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.