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October 13 Information
I previously asked this question here, but did so anonymously and thus I am wondering if I will get more responses here if I openly asked this question. Anyway, here goes:
If a man's epididymis is removed (on both sides, of course) and his vas deferens grows and attaches itself to his rete testis, is this man going to once again have sperm in his ejaculate, but still be incapable of ever impregnating a woman through sexual intercourse as a result of his sperm not being motile enough? Futurist110 ( talk) 02:04, 13 October 2018 (UTC)
- Have you considered handjobs? Sagittarian Milky Way ( talk) 14:38, 14 October 2018 (UTC)
- I see no particular reason why there would not be sperm in the semen, but there might well be less of it, given the loss of its principal storage vessel. So while a modicum of fertility may remain, I have to imagine that it would be reduced. Motility is another question, the answer to which would depend on a number of factors. I am definitely not going to say "incapable of ever" causing pregnancy: far too many "longshots" have resulted in conception in the past. After all, it only takes one sperm. -- Piledhigheranddeeper ( talk) 01:51, 15 October 2018 (UTC)
Our article on negative resistance gives a sourced sentence (and I checked the sources) that "an increase in voltage across the device's terminals results in a decrease in electric current through it." It depicts as an example a fluorescent light bulb, where the current that flows starts an arc that reduces resistance. That too is sourced, but there the sources say "negative resistance characteristics". An online discussion seems skeptical. [1] (The article also lists these in a table as a passive negative resistor)
The problem I have is that I can't picture that a circuit that can actually increase the voltage on a fluorescent light bulb in real time at any point during its cycle would actually observe decreased current flow. Either the flow would increase minimally, or a lot -- I think. The phrasing for the definition of negative resistance would seem to impose a very stringent criterion this way - one met by a tunnel diode, where increasing the voltage (as the independent variable, you might say) really does cause it to choke off the flow of electrons.
This was discussed on the article talk page before I looked at it, but can we get an answer on the overall fact: is a fluorescent light bulb a negative resistor in any meaningful sense at all? Wnt ( talk) 16:28, 13 October 2018 (UTC)
- @
DroneB: Thanks for the figure - I see it is at
neon lamp. Then again, I also see it has had a "dubious-discuss" tag on it there since 2015, so I could use a bit more convincing. More to the point, I still don't know if this chart shows a correlation rather than a causation -- I mean, can this bulb really stop halfway between A and B, and increase current if you decrease the voltage, or decrease it if you increase it?
Wnt (
talk) 20:58, 13 October 2018 (UTC)
- Yes it can. Supply the bulb from a very high voltage in series with a very high resistance, which creates a virtually constant current supply. Of course there won't be any current until after the arc is struck at "A" by some kind of over-current trigger.
DroneB (
talk) 21:13, 13 October 2018 (UTC)
- Wnt, it seems that your confusion about "correlation" or "causation" is because the relation between voltage and current is not simple. More precisely, this graph shows a non-Ohmic relation between current and voltage. The relation is a multivalued function; the physical system has state. We could describe that state as a result of hysteresis or we could describe it as a complete set of other physical parameters; but in such cases, the relation between voltage and current is neither simple nor linear. Nonetheless, it is still true that the applied voltage is the causative agent for the flow of electric current. The magnitude of that current, however, depends on additional factors.
- This is one case where we can't use simple math to describe the physics: if we tried to describe the current-voltage relationship as a single-parameter function, we would just get the wrong answer. The amount of current in a fluorescent light-bulb depends on voltage, but also depends on the time-history of that voltage, and on the ambient temperature, and the pressure of the gas in the bulb, and so on. Even a very simplified model would be a function of many variables, and when you try to address such a problem without using the right mathematical tools, you just get wrong numbers in your answers - and even worse, you may even draw incorrect physical conclusions (such as the confusion about causation that Wnt appears to have described).
- In the case of a light bulb, the volts represent energy applied to the system; the amps represent the system's response to that applied energy, converting it from electric potential energy into the kinetic energy of moving charged particles. The equation that governs how that energy conversion occurs is complicated because there's a non-negligible amount of real-world plasma physics happening in that tube - so the simple "V = I R" linearization just isn't good enough.
