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3

To simplify, let's think of a prismatic volume, as if a planar figure (in $xy$ plane, area $A$) had been extruded in $z$ direction. Let's assume that conductivity ($\sigma$) is dependent on $x$ and $y$, but not on $z$. And let $L$ be the prism length in $z$ direction. We want to compute total resistance between both ($xy$) parallel faces. If we take a small ...


2

A good reference was given in an answer to a related question: Cserti 2000 (arXiv preprint, whose numbers I'll be referring to) solved a number of generalizations of the 2D lattice problem. For a $d$-dimensional lattice, the resistance between the origin and the point $(l_1, \ldots, l_d)$ is given by eq. 18 in that paper: $$ R(l_1, \ldots, l_d) = R_0 ...


2

For your circuit, $V = I\cdot R$. You are plotting (unusually) R along the X axis and $\frac{1}{I}$ along the Y axis, so the slope is $\frac{1}{V}$. Now the fact that this slope is a straight line tells you that the voltage is constant. This means that (over the range of your experiment) your voltage source has a low internal resistance. Imagine for a ...


1

If it's a simple circuit where Ohm's law applies, then we should get $$V=IR$$ so we see that $$V/I = R$$ $$1/I=R/V$$ $$1/I = (1/V) \times R$$ The gradient should then be $1/V$. Seems like a slightly bizarre plot but if you got a straight line then that makes the maths simple at least!


1

You can not find the temperature dependence from this definition. You should look into the microscopic details of the process. For that, you should refer to the Boltzmann transport equation from statistical mechanics. You can find it in a textbook on Statistical Physics or Condensed Matter physics. In short, at high temperatures the resistivity is dominated ...


1

$\frac{2}{a} = 440$ It should rather be $\frac{a}{2} = 440$ So you get $a=880$ for the parallel ciruit. There is still some difference between the two measured resistances. It looks like your ampere meter has an internal resistance of 66 Ohm or your volt meter has an internal resistance of 13 kOhm: If the ampere meter is in series with the volt ...


1

There seems to be some confusion about carrier motion on your part. Carrier diffusion occurs all the time - in a field-free region the net will eventually go to zero in steady state (but carriers are still moving and diffusing around). Drift current is the result of the (slight) bias in charge carrier motion caused by an applied electric field. No field, no ...


1

I agree with both of the two answers above. But I would also suggest that the term resistance should not really be applied at all to a capacitor ( I think it leads the question astray ) Going to the more general notion of impedance ( I would use the complex formulation ) the question makes more sense to me. This immediately brings in the importance of ...


1

To basically summarize and re-organize the linked-to answers: 1) When a charge $q$ is moving (say at velocity v) through a perfect conductor such as an ideal wire, it requires no force to maintain its velocity because it encounters no resistance. This is good, since there can be no electric field inside a perfect conduct and thus no force can be applied to ...


1

Normally, the activation energy is constant. There is a boltzmann type term... $e^{-Ea \over kT}$ where $Ea$ is the activation energy and $T$ is the temperature (and $k$ is boltzmann's constant) so generally the number of free conductors, $N_{free}$, at energy $E$ can be calculated from the number bound, $N_{bound}$, by something like $$N_{free}(E) = ...


1

If you have current flowing one way through a resistor, then the electrons flow through the other way. Since current flows from the high voltage end of a resistor to the low voltage end, then the electrons come in at the low voltage end and come out at the high voltage end. When electrons (which are negatively charged) go from low voltage to high voltage, ...



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