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So I started off with electrostatics and everything seemed nice and mathematical and justified and then "DC circuits" happened!

I just cannot understand the model of electron flow in electrical circuits. Here are my specific doubts-:

1) If potential difference across a tiny cross section of conducting wire is zero, then why on earth does electron flow across that cross section at all? Never mind potential difference across the whole circuit.

2) Is there a constant electric field across a wire connected to a battery? If yes then how is potential difference across a zero resistance wire constant? Shouldnt it be increasing? Doesnt it violate ohms law? If no, then why do electrons flow at all?

Please take time to consider these doubts and relieve me of my frustration. I havs searched through the net for this but every answer seems like beating around the bush. All of the 4 books I have consulted do not address these facts to my satisfaction.

Frankly I think nobody understands this.

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  • $\begingroup$ This must have been asked before. $\endgroup$
    – Qmechanic
    May 31, 2018 at 11:04
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    $\begingroup$ "Doesnt it violate ohms law?" - Ohm's law for a zero resistance is $V = I \cdot R = I\cdot 0 = 0$. Put another way, according to Ohm's law, the voltage across a zero resistance wire is zero for any current through. By what you've written, you seem to be expecting $V$ to be non-zero and increasing according to Ohm's law. Why? $\endgroup$ May 31, 2018 at 11:08
  • $\begingroup$ I did not mean "according" to the ohms law. What I meant was that for constant electric field, V=Ed should increase. This in turn violates ohm law $\endgroup$ May 31, 2018 at 11:51
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    $\begingroup$ To get a current going, a PD is obviously required. Obviously. However for 0 resistance, 0 pd is required to MAINTAIN a current. Ohms law is a STEADY STATE equation. Which tells us, after transients have died down,what is the pd required to maintain a current. The derivation of ohms law, atleast in my experience is skipped over. Leaving lots of people to be confused over this question $\endgroup$ Jan 14, 2022 at 19:59
  • $\begingroup$ Lumped-element model $\endgroup$
    – Roger V.
    Jan 15, 2022 at 14:58

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Consider an electron that enters your piece of wire with some velocity $v_{in}$ and emerges at the other end with a velocity $v_{out}$:

Zero resistance wire

When we say the wire has a resistance what this means is that our electron will lose energy and slow down. That means we have to supply some energy to the electron to keep it moving at the same velocity. That energy is produced by generating a voltage difference $\Delta V$ between the ends of the wire. This gives the electron an energy $e\Delta V$ to replace the energy it loses to the wire.

Now suppose the wire has zero resistance. That means any electrons flowing through it don't slow down, so no extra energy needs to be added to the electrons, so the voltage difference between the ends of the wire can be zero.

This is why electrons will flow through a perfect conductor even though the potential difference is zero. The electrons have a non-zero velocity when they are pushed into the wire by whatever makes up the rest of the circuit, and they simply keep going through the wire at the same velocity.

In fact you could have a potential difference $\Delta V$ between the ends of your zero resistance wire, and what it would do is accelerate the electrons passing through it. You would in effect have built a linear accelerator. An electron would increase its energy by $e\Delta V$ on each pass through the wire. To maintain the voltage you'd need to keep pouring in power at a rate that matched the increase in kinetic energy of the electrons.

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    $\begingroup$ But this works only in the case of straight loop. When the cross section considered is not parallel to the battery how is the potential difference genrated across its ends? $\endgroup$ May 31, 2018 at 11:56
  • $\begingroup$ In the presence of no wire, the battery generates an electric field. The reason the E field in the wire seems to be parallel to the wire at all points is because of "Surface charges" $\endgroup$ Jan 14, 2022 at 23:35
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For a current to flow in a conventional wire (not a superconductor, vacuum, etc.), the potential difference across any segment of the wire and the electric field in it have to be greater than zero.

In most cases, the potential difference in the wires could be approximated as zero, because the resistance of the wires is much smaller than the resistance of other elements in a circuit, including the battery, and, therefore, most of the voltage drops on those other elements.

