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.