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.