- #1
himmellaeufer
- 8
- 0
I have a hard time understanding the concept of electric current and voltage in a circuit. Most of all I do not understand the physics behind the "sudden" voltage drop due to a resistor (although I do know Kirchhoffs and Ohms law).
I think I may have the same problem as discussed here:https://www.physicsforums.com/threads/why-does-an-element-in-a-circuit-cause-a-voltage-drop.823346/
We are probably doing the same thinking mistake. What I don't get is the following example.Lets say we have a + and - pole (i.e. a simple battery) connected by a wire. Now of course, due to the difference in electric potential of the two poles there will be an electric current. According to
E=-∇φ
this is due to the electric field creating a force on the electrons F=qE (which doesn't yield in infinite drift velocity due to scattering - in the simple Drude modell).
If we consider a uniform E-field and a one-dimensional problem, we can also say that φ(x)=-Ex+C. According to this, there would be a linear voltage drop with the spatial coordinate x, no? So even if there was a load there must not be a jump in the electric potential because this is clearly a spatial property.
In this example the E-field should not be altered in x direction due to the flowing current (the E-field components in x-direction of the traveling electrons will cancel each other out and the other don't affect the travel in x direction). I also don't get the analogy of water falling on a wheel - here the potential energy changes due to the distance to Earth's surface. The wheel does not change anything about that. The water just lost some kinetic energy, but it will accelerate again (as would the electron due to F=qE).
Is my understanding of the E-field and the electrostatic potential in a circuit wrong?
Right now I am thinking that if you have a battery and connect the two poles with a wire, the E-field somehow travels through the wire (with c) and the current flows due to φ(x)=-Ex+C. This would also mean that the electric field in the wire would get smaller with longer distances and the same voltage, doesn't it?
I think I may have the same problem as discussed here:https://www.physicsforums.com/threads/why-does-an-element-in-a-circuit-cause-a-voltage-drop.823346/
We are probably doing the same thinking mistake. What I don't get is the following example.Lets say we have a + and - pole (i.e. a simple battery) connected by a wire. Now of course, due to the difference in electric potential of the two poles there will be an electric current. According to
E=-∇φ
this is due to the electric field creating a force on the electrons F=qE (which doesn't yield in infinite drift velocity due to scattering - in the simple Drude modell).
If we consider a uniform E-field and a one-dimensional problem, we can also say that φ(x)=-Ex+C. According to this, there would be a linear voltage drop with the spatial coordinate x, no? So even if there was a load there must not be a jump in the electric potential because this is clearly a spatial property.
In this example the E-field should not be altered in x direction due to the flowing current (the E-field components in x-direction of the traveling electrons will cancel each other out and the other don't affect the travel in x direction). I also don't get the analogy of water falling on a wheel - here the potential energy changes due to the distance to Earth's surface. The wheel does not change anything about that. The water just lost some kinetic energy, but it will accelerate again (as would the electron due to F=qE).
Is my understanding of the E-field and the electrostatic potential in a circuit wrong?
Right now I am thinking that if you have a battery and connect the two poles with a wire, the E-field somehow travels through the wire (with c) and the current flows due to φ(x)=-Ex+C. This would also mean that the electric field in the wire would get smaller with longer distances and the same voltage, doesn't it?