TA: Flavia Filimon

Course: cogs 107B (systems neuroscience)

Quarter: winter 2007 (December – February 2007)

 

 

Clarifications on voltage, current, electrical equivalent circuits, etc ….

 

 

 

Let’s start with the basics:

 

 

·   a simple circuit consists of resistors, capacitors, and a battery (constant current source).

·   A neuron can be viewed as an equivalent electrical circuit:

o                          The inside of the cell/axon/any process has longitudinal/axial resistance - RL

o                          The membrane has resistance – Rm - membrane resistance (current flows through resistors).

o                          The membrane also acts as a capacitor – Cm – membrane capacitance (stores charge – separates +s and –s)

o                          The voltage difference between the inside and the outside of the cell (electrostatic force) and the concentration gradient (different concentrations of ions outside versus inside the neuron) act as the battery, pushing ions (charges, i.e. current Im) in and out of the cell through the membrane.

o                          Channels in a membrane also act as resistors.

 

So let’s have a look at two simple circuits:

 

Several rules will come in handy when trying to understand what happens when current is injected into a circuit (i.e. when we close the circuits above):

 

1). Current in series is the same.

I.e. the same amount of current flows through two things that are arranged ‘in series’ (i.e. in a line).

2). Voltage in parallel is the same.

I.e. the voltage difference across say a resistor and a capacitor arranged in parallel is the same.

3). An uncharged capacitor has no resistance.

Remember that current will flow down the path of least resistance, so if you have an uncharged capacitor in a circuit into which you inject current, all of the current will go straight to the capacitor, and start to slowly charge it up.

It takes time to charge up the capacitor.

4). Once the capacitor is charged, it has infinite resistance – i.e. no current will flow

through it.

5) (very important:) Ohm’s Law: V = IR (voltage = current times resistance).

Large current multiplied by a lot of resistance – e.g. the membrane resistance – results in a lot of voltage difference. E.g. the membrane potential, Vm, is so large (-75mV) because of the high membrane resistance which does not allow current to flow through. The lower the resistance, the more current flow (I = V/R).

6) Q = C*V  (Q = charge, C = capacitance, V = voltage) – the higher the capacitance

and the bigger the voltage difference, the more charge you have.

 

 

Let’s see how these simple rules allow us to predict what the graphs of current flow (I) and voltage change (V) will look like when current is injected into a (I) series circuit and (II) parallel circuit:

 

(Vc = voltage across capacitor; Vr = voltage across resistor, Ic = current through capacitor, Ir = current through resistor)

 

 

 

 

 

 

 

 

 

 

 

I.                           Series circuit:

 

When you inject current, the same amount of current will flow through both the capacitor and the resistor, since current in series is the same. Hence you know that the two curves for Ic and Ir have to look the same. Additionally, you know that most of the current will flow through the capacitor in the beginning, since it is uncharged at first, and less and less current will flow through the capacitor as it gets charged. Hence the curves for Ir and Ic will look like:

 

Ic/Ir

time

 

The voltage curve can be predicted from what we know about capacitors: it takes time to charge a capacitor. Hence initially, there will be little to no voltage difference across the capacitor, but once the capacitor is charged up, we have a maximum voltage difference across the capacitor. The voltage curve for the resistor, on the other hand, can be deduced from the voltage across the capacitor: voltage in parallel is the same, but this is a circuit in series, hence we know that the voltage curve for the resistor has to be complementary to the one for the capacitor:

 

Vc

Vr

time

time

 

 

II.                        Parallel circuit:

 

For a parallel circuit, we already know that the voltage across the resistor and the capacitor (which are in parallel) will be the same, hence the curves look the same. Since it takes time to charge the capacitor, it will start with no voltage difference, after which the voltage increases gradually. For the current, we know that an uncharged capacitor offers no resistance, hence all of the current will initially flow through the capacitor, and none through the resistor. As the capacitor gets charged up, less and less current flows through the capacitor, until all of the current flows through the resistor. Hence, the curves will look like:

 

 

Vc

Vr

time

time

 

 

Ic

Ir

time

time

 

 

 

 

 

 

Let’s look at these pictures “in space” – i.e. let’s look at how current and voltage change along an axon, when we inject either a brief current pulse or some steady-state current:

 

 

 

 

1)  If we inject constant current (steady-state current) into an axon, assuming that there is no capacitance, the current flow or voltage will be highest at the injection site, and as you move away from the injection site, there is progressively less and less current and also voltage. This is because current flowing down an axon encounters more and more resistance, and because current leaks out through the membrane. Hence the inverted V-shaped curve.

2)  If we inject a brief impulse of current into an axon without capacitance, then the curve will look similar to 1) (i.e. there is maximum current/voltage at the injection site and less far away from the injection site), but when you turn the current OFF, that current/voltage IMMEDIATELY dies away.

3)  If we inject constant current in an axon which (realistically) does have capacitance, then current flow or voltage start out small (but are still highest at the injection site) – this is because the current is sucked up by the membrane, i.e. it goes to charging up the membrane capacitor. This is why there is almost no current far away – because all the current goes to charging up the membrane at the injection site. Gradually, as the membrane is getting charged, more and more current can flow (and hence the voltage rises too), and the curve slowly rises until it hits its maximum – the top inverted v-shaped curve.

4)  Finally, if you inject a brief current pulse in an axon with capacitance, then as in 3) the current is initially sucked up by the membrane (hence very little current flows), then more and more current can flow as the membrane is getting charged, until when the capacitor is completely charged, and the current and voltage curves look like in case 1) – the inverted v-shape at the top. However, as soon as the current stops (since it’s a short pulse), the membrane capacitor starts to discharge slowly – and the curve gradually descends back to zero (i.e. time point 1 looks like time point 7).