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I've simulated the very simple circuit shown below:

enter image description here

As you can see, it has two batteries of 1.5 V, and three identical resistors. The electric currents passing through each wire are also indicated. I'm puzzled by the fact that the current on the interior wire is zero (that's why there is no number indicated). I've already verified that this is true by applying Kirchhoff's laws. However, is there any intuitive explanation of why the current of the middle wire has to be zero, without need of doing any calculations?

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    $\begingroup$ A comment: if you swap the left battery and left resistor, the circuit becomes 180° rotationally symmetric. Perhaps it will be intuitive that in a rotationally-symmetric circuit, all the voltage differences and currents will also be rotationally symmetric; then the only current that would keep the middle wire 180° rotationally symmetric is no current. I'm not making this an answer, though; in my head it makes sense that the circuit chunk on the left will behave the same when you swap the battery and resistor as when you don't, but I actually don't know enough physics to be sure of that. $\endgroup$
    – Daniel Wagner
    Jan 2, 2021 at 1:42

7 Answers 7

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Another intuition: both circuits are the same but in reverse from the point of view of the center wire, so each provides the same current but in opposite directions, thus cancelling each other.

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Consider the potential. Let’s call the left upper corner A and right upper corner B. First see upper side. There are two batteries and they’re symmetric. It is easy to imagine that the potential of the mid point of the upper side is the average of A and B. Then see the bottom side. It’s also symmetric. So the potential of the mid point of the bottom side should also be the average of A and B. Since the potential of both ends of the interior wire are equal, there can’t be a current.

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    $\begingroup$ "There are two batteries and they’re symmetric" are they, though? Symmetrical batteries would have + both pointing outwards or inwards, wouldn't they? $\endgroup$
    – Eric Duminil
    Jan 1, 2021 at 4:28
  • $\begingroup$ sorry about the word “symmetric”, since English is not my mother tongue. The main idea here is that two batteries are in series, so the potential is also in the middle. If it’s your configuration then batteries are in parallel. So the mid point will have the highest or lowest potential. $\endgroup$
    – Wenjie Zhang
    Jan 1, 2021 at 9:55
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enter image description here

It's Wolphram jonny's answer with pictures

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Try reversing one of the batteries or changing its voltage. This can help develop intuition.

In the outer wires, both batteries try to push electrons in the same direction but on the inside they push on the opposite sides and because they are both $1.5V$, they will cancel each other's effort. If one of them was stronger you would have current in the middle.

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You can connect a resistor (or a solid short for that matter) between any 2 equipotential points and it will have no effect on the circuit. Because both resistor terminals are at same potential no current flows.

Remove your middle resistor, analyze the circuit, and you will see it clearer.

p.s. Voltages are meaningless unless measured to some reference point. This can be any point in the circuit. It is typical to make this point the ground reference - in your case, putting it between the batteries would be a good spot.

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    $\begingroup$ affect -> effect $\endgroup$
    – Acccumulation
    Dec 31, 2020 at 22:15
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Imagine voltage/potential at some point as density of electrons passing on a road. They all need space, so if there's a place near them with lower density, part of the electrons move there.

Now, let's say on the bottom end of the wire in question there is a potential, so the density of electrons is higher than usual. They see they can go up to the middle wire, and they do move there, but at some point they meet the electrons from the upper end of the wire, which has the same potential and same density of electrons. So, after they meet, they cannot go anymore, there's basically no direction they'd want to move to, because the density is now equal at all points, and therefore new electrons also no longer come from any end.

That way, we had a quick movement of eletrons which then established a balanced state, and no movement after that. But current is constant movement, so it doesn't count. Constant movement can only be created by constant difference in density and we don't have that.

So, you had a right feeling that some movement from the ends to the middle part should happen, it's just that it lasts very short.

Also, you can imagine some gas moving through the pipes, with the same result. Imagine that batteries emit gas from one end (high potential) and suck it from the other end (low potential). That way you can clearly see how density of the gas smoothly decreases from high potential to low, therefore creating equal current across all the outer wire.

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Kirchhoff's junction rule:

The sum of all current leaving a junction must be equivalent to the sum of all current entering it.

Take the bottom-middle junction for instance. We know that $1.51$ A enters the junction from the left and that $1.51$ A leaves the junction from the right. As such, by Kirchhoff's junction rule, there is no current remaining to run upwards.

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  • $\begingroup$ Isn't that circular logic, though? How do we know that 1.51A flow in both directions? Because no current is flowing in the middle connection? $\endgroup$
    – Eric Duminil
    Jan 1, 2021 at 17:13
  • $\begingroup$ Correct me if I'm wrong, but isn't the current values that I mentioned in my response labeled on the diagram on top of the respective wires? $\endgroup$
    – Spyre
    Jan 1, 2021 at 19:37
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    $\begingroup$ OP is trying to understand why the results are the way they are. So you cannot use the results in your answer. $\endgroup$
    – Eric Duminil
    Jan 1, 2021 at 19:40

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