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Imagine a circuit with a constant voltage and no resistance. Every electron would gain the same kinetic energy as it moves from the negative to the positive terminal. If we add a resistor to the circuit, some of the electrons energy is given to the atoms of the resistor due to constant collisions and re-accelerations. Usually a resistor has a smaller free electron density. So, the only way a high current could be maintained would be a higher drift velocity in the resistor. However, this isn’t possible in this case because we assumed the voltage to be constant. Therefore, the current decreases as a result of adding a resistor.

When the resistor is exchanged for a longer resistor, the current in the circuit drops even further, due to the higher resistance. That would mean the drift velocity of electrons in the resistor is reduced. The free electron density and the acceleration the electrons experience shouldn’t change if the new resistor consists of the same material and if the voltage stays constant. So why would the drift velocity decrease when the length of the resistor increases?

Qmechanic
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QuantumQuasar
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1 Answers1

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There is no such thing as ideal wires - wires always have resistance that limits the current. What may mislead here is that in the [lumped element model][1] we assume that all the resistance is concentrated in a resistor: this simplifies thinking about circuits and making calculations, but one should be aware that ideal wires between lumped elements are not real physical wires.

Thus, the electrons flowing in a circuit always experience resistance, that is they can be though of as being under influence of two forces:

  • the electric force due to the potential difference, which accelerates the electrons
  • resistance/friction force that slowers the electrons As the result the electrons move with a nearly constant drift velocity. One could write a very crude version of [Drude model][2] as a Newton's second law $$ m\dot{\mathbf{v}}=q\mathbf{E} - \gamma m\mathbf{v}, $$ where $\mathbf{E}$ is the electric field, whereas $\gamma$ is the "friction coefficient". The stationary solution of this equation is $$ \mathbf{v}_d=\frac{q\mathbf{E}}{m\gamma}, $$ which becomes the current density when multiplied by $qn$: $$ \mathbf{j}=qn\mathbf{v}_d=\frac{q^2n\mathbf{E}}{m\gamma}=\sigma\mathbf{E}, $$ where $\sigma$ is the conductivity.

The total current flowing through the wire is then proportional to the current density times the cross-section, $I=jA$, whereas the electric field is proportional to the voltage divided by the length of the wire, $E=V/l$, which leads to $$ I=jA=\sigma E A =\frac{\sigma A}{l}V=GV=\frac{V}{R}, $$ where $$ \frac{1}{R}=G=\sigma\frac{A}{l}. $$ The short answer is thus: the potential difference in a longer wire falls across greater length and results in smaller electric field.

Roger V.
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