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As far as I understand, you need potential difference to draw current.

enter image description here

Assuming ideal conditions, we can state that there is no resistance in the path, and therefore the potential difference between the resistor R1 is V1. This potential difference results in a current of V1/R amperes flowing through the circuit. If there were no potential difference, the current would be 0/R amperes, which is equivalent to 0 A.

What I don't understand is this:

enter image description here

I marked two points on the path, P and Q. Assume ideal conditions with no resistance in the wire. In such a scenario, there should be no reason for the potential to drop between P and Q. If there was a potential drop, it would indicate the presence of some resistance, contradicting our initial assumption. Consequently, we conclude that P and Q are at the exact same potential.

What confuses me is how there can be any current flow between P and Q when there is no potential difference between them. I am aware that this model is an oversimplification and represents an idealized scenario. I am not questioning the real-world implications; instead, I am specifically asking about the occurrence of current under these ideal conditions and how it manages to pass through in order to reach the load. How is this explained in the theory itself?

EDIT

I want to clarify something:

People have been saying that P and Q are actually the same point in space, so there cannot be any current between them (no area, no current; makes sense). This might be the explanation from a circuit theory point of view. However, Ohm's Law doesn't state that. People have also been saying that Ohm's Law defines current in a non-zero resistive path (which I've never seen as a constraint in Ohm's Law before). If we include that condition, then we are good. Otherwise we have a problem. Any theory (whatever it may be) should be valid within its realm of validity (otherwise, it is logically contradictory, let alone mathematically). Can you think of any violation of Newton's Law or any place where Newton's Law cannot explain the phenomenon under sufficient assumptions and within the realm of its validity? I don't think so. The same must be true for Ohm's Law. As far as I know the law, there is nothing that lies outside the definition of the law. There is a resistive path with zero resistance, a load, and a voltage source, so tell me the current between these two points? Better question is "Why is there a current between these two points? Can your theory explain that?" People have also been suggesting checking out superconductors (which are fun and I like them), but I don't think that answers the question.

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    \$\begingroup\$ I almost never move comments to chat - I have done so here. The question has a number of good enough answers (including mine :-) ) and enough material in comments (now in chat) to resolve the question. See chat here.; Please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. \$\endgroup\$
    – Russell McMahon
    Jun 7, 2023 at 12:34
  • \$\begingroup\$ Comments that do not request clarification or suggest improvements usually belong as an answer, on Electrical Engineering Meta, or in Electrical Engineering Chat. Comments continuing discussion may be removed. \$\endgroup\$
    – Russell McMahon
    Jun 7, 2023 at 12:38
  • \$\begingroup\$ This is sort of like asking "how can X be non-zero if X times zero is equal to zero?" (X is the current; the zero it's multiplied with is resistance between P and Q; the resulting zero is the potential difference.) The answer is that solving for X entails dividing by zero, so any result is possible (rather, we have no information). \$\endgroup\$ Jun 9, 2023 at 6:16

10 Answers 10

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If you are talking about an ideal schematic, ideal circuit theory, then P and Q are two co-incident points on the same node, there is no wire between them. The V1 +ve terminal and the top of R1 are also points on the same 'top wire' node.

An ideal node is an equipotential connection between ideal components, which has no size itself. It just conveys any current entering from any one connection to any other, while obeying Kirchoff's Law that charge doesn't build up on a node.

We just connect points on the node with 'wires' for drawing convenience, so that everything doesn't end up on top of everything else.

If we did want to model a real wire between V1 and R1, then we would give it some finite resistance, and then V1 and R1 would be connected to two separate nodes, connected by the wire's resistance.

So there is no wire PQ, there is no spoon.

If, on the other hand, you want to talk about reality, split that top node, and place a real wire between P and Q, then it's a different argument.

The problem is with your expectation that you need a potential difference to have a flowing current. That's false. Just look at superconductivity, charge flows along a wire without any voltage drop.

Resistance is just a description of what we observe. We see a current in a wire, we measure the voltage drop between the ends. The ratio we call resistance. If it's a small voltage drop, then we say the resistance is low. If it's a zero voltage drop, then the resistance is zero.

