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In an ideal circuit, How can there be a current b/w points a & b, when there is no potential difference and thus no electric field between a & b? If there is no current, then where does current across the resistor come from, because that means no charge is coming from the battery (?).

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    $\begingroup$ Related: physics.stackexchange.com/q/81536 duplicate: physics.stackexchange.com/q/28660 $\endgroup$ – user6972 Nov 16 '13 at 5:46
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    $\begingroup$ Your question from a different perspective answered here: physics.stackexchange.com/q/80400 $\endgroup$ – Alfred Centauri Nov 16 '13 at 14:25
  • $\begingroup$ What evidence do YOU have, that there IS a current flowing, under the conditions you cite. Why not do the experiment, and report what current you observe with zero Voltage. $\endgroup$ – user26165 Nov 18 '13 at 6:36
  • $\begingroup$ @GeorgeE.Smith Due to the current state of industry, no ohmic resistance is available having $0$ resistance, however there is no law of physics in electromagnetism that would stop such a conductor from existing. $\endgroup$ – Isomorphic Nov 18 '13 at 17:59
  • $\begingroup$ @GeorgeE.Smith Also, if there was no current flowing between a & b, I do have evidence, that electrons are flowing the resistance R. I do not know then where they came from and are coming again and again. $\endgroup$ – Isomorphic Nov 18 '13 at 18:02

An electric field isn't necessarily required to sustain a current. Remember electric charge is accelerated by an electric field.

In the case of an ideal conductor, which is assumed to connect the source to the resistor, the current can be any value and the voltage across the conductor is identically zero.

This isn't a contradiction. Consider the motion of an object in the absence of friction. No force is required to sustain that motion (only to change it).

Analogously, in the absence of resistance in the ideal conductor, no electric field is required to sustain a current through.

If it helps, consider a non-ideal conductor with some total resistance R. The voltage across to sustain a current $I$ through is:

$$V = I\cdot R$$

Now, let $R$ go to zero and see that, for any value of $I$, the voltage across is zero.

I reluctantly add this because, after some discussion in the comments, I think there is some confusion over the meaning and purpose of ideal circuit theory.

When the OP opens the question with "In an ideal circuit", he sets the context as ideal circuit theory which is a well known, well understood, widely used branch of electrical engineering. Perhaps the OP isn't aware of this context. Perhaps some of those that answered and/or commented aren't aware. Thus, this addendum.

What needs to be made clear is that ideal circuits and circuit elements are used to model physical circuits and physical circuit elements. The ideal circuit elements are meant to correspond to mathematical terms in the equations for the solution of the circuit. They do not represent physically realizable electric circuit components.

Thus, any answer along the lines of "there are no ideal circuits" entirely misses the point.

And, any complaint along the lines of "there must a voltage across because of Ohm's Law" entirely misses the point.

The confusion lies, I think, with the distinction between a physical schematic or, if you will, a "wiring diagram", and an ideal circuit schematic.

What's the difference?

The first represents the physical components and their connections. Useful for technicians, test engineers, etc. etc. but not for calculations and/or simulations.

For that, an ideal circuit schematic is used either explicitly or implicitly to translate the physical circuit into a mathematical model that can be used to calculate and simulate.

For example, here's the schematic symbol for an ideal transformer with the secondary connected to a load:

enter image description here

Unlike a real, i.e., physical transformer, the ideal transformer is lossless and has infinite bandwidth. How would one calculate or simulate a real transformer? By augmenting the ideal circuit schematic with additional ideal circuit elements that model the non-ideal characteristics.

For example, an ideal circuit model of a real transformer looks like this:

enter image description here

Note that every circuit element in that diagram is ideal and thus, isn't physically realizable but the entire ideal circuit corresponds to a good mathematical model of a real transformer that can used for calculations and simulations.

To further drive this point home, let's consider the OPs schematic as a "wiring" diagram for a physical battery connected with wires to a physical resistor.

Since this a DC circuit, a simple model of a battery is an ideal voltage source in series with some small value ideal resistor. A simple model of a physical wire is small value ideal resistor. Thus:

enter image description here

But, and again, each circuit element above is ideal including the wires that connect the ideal circuit elements.

