Using a simple lattice model of conduction, where electrons are accelerated by an electric field, and are slowed down by bumping into the lattice, you get the following equation for current density:
$\vec{J}_n=nqμ_n\vec{E}$
Let's imagine an ideal DC voltage source connected with perfectly conducting wires to a resistor. Just from intuition, as the electrons reach the lattice of the resistor, you'd think that there'd be a pileup of electrons, since they don't have as much mobility in the resistor (almost like a traffic jam at tight roads). Do electrons or other charge carries collect at the end of resistors? If they do, is this what creates a voltage drop across resistors (or equivalently, an electric field across a resistor)? This idea of a collection of charge seems to imply a capacitance to the resistor. Do real resistors display any in-built capacitance?
Answer
Do real resistors display any in-built capacitance?
Yes, and series inductance too. In fact, physical resistors have self-resonance frequency and a Q. At "high-enough" frequencies, these non-ideal contributions must be taken into account.
See, for example, this lecture: The Hidden Schematic
If I recall correctly, there is a charge density gradient through the length of the resistor. I remember seeing a nice graphic of this but I haven't found it online as yet. I will post as soon as I do.
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