That relation is often inverted to say that current = voltage/resistance. That's useful when resistance is finite, which is the case when superconductors aren't around, and reasonably useful when resistance is more or less constant, as in metals, 'resistors' etc, but crucially not useful in diodes and other non-linear components.

What 'makes' the current flow? In a real world wire, starting from zero, we would need an initial voltage, to get a rate of change of current through the wire's inductance. Once the current is flowing steadily, zero rate of change, zero volts is needed across the inductor. Now we see whether a voltage is needed across the resistance of the wire to keep it going.

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    \$\begingroup\$ I see what you did there, with the spoon, and the username! \$\endgroup\$ Jun 7, 2023 at 5:28
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    \$\begingroup\$ Well, the problem is still the same: zero ohm resistors do not exist, they are just a graphical aid to represent something, pretty much like the ideal wire. You can draw a dragon on a piece of paper, but cannot be worried whether it'll burn you alive or not. \$\endgroup\$ Jun 7, 2023 at 10:09
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    \$\begingroup\$ @VladimirCravero Of course zero ohm resistors exist, superconductors. If it happens in the real world, then it exists, whatever convention you use to draw or imagine it. We observe continuous current flow round a loop of superconductor, undiminished with time, hence zero resistance. \$\endgroup\$
    – Neil_UK
    Jun 7, 2023 at 10:43
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    \$\begingroup\$ @Neil_UK While superconductors have zero resistance, they don't have zero inductance, and so they act more like an ideal inductor than a zero ohm resistor. \$\endgroup\$
    – Chris
    Jun 7, 2023 at 16:37
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    \$\begingroup\$ Here is my comment on the question, which was moved to chat. But I think it adds something here. It might also be useful to consider that a wire, even with no resistance, will present an inductance. So, when the connection is initially made, there will be a brief period of time where a voltage will appear across the two ends of the wire. As the magnetic field increases, it will reach equilibrium where the voltage will be zero. But if the current tries to decrease, the collapsing field will try to keep the same current flowing by generating an opposing voltage. \$\endgroup\$
    – PStechPaul
    Jun 8, 2023 at 2:01
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Your assumed rule ("a potential difference causes current") is missing a crucial clause. The complete rule is "a potential difference causes current in a non-zero resistive path". Since there is no resistance in the path between P and Q, then that rule is not applicable. As the rule is not applicable, it is not violated.

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    \$\begingroup\$ So what causes current in a conductive path? \$\endgroup\$ Jun 6, 2023 at 15:46
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    \$\begingroup\$ That is not your original question. I answered your original question. You already know the answer to this new question: the fact that a load (on the right) is connected across the voltage source (on the left). More generally: if something in the past started a current flow in a zero-resistance circuit, current will continue to flow forever, even though the original cause is gone. Similar to how the Earth rotates around the sun. \$\endgroup\$ Jun 6, 2023 at 15:52
  • \$\begingroup\$ "Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points." This definition is from Millikan, Robert A.; Bishop, E. S. (1917). Elements of Electricity. It doesn't include the additional constraint you've added. \$\endgroup\$ Jun 6, 2023 at 16:03
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    \$\begingroup\$ Ohm's law only applies to resistive circuits. OP's ideal conductor is not a resistive circuit, hence Ohm's law doesn't apply. I can't help it if our university professors glossed over that critical detail: Ohm's law only applies to resistive circuits. \$\endgroup\$ Jun 6, 2023 at 16:27
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    \$\begingroup\$ @DavideAndrea I would even go a bit further: Ohm's law only applies to particular electrical conductors under particular physical conditions, and is then only extrapolated to wide array of circuits as a result of an empirical generalization. Ohm's experiment was for a very specific galvanic circuit, he never stated the "law" as we know it. He basically only said that galvanometer's reading is proportional to thermocouple temperature and inversely proportional to wire length+const... and that's where the Wittgenstein's ladder begins IMO. \$\endgroup\$
    – user213769
    Jun 9, 2023 at 11:24
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The short answer is that in the lumped-element model, you don't need a potential difference in order to draw current.

In case you're not familiar with it, the lumped-element model is the "usual" model of electric circuits, which makes various simplifying assumptions such as "the voltage between any two points on a wire is always 0" and "the current in a wire is equal at all points on it."