And, again, for the ideal wire, there is no voltage across for any value of current through. This defines the ideal wire and that's really all that needs to be said about this.

  • $\begingroup$ Ok, then what starts the flow in the first place ? Analogous to motion of mass, to move on a flat surface one needs to start the motion. $\endgroup$ – Isomorphic Nov 15 '13 at 18:09
  • $\begingroup$ I'm having trouble with your statement: "An electric field isn't necessarily required to sustain a current." While true for a superconductor which is a QM effect, is this true for ohm's law cases where current density is proportional to E field. $\mathbf{J} = \sigma \mathbf{E}$ $\endgroup$ – user6972 Nov 15 '13 at 18:11
  • $\begingroup$ @Iota When the source is first applied there will be a potential difference that must travel at close to the speed of light across your ideal wire assuming some physical length exists. This voltage wave will travel to the load and back to the source. It might do this several times before reaching equilibrium with the entire circuit, thus "starting the flow". However I think once reaching the static case there has to be some conductivity and a very small field to keep the current flowing through it. In an "ideal" wire the resistance isn't non-zero, just too small to matter. $\endgroup$ – user6972 Nov 15 '13 at 18:20
  • $\begingroup$ @iota, you wrote "this is an ideal circuit", i.e., the conductors connecting the resistor and source are ideal, i.e, perfect. From "Electromagnetic Theory for Microwaves and Optoelectronics", pg 21: "The charges inside a perfect conductor are assumed to be so mobile or the relaxation time so small that charges move instantly in response to changes in the fields, no matter how rapid. Then just as in the static case, there must be neither a time-varying electric field nor electric charge inside a perfect conductor..." $\endgroup$ – Alfred Centauri Nov 15 '13 at 18:32
  • $\begingroup$ @user6972, see my response to iota. $\endgroup$ – Alfred Centauri Nov 15 '13 at 18:33

There's a limit to how an "ideal model" can be used. Actually, that was an unintended joke: sometimes you need to consider how an ideal model behaves in the limit, not at the limit.

So, consider a voltage source with $V = \epsilon$ with zero internal resistance and look at the current flow through a resistor as the resistance $R$ approaches zero. The current increases without bound.

Next, what's the current as both $V$ and $R$ approach zero? The answer is, of course, "it depends" on how they approach zero.

  • $\begingroup$ +1 for "how they approach zero" because I don't think ohm's law really allows them to be zero because then there would be no current density. $\endgroup$ – user6972 Nov 15 '13 at 18:44
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    $\begingroup$ @user6972, the circuit is either ideal or it isn't. The OP has specified that the circuit is ideal but you seem to be insisting that it isn't ideal because that's not physical. But ideal isn't physical. It's an approximation. $\endgroup$ – Alfred Centauri Nov 15 '13 at 19:24
  • $\begingroup$ @AlfredCentauri I tried to elaborate in my own answer. $\endgroup$ – user6972 Nov 16 '13 at 4:57
  • $\begingroup$ If $\vec E=0$,R doesn't matter. Moreover R is a constant for a conductor. Ohm's Law is I is proportional to V. So if V=0, i=0, R doesn't matter. $\endgroup$ – Isomorphic Nov 16 '13 at 12:10
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    $\begingroup$ @Iota You're mixing ideal elements (constant R) with a denial of other ideal elements (R=0). Also see "superconductor" $\endgroup$ – Carl Witthoft Nov 16 '13 at 14:42

Your question is difficult to answer exactly because of the assumptions you have to make for the ideal case. In the real world you can not have current flowing without an electrical field due to Ohm's Law where $\mathbf{J} = \sigma \mathbf{E}$.

You can try to argue in a perfect conductor $\sigma \rightarrow \infty$ so it doesn't matter what $\mathbf{E}$ is, but this is just the type of hand waving you can get away with only when you are talking "ideals". When you ask:

How can there be a current b/w points a & b , when there is no potential difference and thus no electric field between a & b ?

There is a subtlety in your question that I think is important to understand conceptually when learning circuits. And in fact you are correct to question it because current could not flow without a field, this is a requirement. However to simplify the model we often ignore it and this assumption works quite well for most cases however it can also lead to gross misunderstandings of conductivity in more complex situations.

