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?

  • $\begingroup$ 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. $\endgroup$ – Pieter Feb 13 '18 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).


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 $A$).


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.

  • $\begingroup$ No, electrons won't normally get slowed down in a motor. $\endgroup$ – Pieter Feb 13 '18 at 11:02

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