in order to confirm whether I understand the relation ship between amps, columbs and electrons I decided to try out the following.

my laptop adapter's label reads 3.34 amps. I assume that:

  • amps are the rate of flow of charge (i.e speed of flow measured in coulombs per second) at a point in a circuit.
  • that elementary charge (e) is the charge of one electron
  • that one coulomb is the sum of the charge given by 1/e electrons

so the number of electrons that will pass through my laptop (or at least the adapter) per second are:

3.34 x (1/e) = 2.084664255E19 electrons

is this correct?

EDIT: having seen the answers I have seen that there's a lot of information missing here.

The point of the circuit I would like to focus on is any cross section of the cable that leads from the adapter to the laptop.

I also assume that the laptop is at full power and drawing the following as labelled on the adapter:

  • 19.5 V (since this is out of the adapter I'm assuming DC current)
  • 3.34 A
  • 1
    $\begingroup$ yes.. Ampere is Coulomb (a unit of charge) per second so you just have to navigate the units correctly (like you did). $\endgroup$ – BjornW Jul 24 '11 at 16:06
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    $\begingroup$ One small problem: The label of such an adapter does not give the current which flows actually, this is always the maximal current the adapter is dimensioned for. So to get the real charge per second, You had to measure the current. $\endgroup$ – Georg Jul 24 '11 at 19:29
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    $\begingroup$ Also note that the input electricity is AC, so the number of electrons passing through any circuit is time-variable. $\endgroup$ – Tim Jul 25 '11 at 8:09
  • $\begingroup$ @Tim's suggestion would imply that the device has a power output of 400 W if we assume a 120 V outlet. This sounds the most plausible since that's what many computer power supply units are rated for. Indeed, the # of net electrons passing along the length of wire is different at both sides of the converter, and the definition needs to be more carefully considered anyway. $\endgroup$ – Alan Rominger Jul 25 '11 at 13:28
  • $\begingroup$ I've clarified some more of the question now but, I see the main barrier now is drift. $\endgroup$ – code shogan Jul 26 '11 at 7:17

I would start from power consumption. It's a little unclear to me if you're talking about electron use in all of the components or the CPU, but since 3 Volts was used, I'll take the discussion to be limited to CPU. Let's say 30 W of power consumption.

$$I=\frac{P}{V}=\frac{30 W}{3 V} = 10 A$$

Now, for the electron flux (denoted $\phi$), we can just divide by the electron charge.

$$\phi = \frac{I}{e} = \frac{10 A}{1.9 \times 10^{-19} C} = 5.263 \times 10^{19} \frac{1}{s} $$

This is the number of net electrons passing the power supplying terminals. I've heard that the drift velocity of electrons is 1000 to 10,000 times less than the actual velocity of electrons. So... you could just multiply the above number to get $5.3 \times 10^{22}$ electrons participating in current passing through a cross section of the terminal wire, but this is unsatisfactory for a number of reasons.

Firstly, at any given point in your circuit there are lots of electrons passing every which way. Why bother counting these unless we're talking about the role of conduction band electrons in temperature itself? Additionally, even if we restrict ourselves to the net electron flux in a wire - there are LOTS of wires in the CPU, and the passing of electrons through the terminal wire is not exclusive with electrons passing through other points, and a CPU is a very complicated machines with many many routes. In order to obtain more objective measures, you might want to look at the rate at which electrons bleed off, or loose voltage in some way (this would be the most analogous to a computer "using" those electrons). Otherwise, there is a virtually unlimited number of net electrons passing through all points in the circuits.

I also don't mean to present this as a very thorough answer, but at least I've taken a different approach to get the same order of magnitude of electrons that the question had. These numbers are just generally representative of a few Amps, but one needs to be specific in labeling what that number represents.


As explained here, "Electric current is the rate of charge flow past a given point in an electric circuit, measured in Coulombs/second which is named Amperes."

The charge in a normal conductor is essentially all due to electron flow. Therefore, the rate of electrons past a given point in a circuit (your computer) can be calculated based only on the amount of current and the charge of each electron.

So yes, your calculation is correct: 3.34 amps divided by the elementary charge (~1.602 × 1019 coulombs per electron) is around 2.08 × 1019 electrons per second.


most wall outlets in the united states run at 60HZ. each cycle gives two peaks so if you multiply your result by 120 you should get the number of electrons. though to be fair I think if you actually study conductors you would know that the mean drift velocity is super slow--- so that it's not really correct to say that electrons are passing through the laptop... though I don't really know enough about this to go any further.

  • $\begingroup$ this is actually a much more complicated question than it appears to be... I'm gonna leave the rest to someone who really understands electrical current + conductors in great detail $\endgroup$ – Timtam Jul 25 '11 at 10:05
  • $\begingroup$ I thought about the HZ but figured that since the adapter functions as an ac~dc converter that I wouldn't have to take that into account. $\endgroup$ – code shogan Jul 26 '11 at 6:38

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