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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

Exactly what is this "circuitous route"? Does the thing I touch also have to be touching the carpet? Though I'm not a native English speaker I am pretty sure that a circuitous route is a path that combines you shoes and the carpet as were they a part of a circuit. The thing you touch has to be connected to the carpet (by touching the carpet itself or ...


2

I am not really sure what equalize surface charges means, but yes, charge gets redistributed and shared in this new part of the circuit as there will be a new equilibrium potential in between the two original values in the seperated systems. Remember though that their potential must be measured with the same reference (or else you could not know if one was ...


2

The answer is not quite simple, to show this we need some graph theory and matrices. There is a beautiful document explaining this relation in detail http://www2.math.uu.se/~takis/L/Circuits/2000/handouts/graphsandckts/graphsandckts.pdf I think the "fundamental reason" of this is related with the fact that every loop have different variables, if we can ...


2

Here's a counterexample: Suppose two identical, ideal batteries (with zero internal resistance) are both connected in parallel across a single resistor; equivalently, replace one of the resistors in your diagram with a second, identical battery. Also assume the conducting wires are ideal (again, no resistance). Kirchhoff's laws in this case result in an ...


2

A battery is no capacitor, and the actual charge stored in the battery terminals is very low. When you connect the anode of one battery to the cathode of another, that charge is transferred very quickly, and the voltage drops to zero. When you connect anode and cathode of the same battery, a chemical reaction takes place, and charges flow inside the battery ...


2

Okay, so capacitors allow current to flow, but not electrons to flow, between the terminals. As a consequence, when they are operational, a density of electrons builds up on one side and a density of missing-electrons builds up on the other side, until the charge on either plate is given by $|Q| = C |V|$ where $C$ is the capacitance and $V$ is the voltage ...


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

Anything (charges or any objects) can move at constant speed in vacuum without requiring any force exerted on it. That's just Newton's first law of motion. Here, the charge is not in a vacuum but in a wire where you assume that the resistance is negligible. In reality that is impossible in a regular metal wire at normal temperatures : there MUST be a very ...


1

Consider a circuit with a capacitor, a voltage source, and a switch. Suppose the voltage source is DC and we flip the switch. If the capacitor is initially uncharged, then at the instant you close the switch current will flow as if the capacitor was not there. Instead of an electron crossing the capacitor, an electron will arrive at the negative capacitor ...


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

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, ...


1

For a capacitor with closer plate spacing (all else being equal), the electric field in the dielectric between the plates is stronger. Assuming a uniform electric field, the potential difference is given by the product of the spacing and the strength of the electric field: $$\Delta V = E\cdot d $$ So, the potential difference can be the same for both ...


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

Here the use of DC in your question is a bad choice. DC does not imply constant current, it means that the current have same polarity in a period that we refer to. So you would like to use steady state in your question. That said, in steady state the the current is not changing with respect to time so the flux is not changing and hence there is no emf. But ...


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

There are a few key bits of physics and math to understand here, but one does need to be very careful to avoid producing a circular argument. The key, concept I think is that of a linear circuit element, for which the output is precisely proportional to the input. The precise meaning of 'input' and 'output' depend on the precise device, but this does ...


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

The electron gun produces electrons by heating the cathode; this shakes out electrons from the metal ("boils them off"), and as soon as they're out, they are repelled and accelerate away to do your bidding. This is called Thermionic Emission. When an electron is emitted, another one comes in from the cable connecting the cathode to the power source. The ...


1

The emf is not the work done per unit charge integrated along a closed loop by a source that is not electrostatic, it's just: $$ \mathscr E =\oint \vec{f}_s \cdot d\vec{\ell},$$ where the integral is around the circuit, and $\vec{f}_s$ is the net force per unit charge on the conduction charges that move about within the circuit element. This might seem ...



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