what important elements can relate to electrical resistance when considering about the resistance effect and force? i find some equations may be concerned,like the Ohm's law,R=V/I.V is voltage and I is currents.Also R=gL/A g is resistivity and L is length, A is cross section also how to measure the resistance in experiments?


closed as unclear what you're asking by Neil_UK, Bruce Abbott, Daniel Grillo, Dmitry Grigoryev, JIm Dearden Aug 30 '16 at 16:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


The main four things are the mobility of the charges, the charge density, the average cross section area of the conductor, and the length of the conductor.

The essential idea is that free electrons can exist as a kind of cloud within a material. (The so-called 'electron-gas' theory.) When there is an applied electric field between the two ends, this field accelerates the cloud of charges in a particular direction. As they accelerate under that force, they eventually collide with an atom (electrons are very much lighter than heavy metallic atoms), where it is presumed that they simply come to an abrupt stop. They then start back up again under the electric field accelerating potentials, accelerating until the next impact happens. On average, there is a certain average velocity called the "drift velocity" and given the time between collisions you can then get the "mean free path" that they achieve before hitting another atom, again, and stopping. The average speed reached is \$v = \mu \mathscr{E}\$, where \$\mathscr{E}\$ is the electric field usually in \$\frac{volts}{meter}\$ and \$\mu\$ is the mobility in \$\frac{meter^2}{volt\cdot second}\$. You can experimentally approach getting the mobility of a material using one of several methods, which include time of flight (using thicker materials) and using an impulse voltage and observing the resulting current impulse response function (thinner materials.) Similar concepts apply to positive charges.

Energy is lost into increasing the vibrational energy of the impacted atoms. This becomes heat.

The reason why length would affect the total resistance (not resistivity) should be pretty obvious, as more length means more impacts and more lost energy. A larger charge density (number of charges within a volume times their charge and then divided by the volume they occupy, or \$\frac{N\cdot q}{L\cdot A}\$) clearly leads to more charges per unit time crossing some chosen cross-section and therefore increases the net current given some electric field. So clearly a larger charge density should reduce the resistance. Finally, the mobility itself is directly related to drift velocity (scaled by the applied electric field) and so better mobility (higher average velocity given a specific electric field) means lower resistance. The resulting equation is something like \$R = \frac{L}{p\mu A}\$, where \$L\$ is the length, \$A\$ is the cross section area, \$p\$ is the charge density, and \$\mu\$ is the mobility of the material.

It's actually a pretty simple idea to apply, when you think about it. The mobility is the tricky measurement to get right. But if you have it, you can figure out a lot.

For example, say you have a bit of silicon that is \$2cm\$ long and has a square cross-section of \$2mm \times 2mm\$. The mobility of silicon is already tested in the lab and is \$\mu = 1300 \frac{cm^2}{V\cdot s}\$. You now impress \$V= 10V\$ between the ends of the bar and measure \$I=80mA\$. You can now get the conductivity as:

\$ \sigma = \frac{L\cdot I}{A\cdot V} = \frac{2cm\cdot 80mA}{4mm^2\cdot 10V} = 40 (\Omega\cdot m)^{-1}\$

And can now compute the drift velocity under a \$10V\$ accelerating field between the ends as,

\$v = \frac{I\cdot \mu}{A\cdot \sigma} = 65 \frac{m}{s} \$

Just think of it as a cloud of electrons with equal spacing between them like a gas, sitting in a volume of material. There will be an effective charge density as a result of that idea. And if you apply an accelerating force (voltage), then there will be an electric field potential set up, measured in \$\frac{volts}{meter}\$, and this will now accelerate the cloud. These charges will periodically smack into atoms along the way, coming to a stop and depositing some equivalent thermal energy, then will accelerate back up to speed again before the next impact. It's a very simple concept that works pretty well.


Not the answer you're looking for? Browse other questions tagged or ask your own question.