Having some difficulties with the concept of pd in a circuit.

Potential difference is a measure of how much energy is used up between two points in a circuit.

However, see the picture I have drawn. In a circuit, there is a potential difference between the negative and positive terminals of the battery. This causes the electrons to flow around the circuit and the movement of charge transfers energy through the circuit.

Now, we expect the pd across the bulb to be 12V in an idealised circuit (no internal resistance, no resistance in the wires). This pd of 12V represents the potential energy lost by the electron (?). As such, when the electron exits the bulb, I can then not explain why the electron continues to move in the circuit without any potential energy.

I have a few ideas:

  • Electrons themselves do not carry energy, this energy is carried in the electric field. So perhaps the energy we see on the voltmeter is some sort of measure of the energy lost by the electric field (if that is a thing?)
  • Wires have resistance to some degree. The resistance of these wires means that the electron does not lose all of its potential energy across the bulb.

But as you can see, it's all quite imprecise and I have no real certainty in any of the ideas. I also know I am wrong since I can pick out even more flaws in my argument. So help in understanding would be appreciated. enter image description here


A potential energy difference due to an electric field is only required to accelerate charges. In the region of the resistor the electrons are not able to move in straight lines. They collide with ions in the resistor and lose some energy (This energy appears as heat in the resistor). So an electric field is required to accelerate them once again. The average velocity attained by the electrons is called drift velocity. The electrons exit the resistor with this velocity.

In the region of the ideal wire the electrons are able to move in straight line paths without any collisions because an ideal wire has no resistance. So even in the absence of an electric field the electrons maintain their velocity.

A good analogy for this is a block moving with a constant velocity. It needs an external force acting on it to move with a constant velocity on a surface with friction, but on a frictionless ideal surface it moves with a constant velocity even in the absence of an external force.


I have not considered thermal motion of electrons but this is because net velocity due to thermal motion alone is zero.

| cite | improve this answer | |
  • $\begingroup$ Is the electric field supplying energy to the electrons to get them moving as they lose energy through collisions in the bulb or to the bulb to illuminate? Because as far as I am aware, collisions is not the mechanism for energy propagation in a circuit (?) - energy propagates through EM waves. $\endgroup$ – PhysicsMathsLove Jul 13 at 15:58
  • $\begingroup$ physics.stackexchange.com/a/143945/269321 this answer addresses this question. The energy transfer can just be viewed in two different ways. Either as through collisions or as through the electromagnetic field. $\endgroup$ – trinitrotoluene Jul 13 at 16:03
  • $\begingroup$ Thanks that's useful - one more weird question: how does the electric field 'know' where to do the work? For example, it is knows that the electrons have to be accelerated inside the resistor but not elsewhere... Probably some quantum mechanical stuff but I'd be interested if you do know :) $\endgroup$ – PhysicsMathsLove Jul 13 at 16:19
  • $\begingroup$ The electric field does not accelerate electrons in the wire because it does not exist in an ideal wire. It only exists in resistor. This is because of the induction of a very small charge on either end of the resistor. This website explains this concept: phys.libretexts.org/Bookshelves/University_Physics/…. $\endgroup$ – trinitrotoluene Jul 13 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.