4
\$\begingroup\$

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?

|improve this question
\$\endgroup\$
  • \$\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 '15 at 16:49
  • 2
    \$\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\$ – mkeith Dec 13 '15 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\$ – mkeith Dec 13 '15 at 17:00
7
\$\begingroup\$

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.

|improve this answer
\$\endgroup\$
1
\$\begingroup\$

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.

|improve this answer
\$\endgroup\$
0
\$\begingroup\$

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.

|improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.