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  • $\begingroup$ Understood. But what about the electric field? is it constant in any configuration across a wire? How is it generated anyway? $\endgroup$ May 31, 2018 at 10:36
  • $\begingroup$ It is not constant and it is generated automatically through redistribution of electrons, which will accumulate in front of sections of high resistance to support whatever current ends up flowing in a circuit. You can check out this post for more details: physics.stackexchange.com/q/407558 $\endgroup$
    – V.F.
    May 31, 2018 at 10:48
  • $\begingroup$ If the charges redistribute and accumulate in regions of circuit then this violates Kirchoffs junction law. The fact that field is constant is used to derive J=§E which is another form of ohms law from drift velocity $\endgroup$ May 31, 2018 at 11:53
  • $\begingroup$ The redistribution/accumulation is not continuous, i.e., the charges redistribute themselves as needed and stay that way until some conditions change. So there is no violation of Kirchoff''s law, which is applicable to the steady state. $\endgroup$
    – V.F.
    May 31, 2018 at 13:19
  • $\begingroup$ @V.F., KCL also applies to dynamic situations, but to apply it to this situation you'd have to consider displacement currents (aka parasitic capacitance). $\endgroup$
    – The Photon
    May 31, 2018 at 15:45
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In ordinary circuit analysis (no superconductors involved) when we say the resistance of a wire is zero, we really mean it's close enough to zero that the voltage drop across it doesn't significantly affect the circuit behavior.

If you want to understand exactly how this could work, you could model each wire as a low-valued resistor, and then take the limit as the resistance value goes to zero.

If you want to know the effect of the actual resistance of the wire, you can easily calculate the resistance from the resistivity of copper (~$1.72\times 10^{-8}{\rm\ \Omega\ m}$) and the geometry (cross-section area and length) of the wire. If you design high power circuits, you will certainly run into situations where wire resistance must be considered in the design.

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  • $\begingroup$ Thanks. This is the part I have now grasped. What i cannot understand is why "if potential difference between two points is zero then no current flow takes place b/w them". According to me zero field should be the basic condition not pd, for in some cases work by field across some distance cancels out work by another field between the points considered. So charge can actually move between those points. $\endgroup$ Jun 1, 2018 at 17:58
  • $\begingroup$ @IncludedExcluded, if there are two fields in a conductive material exactly cancelling each other out, then the net current in that region will be 0. But when we talk about zero-resistance wires in a lumped circuit analysis, we're working at a much higher level of abstraction, and we shouldn't be thinking about the details of current density distribution on the wire, etc. $\endgroup$
    – The Photon
    Jun 1, 2018 at 18:06
  • $\begingroup$ My doubt was not that fields cancel each other but there work done does. I get the part where you say Is shouldnt be too picky about the field part $\endgroup$ Jun 1, 2018 at 18:35
  • $\begingroup$ How can the work cancel if the field contributions aren't equal and opposite? $\endgroup$
    – The Photon
    Jun 1, 2018 at 18:38
  • $\begingroup$ If the fields act across different distances $\endgroup$ Jun 1, 2018 at 18:40
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  1. If potential difference across a tiny cross section of conducting wire is zero, then why on earth does electron flow across that cross section at all? Never mind potential difference across the whole circuit.

The elements of circuits are ideal, just like massless springs in mechanics for example. So, if the wiring from positive to negative pole of a battery have some resistance, it can be represented by a resistor connected by wires without resistance. Real wires have some resistance even between close points, but usually orders of magnitude smaller than a resistive load like a lamp.

  1. Is there a constant electric field across a wire connected to a battery? If yes then how is potential difference across a zero resistance wire constant? Shouldnt it be increasing? Doesnt it violate ohms law? If no, then why do electrons flow at all?

$$E = \frac{\partial \phi}{\partial x}$$ so, if the potential $\phi$ is constant there is no electric field. In pratical, real wires have some resistance, that is the ratio between a small potential difference and the current. The gradient of that potential is the electric field. It is not constant for the circuit however. It is much greater across a lamp than across the wires between the lamp and the source.

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Addressing your first doubt, I assume here you are talking about superconducting wires.

If electrons are at rest and 0 potential difference is applied across a section of the wire. Then obviously there will be 0 current. If there is no force acting on them they will continue to move at a constant velocity (in this case, 0 velocity)

now alot of people naively apply ohms law to get the conclusion that in superconducting wires from this rest state, where the electrons are at rest, 0 potential is needed to generate a current in the wire. This is absolute NONSENSE for a current to be GENERATED a force must be applied meaning a potential difference MUST be present to GENERATE a current.

