# Is current the speed of the electrons?

My teacher got me really confused this time. He said that current is the rate of flow of charge. If this is true, then current could be the speed of the flow of charge (ie. the kinetic energy of the electrons). I decided that this was the case, and didn't ask him any questions. I thought I was all clear, until he said that the current is the same in all parts of a circuit. I asked him further questions, like what happens to the current if it passes through a motor or something, and the answer was 'the current stays the same, but the electrons slow down'. How is this even possible?

• Sounds as if the teacher was wrong or you may have misunderstood. The drift velocity is generally lowest in the leads. The larger the diameter, the slower the drift velocity.
– user137289
Commented Feb 13, 2018 at 10:14

In the most general case the particle current density is defined as particle density $n$ times the mean particle velocity $\vec v$ (drift velocity): $$\vec j = n \cdot \vec v$$ If the particles carry electrical charge ($e$ for electrons) you get an electrical current density: $$\vec j_{el} = e \cdot \vec j = e \cdot n \cdot \vec v$$ (the same with mass and mass current, etc.).

The electrical current used in circuits is the current density based on the traversed area: $$I = \int \vec j_{el} \cdot d\vec A$$

As you can see, the electrical current density is indeed something like the rate of flow of charge, but there is also the particle density. If the electrons slow down somewhere the current density stays the same because the particle density increases. More particles with less speed give the same current density as less particles with more speed.

I used the electrical current density to explain this but it is also valid for the electrical current itself, unless the traversed area changes (in a cable the cross section is everywhere the same, also in a motor).

• I don't even know what electrical current density is ._. I'm still 15, still studying my IGCSEs. This cleared a lot of my doubts though Commented Feb 13, 2018 at 12:44
• @iNehMan have a look here hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html Commented Feb 13, 2018 at 19:00

The current in a conductor is the charge passing through a cross-section of that conductor per second.

To get a mental picture think of electrons flowing in a metal wire. Imagine you could count the number of electrons going through a chosen cross-section per second, as if you were counting the number of vehicles passing under a bridge on one carriageway of a road. In the case of the electrons you'd get the current by multiplying the number per second by the charge ($1.60\times 10^{-19}$ C) on each one.

So the current, $I$, isn't at all the same thing as the (mean) velocity, $v$ of the charge carriers (electrons), though $v$ is one of the factors on which $I$ depends. In fact$$I=nAve$$in which $n$ is the number of charge carriers (each of charge $e$) per unit volume of the conductor (cross-sectional area $A$).

Current is related to the speed of electrons but it is not the speed of electrons! Current is the amount of charge that passes through a cross sectional area in one second. Relating current solely by the speed of free electrons is wrong.

Current is given as $$N \times A \times V \times E$$ where:

• N is the number of free electrons per unit volume
• A is the area of cross section
• V is speed of free electrons and
• E is electron charge

There is no formal derivation for this equation because it is basically the definition of current.

Imagine there are some let's say 500 electrons in unit volume of a conductor with some cross section. Through that cross sectional surface in one second $$A \times V$$ volume passes and in that volume $$N \times A \times V$$ electrons are present and the number of electrons times the charge of an electron is the current passing through that surface and hence the equation!

Now in a series circuit we assume there are no charge accumulation at any cross section, hence by conservation of charge, currents throughout the conductor must be the same!

Thinking of current as the "speed of the charge" like you do is a good way to think about it. And it's correct that the speed is the same in the entire circuit (if there is only one path for the current to take) like your teacher said.

It's also true that current through, for example, a motor will slow down the charges, or "reduce the speed".

But the thing is that if the charges slow down in the motor, they also have to slow down everywhere else!

Just look at the animated gif below. It represents the charges flowing in a circuit. As you can see, the speed is the same everywhere.

Every time an electron goes out of the battery on one side, an electron goes into the battery on the other side. This is how it works. So it's not possible that the charges stack up anywhere in the circuit.

This means that if you introduce a motor in the middle of the circuit that slows down the charges, the charges will have to slow down everywhere in the circuit, not just at that point:

• why does a motor in the middle of the circuit slow down the charges?
– john
Commented Jul 19 at 2:12
• @john Because a motor has resistance in it's wires. Resistance slows down the charges. Commented Jul 22 at 7:46

If current had depended only on the speed of its electrons,then current would be equal to speed.But that's not the case current depends on the number of electronic particles passing through a cross sectional area of a conductor. This means current depends on the amount of electrons,Area of the conductor and speed at which the electrons travel. I=N.A.V If the electrons are charged electrically then its called electrical current thus I=N.A.V.e

The electric current is defined as the Change of Charge with time. The Change of charge $\Delta Q$ can be expressed by the total number of electrons $\Delta N$ that are flowing by the relation

$\Delta Q = e \Delta N$

with the elementary Charge $e$ (this is constant). Consider a wire with $n$ electrons per unit volume. During a time interval $\Delta t$ the volume that electrons will sweep out during the flow is

$\Delta V = A \Delta s = A v \Delta t$

with the cross section area $A$ of the wire. Also we have $\Delta N = n \Delta V = nAv \Delta t$; hence the current is given by:

$\frac{\Delta Q}{\Delta t} = I = enAv$.

From this equation you see that the current will NOT ONLY depend on the drift velocity of the electrons. It will also depend on the number of electrons per unit length $\sigma:=nA$. When the current passes through some devices (e.g. motor), electrons will not only flow through it; the electrons will be slown down (e.g. due to collisions with other particles or external magnetic fields acting on it). However, the current will remain the same; the value $\sigma$ must behave as $\sigma = \frac{I}{ev}$ for equal $I$; thus this value must increase if the electrons slow down.

If there is a brancing in a circuit into 2 pathways, the current will split as follows:

$I = I_1+I_2$

where subscripts denote the quantity belonging to the corresponding pathway (Kirchhoff law).

According to electrodynamics, the total Charge is conserved in a closed System. This also implies that the incoming current in some control volume must be equal to the outgoing current.

• No, electrons won't normally get slowed down in a motor.
– user137289
Commented Feb 13, 2018 at 11:02