How do Electrons perform work in circuits?

[nuby]

Do electrons flowing in a circuit perform work with their mass (or kinetic energy)? For example like a water flowing in a creek through a turbine. If so does E = 1/2 mv^2 apply.

 

[pmb_phy]

Do electrons flowing in a circuit perform work with their mass (or kinetic energy)? For example like a water flowing in a creek through a turbine. If so does E = 1/2 mv^2 apply.

The kinetic energy of a non-relativistic electro, such as those in an electric circuit, is K = mv^2/2. A deeper analysis requires quantum mechanics where the concept of force, and thus work, is meaningless.

Pete

 

[nuby]

What does the "/2" represent in that equation (in reality)? K = mv^2/2

Thanks for your response.

 

[pmb_phy]

What does the "/2" represent in that equation (in reality)? K = mv^2/2

Thanks for your response.

Division. E.g. the expression x/2 means that x is divided by two.

And you're very welcome. Glad to be of service.

Best wishes

Pete

 

[nuby]

i knew it was divided by 2 :P .. but I was wondering more why it is divided by two. :)

 

[Snazzy]

$$KE=\int_0^Vmv dv = m\int_0^V v dv = \frac{1}{2}mv^2 \ |_0^V =\frac{1}{2}mV^2$$

 

[nuby]

I'm guessing the /2 means it is only 1/2 the force used to put an object in motion. Is that correct?

 

[chroot]

As Snazzy showed, the 1/2 is the result of taking an integral of a first-order quantity, velocity.

- Warren

 

[Dale]

Whenever you see constants like 1/2 in an equation you can be pretty sure that the equation was derived from some more basic calculus expression. Once you get a few weeks of calculus those terms will make sense.

 

[nuby]

Any other opinions on this topic? The basic question is: Are moving electrons' mass the only source of energy in an electronic circuit.

 

[Dale]

I would think their electric field is much more relevant than their mass.

Your reference earlier to water in a turbine is close. I would actually relate it to water in an enclosed pipe. Hydraulic energy is pressure times volume. The analogy would be of "electricity" being like an incompressible fluid where the energy is voltage times charge. The mass doesn't really factor in explicitly for either case.

 

[nuby]

An electric field in my opinion is just a force that makes electrons move (measured in volts), and not necessarily the main 'work' force in a circuit.

If you have two magets sitting on a table (N/S), with say a unipolar magnet between them (o), and a non-magnetic barrier (I).

like this:

..N<...I...o...<S..

From a simple point of view, the mass of the 'o' will be the main factor of the work actually done to push over (I). I believe this is the same concept occurring in most electronic circuits, and the electron push/pull force then would be the energy 'transmitters' and the mass of the electron, which changes due to electrical resistance (giving off heat/light) is doing the actual 'work'.

If this is true, then voltage is actually a measurement of the potential speed of the electrons in a circuit.

 

[Dale]

If you have two magets sitting on a table (N/S), with say a unipolar magnet between them (o), and a non-magnetic barrier (I).

like this:

..N<...I...o...<S..

From a simple point of view, the mass of the 'o' will be the main factor of the work actually done. I believe this is the same concept that is occurring in most electronic circuits.

First, there is no such thing as a unipolar magnet.

Second, the mass of 'o' is irrelevant to the work done in this example. Remember, work is the (scalar) product of force and distance, mass does not enter in. If 'o' experiences 1N of force and travels a distance of 1m then the work done on 'o' will be 1J regardless of the mass of 'o'.

 

[nuby]

Ok, I understand there is no such thing as unipolar magets, and this is probably a bad example to show my thought... which is, electron mass contributes to most of the work done in electronic circuits

DaleSpam,

So if you throw an object in space, and it travels 1M at 1N force, is the work still 1J regardless of the mass?

 

[Dale]

So if you throw an object in space, and it travels 1M at 1N force, is the work still 1J regardless of the mass?

Correct. An object with 1kg mass will end with a velocity of 1.41 m/s and an object with 2kg mass will end with a velocity of 1 m/s, but the work is 1J in each case and the KE is also 1J in each case.

 

[nuby]

Correct. An object with 1kg mass will end with a velocity of 1.41 m/s and an object with 2kg mass will end with a velocity of 1 m/s, but the work is 1J in each case and the KE is also 1J in each case.

DaleSpam,

Can you give me the simple equation for that, as well as the equation for energy required to slow down that same object?

Thanks

 

[Ulysees]

Sorry to interrupt you here, but reading your initial question I wonder something has been misunderstood. You said:

Do electrons flowing in a circuit perform work with their mass (or kinetic energy)?

They'd have to slow down to perform work out of their kinetic energy. But they don't slow down to a halt, as they move from atom to atom. Instead they gain some kinetic energy under the influence of the field, they lose some (turned into heat), they gain some, they lose some, and so on. Gaining and losing overlap. But they don't stop, so any expression like 1/2 mv^2 is irrelevant to the work actually done to turn their potential energy into heat and warm up the conductor.

 

[Dale]

Can you give me the simple equation for that, as well as the equation for energy required to slow down that same object?

W = f.d
KE = 1/2 m v^2
dW = dKE

 

[nuby]

Can KE = 1/2mv^2 and F=ma or be applied to electrons.

 

[Snazzy]

In some circumstances, it can. (cathode ray tube questions and sorts).

 

[Dale]

Can KE = 1/2mv^2 and F=ma or be applied to electrons.

Yes. However, remember that in most circuits (other than cathode rays etc. as mentioned), the electrons never move with any significant velocity, never attain any appreciable amount of KE, and give up any energy imparted to them almost as soon as they get it.

 

[nuby]

I've heard electrons travel very slowly in standard circuits before (cms per hour), however, I've never seen any proof of this. How was this measurement taken?