Notice how I said GENERATE a current.

Ohms law is stated as

V=IR

plugging in 0 to this equation gives that for a specific current "I" the potential across this wire is 0

Now if you understand what ohms law is actually saying, then this IS technically true. HOWEVER this potential difference does not represent the potential needed to GENERATEA a current, per say. It actually represents the potential difference needed to MAINTAIN a CONSTANT current.

With zero resistance forces opposing the motion of the electrons in the wire, the electrons will move at a constant speed. Lets say I have a superconducting wire, and I apply a potential difference across this wire. Clearly, starting from rest, the electrons are going to be ACCELERATING and the current is going to increase.

Now what potential difference is needed to MAINTAIN this constant current, not generate this current. Clearly in the absence of resistive forces , zero potential is needed to maintain this current as if I stop applying a potential to my wire, the electrons are going to keep on moving at that constant speed.

Why does ohms law represent the potential needed to maintain the current?

To see this clearly , let's derive the microscopic ohms law.

(Search up the drude model of conductivity.)

The equation of motion of an electron is:

$ma = Eq -(m/T) v$

Here there is an applied E field, with a resistive term proportional to the charges velocity.

Clearly the acceleration of the electron is going to be high at first when the V term is low. But as the electron accelerates, V is going to get bigger and bigger until it is exactly equal to the applied electric field, once this happens, the electron is going to move at a constant velocity

Ohms law is a STEADY STATE solution to this differential equation, and represents the FINAL velocity of the electron once the resistive force equal the applied electric field .clearly to find the final velocity of the electron we set a to be equal to 0,

Set a=0:

Doing so gives

Meaning

$ Eq-m/T v =0$

$v= (qT/m) E$

Subbing in the definition of current density

by definition J = $\rho v $

or $J= nq v$ where n is some number density. Plugging back into our equation

$$J = (nq^{2}T/m)E$$

$$J = \sigma E$$

which means current density J in the steady state is proportional to the electric field at a point. so for a constant electric field, J everywhere will reach some constant value

So now we can understand how ohms law is derived, by taking the steady state solution of the above differential equation, we can see why ohms law gives V=0 for R=0, as the equation represents the conditions needed to have ZERO acceleration. Which obviously comes from zero potential.

Obviously to get the current going , a potential difference IS needed

Your second point is answered above, as ohms law doesn't actually say there IS zero potential, rather the PD to maintain a constant current. If there is a battery connected to a superconducting wire, there most certainly IS a potential difference across the wire, however the current won't be constant, it will be increasing. Hence using ohms law in this way is just nonsense

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  • $\begingroup$ If there's an E field in a wire there is a pd, regardless of conductivity/resistance $\endgroup$ Jan 15, 2022 at 12:13
  • $\begingroup$ If there's a conservative E field in a wire there is a pd $\endgroup$ Feb 6, 2022 at 4:40
  • $\begingroup$ This can also be extended to NON Conservative $\vec{E}$ fields aswell. Hence my generality. $\endgroup$ Feb 6, 2022 at 10:03
  • $\begingroup$ Yes, Physicists sometimes like to use the magnetic vector potential to create a "potential difference". However, for Electrical Engineers, the voltage along a path is proportional to the work associated with moving a unit charge along that path, and a potential difference (for them) is a path-independent voltage in that sense. In the case of a time-varying magnetic field, the work associated with moving a charge depends upon the path. $\endgroup$ Feb 6, 2022 at 12:44
  • $\begingroup$ But this is a physics stack exchange and would not be accurate to say that only Conservative fields can cause a potential difference. This is undeniably false. $\endgroup$ Feb 6, 2022 at 15:51
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  1. If potential difference across a tiny cross section of conducting wire is zero, then why on earth does electron flow across that cross section at all? Never mind potential difference across the whole circuit.

The potential difference between two points in a wire is defined as the work per unit charge required to move the charge between the two points. If there is no resistance to the movement of charge in the wire (i.e., a zero resistance wire) then no work is required to move the charge between two points in the wire, and thus the potential difference is zero.

