# Why does resistance reduce current in whole circuit?

• I
• bmhiggs
bmhiggs said:
no wire is idea, so all wires put up a certain amount of resistance, "using" a small amount of the energy of the field, contributing to the overall resistance and creating a field throughout the wire
There are superconducting wires, but yes, for ordinary copper wiring there is a small resistance

bmhiggs said:
...what is the mechanism for the energy transfer in a circuit - say a basic filament lightbulb?
Perhaps my response in this thread from June: Power flow outside a wire - how close? helps to answer this question.

Other than a trivial amount of kinetic energy carried by the low-speed flow of electrons, the vast majority of energy flows between the battery and the light bulb via the electric and magnetic fields outside of, and well removed from, the wires. The electron current in the wires simply serves to establish and guide these fields in the space outside of the wires. Since changes in these fields propagate at the speed of light, the bulb appears to instantly illuminate when the circuit is completed.

Dale
bmhiggs said:
Ok:But:

How can there be a current without an electric field?? Or is that vector E referring to something more specific?
[Keep on exercising patience with me here if that should be obvious...]
Of course, if there's no force on the electrons they come to rest due to friction. There's only a current if there's a force due to an electromagnetic field. If you just have a static electric field along the conductor, after a short while after switching this field on, the electric force, ##\vec{F}_{\text{el}}=-e \vec{E}##, is compensated by the friction force ##\vec{F}_{\text{fric}}=-\gamma m \vec{v}=-\vec{F}_{\text{el}}=e \vec{E}##, i.e., you have ##\vec{v}=-e/(\gamma m) \vec{E}##
$$\vec{j}=-n e \vec{v}=+\frac{n e^2}{\gamma m} \vec{E},$$
where ##n## is the number of conduction electrons per unit volume, i.e., ##\rho=-e n## is the charge density of the conduction electrons.

Now the above is nothing else than Ohm's Law in local form, i.e., you get for the electric conductivity for DC
$$\sigma=\frac{n e^2}{\gamma m}.$$

Dale said:
There are superconducting wires, but yes, for ordinary copper wiring there is a small resistance
Superconductors cannot described by the usual constitutive relations, i.e., not be Ohm's Law, ##\vec{j}=\sigma \vec{E}## and not with a Drude model as sketched in my previous posting, because obviously you cannot simply set the friction coefficient ##\gamma=0## or the relaxation time ##\tau=1/\gamma \rightarrow \infty##.

The reason is that superconductivity is a generic quantum effect, leading to an effective classical theory, which is described by the London equations. For a nice treatment, see

https://www.feynmanlectures.caltech.edu/III_21.html

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