-
Nimur (
talk) 00:55, 14 October 2018 (UTC)
- @ Nimur: I really do understand all of that. I understand that a light bulb at the point shown on the graph at right that has an x-value halfway between A and B could have either value for the current depending on circumstance. Other things like the tunnel diode have a similar situation. The key issue for me is still that whereas the tunnel diode actually can be shown to have a true negative differential resistance, where turning up the voltage reduces the current, I still strongly doubt that the fluorescent or neon light bulb can be made dimmer as a consequence of turning up the voltage at a given moment. I mean, turning up the voltage can't reduce the temperature, it can't reduce the amount of current flowing through an existing plasma arc, what can it possibly do (aside from interacting with the ballast...) that would dim the bulb? But that's the definition the article gives for negative resistance. Wnt ( talk) 02:23, 14 October 2018 (UTC)
- Yes it can. Supply the bulb from a very high voltage in series with a very high resistance, which creates a virtually constant current supply. Of course there won't be any current until after the arc is struck at "A" by some kind of over-current trigger.
DroneB (
talk) 21:13, 13 October 2018 (UTC)
- Does this paper help? [2]
- One thing that helps people to think about negative resistance is to consider an ordinary fuse. Up to a point, an increase in voltage causes an increase in current (fuses do have a small resistance, and would not function is the resistance was exactly zero). Past that point, an increase in voltage results in a dramatic decrease in current. And every wire is a fuse. -- Guy Macon ( talk) 06:49, 14 October 2018 (UTC)
- @ Guy Macon: I don't think a fuse is a good example, because the decrease in current with a fuse is a function of time more than anything else. Put a very rapid spike of voltage across it (enough not to induce heating...) and it would be just a conductor.
- However, the paper is great. I may not have really understood the step between equations 7 and 8, let alone the heavy math at the end, but I think I can fairly say that "a plasma discharge has a negative differential resistance when the slope of the filament characteristic curve exceeds the slope of the plasma production curve at their intersection." That said, the discharge curve they show is discharge current as a function of electron probe current; then they show both discharge current and plasma production as functions of plasma density. In figure 7 it is clear that in the negative resistance region (C to D), the filament characteristic is increasing faster than the plasma production. The problem is, I still don't really understand why that means the current drops from increased voltage. I have a sense that, perhaps, the flow is being "choked off at the filament" because the emission of electrons is somehow saturated, but as best as I can tell that seems backwards from the increased "filament characteristic"? I don't feel like I understand well enough to explain it in an edit, beyond possibly the quote at the beginning of this paragraph. But it's progress. ;)
Wnt (
talk) 13:41, 14 October 2018 (UTC)
- Re: "the decrease in current with a fuse is a function of time more than anything else", that's a really useful way of looking at it. The plasma discharge also involves time -- it doesn't create new ions and electrons instantaneously, nor do they revert to ordinary atoms instantaneously. But then again, a resistor does not act as a resistor unless you give it some small amount of time after a sudden change in voltage -- because every resistor is also an inductor. I would think it fair to say that both the fuse and the plasma show a negative differential resistance when subjected to a slowly-changing voltage, and that pretty much everything acts differently at very short time scales. One could say that everything is a perfect insulator if the time scale is fast enough that the speed of light is the limiting factor. -- Guy Macon ( talk) 15:08, 14 October 2018 (UTC)
- With the neon lamp biased at a point midway between A and B by a current through a high value resistor as I described [3], apply a small voltage through a capacitor. That should demonstrate that the current and voltage swings are in the opposite direction consistent with the negative I/V slope. Also you may see the lamp brightness modulated with decreasing brightness with increasing voltage swings and vice versa. However a constant voltage supply is not good for testing the lamp because of the danger of the current leaping into the possibly destructive segment B->C. It's better to make a constant current supply be the controlling variable, which enables straightforward plotting of the whole V = f(i) characteristic. Note that this is not an efficient way to create a fluorescent lamp Dimmer. This is better achieved by a solid-state circuit that delivers a variable pulse-width voltage. DroneB ( talk) 20:25, 14 October 2018 (UTC)
