Electricity. What kind of current should I use?

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1. Apr 8, 2015

MarcusAu314

Look, I found that there exist two kind of electric current: conventional flow and electron flow. I have read that in every situation I have to use the conventional flow, however I'm going to work with electric circuits using copper wires. In solids, only negatively charged particles (electrons) flow across an electric conductor.

So, what kind of current should I use?

Is it the same thing after all? If so, does everything remain the same or something changes?

2. Apr 8, 2015

rootone

Current together with the voltage applied, produce a measure of the amount of power which the circuit is using.
Conventionally current is measured in Amps, there would be very few situations where it would be preferable to describe current in units of electron flow.

3. Apr 8, 2015

nsaspook

I'm an old Navy guy from the vacuum-tube era so we used electron flow in training.

4. Apr 8, 2015

nasu

It will depend on what do you mean by "use". What do you want to do?

5. Apr 9, 2015

vanhees71

If you use vector calculus and the charge density, there's no such confusion left. Sometimes a slightly more advanced mathematics helps to resolve such issues. A current is then easily understood as the flow of electric charge. If you have a particle density $n(t,\vec{x})$ (number of particles per unit volume around $\vec{x}$), its flow $\vec{t,\vec{x})$ (collective velocity of the particles around $\vec{x}$) and if the particles carry a charge $q$, the current density is given by
$$\vec{j}(t,\vec{x})=q n(t,\vec{x}) \vec{v}(t,\vec{x}).$$
This gives the charge flow (current density). The meaning is as follows: If you have an arbitrary surface $F$ with surface-normal elements $\mathrm{d}^2 \vec{F}$, the total current (charge per time) flowing through the surface is given by
$$I=\int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{j}.$$
Note that the sign of this current depends on the choice of the direction of the surface-normal elements (i.e., on the orientation of the surface, which is of course arbitrary).

The current density $\vec{j}$ is easier to understand: It's direction at any point is in direction of the flow velocity of the charges, if $q>0$. If $q<0$, the current density points into the opposite direction. So for electrons, which are the usual conduction-current carriers, the current density is oppositely directed to the flow velocity. It's as simple as that!

6. Apr 9, 2015

7. Apr 9, 2015

DaPi

The theoretical answer is that as long as you consistently stick to whatever convention you choose, it doesn't matter.

In real life, the heart-ache arises when you need to talk to someone else or, worse, incorporate their results in yours. You'll come across this when you need to consider the "direction" of the magnetic field produced by your current.

Vacuum-tube designers will have their own "trade" conventions and be well aware that the rest of the world does it differently. How about transistor designers? They have to consider holes(+) as well as electrons(-). And when you look at the battery that powers your experiment or electrolysis in the chemistry class, you'll also have to consider charge carriers of both signs.

8. Apr 9, 2015

Staff: Mentor

We should nominate Ben Franklin's arbitrary choice as the most calamitous accident in the history of science.

1000 or 100000 years from now, is there any chance that society will have flipped the sign convention for electron charge to be positive? Probably not.

9. Apr 9, 2015

DaPi

Plank not using E = h * omega has probably spilt just as much ink.

10. Apr 9, 2015

vanhees71

No, Ben Franklin didn't do anything wrong. What mistake should he have made?

11. Apr 9, 2015

Staff: Mentor

He chose which polarity was positive and which negative. See "Benjamin Franklin (1706–1790)", Science World, from Eric Weisstein's World of Scientific Biography.

His choice was arbitrary, so it wasn't wrong, it was unfortunate. This very thread is an example. Centuries later, millions of students struggle with the idea that electrons flow in the opposite direction of current.

12. Apr 9, 2015

MarcusAu314

So it doesn't matter at all which one I use?

13. Apr 10, 2015

Staff: Mentor

I think you should use conventional current so you are speaking the same language as everyone you interact with.

14. Apr 10, 2015

nsaspook

Understand what's happening and go with the flow.
Wiring: Conventional current is what almost everyone uses in diagrams.
Tubes: ?

15. Apr 13, 2015

vanhees71

First of all, what's "unfortunate" in Franklin's choice? It's simply arbitrary and became a convention.

Again, I can only advise not to use confusing ideas about a "technical" and "true" direction of current, but to start from the local fields, which are the fundamental quantities and have the most simple and intuitive meaning. If $n(t,\vec{x})$ is the number density of charges under consideration (particles per volume) and $\vec{v}(t,\vec{x})$ is the flow field of the charges and $q$ the charge of a single charge, then the current density is given by
$\vec{j}(t,\vec{x})=q n(t,\vec{x}) \vec{v}(t,\vec{x}).$
The current involves then another artibrary convention, namely the direction of the surface normal vector used to define its sign. It's defined by
$$i(t)=\int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{j}(t,\vec{x}),$$
and gives the charge per time running through the surface $F$, with the sign given by the direction of both the current density and the surface normal vectors $\mathrm{d}^2 \vec{f}$.