# Would an electron use up its kinetic energy after competing a circuit?

1. Apr 23, 2006

### Twukwuw

I was told that, in an simple electronic circuit (consist of resistor and battery), the power supplied by the battery must be equal to the power (heat) dissipated by the resistor.

In form of equation:
VI = RI^2 ,where V=voltage supplied by the battery, I=current, R=resistence.

I find that, this equation is based on 1 assumption:
on average, the electron will end up with ZERO kinetic energy when it reach the terminal of positive. In other words, all the energy it gains from the battery will be
used up competely to overcome the collision and other resistence along the circuit.

My qeustion is, has the assumption I mention here been made?
If yes, what GUARANTEE that on average all the kinetic energy must be used up after completing a circuit?

I think, there is a possibility for the elctrons that, after several collisoon, they may have some "residual" kinetic energy and they continue to reach the positive terminal.The consequesce is, these powerful electron will collide with the positive terminal of battery and hence stop there by losing their kinetic energy in form of heat.
This is supported by the fact that, after some time, the battery will be heated up.

2. Apr 23, 2006

### vanesch

Staff Emeritus

What you write would be correct, if the electrons were moving in a vacuum. In fact, what you write is what happens in a particle accelerator, and indeed, when one "dumps" the beam, there's a lot of dissipated heat generated.

However, the kinetic energy of electrons IN A CONDUCTOR is extremely low, because of the successive collisions. This is BTW exactly the reason of electrical resistance.
The heating up of the battery has nothing to do with the remnant kinetic energy of electrons completing the circuit, but with losses during the electrochemical process that produces the potential difference (in other words, not 100% of the chemical energy goes into restoring the potential difference ; some percentage is lost in heat during the chemical reaction).

3. Apr 23, 2006

### Twukwuw

oh I see....

That means, along the way, the elctrons actually always maintain their kinetic energy at "small magnitude" due to collisions. In other words, the gain in kietic energy (from the battery) is always balanced out by the loss of heat in collisions, roughly speaking.

Moreover, the change in speed of an electron is not significantly large when it reaches the posotive terminal....so, when the electron complete a circuit, the change in kinetic energy is always negligible.

This is why I understand...is it correct?

Twukwuw.

4. Apr 23, 2006

### vanesch

Staff Emeritus
Yes. Although extremely elementary, look up the "Drude model"
http://en.wikipedia.org/wiki/Drude_model of conduction in a metal.

In fact, the Drude model is essentially correct concerning electron transport in gasses, but less so in metals.

5. Apr 24, 2006

### michaeltorrent

typical drift velocity of electrons in a small circuit is a few mm per second. while their rms speed due to thermal energy is several thousands meter per second.

6. Apr 24, 2006

### Gokul43201

Staff Emeritus
The kinetic energy of the elctrons remains the same; it is the potential energy that is decreasing.

The analogy to a river flowing downhill can be helpful. The battery provides the height of the (for simplicity, conical) hill and the series resistance determines its base diameter (or the horizontal distance traveled by the river). Increasing the height of the hill (battery voltage) or decreasing the diameter (resistance) makes the river flow faster (increases the current).

During the course of the journey, however, due to dissipative forces along the way, there is no change in the average flow velocity.