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I'm reading a book on computing that has a couple of introductory chapters on electronics. The author imagines a situation in which you and your next door neighbor set up a bidirectional telegraph to communicate with each other at night via Morse code: enter image description here

And then you and your neighbor realize that a common part of the circuit can be shared to reduce wire costs: enter image description here

Where I get confused is how the author analyzes the circuit when one or both switches are closed. The author depicts the closing of the switch at battery 1 like so: enter image description here

He says that no electricity flows in the other part of the circuit because there’s no place for the electrons to go to complete a circuit. My question is, why can't electrons also flow from the negative terminal in battery 2, through the right bulb, and into the positive terminal of battery 1?

He also analyzes the situation when both switches are closed: enter image description here

The author states (with absolutely no explanation) that "No current flows through the common part of the circuit." OK, but why? Why doesn't current flow form the negative terminal of battery 1, through the right bulb, and into the positive terminal of battery 1. And why doesn't current flow form the negative terminal of battery 2, through the left bulb, and back to the positive terminal of battery 2?

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  • \$\begingroup\$ Do you know Kirchoff's current law? What happens if you try to calculate the current in that wire? \$\endgroup\$
    – Greg d'Eon
    Dec 13, 2015 at 16:49
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    \$\begingroup\$ When there is a voltage gradient, and assuming there is a conductive path (such as a wire) current will flow from positive to negative potential. When a switch is open, there is no conductive path through the corresponding light bulb. So no current flows through that bulb. When both switches are on, the author is suggesting that the currents are exactly equal and opposite, thus net current in the common segment is zero. In the real world, if the currents are not exactly equal in magnitude, a small net current WOULD flow in the common segment. \$\endgroup\$
    – user57037
    Dec 13, 2015 at 16:54
  • \$\begingroup\$ The other thing you are not gettting is that batteries cannot supply or sink current without a completed circuit either. And the current at the positive terminal of a battery is exactly equal to the current at the negative terminal in magnitude. If one electron flows into a battery at one terminal, an electron MUST flow out at the other terminal. This is a basic rule of circuit analysis (and the universe, for the most part). So if one terminal of a battery experiences an open circuit, then the current in the other terminal will always be zero. \$\endgroup\$
    – user57037
    Dec 13, 2015 at 17:00
  • \$\begingroup\$ EVERY time you have two circuit nodes connected with nothing but a wire, that's just one node. Try redrawing your last figure the way to see if it helps your understanding \$\endgroup\$ Sep 10, 2023 at 14:05

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why can't electrons also flow from the negative terminal in battery 2, through the right bulb, and into the positive terminal of battery 1?

Because the only way to get to the negative terminal of battery 2 is to come from the body of the battery, and the only way to get there is from the positive terminal of the battery. And the only way to get there is through the switch in the friend's house, which is open so no current can flow through it.

Batteries don't create electrons from nothing. They "pump" them from one terminal to the other via a chemical reaction, causing current to flow (through the battery) from a low potential to a high potential.

Current only flows in complete circuit means the current has to flow through the battery just as much as it has to flow through every other circuit element.

As an aside, it's also why it's silly to say that current always flows from high potential to low --- the battery is a trivial example of when it's the other way around. And it's also silly to say that all current is a flow of electrons in the opposite direction from conventional current --- the battery is a trivial and everyday example of when ionic currents should be considered.

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In the first case, since sw2 is open, battery2 does not have a return path for electrons, so it cannot contribute to the current.

In the second case, when both switches are closed, the "common part" is at zero volts. A difference in potential is required for the current to flow. Think of current flow as water flow. If there is no difference in height (potential), water will not flow.

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taking problem solely on circuit theory there is an existence of a shorted path i.e. "potential difference in between the two nodes of the path is zero" which implies no current from battery 2 requires to flow through battery 1 line since already a path with zero resistance is available. Same goes for battery 1. Thus circuits though physically connected remain electrically isolated.

To simplify and understand my above said statement just assume ground on the interconnected nodes and you will get my point pretty easily.

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I think some answers are getting a little complicated just because of the drawing tools for circuits that are easily available. This answer matches this already posted answer, but adds an image to explain

enter image description here

The above image is simply a redrawing of the circuit you provided, just twisted around a bit differently. Everything drawn in black is circuit. I provided a frame of red dotted line to make a point. Everything in that frame is the equivalent of one node in this circuit. Current can certainly be flowing through this node, but it doesn't make much sense to be talking about current flow from one node to the same node.

This, of course, is part of an ideal assumption -- that wires have zero resistance. In reality, wires do have resistance, and that resistance can have shockingly big impacts on some circuits. In this case, though, those impacts will be pretty benign.

