Does current decrease with frequency increasing?

Question

Electric field accelerates electrons in wire.

Because of at least inertia, when the electric field (voltage) is applied, electrons will not instantly have some maximum speed, but will continuously accelerate to maximum value that can be calculated by Ohm’s law.

I am wondering: if electrons need time to achieve some speed, will the current (amplitude speed of electrons) decrease with high-frequency AC voltage applied?

Say, for 100V (DC) electrons need 1ms to accelerate to maximum speed. So, if we apply, for example 251 Hz AC square-waved source, the voltage will be decreased before 1 ms, and will be applied to opposite direction, damping electrons. So they will not accelerate to maximum, and, hence the resulting current will be decreased?

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I actually forget about Ohm’s law for AC voltage, and yes, current inversely depends on frequency, so higher frequency, lower current, but I want to know is my understanding correct. I want to understand the fundamentals of the phenomenon. People in comments started to answer kinda “it is not because of inertia, it is because of inductance”, the same as to state, for example, that ballon inflates not because outside air atoms will bump it insides, but because of “pressure”, although it is the same, just a bit more fundamental explanation.

Answer 1

It could be instructive to work out yourself the ac response of the Drude model or its simplified version (see my own posts here and here). The key point is that electron acceleration by the electric field is already taken into account in the electron mobility/conductance $$ \sigma_0 = \frac{nq^2\tau}{m}, $$
since the electrons are constantly reaccelerated by the field after being scattered from the lattice (impurities and phonons). In particular, the ac conductance is predicted to be $$ \sigma(\omega)=\frac{\sigma_0}{1-i\omega\tau}= \frac{\sigma_0}{1+\omega^2\tau^2} + \frac{i\omega\sigma_0}{1+\omega^2\tau^2} $$ As one can see, the conductance has an imaginary part, characterizing the current retardation in respect to the electric field. Moreover, it decreases in magnitude at higher frequencies - that is, the cvurrent decreases with increasing the frequency, as $$ \mathbf{j}(\omega)=\sigma(\omega)\mathbf{E}(\omega) $$

Answer 2

In the classical theory the free electrons are in thermal equilibrium with the atoms of a conductor, and are bouncing at random at a very high speed from one atom to another. An electric field will cause an acceleration between each bounce and cause a very small drift. The mean free time between collisions is small compared with the period of most achievable frequencies.