This is, of course, a situation where the model differs from reality. In reality, as far as I know, you need an electric field in order to produce a current. But you asked about the theory, not about reality, so let's go back to the unreal world.

In the lumped-element model, the current through a wire is determined only by the following rules:

  • The current into one end of the wire is the opposite of the current into the other end of the wire.
  • At a junction, the sum of the current into each of the leads connected to that junction is zero. (By "lead," I mean one end of a wire, or one of the leads or pins of a device.)

By using these rules, Ohm's law, and a handful of other rules, we can calculate how much current flows through each of the wires in your diagram.

At this point, you're probably still wondering, "What actually causes the current to flow? What is the explanation for how that happens?"

The answer, unfortunately, is that there is no answer. The lumped-element model gives us a system of equations that we can solve to predict what will happen, but the lumped-element model doesn't say anything about how or why it will happen. Causation and explanations simply are not a part of the lumped-element model.

Of course, the lumped-element model is not the only model out there. There are other models that do provide some explanation for why current starts to flow in this circuit. However, I'm not aware of any model that provides an explanation other than "because there's a potential difference" or "because there's an electric field." In other words, I'm not aware of any model in which ideal wires exist and there is an explanation for why current flows.

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    \$\begingroup\$ That was a good explanation. Thanks. \$\endgroup\$ Jun 7, 2023 at 11:08
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    \$\begingroup\$ You mention other models of why current flows: Veritasium made an interesting video about electric fields outside (real) wires as a realistic explanation for why current flows. It makes correct testable predictions, such as current starting to flow in a physically nearby but electrically distant part of a long loop of wire when the switch is closed, before a "signal" would have time to propagate through the wire at c. How Electricity Actually Works - Veritasium. He compares it to the lumped-element model. (ping @merovingian) \$\endgroup\$ Jun 8, 2023 at 4:11
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    \$\begingroup\$ That model I think requires wires to have at least inductance (which even superconductors have), if not resistance. So not ideal wires. The video spends some time explaining the surface-charge model for making electricity flow in them. \$\endgroup\$ Jun 8, 2023 at 4:20
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Your issues are:

  • Dealing with ideal objects and then trying to assign them non-ideal attributes.

  • Mixing circuit diagrams and real-world circuits without attention to the fact that they are NOT identical.

You are attempting to consider something ideal in one context and non-ideal in another.

Either

  • The line on the circuit diagram is a real component with real parameters and it does NOT have zero voltage drop if current flows in it.

or

  • It is not a real component, and its size, thickness, shape, colour or anything else on the diagram is of no relevance to the circuit. It's ONLY attribute, if ideal, is showing that all points on the line are the exact same point in physical reality.

If a real world wire or bus connects two or more components but it has no parameters on the circuit diagram then either it must be close enough to ideal to be able to be ignored in real life OR the diagram is not a true depiction of reality.

An ideal wire has zero resistance per length.
When depicted on a circuit diagram it can show an ideal wire of anywhere from zero length to infinite length. The length of the line on the diagram has no real world significance.

Insisting that it can have zero resistance but that the length is meaningfully represented on a diagram is erroneous.

A circuit diagram depicts how components are connected.
If a wire has relevant electrical parameters then these MUST be included if the wire is to be seen as a component in its own right. Typically these may be resistance inductance and capacitance, and perhaps mutual inductance and capacitance relative to other components. IF it has none of these then it is NOT a "real" component, and showing it on a circuit diagram is simply a convenience to avoid drawing components in immediate proximity.

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As far as I understand, you need potential difference to draw current.

Yes this is true, but a potential difference is not required to pass current. Water in a river can pass through a horizontal section of the river where there is no change in height (potential difference) where it is being pushed by the water from upstream that does have a change in height.

If there were no potential difference, the current would be 0/R amperes, which is equivalent to 0 A.

This is not true in general, but is true for a load on a voltage source.

What confuses me is how there can be any current flow between P and Q when there is no potential difference between them

Ohm's law states that \$I=V/R\$. If \$R=0\$ then the result is undefined, although some might say the current approaches infinity, which clearly does not make sense.