- The question has been answered by now, but I will still leave a link to electrical breakdown. (Though not exactly the same thing, it helps understand how resistance can decrease by the application of strong voltage.) Tigraan Click here to contact me 11:46, 16 October 2018 (UTC)
| Science desk | ||
|---|---|---|
| < October 12 | << Sep | October | Nov >> | October 14 > |
| Welcome to the Wikipedia Science Reference Desk Archives |
|---|
| The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
October 13 Information
I previously asked this question here, but did so anonymously and thus I am wondering if I will get more responses here if I openly asked this question. Anyway, here goes:
If a man's epididymis is removed (on both sides, of course) and his vas deferens grows and attaches itself to his rete testis, is this man going to once again have sperm in his ejaculate, but still be incapable of ever impregnating a woman through sexual intercourse as a result of his sperm not being motile enough? Futurist110 ( talk) 02:04, 13 October 2018 (UTC)
- Have you considered handjobs? Sagittarian Milky Way ( talk) 14:38, 14 October 2018 (UTC)
- I see no particular reason why there would not be sperm in the semen, but there might well be less of it, given the loss of its principal storage vessel. So while a modicum of fertility may remain, I have to imagine that it would be reduced. Motility is another question, the answer to which would depend on a number of factors. I am definitely not going to say "incapable of ever" causing pregnancy: far too many "longshots" have resulted in conception in the past. After all, it only takes one sperm. -- Piledhigheranddeeper ( talk) 01:51, 15 October 2018 (UTC)
Our article on negative resistance gives a sourced sentence (and I checked the sources) that "an increase in voltage across the device's terminals results in a decrease in electric current through it." It depicts as an example a fluorescent light bulb, where the current that flows starts an arc that reduces resistance. That too is sourced, but there the sources say "negative resistance characteristics". An online discussion seems skeptical. [1] (The article also lists these in a table as a passive negative resistor)
The problem I have is that I can't picture that a circuit that can actually increase the voltage on a fluorescent light bulb in real time at any point during its cycle would actually observe decreased current flow. Either the flow would increase minimally, or a lot -- I think. The phrasing for the definition of negative resistance would seem to impose a very stringent criterion this way - one met by a tunnel diode, where increasing the voltage (as the independent variable, you might say) really does cause it to choke off the flow of electrons.
This was discussed on the article talk page before I looked at it, but can we get an answer on the overall fact: is a fluorescent light bulb a negative resistor in any meaningful sense at all? Wnt ( talk) 16:28, 13 October 2018 (UTC)
- @
DroneB: Thanks for the figure - I see it is at
neon lamp. Then again, I also see it has had a "dubious-discuss" tag on it there since 2015, so I could use a bit more convincing. More to the point, I still don't know if this chart shows a correlation rather than a causation -- I mean, can this bulb really stop halfway between A and B, and increase current if you decrease the voltage, or decrease it if you increase it?
Wnt (
talk) 20:58, 13 October 2018 (UTC)
- Yes it can. Supply the bulb from a very high voltage in series with a very high resistance, which creates a virtually constant current supply. Of course there won't be any current until after the arc is struck at "A" by some kind of over-current trigger.
DroneB (
talk) 21:13, 13 October 2018 (UTC)
- Wnt, it seems that your confusion about "correlation" or "causation" is because the relation between voltage and current is not simple. More precisely, this graph shows a non-Ohmic relation between current and voltage. The relation is a multivalued function; the physical system has state. We could describe that state as a result of hysteresis or we could describe it as a complete set of other physical parameters; but in such cases, the relation between voltage and current is neither simple nor linear. Nonetheless, it is still true that the applied voltage is the causative agent for the flow of electric current. The magnitude of that current, however, depends on additional factors.