Is it possible that conduction band ("free") electrons move rapidly through a conductor in a spiral path, but only 'outputs' as much current is being drawn... This could give the appearance of slow moving electrons.

 

[awvvu]

I've heard electrons travel very slowly in standard circuits before (cms per hour), however, I've never seen any proof of this. How was this measurement taken?

Yes, individual electrons (or rather, I mean the drift velocity) actually move very slowly. See this:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

 

[Snazzy]

The current in a circuit is defined by the amount of charge moving past a certain point each second:

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


Where:

$$\Delta Q=nA\Delta x q$$

Where n is the density of charge carriers (a very large number) and is a property of the material you are running a current through, ie. a constant. Therefore:

$$I=\frac{nA\Delta x q}{\Delta t}={nAv_dq$$

Where v_d is the drift velocity.

 

[nuby]

Is it possible that conduction band electrons, and valance, etc. move a difference speeds?

 

[nuby]

It seems ohms law, V = IR has some similarities to F = ma

Couldn't voltage be considered a force that moves electrons, similar to Newtons, a force that moves mass.

Current is a quantity of electrons (mass) over time.

Resistance will cause a deceleration of the electron flow, same as the deceleration (a) of mass (electrons) due to friction or an opposing force.

How fast would an electron need to travel in order to fit in the equation F = ma or KE=1/2mv^2 in a conductor?

 

[Dale]

Here is an example of the kind of calculation that Snazzy and awvvu alluded to:

My hair dryer operates at 125 V and 15 A. The cable is about 2 m long plus the wiring inside the dryer itself, so let's say that the total path length is 5 m. The cable is 15 AWG which is about 1.7 mm². The molar volume of copper is 7.11 cm³, and there are Avogadro's number of copper atoms in a mole, and one free valence electron per copper atom. Combining these numbers with the mass and charge of an electron we get:

a drift velocity of:
v = 6.5E-4 m/s

and a total mass of electrons of:
m = 6.56E-7 kg

So the total KE of the electrons is:
KE = 1.39E-13 J

If it takes 3 minutes to dry your hair that is a total energy consumption of:
E = 3.4E5 J

So the kinetic energy of the electrons themselves is a miniscule portion of the total electrical energy delivered. In fact, it amounts to only one part in 2400 trillion. The point is that the kinetic energy of the charge carriers is practically irrelevant, what matters is the electric charge of the electron, not its mass, the energy is transferred in the electric field, not in the KE.

 

[Dale]

Couldn't voltage be considered a force that moves electrons, similar to Newtons, a force that moves mass.

The units aren't quite right for that, also voltage is a scalar and force is a vector. Do you know what the gradient of a scalar field is? The E-field is the gradient of the voltage. So what I would say is "the E-field is the field that moves electrons, similar to the gravitational field, a field that moves mass". In this manner the voltage is similar to the height, a measure of the potential energy in the mass or charge and only indirectly related to the force on a mass or charge.

 

[zarbanx]

the work done by electrons in a circuit is through change in its potential.the current in the circuit can be viewed as a stream of positive charges moving from +ive plate to -ive plate.thus when these +ive charges pass through the wire(a non ideal wire which has resistance) there is a fall in potential of the +ive charges and it is this energy tht reappears as heat energy.

i hope it was of some help. i really could not get wht u want to know so i wrote wht i know tell me if it was of any help or not

 

[nuby]

the work done by electrons in a circuit is through change in its potential.the current in the circuit can be viewed as a stream of positive charges moving from +ive plate to -ive plate.thus when these +ive charges pass through the wire(a non ideal wire which has resistance) there is a fall in potential of the +ive charges and it is this energy tht reappears as heat energy.

i hope it was of some help. i really could not get wht u want to know so i wrote wht i know tell me if it was of any help or not

What do you mean work is done through change in potential? What happens to actually make 'work' occur, as the potential changes?

 

[marmoset]

About the integral explanation of the 1/2 in the expression for kinetic energy, I understand the integration, but why does integrating mass x velocity with respect to velocity give kinetic energy?

I had always explained the half by saying that to stop an object mass m traveling at a velocity u, you apply a force which gives the object an acceleration of -F/m. Then I use v2 = u2 + 2as to work out the distance (s) this object travels before it stops. So 0 = u2 - 2Fs/m, and so Fs = mu2/2. Fs expresses the work done to stop the object, and so from conservation of energy the object's initial KE must have beeen mu2/2.

 

[nuby]

When one object(a) is pushing on another object(b) (accelerating it). Isn't 1/2 of the force pushing object (a) from (b), and the other 1/2 (of the total force in system) pushing (b) from (a)? This is how I see it. Similar to centrifugal and centripetal force.

 

[Snazzy]

About the integral explanation of the 1/2 in the expression for kinetic energy, I understand the integration, but why does integrating mass x velocity with respect to velocity give kinetic energy?

I had always explained the half by saying that to stop an object mass m traveling at a velocity u, you apply a force which gives the object an acceleration of -F/m. Then I use v2 = u2 + 2as to work out the distance (s) this object travels before it stops. So 0 = u2 - 2Fs/m, and so Fs = mu2/2. Fs expresses the work done to stop the object, and so from conservation of energy the object's initial KE must have beeen mu2/2.

$$F \cdot dx = F \cdot vdt=\frac{d(mv)}{dt}vdt=v\cdot d(mv)=mv\cdot dv$$

$$KE=\int F\cdot dx = \int mv\cdot dv = \frac{1}{2}mv^2 + C$$

$$KE(0) = 0$$

$$\frac{1}{2} m0^2 + C = 0$$

$$C = 0$$

$$KE = \frac{1}{2}mv^2$$