Voltage drop in Drude's model and the electromotive force I'm trying to understand the meaning of the voltage drop and of the electromotive force in view of Drude's model. This model assumes a drift speed for electron equal to:
$\mathbf{v_d} = \frac {-e \mathbf{E} \tau } m$
where τ is average free path time.
The field E is a electromotive field thus around the circuit we have:
$ R\cdot i= \int_C \vec{E} \cdot d\vec{l}$
When electrons strike the ions in the lattice their velocities is zeroed and then accelerated again by E to reach $\mathbf{v_d}$ before a second collision. In this way, isn't the energy lost during the collision regained by the electrons? 
By considering a led + resistor circuit the resistor protects the led by the excess of energy. If the current is stationary we have the same $i$ inside and outside the resistor. If the electrons are accelerated to the same $\mathbf{v_d}$ by the field the resistor shouldn't be of any use; There shouldn't be any voltage drop at all.
The only possible explanation I can give is that the electromotive field is decreasing along the circuit. If it is so I'd like to better appreciate how E varies inside the circuit. I have never found any expression for E and in Drude's model it is never said that is varying.
 A: It sounds like you have a pretty good understanding of the Drude model.
What you need to remember is that electric field provides a force on a charge. So for a free charge, electric field will be proportional to acceleration, not velocity (because $F=ma$). If applying a field results in a constant velocity, then there is no net force on the charge. This can occur if there is a friction-like force proportional to velocity (as in the Drude model, and consequently Ohm’s Law). 
So in a resistor, the collisions (phenomenologically described by $\tau$) evidently result in an effective force equal and opposite the applied field.
This “scattering” force, which is friction-like, occurs over some distance and corresponds to a potential drop over that distance. So you could think of the voltage drop as the potential the electron would have had if it had been allowed to accelerate. Ohm’s Law just says that since the opposing force is proportional to velocity, you can calculate the potential drop from the final velocity ($\propto$ current) and scattering rate ($\propto$ resistivity).
