And this is what we call the synaptic battery.

E, E synapse over here. Okay?

So, this is a full circuit. And now you can realize that whenever I

add into a given circuit, the passive circuit, another path for current flow

that will be probably voltage change. Just because you have a battery, which is

different than the resting battery. If this battery would of been exactly the

same as the rest battery, then changing the conductance here will not make a

difference because this battery and this battery are the same.

But typically they are not the same. For example, this batter is more positive

inside than the resting battery, then you can ex-, you can expect that if I open

these conductance. Because transmitter was interacting with

the receptor and I open this path, I would expect now positive current to flow

inside the cell and the cell will become more positive than before.

Why? Let's write now the equation that

describes the circuit as we did before. So, what is the equation?

This is now almost the same as before when I had passive membrane.

But not exactly. Before, there was Cdvdt as before, also

here. The current that is in this [UNKNOWN], in

this, in this point may flow this direction and will become a capacitative

current. That is before.

As before, we have now g rest, is this conductance, multiply by V minus E rest.

This is exactly Ohm's law. What is written here, exactly, is that

the current that flows here, depends on the difference between the voltage V in

one side of the resistor of the conductance, to the other side of the

conductance. So, it's V minus ER.

This is the voltage difference between this two side of these conductance

divided by R or multiplied by g. So, this is the passive current.

This is the passive current. This is the capacitative current.

It's like before. But now you have a new current.

This is a new current. This is the red current.

The current that flows through new channels that are being opened.

And it's exactly like the passive current, but in this case it's the, if

the synaptic current g synapse multiply by the difference of voltage V minus the

synaptic battery E synapse.

[SOUND].

Okay? So, this is exactly V, the membrane

voltage at this side minus the synaptic battery at this side, divided by the

resistance of the, of the synapse or multiplied by the conductance of the

synapse. So, this is [SOUND] the synaptic current.

[SOUND]. So, you see that in this circuit you have

three types of, of currents. The capacitive current, the passive

current or the resting current, and the synaptic current.

The sum of these three must be 0. There is no extracellular, there is no,

sorry, there is no external current going into the cell.

I don't inject any current. I just open a new path.

So, at any given point here. According to Kirchhoff's Law, the sum of

this one plus this one plus this one, should sum to 0.

Okay? So, this is the equation.

Capacitative current, plus passive current, plus synaptic current equals 0.

So I need to solve this equation in order to get this V, which is the voltage being

generated by the synapse, by the fact that there is a synaptic conductance.

There is a voltage change and I'm interested in this voltage change.

I want to know what is the change due to the vol-, due to the opening of the

conductance of the synapse, that I will call the post-synaptic potential.