Nerves, the heart, and the brain are electrical. How do these things work? This course presents fundamental principles, described quantitatively.

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Bioelectricity: A Quantitative Approach

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Nerves, the heart, and the brain are electrical. How do these things work? This course presents fundamental principles, described quantitatively.

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Electricity in Solutions

This week's theme focuses on the foundations of bioelectricity including electricity in solutions. The learning objectives for this week are: (1) Explain the conflict between Galvani and Volta; (2) Interpret the polarity of Vm in terms of voltages inside as compared to outside cells; (3) Interpret the polarity of Im in terms of current flow into or out of a cell.; (4) Determine the energy in Joules of an ordinary battery, given its specifications; (5) State the “big 5” electrical field variables (potentials, field, force, current, sources) and be able to compute potentials from sources (the basis of extracellular bioelectric measurements such as the electrocardiogram) or find sources from potentials.

- Dr. Roger BarrAnderson-Rupp Professor of Biomedical Engineering and Associate Professor of Pediatrics

Biomedical Engineering, Pediatrics

Hello, again.

This is Roger Cook Barr for the Bioelectricity Course.

In this section, section 1 subsection 9, we're going to talk about and

compute the values for some axial currents in our core conductor nerve fiber.

So this easy.

We could do it with Ohm's Law.

We assume first that the current is uniform in between this disc on

the left and this disc on the right.

Uniform, I mean, is not coming into the fiber or going out of the fiber.

It's just traveling along the fiber from one place to another.

We can use Ohm's law, just regular standard Ohm's law.

See as the current is the voltage divided by the resistance.

We know the voltage from the previous section.

We know the resistance from two sections back, two subsections back

to 40 millivolts and 636,000 Ohms, approximately.

If I substitute those values into Ohm's law, I get

the axial current 62.8 nanoamperes.

Nanoampere is one billion of an ampere.

So a very tiny current, relative to an ampere,

which might be the current required to light a lightbulb.

So, it's very tiny current in comparison but none the less sufficient.

There's another way to get the axial current from the electric field.

Let's do it that way too, just for comparison.

Again, we assume that the current density is uniform going from A to B.

Now let's find the electric field.

The electric field is defined as the change in potential,

divided by change in position.

So going from A to B, we'd say the electric field,

the magnitude of the electric field, potential of B minus the potential of A,

divided by the position of B, minus the position of A.

So, that will be equal to minus V AB, note that is a negative sign for

V AB because we are saying B minus A, divided by the length in between.

EX you noticed has a direction.

The formula gives the magnitude,and the sign will indicate the direction.

VBA and L come from our earlier work, so, we compute the electric field,

4 volts per centimeter, and in the positive X direction.

Once again, please repeat this calculation for

yourself, so you see exactly where it came from.

Once we know the electric field, we can find the axial current density and

then the axial current.

First, to get the axial current density, we use this expression,

and substitute in numbers.

The axial current density J in the X direction is equal to

the interior conductivity times the electric field.

You notice the result is current per centimeter square.

That means the amount of current per each centimeter square of cross section.

If we now multiply the axial current density, Jx by the cross sectional area,

then we find the current flowing along x.

Here the cross-sectional area is quite small, so,

we multiply the axial current density Jx times this

cross-sectional area and we come up with the axial current.

And of course, it comes out to be the same as that before.

Now, whether to do it one way or

whether to do it the other way, it's here just a matter of arbitrary choice.

But in other problems, it is advantageous to do it one way or

do it the other way because it simplifies the problem.

So that's why I bring it up.

Also, it's interesting to think about doing it this way,

because the electric field is a measure of

the force that will be exerted on each of the charged particles.

So knowing the electric field gives you some insight that you

don't have unless you calculate how much that is.

[SOUND] Well, once again, thank you for watching this section.

I'll see you again soon.