Okay. So I need to ver-, to show you what I promised to show you. Is, when I take a patch of membrane. [SOUND] . [UNKNOWN] representing it, representing it as a spherical, small spherical cell, which I will call soon iso potential sphere. Meaning that there is no change in, in voltage across this membrane. So there will be the voltage difference between the inside and the outside will be V at any location. So this will be iso potential. No change, no drop in voltage across this membrane will be an iso potential cell. I showed you that when I inject current here. I, I get voltage response V, that looks like this. This will be my I, this will be my V. And I told you that the representation of this, is a first approximation is an RC circuit, like this. This will be my inside, this will be my R, this will be my C, this will be out. So, why do I say that? A good electrical circuit, representing this behavior of a cell, is an RC circuit. And this is because of the mathematics of this circuit. And I can write down the equation that describes what happens when I inject current to the two sides of this circuit, I, as I did before. What will be the voltage development between these 2 sides V. So I want to measure V, in response to I. So V is the voltage difference between outside and inside. And I is the injected current and I can write the following, basic equation. I am saying that the total current that I inject, I, the total current that I inject into this location. Can either go and become a capacitative current. Or can go and become a resistive current, because. According to Kirhoff law, everything that inj-, is injected here must split either in this branch or in this branch. So that's exactly what I'm writing now. I'm saying that the, the capacitative current, which is c dividity[SOUND]. This will be my capacitive current. Plus the resistive current, which is just regular. Ohm's law, V divided by R must equal I, the injective current. So this is my resistive current, current, this is my injected current. Injected current. So what I wrote here actually, is Kirchoff's law. Saying that the current that I inject splits into capacitive current, here. Charging the capacitance, and this, the resistive current. Is the current that goes through the resistance, and leaks out. Actually, current can only flow outside through the resistance. And it can charge the capacitance. If I inject a positive current, then the capacitance is charged, inside. It's charged inside. Inside the cell, inside the cell, you charge the capacitance with positive charges. So the inside of the cell become positive, and that's what you see here. The inside of the cell becomes positive, and that's our agreement. If I inject positive current inside the cell, I get the positive voltage inside the cell. Which I call depolarizing. Okay, so now I have to take this equation. And show you that when I solve this equation for V, I get such a behavior. So let's see, so I'm trying now to solve this and this is actually very easy. Because now you have a very simple linear, because this is constant, the capacitance. This is constant, the resistance. There is a constant current I. It's a linear, 1 dimensional partial equation. So you only have dv, dt. There is only time here, t is time. As I said before, this is now here, I injected in time. So c, dv, dt plus vdivided by R is equal I. And I need to solve this equation. It's a very simple solution. So let me write to you the solution. But before writing the solution for such I thing, I have to define initial conditions. What do you start with? So we already said that we start with v at t equals 0 equal lets call it 0. So I start with v equal 0here. V equal 0. And I get to some v with time. So I want to solve for v, I want to solve this equation for v as a function of time. So the solution looks like this. V as a function of time, is equal I which is my current injection multiplied by R which is the resistance. Multiplied by 1 minus e, exponent to the power of minus t time divided by R multiplied by C. This is the general solution, for this equation. Assuming that v at t equals 0 is indeed 0. So let's check it. Let's say the t is equal to 0. What happens then? When t is equal to 0, e to the power of 0 is 1. 1 minus one is 0. So v, so v, at t equals 0, is 0. So this means that we satisfied our intial condition, that's fine. Let's look at the other extreme, which is the case when I inject my current for a long, long time. For an infinite time. So, I inject my current for long, long, long time. What happens then? So, if t goes to infinity, e to the powers of minus t goes to 0. This goes to 0. And you are left with v. At t equal infinity is equal IR. Because this is, this remain, this goes to 0 and you have then, IR. Okay, so these are the two extreme cases. You start with v equals 0. And if you inject this I current for infinite long time, you will get at the steady state. This means steady state because you inject so low. And the voltage remains at the steady state and, the steady state value is I multiplied by R. So that means again, I'm repeating this drawing. I inject current I. I look at voltage v of t. Okay, at 0, here, at time 0. I have voltage 0. And at infinite time, if I inject it very very long time, this becomes I equal to I, I multiplied by r. So this is the steady-state value, the steady-state value of this voltage, development, if I wait enough time. And this comes from this equation.