you have no change in voltage.
And what that means is that we can take this inductor, and if we're in steady
state mode, we can replace it with a short circuit.
So knowing that, let's kind of start tackling this problem, so our first step
is before the switch is flipped. So let's redraw the circuit with the
switch flipped. So we have our current source, our
resistor, a a closed switch and our inductor.
Now, the inductor, the 100 millihenry inductor, since this circuit has been
running for a really long time, and that switch has only been flipped at t equals
0. This inductor is fully charged, and for
inductors, the, that means that we have reached a steady state position.
And we can replace the inductor with a short circuit, so let's go ahead and
simplify this circuit. So now, we replace this with the, a short
circuit and we can move our resistor over, and so we have a 5 ohm, a 30 ohm
and a 30 ohm resistor here, and a 3 amp current source.
So now, we want to solve the circuit, we want to solve for i1, i2, and iL.
So you know, there's isn't an iL, we're just calling this branch current iL.
So how do we go about doing that? Well, we can you know, combine all of the
resistors together and then solve for the voltage across them.
You know, and then do v equals i r. You can also do a current divider rule.
So the current divider, if you remember, the current divider tells us that the
current splits between two resistors by the weight of the second branch.
So if this is i1 and this is i2 and this is an i input, this is R1 and R2.
i1 is equal to i times R2 over R1 plus R2, and similarly, similarly i2 is equal
i times R1 over R1 plus R2. So, let's go ahead and calculate the i1
current. So, i1 is the 3 amps times the second two
branches, the 30 and 30. Parallel over the sum of the branches,
the 5 and the 30s, and this notation, the parallel bars, it stands for a parallel
combination. So just as we use a sum to indicate a
series connection, the parallel bars help us reduce notiation for a parallel
connection. And so, when you see 30 in parallel with
30, you should immediately think of doing the operation 1 over 30 plus 1 over 30
all inverted, so this is 15 ohms. So i1 is 2.25 amps.
And we can repeat the process for i2. i2 is the current in this branch.
So the current in this branch is 3 amps times the second two branches combined,
the 5 and the 30, divided by the sum of the branches, 30 plus 5 in parallel with
30. And so, if you do the same operation with
5 in parallel 30, you should get 0.375 amps.
So we take we're looking at this branch, and we write out the same equation and we
should get 0.375 amps. And you can check this by summing the
currents at using a KCL equation. So, for this KSL equation we have 3 amps
coming in, 0 2.25 amps in this first branch, and 0.375 in the second branch,
and 0.375 in the third branch, and that all sums up to be 3.
So we, we have our currents for this, the first part of the problem, for the
pre-switch flip. Now we're going to be looking at the
post-switch flip. So we are, we switch our flip we flip our
switch, and now, we have a kind of new set a new circuit.
So let's go ahead and draw out that circuit.
We have a current source and our first resistor.
And we have an open circuit at our, where our switch used to be.
And that come the bottom comes over still, and we have our second branch
resistor, and our third branch resistor, and our inductor.
And this is a 5 ohm, 30 ohm, 30 ohm, and 100 millihenry conductor.
So, what is going on at this bottom node? Is there, is there current in this
branch? Well, we can do a KCL to figure that out.
So, let's look at the KCL at this bottom node.
So the KCL here well, this, this current source tells us we have 3 amps in this
left wire. So we have 3 amps.
And in this top top wire at the node, we have also 3 amps, right?
The source this three amp current source tells us there's a constant 3 amp current
on this on this branch. And by KCL, that tells us we have 0 amps
in this connecting branch. And that means that these two circuits
are actually separated, they're independent from each other.
So, right away, we can say that i1 is 3 amps because of this current source.
So now, let's take a look at this, at the right half circuit.
The right half circuit has you know, two resistors, and an inductor and there's
two there's two states to this circuit. There's a transient, right when the
switch gets flipped, and a steady state, where this inductor has smoothed out,
it's evened out. And remember for steady state, we have
all of our time derivatives go to 0. And so, we can replace that inductor with
a short circuit, just like in the first part of the problem.
And if we calculate for the two branch currents, iL and i2, we'll find that
they're 0, because there's no source here.
So i2 is 0 and iL is 0. So now, let's look at the transient
response, where we have, we have the full the full circuit including the inductor.
We have our resistor, our second resistor, and our inductor, 30 ohms, 30
ohms and 0.1 henries. What's so now, let's, to solve this
circuit, let's do a KVL around this loop. So the KVL starting with the kind of left
part, the KVL is 30, the resistance times the current.
So this, the, the voltage drop across this resistor, the voltage drop across
this resistor is the same, 30 times iL. And then the voltage drop across an
inductor is equal to L times diL dt, and all of that equals 0.
So, this is our, our KVL number. Now, L is 0.1 henries.
So, you remember from lecture 3.6. We found that the derivative this, this
differential equation, dy of t dt plus ay of t equal to k.
It has the solution y of t equal to k over a times 1 minus e, 1 minus e to the
negative at plus y at 0, the initial condition, times e to the negative at.
And this is valid for t greater than, or, for, for all time after the, the switch
flips, or basically the time after this this starting condition.
So using that previous work, we can rewrite our KVL equation in the similar
form. So,let's go ahead and reorganize that
equation. diL, diL of t plus 630, and then this 0.1
gets moved over to this constant plus 600 iL equals 0.
And this is actually, this is in the same form as our known solution, so we can
just plug in our numbers. So our iL of t has the solution k in this
case, our k is 0, and our a is 600. So, iL of t is going to be 0 times, you
know, all of this plus our initial condition.
From the first part, we found that iL at 0 is 0.375 amps.
So plus 0.375e to the negative 600t for t greater than or equal to 0.
So we have our, our solution for kind of all time.
Now, what is, what is the current in i2? So i2, remember, i2 is pointing down
through the 30 ohm, and iL is pointing down through the inductor.
And they're, so since they're their reference directions are opposite, we can
say that i2 is equal to negative iL. So let's summarize kind of what we have
now. So we have i1, where t less than 0, and
that is zero point 2.25 amps. i2 for t less than 0 is 0.375 amps.
And i3 for t less than 0 is 0.375 amps. And our i1, when T, f, for the, the
switch is flipped turns into 3 amps or i2 after the switch is negative 0.375e of
the negative 600t. And our i3 amps, and our i3 is 0.375e to
the negative 600t amps. So, let's, let's graph that solution and
see what it looks like. Whoops, sorry, these are L's.
So iL of t starts right at t less than 0. We know that it's kind of a constant
0.375, and at infinity kind of in the steady state, it approaches 0.
So right when this switch flips, we turn from steady state to transient.
And the transient here, the 0.375e to the negative 600t, it's connecting these two
points together. It's connecting the steady state before
the switch was flipped and the steady state when the inductor fully discharges.
And that's the, this is the solution of the problem.
So what do we, what do we do? We started with a we started with a
circuit [NOISE ]
[COUGH].
Okay. So we started with our RL circuit, and we
solved for the circuit where, before the switch was flipped in a steady state, and
we calculated each of those currents. And then after we looked at a new
circuit, after the switch was flipped. So after the switch was flipped, the we,
we had two separate independent circuits, one with a current source and one with
just Rs and Ls. And then we proceeded to solve, set up a
differential equation for that, for that RL circuit, and we found that the
inductor slowly dissipates its energy and ends up fully discharged, kind of in the
steady state.