What is happening in those wires?

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
Given the circuit below, with the capacitor previously charged to 100 volts and the resistor large enough so that full discharge will require about an hour:

cap 4.jpg

Please describe the physical changes that take place in wire A and wire B during the discharge period.
 

Papabravo

Joined Feb 24, 2006
12,514
There are no observable physical changes taking place in the wires that I am aware of. Full discharge is of course an impossibility since the exponential decay to zero takes an infinite amount of time. You should phrase the proposition as: it takes an hour for the voltage to drop to 0.1% of it's fully charged value.
 

MrChips

Joined Oct 2, 2009
19,408
Since the wires A and B are subjected to the Earth's magnetic field, the wires will experience two diminishing forces in opposite directions given by the cross product of current times magnetic field.
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
There are no observable physical changes taking place in the wires that I am aware of.
The changes may not be observable, but we can measure their effects (say, with a voltmeter). So what are the theoretical changes in those wires during the discharge period?

Full discharge is of course an impossibility since the exponential decay to zero takes an infinite amount of time. You should phrase the proposition as: it takes an hour for the voltage to drop to 0.1% of it's fully charged value.
You're correct. I should have said something like, "discharge to a state of approximate equilibrum", since the exact figures are not critical to the question.

Extremely slight increase in length and diameter due to a miniscule amount of heat release due to resistance?
Possibly, but I'm thinking smaller. I was hoping for something like, "At the start of the discharge period, wire B will have ["more electrons" or "electrons that are vibrating more quickly" or "electrons further away from their protons"] than wire A".
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
Since the wires A and B are subjected to the Earth's magnetic field, the wires will experience two diminishing forces in opposite directions given by the cross product of current times magnetic field.
That may be true, but I don't think the Earth's magnetic field is relevant to the question, since I believe the circuit will behave the same in space, far away from such fields.
 

dl324

Joined Mar 30, 2015
9,142
Given the circuit below, with the capacitor previously charged to 100 volts and the resistor large enough so that full discharge will require about an hour:

Please describe the physical changes that take place in wire A and wire B during the discharge period.
Is this school work?
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
Is this school work?
No. Well, sort of. I'm a teacher and for several years I've been trying to write an illustrated, cartoon-style book about electricity and electronics for middle- and high-school kids, but I can't seem to pin down a satisfactory answer to this question (which is fundamental to the "picture" of electricity that I want to plant in their minds). One of my inspirations is this book: https://www.amazon.com/dp/0962781592/
 

MrChips

Joined Oct 2, 2009
19,408
You're correct. I should have said something like, "discharge to a state of approximate equilibrum", since the exact figures are not critical to the question.
There is no state of equilbrium until infinite time.

Possibly, but I'm thinking smaller. I was hoping for something like, "At the start of the discharge period, wire B will have ["more electrons" or "electrons that are vibrating more quickly" or "electrons further away from their protons"] than wire A".
Sorry, electrons in a wire don't flow like that. Wire A and wire B will experience the same flow of electrons.
 

MrChips

Joined Oct 2, 2009
19,408
No. Well, sort of. I'm a teacher and for several years I've been trying to write an illustrated, cartoon-style book about electricity and electronics for middle- and high-school kids, but I can't seem to pin down a satisfactory answer to this question (which is fundamental to the "picture" of electricity that I want to plant in their minds). One of my inspirations is this book: https://www.amazon.com/dp/0962781592/
Then I sincerely suggest that you teach some other subject and not electricity and electronics.
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
Wire A and wire B will experience the same flow of electrons.
Then what is physically different about wires A and B during the discharge period? How do we account for the measurable differences in potential between the two wires during the discharge period? A voltmeter connected to two different points on wire A will read very close to zero potential difference the whole time; but the same meter connected from wire A to wire B will reveal a significant potential difference at the start of the discharge period. What is the physical cause of this potential difference? There must be something different about wires A and B during the period in question. What is it?
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
There is no physical change in the wire. Electrons flow, but the density isn't sufficient to alter the physical characteristics of the wire.
It seems to me that a wire with electrons flowing in a given direction, and a wire with electrons that are not flowing in any particular direction, are physically different. But if the flow is the same in both wires, how do we account for the potential difference between the two?
 

dl324

Joined Mar 30, 2015
9,142
What is the physical cause of this potential difference?
The capacitor.
There must be something different about wires A and B during the period in question. What is it?
Considering the fact that no two wires can be exactly the same, by definition, they're different.

For a typical analysis, we don't deal things at an atomic level. The wire is the same before, during, and after the flow of current. So long as the current density isn't sufficiently high to cause something undesirable, such as electromigration.
 

dl324

Joined Mar 30, 2015
9,142
It seems to me that a wire with electrons flowing in a given direction, and a wire with electrons that are not flowing in any particular direction, are physically different. But if the flow is the same in both wires, how do we account for the potential difference between the two?
I hate the water analogy, but it can be illustrative in this situation...

When an electron leaves the capacitor, it displaces an electron in an atom of the metal in the wire. That causes a "chain reaction" where other atoms give up an electron and almost immediately gain one. So the net number of electrons in any atom of metal remains constant.
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
When an electron leaves the capacitor, it displaces an electron in an atom of the metal in the wire. That causes a "chain reaction" where other atoms give up an electron and almost immediately gain one. So the net number of electrons in any atom of metal remains constant.
And all that happens very fast, close to the speed of light, yes? So if the flow of electrons is steady and virtually the same in both wires, how do we account for the measurable potential differences in the wires throughout the hour in question?

Perhaps it will help if we back up a little. You also said that the "the capacitor" was the physical cause of the potential difference. If so, we could think of wires A and B as temporary "extensions" of the capacitor's plates, and the question would then become, What accounts for the potential difference between the two plates inside -- and, by extension, outside of -- the capacitor?
 

dl324

Joined Mar 30, 2015
9,142
You also said that the "the capacitor" was the physical cause of the potential difference. If so, we could think of wires A and B as temporary "extensions" of the capacitor's plates, and the question would then become, What accounts for the potential difference between the two plates inside -- and, by extension, outside of -- the capacitor?
In your circuit, you have two electrical nets. One for the capacitor anode which includes the wire used to connect the anode to one end of the resistor. And one for the cathode of the capacitor. We don't think of the wires as extensions of anything.

We don't think of things happening at the atomic level. That would be an unnecessary complication.
 

Thread Starter

Gerry Rzeppa

Joined Jun 17, 2015
170
In your circuit, you have two electrical nets. One for the capacitor anode which includes the wire used to connect the anode to one end of the resistor. And one for the cathode of the capacitor. We don't think of the wires as extensions of anything.
Okay, then what accounts for the potential difference between the capacitor's anode and cathode? In other words, what is physically different about the capacitor's anode and cathode when the capacitor is charged to 100 volts?
 

MrChips

Joined Oct 2, 2009
19,408
Perhaps the answer you are seeking is the electron concentration at wire B is higher than that at wire A.

Wire A and wire B are practically equivalent. We do not consider charge concentration as a physical difference.
 
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