Okay, now that you've been introduced to inductors, we're going to look at a

simple circuit with a resistor and an inductor in it.

And look at the transient behavior in that circuit, just like we did for the

resistor-capacitor circuit. So, here's the circuit that we're going

to analyze. It just has a resistor in series with an

inductor. And a battery, and then there's a switch.

And we're going to assume that the switch is closed at t equals 0.

And what we want to find is the current, I, as a function of time that flows

through this circuit. So, to do that, we'll do the same

approach that we used before and start off by using Kirchhoff's voltage law.

So, right after the switch is thrown, the circuit is complete, and I can write a

Kirchhoff's voltage law, I go, I start at ground.

I pick up V0 when I, when we go through the battery.

Then, there's a voltage drop across the resistor and a voltage drop across the

inductor. So, rearranging this equation, I get VR

plus VL equals V0, the source voltage. Now, VR is just the current times R, from

Ohm's law. And the relationship between the voltage

and the current, time rate of change of the current for an inductor, that we just

introduced is l dI/dt. That's the voltage drop across the

inductor. So, these two add up to V0, the battery

voltage. Now, we have to specify the initial

condition, too. The initial condition is that I at t

equals 0 is 0. So, we start off with no current flowing

this circuit. Okay, now we're going to use the same

method of solving this equation that we used previously.

So that means, the first thing we do is we find a solution we're looking for this

function I as a function of time. And we have to find any old solution that

will satisfy this equation. Well, if I assume that I is a constant,

is V0 over R, then the derivative term goes away.

And if I put V0 over R in here, then the RS cancel and I just get V0.

So, this indeed, if I plug it into this equation does solve it.

So, that's the particular form of the solution.

Now, the, I need then the general form. When I set this equal to zero, the right

hand side equal to zero, I have to find the function of time that will solve that

equation. And again, that's this exponential

function. When I take a derivative, I bring down a

minus R over L, and so I get minus R over L in front of this term.

The Ls cancel and I get a IR minus IR. And if I have an A, just any old

constant, unspecified constant in front of each term then those would cancel out.

So, this then solves the related homogeneous equation where all of this is

equal to zero on that side. So, now the method says that I take the

particular solution and the general solution.

Add them together then there's my solution.

So now, I have the solution of this equation, but I have unspecified

constant, A, that we have to find. And the way we find that is we apply the

initial condition. The initial condition, to remind you, was

that t equals 0, the current was 0, so if I plug t equals 0 in this equation, E to

the 0 is 1. And, so I have V0 over R plus A has to

equal 0, so that means A is minus V0 over R.

So here's the final full solution. I factor out, here's a V0 over R minus V0

over R times that exponential function. So, I factor the V0 over R out, and there

it is. So, there's a solution that satisfies the

initial condition. So, here's what a graph of the solution

looks like. plotting I of t divided by V0 over R.

So, I'm dividing the V0 over R. And to this side, so I'm just looking at

this factor, in this plot. And, we're measuring time in units of L

over R. L over R is the so, so-called LR time

constant of the, of this circuit. So, I'll measure time in those units.

And so, in this case, the current starts off at zero, and it's going to reach 63%

of its final value after one time constant, L over R.

And then the final state, the steady state condition.

As time gets large is the current just goes to V0 over R, so that's just the

current that would go through the resistor.

And like the inductor looks just like a short at that point.

So, that's the response of the RL circuit through a transient.