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  • \$\begingroup\$ Why do you say it doesn't make sense to talk about current flowing within a node? \$\endgroup\$
    – Hearth
    Sep 11, 2023 at 0:04
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    \$\begingroup\$ @Hearth. A node has no geometry, area, or volume. \$\endgroup\$ Sep 11, 2023 at 0:22
  • \$\begingroup\$ Scott S. - thank you. This redrawing is very helpful to clarify things. \$\endgroup\$
    – LvW
    Sep 11, 2023 at 13:36
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    \$\begingroup\$ @Hearth Sure, you can talk about currents flowing through a node, or into and out of the node, but not within a node. Nodes have no extent - they are not points in a metric space. They are parts of a graph that represents the circuit. They are entirely abstract concepts with no physical equivalent - only physical approximations. \$\endgroup\$ Sep 11, 2023 at 14:16
  • \$\begingroup\$ I cannot agree with this presentation. A "single node" is a special case where the resistance is zero. This is just one of many cases where an arbitrary element is included between two nodes. For the purposes of understanding, it is more useful to draw the circuit with two nodes and then connect various elements between them (resistors, lamps, voltmeters, ammeters, and finally even a short circuit) to show that this does not change the picture of the currents. Intuitive understanding is something different from strict circuit theory and it gains nothing by shrinking a long wire to a point. \$\endgroup\$ Sep 11, 2023 at 14:50
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Basic idea

Perhaps at one time this phenomenon was really a mystery, but now it is not because it is the well-known virtual ground. Let's transform the schematic in a few steps to see it.

Original circuit

schematic

simulate this circuit – Schematic created using CircuitLab

"Inverting amplifier"

If we swap V1 and R1 (or V2 and R2), we will easily recognize the familiar configuration of two resistors in series (aka voltage divider, summing resistor circuit, etc.) supplied by a split power supply (note that one of the voltage sources must be reversed). Its most famous application is, of course, the op-amp inverting amplifier. Now the explanation is easy.

schematic

simulate this circuit

There is a virtual ground between R1 and R2, and a real ground between V1 and V2. The two grounds are connected by the common line (the ammeter). Since there is no voltage difference between them, there is no current flowing.

From another viewpoint, two (zero) voltage sources - one imperfect and the other perfect, are connected in parallel. So there is no conflict between them, and no current flows through them.

Also, this configuration can be thought as a kind of bridge circuit whose output is shorted (by an ammeter here).

Visualized currents

Before to continue, let's simplify the schematic by replacing the "dead" resistors by "live" ammeters with intentionally increased internal resistance of 100 Ω. Thus they will function both as resistors and ammeters. To do it, open the ammeter parameters window and set the resistance.

Connected grounds

As you can see, the current travels along the loop of four elements - V1, R1, R2 and V2, but there is no reason for it to go through the bridge element in the middle.

schematic

simulate this circuit

Disconnected grounds

So we can easily remove the connection between the two grounds and the current (limited by the two ammeter internal resistances) continues to circulate around the periphery.

schematic

simulate this circuit

A reversed story

@nani chan wrote:

Circuit fantastist why the electrons don't flow from the middle wire because they can also complete the circuit by going from the middle wire

I put together a series of schematics that reveal the idea in reverse order to answer the question, "Why does the ammeter show a reading of 0?" (for some reasons, closed). But what was that idea?

In circuits, it is known as "bootstrapping" and consists of the following:

If we connect two identical voltage sources through some element, no current will flow through it because there is no potential difference between the two points. Figuratively speaking, each of the sources stops the current of the other by a "counter-voltage". This gives the impression to each of the sources that an element of infinite resistance is connected to it.... but actually the resistance can be very low, even zero.

Let's first see the simpler single-supply arrangement...

schematic

simulate this circuit

Open circuit

... and then the more sophisticated split-supply version. Here the current travels along the loop of four elements - V1, R1, R2 and V2, but does not try to go through the middle because there is no path.

schematic

simulate this circuit

Resistor

Now there is path in the middle but still the current does not try to go through it. Why?

Imagine that the current is water that descends from above and reaches the "junction" (the middle point between the two resistors). There it has the option to turn left (through the middle) or continue straight (down). However, to the left is flat (horizontal); that is why the water (current) goes down the slope.

schematic

simulate this circuit

Now insert all sorts of elements into the middle to convince yourself more and more that no current is flowing...

Capacitor

schematic

simulate this circuit

Ammeter

schematic

simulate this circuit

Short connection

schematic

simulate this circuit

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    \$\begingroup\$ That's a real ground between R1 and R2. \$\endgroup\$ Sep 10, 2023 at 13:18
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    \$\begingroup\$ No, it's just a ground. \$\endgroup\$ Sep 10, 2023 at 13:53
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    \$\begingroup\$ With SW1 and SW2 closed, it's just a short circuit, and an open circuit with either switch open. \$\endgroup\$ Sep 10, 2023 at 13:58
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    \$\begingroup\$ An ideal ammeter is a short, thus you have +10 shorted to -10. \$\endgroup\$ Sep 10, 2023 at 14:23
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    \$\begingroup\$ Then you don't have a real ground or a virtual ground. You have a node in a circuit. \$\endgroup\$ Sep 10, 2023 at 14:42

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