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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 ...


0

Each branch containing resistors acts as a potential divider, dividing the voltage across the whole branch in the ratios 40:50 and 50:40 (top and bottom respectively), ie the ratios 4:5 and 5:4. This means that each resistor will have either 4/(4+5)=4/9 or 5/(4+5) = 5/9 of the total voltage (18V), which as you have correctly calculated is 8V for both the ...


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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There may be a confusion between enery and power. While the current, which is the number of charges per second, and the energy of the ,electron do'nt change at the output of the resistor, the power does. Power is the amount of energy per unit time, and that does not affect the current which, again, is the amount of charge per unit time.


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See picture. The current equal -1 Ampere. It is easy to obtain, if we determine potential in each point. I will set $\phi=0$ in up right corner of your picture. I did mistake in a sign of current on the picture, but I lazily to remake picture.


0

I don't have the facilities at hand to draw a circuit diagram, so I'll just describe it. A conventional electromechanical current meter consists of these parts: M1 -- the meter movement itself (which we idealize as having zero internal resistance). R1 -- the innate resistance of the meter movement. This is packaged inside the meter, but we can consider ...


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You can find a proof of Thevenin's theorem at the following site. https://www.researchgate.net/publication/262764661_Thevenin%27s_Theorem__An_Easy_Proof_Suitable_for_Undergraduate_Teaching


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) = ...


0

Let us assume that the instrument design is the same as in http://en.wikipedia.org/wiki/Ammeter (the words "spring constant / torsional constant", "flux, number of loops" seem to imply that). Then, to get the same deflection for lower current you need more loops, therefore, the resistance of the milliammeter will be higher. It looks like the same ...


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Answer #1 uses an invalid argument. The conclusion that a milliammeter has lower resistance does not follow from the observed fact that the milliammeter has higher deflection. Consider that some work must be done to deflect the needle. Because of the assumption that the physical constants are the same, the work is equal for equal deflection. Also ...


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

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 ...


0

A capacitor has an infinite resistance (well, unless the voltage gets so high it breaks down). The simplest capacitor is made from two parallel plates with nothing but space in between - as you can guess from its electronic symbol. In a DC circuit, a capacitor acts as an open circuit and does not permit current to pass. In an AC circuit a capacitor has an ...


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Actually, the capacitor is an open circuit. It has infinite resistance. (I will assume a DC circuit.) But since the plates are so close, charge build up on one plate induces charge in the opposite plate of the opposite sign. This means that putting a battery across a capacitor (see the picture below) will let negative charge (in the form of electrons) flow ...


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Any electrical component has some kind of "resistance" - that is, it takes effort to move a current through it (leaving aside for a moment superconductors). We talk about "resistance" as a way to quantify this effort; for a simple resistor, the resistance has a constant value (at a given temperature) which means that the relationship between voltage and ...


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 ...


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 ...


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 ...


0

Olly, the first part of your thinking is correct, as the atoms receive more energy, the electrons do collide more energetically, but they also move "away" from the atom's center. The further they are from the center, the easier it is for an electric field to "move" them. This means that for the same effort (voltage), more electrons are moved (larger ...


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 ...


0

Yes ... but you may not like it. It can be very math intensive, especially when a little practice or other methods tell you the same thing with less effort. Let me first suggest some other methods. For most undergraduate and high school physics courses, you can usually identify this by drawing a circuit diagram, and use the smallest sections of circuit to ...


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If I had to solve this problem on a paper with a pen but without the Kirchhoff equations then I would do it in four stages in each replacing the voltage sources with a short circuit and the current source with an open circuit. Then one by one putting them back (3 sources - 3 steps). In step 0 without the sources I would calculate the resistance between the ...


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You could use a freeware like pspice student or ltspice which allow you to model electric circuits like the one you descirbed. You can easily calculate currents and voltages in any element in your circuit.


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You can clearly move lines (which are assumed to have no resistance) around by continuous movements. Try to see how to go from the first picture to the second one by this sort of operations, that is, stretching/squeezing/identifying lines with same potential


15

Start with the initial diagram, but let's color code everything: Now move some wires around, without actually changing the connectivity: Finally, rotate the left and right blocks while again not changing the connectivity:



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