Rearrange Ohms'Law to \$V=RI\$. Now, if \$R=0\$, then \$V=0\$, while \$I\$ takes on any value, which is what is would be measured for an ideal wire.

The way I understand:

It is the flow of charge through the resistor that creates the voltage drop across it. The applied voltage accelerates the charge. In order for the current to become constant the reaction of the electrons with the lattice of the resistor creates a counter voltage that stops the acceleration. This happens very fast. We know that this happens by applying a current source to a resistor. A voltage will appear across the resistor.

So I treat the voltage source that as an accelerating voltage, and the voltage across the passive elements as resulting from current through them, except where connected directly across a voltage source.

So when one form of the Ohm's Law equation doesn't fit, rearrange it.


UPDATE #1

Including inductance.

The wires connecting the source with the load are ideal inductors, unless the “ideal conditions” expressed in the OP include zero inductance as well.

At the instant that the voltage is applied, all the voltage appears proportionally across the two inductors. Then, the inductor voltage will decrease to zero as the current becomes steady.

So a transient voltage can appear across a length of ideal conductor if the current is changing.

So the V-I relationship, with zero initial conditions, for the connecting wires is according to the defining relation for the inductor.

$$i_L=\frac{1}{L}\int_0^t v_L(\tau)d\tau$$

Further, the alternative form for the inductor’s definition is

$$v_L=L\frac{di_L}{dt}\text{ ,}$$ which indicates that the voltage is zero for any nonzero constant current.

My first answer still applies under steady state conditions.


Update #2

Can you think of any violation of Newton's Law or any place where Newton's Law cannot explain the phenomenon under sufficient assumptions and within the realm of its validity?

Yes I can. Newton's second law, \$f=Ma\$ can be written:$$a=\frac{f}{M}\text{, } M\ne0$$ So Newton's second law cannot describe the phenomenom of acceleration when the mass \$M=0\$.

The same must be true for Ohm's Law.

Comparing an energy dissipating element with an energy storage element is questionable. However I take your point.

So I answered this in my original post. Comparatively:

Ohm's law is $$i=\frac{v}{R}\text{, }R\ne0$$ So Ohm's law cannot describe the phenomenom of charge flow when the resistance \$R=0\$.

In both the mass and resistance cases, Division-by-zero is a mathematical limitation of both laws when trying to evaluate acceleration and current respectively.

So for zero mass and any acceleration, the force is zero.

And for zero resistance and any current, the voltage is zero.

"Why is there a current between these two points? Can your theory explain that?"

It's not my theory.

When the source pushes an electron into one end of the series circuit, all the other electrons in the circuit move away like a pressure wave.

The electron at the other end is pulled from the circuit. All the other electrons in the circuit move toward like a pressure wave. The wave moves at the speed of light. The electrons not so much.

It is this action that creates the observation that elements in series must have the same current regardless of the resistance of the component parts.

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    \$\begingroup\$ With all respect - but I think the first two sentences of your answer are very misleading (wrong?). We must consider resistors and pices of wire in a closed current loop simply as a series connection of several very low resp. larger resistive elements. And a current through this loop is, of course, driven by a voltage (potential difference). This voltage causes an E-field within each individual part which allows the movement of electrons within each part. \$\endgroup\$
    – LvW
    Jun 7, 2023 at 6:54
  • \$\begingroup\$ @LvW: An electric field cannot exist in a metal with 0 resistivity. The OP is asking, “How can these electrons move without the electric field?” The first two sentences are correct in making a distinction between drawing current and passing current. \$\endgroup\$
    – RussellH
    Jun 7, 2023 at 10:22
  • \$\begingroup\$ Question: Does a metal exist with zero resistivity? \$\endgroup\$
    – LvW
    Jun 7, 2023 at 13:28
  • \$\begingroup\$ For simplicity let us discuss the second sentence of your answer: "Water in a river can pass through a horizontal section of the river". Of course, this is correct - however: Which force causes the water to flow? From where it comes and where does it go? Is there really no "change in height" at the beginning resp. the end of the horizontal path? \$\endgroup\$
    – LvW
    Jun 7, 2023 at 13:54
  • \$\begingroup\$ Zero resistivity is the constraint (speculation) of the question. Mentioned elsewhere, superconductors do exist. If you want to discuss please set up a chat room. \$\endgroup\$
    – RussellH
    Jun 7, 2023 at 15:01
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Your mistake is just assuming that Ohm's law talks about cause and effect. It doesn't. It puts into relation resistance, current, and voltage, with the latter being the product of resistance and current. If either current or resistance are zero, the voltage is zero. And that's what happens here.