It is important to understand that ideal conductors don't exist and those lines drawn between elements can have a profound impact on how a circuit works. There are 4 major conditions where the mentality of an "ideal wire" burns people because they ignore the fields:

  • When the wavelength approaches 0.1 * circuit size or smaller.
  • When the current inrush is very large
  • When large or fast transients occur
  • When the wire is significantly long in relation to its current load.


And what about those superconducting rings that don't loose their circulating current for 3+ years in the lab and won't until long past the age of the Universe? Often people are confused by the fact that superconductors can conduct current without an electric field. This is unrelated to the idea of an ideal conductor. In an ideal conductor based on ohms law, a superconductor could never do this. So we have to look for another mechanism. This is the Meissner Effect which only happens in a special case of superconductivity while the material is exposed to a magnetic field during the superconductivity transition:

Any perfect conductor will prevent any change to magnetic flux passing through its surface due to ordinary electromagnetic induction at zero resistance. The Meissner effect is distinct from this: when an ordinary conductor is cooled so that it makes the transition to a superconducting state in the presence of a constant applied magnetic field, the magnetic flux is expelled during the transition. This effect cannot be explained by infinite conductivity alone.

Zero resistance would imply that if you tried to magnetize a superconductor, current loops would be generated to exactly cancel the imposed field (Lenz's law). But if the material already had a steady magnetic field through it when it was cooled trough the superconducting transition, the magnetic field would be expected to remain. If there were no change in the applied magnetic field, there would be no generated voltage (Faraday's law) to drive currents, even in a perfect conductor. Hence the active exclusion of magnetic field must be considered to be an effect distinct from just zero resistance.

  • $\begingroup$ I don't think you need potential difference for flow of current, let's get an analogy from gravity. Yes to start motion of a mass, you need a height drop but once a body has achieved that height drop, it can move indefinitely on a smooth surface. Also saying, that an ideal circuit doesn't exist, so we won't answer about ideal circuits, is like saying a perfectly smooth surface doesn't exist, so newton's second law is an approximation only. Now it may be an approximation, but these laws form a mathematical system and we should be able to deduce things from these laws ultimately. $\endgroup$ – Isomorphic Nov 16 '13 at 6:13
  • $\begingroup$ (contd.) If I would have asked to find acceleration of a body for an object on a smooth surface, then an answer like a perfectly smooth surface does't exist may be true, but that isn't a law of physics as we know it. I would be happy to accept and up vote your answer if you could please consider these points. $\endgroup$ – Isomorphic Nov 16 '13 at 6:16
  • $\begingroup$ @iota, I do believe you grok the context perfectly. As you might already know, to model a real conductor, with finite conductivity, we simply add a resistor to the ideal circuit schematic that models the total resistance over the length of the conductor. This is how we simulate real conductors in circuit simulators such as SPICE. $\endgroup$ – Alfred Centauri Nov 16 '13 at 12:53
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    $\begingroup$ "but this is just the type of hand waving you can get away with only when you are talking "ideals"." But that is precisely the context, ideal circuit theory. While your at it, please do point out that that the resistor in the circuit diagram doesn't exist since any real resistor has associated inductance and capacitance. And do point out that the voltage source in the schematic doesn't exist since any real voltage source has internal resistance, limited energy etc. All of the symbols in that diagram are "hand waving ideal approximations". But, I don't think that's news to the OP. $\endgroup$ – Alfred Centauri Nov 16 '13 at 13:20
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    $\begingroup$ @iota, user6972 wrote above "you need a field to move charge". Well, no. An isolated charge at rest in one frame is a moving charge, i.e., a current, in another relatively moving frame. Another example is diffusion current: "Diffusion current occurs even though there isn't an electric field applied to the semiconductor. It does not have E as one of its parameters." ece.utep.edu/courses/ee3329/ee3329/Studyguide/ToC/Fundamentals/… $\endgroup$ – Alfred Centauri Nov 16 '13 at 21:40

There is some small potential difference between a and b, because conductors always have nonzero resistance (at room temperature).


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