# Steady current and logical inconsistencies that Im struggling to resolve

1. Apr 28, 2007

### O.J.

Dear all

i've thinking, how can a circuit have a steady current?
the reason for this seemingly naiive question is:

if I is contant it means the velocity at which the charges move is constant. But if the velocity is contant that means their KE is constant. So in a circuit with a resister for example, how can electrons before the resistor have the same velocity as the ones coming out of it? doesnt that mean they havent lost any energy?

2. Apr 28, 2007

### Staff: Mentor

The electrons don't gain energy (at least not in the direction of the field).

Loosely speaking, the electrons are continually banging into the lattice of the conductor/resistor. Every bit of energy they gain from the driving electric field gets dissipated as heat. Realize that the net speed of the electrons through the circuit (the drift speed) is quite small--they never get the chance to build up speed. You might find this discussion helpful: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html" [Broken]

Last edited by a moderator: May 2, 2017
3. Apr 28, 2007

### O.J.

so my question is
suppose u have a number of electrons approaching a resistor. they'd be moving at a net drift velocity. Now they've went past that resister. is their drift velocity the same or less?

4. Apr 28, 2007

### Staff: Mentor

I'd say it would be the same.

5. Apr 28, 2007

### daniel_i_l

And just to be precise, a stronger current doesn't mean that the electrons are moving faster, it means that more electrons are passing through the wire every second. (I = dq/dt)

6. Apr 28, 2007

### O.J.

if its the same, then how come the resistor has energy dissipated through it? where did it come from if not from the KE of the electrons?

7. Apr 28, 2007

### Staff: Mentor

It comes from the electrical potential energy that the electrons lose as they travel through the resistor: $q_e \Delta V$ for each electron, where $\Delta V$ is the potential difference (in volts) between the ends of the resistor.

8. Apr 29, 2007

### cesiumfrog

It maybe helps to consider this analogy: Water pumped at high potential (pressure, voltage) through a hose which has a section of high resistance (a narrow kink, resistor). The current flow rate will be the same before and after the section of high resistance (obviously the water/electrons can't just pile up), however, the potential (the pressure relative to the atmosphere at the end of the hose, or the voltage relative to ground) will be much higher in front (compared to behind) of the section with high resistance.

If you put pinholes in the hose (on both sides of the narrow kink), the water would spray much higher from the hole in front of the kink. This demonstrates that the water molecules have lost energy through the highly resistive section (and indeed, the flow's friction would have heated that section more than the rest of the pipe). Nonetheless, this doesn't contradict that the current is constant along the hose. The water molecules have the same average net drift velocity on both sides, but the individual velocities also have a random component (which is larger on one side of the kink and observed on that side as exerting larger pressure on the containing walls of the pipe).

Likewise, there is no contradiction in the fact that the current (or average drift velocity) remains constant despite electrons each individually loosing kinetic energy in a resistor.

Last edited: Apr 29, 2007
9. Apr 29, 2007

### O.J.

so every point on a circuit closer to the attery is at a higher potential than the ones following it?

Last edited by a moderator: Apr 29, 2007
10. Apr 29, 2007

### Xezlec

No, the closer a point is to the positive side of the battery, the higher its potential. The closer to the negative side, the lower the potential. Potential just means voltage. Voltage just means the pressure of the electrons in a certain segment of metal. See https://www.physicsforums.com/showthread.php?t=166904".

I'm not sure it's helpful to talk about the electrons themselves losing energy when they pass through the resistor (I mean I guess they do, but only in they sense that they go from a higher pressure to a lower pressure). It's more useful to think about the electromagnetic wave delivering energy to the resistor.

As I told someone recently, electrons only drift at a few inches per hour in normal circumstances. But when you flip a switch, the light turns on long before the electrons get there. The electrons didn't have time to carry "kinetic energy" all the way to the light. That isn't how it works.

The energy is carried by an electromagnetic wave, which carries an electric field to the light at nearly the speed of light. The electric field instantly makes the electrons in the light start moving. These electrons encounter resistance, but they are pushed onward by the electric field. It is the field that delivers the energy. Kinetic energy of the electrons has nothing to do with it.

Last edited by a moderator: Apr 22, 2017
11. Apr 30, 2007

### NoTime

Yes, barring the case of a zero resistance conductor.
The component resistor just represents the major potential drop in the circuit.
A plain piece of wire shorting a battery might make this simpler.

The KE of the electrons is insignificant in a circuit like this due to the generally low drift rate.
However, the KE can not always be ignored.
For example, the KE of the electron is responsible for light output of a computer display (CRT) as it strikes the phosphor screen.