When the resistance is zero, you can deduce that the voltage is zero without knowing anything about the current (and indeed, without being able to deduce anything about the current in spite of knowing both voltage and resistance).

That you can deduce B from A does not mean that B is caused by A. Cause and effect are metaphysical interpretations, not physical or mathematical relations.

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It's important to understand the difference between lumped circuit theory and physics. Many beginners attempt to use a SPICE-like circuit simulator to find truths about physics, without realizing what they're looking for is not there. The map is not the territory, and lumped circuit theory is not even designed to be a realistic map.

This question can be interpreted in two ways:

  1. In lumped circuit theory (or SPICE-like circuit simulators), how can current flow in a ideal "wire" without a potential difference across it?
  2. In physics, how can (DC) current flow in an ideal conductor or superconductor without a potential difference across it?

Both are good questions. The difference is that answering the first question only requires explaining how circuit simulators are programmed, while answering the second question requires explaining the physical concepts in electrostatics, the structure of atoms, etc,.

What is not correct, however, is attempting to use the circuit simulator to answer physics questions. Circuit simulators do not model physical objects, so it's not appropriate to measure the voltage at a floating node, measure the voltage or current across an open circuit, or measure the voltage across an ideal wire. The results are not physical (although the results can be the same as an ideal result from a physical point of view, but it happens as a mathematical limiting case, not because of modeling any physics).


Since I'm not a physicist, I'm only going to answer the first question.

In lumped circuit theory, visual locations of wires in the circuit diagram doesn't matter, the "wires" are not physical objects, but just a way to say "located at", "equal to", "is the same node as".

To describe connections in a circuit schematic diagram in a computer, lumped circuit theory is usually formulated in terms of netlist and nodes. If you draw a schematic diagram on a computer, internally, the CAD represents this circuit using a language called the "netlist" - which is the list of connections between nodes. If two resistors are connected in series with a voltage source, in this schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

In a netlist, each "terminal" is a node, and a node is the most basic object of any calculation in the software. In software, the netlist would be written as something like this:

  1. The right terminal of the voltage source is connected to the left terminal of resistor R1.
  2. The right terminal of R1 is connected to the left terminal of R2
  3. The right terminal of R2 is connected to the left terminal of the voltage source.
  4. The left terminal of the voltage source is the ground reference.

This is the SPICE netlist for this circuit, in computer language:

V1 _net0     0  DC   1
R1 _net0 _net1      50
R2 _net1     0      50

.control
exit
.endc

.END

As we can see, the only purpose of a wire is declaring the relationships between nodes. A wire plays the same role of an "equal sign" in algebra.

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  • \$\begingroup\$ That's a good illustration of Neil_UK's comment. Thanks. \$\endgroup\$ Jun 7, 2023 at 20:48
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Besides all the good answers before, there's a missing point in OP assumptions.

An electrical current is the flow of charges through a surface, there's no mention of voltage potential in this.

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Ohm's law applies to this circuit as well - there is a voltage Vqp applied by the input voltage V1 through the resistance R1 to the resistance Rqp but both are negligibly small (tend to zero).

We can also think of section QP and resistor R1 in series as forming a voltage divider with a very small ratio. The output voltage of this voltage divider is applied to the "resistance" Rqp.

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In circuit analyses, circuits are represented in LUMPED-PARAMETER form, where actual wires are simply nodes which do not have any properties. So P and Q are topologically one entity in a lumped-parameter model and it can only represent a singular point (like a voltage reference), but it has no other properties (like current) of its own since it doesn't represent an actual component for the circuit (e.g. an actual wire that minutely draws some current due to its own resistance).

To represent an actual wire component with voltage drop due to wire resistance, you have to model the resistance of that wire itself as a lumped model (i.e. resistor).

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