The mechanical analogy is an object (analogous to charge) moving at constant velocity on a perfectly frictionless surface (analogous to a zero resistance wire). No work (potential difference) is needed to keep the object moving at constant speed on the frictionless surface. However, all conductors except superconductors have resistance.

  1. Is there a constant electric field across a wire connected to a battery? If yes then how is potential difference across a zero resistance wire constant? Shouldnt it be increasing? Doesnt it violate ohms law? If no, then why do electrons flow at all?

The electric field $E$ along a uniform wire equals the voltage gradient along the length $l$ of the wire, or $E=V/l$. So for a zero resistance wire, $E=0$ in the wire. Such a wire would be considered a super (zero resistance) conductor. For a superconductor, electrons can flow continuously without an electric field. However, to initiate the flow of electrons, they need to be accelerated by the force of an electric field.

That field could be supplied by a battery. But once the electrons begin to flow, all of the emf (and field) of the battery would be across the internal resistance that is always present in real batteries, and not across the zero resistance wire.

There is no violation of Ohm's law $V=IR$. If $V=0$ and $R=0$, $I$ can theoretically take on any non zero value.

Hope this helps.

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Wire is a term closely related to a piece made of metal. Metals are not superconductors that pose zero resistance. So the question is somewhat ill posed.

There are some great definitions already found by google.com that answer to some degree the questions subsumed to the main questions. For example:

A circuit with no resistance (0) would indicate a complete circuit, or one that has no short. A circuit with total resistance would indicate a completely open circuit -- one through which no current was able to pass.

Here another one:

An ideal wire has zero resistance, to deliver power with no loss. According to Ohms law, Voltage drop across the wire is V = I x R, if R is zero, no matter what the current the voltage drop across the wire is zero.

There is another way to write the formula due to Siemens that is opposing Ohms lay.

Their process of conduction is in both interpretation statistical and based on models like the Brownian motion. If no such process that can be described with Brownian motion is present than the resistivity or conductivity changes.

There was lately a new discovery on that topic additional to supraconductivity, superconductivity or Cooper pairs. This made use of the fact that pairings are favored for delivering changes in resistivity or conductivity.

That is naming the limits somewhat in more detail:

The superconductor does not shield or change an electric field to measurable extents. The potential difference is the one that is applied. This results in a current which is infinite because of zero resistivity. The current is limited by inductance. It does not go instantly to infinity but increases indefinitely. In the real world, the current is limited by the critical current density or the critical magnetic field strength of the material.

One of the famous phenomenons are the Cooper pairs. These are verified in supraconductors. This is under recent experimental development including discoveries like this discovery segmented fermi surface cooper. They name this supercurrent-induced quasiparticles. This page shows some older models: Cooper pairs. These groups work on raising the temperature for which Cooper pairs appear: cooper pairs spotted above critical temperature for superconductivity.

This deals with

"“Two paradigms” The researchers found clear evidence that, though the sample ceased to be a superconductor at 2.95 K, the shot noise only dropped from that of Cooper pairs to that of electrons when the temperature rose above 7.2 K. The bizarre aspect is that titanium nitride does not have a pseudogap. “From a physics point of view, I find this very fascinating, because there are these two paradigms – one is electron pairs in a superconductor, the other is a good metal made out of fermions,” says Allan. “Ours is a good metal: it doesn’t have a gap. But it looks like our metal is made out of pairs, which are more like bosons than fermions. We didn’t discover anything new about the pseudogap, but we put some of these discoveries in context."

So to understand something is not only being able to deal with an advanced temperature concept compared to Brownian motions but to be firm with Fermi spheres in solids, to known all lot about shot noise like how it is generated, maintained, and measured, and a lot more. Besides Thermodynamics, Quantummechanics a lot more has to be known and dispositioned for gathered some competencies for understanding and describing this zeroness of the resistance or conductance in solids.

These groups use Graphene a somewhat simpler solid offering simpler answers: Cooper-pair propagation and superconducting correlations in graphene.

There is already a type-II superconductors really superconduct – they transmit electricity without dissipating energy.. These extend this pairing wonder to photons: Photons Couple Like Cooper Pairs which is highly required due to the statistics valid for Cooper pairs.

So the limits are not reached and will not be reached since we have the earth's environment, are humans, and so on. Finiteness is physical, infiniteness is not. To humans, accessibility is always finite.

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