- This is one case where we can't use simple math to describe the physics: if we tried to describe the current-voltage relationship as a single-parameter function, we would just get the wrong answer. The amount of current in a fluorescent light-bulb depends on voltage, but also depends on the time-history of that voltage, and on the ambient temperature, and the pressure of the gas in the bulb, and so on. Even a very simplified model would be a function of many variables, and when you try to address such a problem without using the right mathematical tools, you just get wrong numbers in your answers - and even worse, you may even draw incorrect physical conclusions (such as the confusion about causation that Wnt appears to have described).
- In the case of a light bulb, the volts represent energy applied to the system; the amps represent the system's response to that applied energy, converting it from electric potential energy into the kinetic energy of moving charged particles. The equation that governs how that energy conversion occurs is complicated because there's a non-negligible amount of real-world plasma physics happening in that tube - so the simple "V = I R" linearization just isn't good enough.
-
Nimur (
talk) 00:55, 14 October 2018 (UTC)
- @ Nimur: I really do understand all of that. I understand that a light bulb at the point shown on the graph at right that has an x-value halfway between A and B could have either value for the current depending on circumstance. Other things like the tunnel diode have a similar situation. The key issue for me is still that whereas the tunnel diode actually can be shown to have a true negative differential resistance, where turning up the voltage reduces the current, I still strongly doubt that the fluorescent or neon light bulb can be made dimmer as a consequence of turning up the voltage at a given moment. I mean, turning up the voltage can't reduce the temperature, it can't reduce the amount of current flowing through an existing plasma arc, what can it possibly do (aside from interacting with the ballast...) that would dim the bulb? But that's the definition the article gives for negative resistance. Wnt ( talk) 02:23, 14 October 2018 (UTC)
- Yes it can. Supply the bulb from a very high voltage in series with a very high resistance, which creates a virtually constant current supply. Of course there won't be any current until after the arc is struck at "A" by some kind of over-current trigger.
DroneB (
talk) 21:13, 13 October 2018 (UTC)
- Does this paper help? [2]
- One thing that helps people to think about negative resistance is to consider an ordinary fuse. Up to a point, an increase in voltage causes an increase in current (fuses do have a small resistance, and would not function is the resistance was exactly zero). Past that point, an increase in voltage results in a dramatic decrease in current. And every wire is a fuse. -- Guy Macon ( talk) 06:49, 14 October 2018 (UTC)
- @ Guy Macon: I don't think a fuse is a good example, because the decrease in current with a fuse is a function of time more than anything else. Put a very rapid spike of voltage across it (enough not to induce heating...) and it would be just a conductor.
- However, the paper is great. I may not have really understood the step between equations 7 and 8, let alone the heavy math at the end, but I think I can fairly say that "a plasma discharge has a negative differential resistance when the slope of the filament characteristic curve exceeds the slope of the plasma production curve at their intersection." That said, the discharge curve they show is discharge current as a function of electron probe current; then they show both discharge current and plasma production as functions of plasma density. In figure 7 it is clear that in the negative resistance region (C to D), the filament characteristic is increasing faster than the plasma production. The problem is, I still don't really understand why that means the current drops from increased voltage. I have a sense that, perhaps, the flow is being "choked off at the filament" because the emission of electrons is somehow saturated, but as best as I can tell that seems backwards from the increased "filament characteristic"? I don't feel like I understand well enough to explain it in an edit, beyond possibly the quote at the beginning of this paragraph. But it's progress. ;)
Wnt (
talk) 13:41, 14 October 2018 (UTC)
- Re: "the decrease in current with a fuse is a function of time more than anything else", that's a really useful way of looking at it. The plasma discharge also involves time -- it doesn't create new ions and electrons instantaneously, nor do they revert to ordinary atoms instantaneously. But then again, a resistor does not act as a resistor unless you give it some small amount of time after a sudden change in voltage -- because every resistor is also an inductor. I would think it fair to say that both the fuse and the plasma show a negative differential resistance when subjected to a slowly-changing voltage, and that pretty much everything acts differently at very short time scales. One could say that everything is a perfect insulator if the time scale is fast enough that the speed of light is the limiting factor. -- Guy Macon ( talk) 15:08, 14 October 2018 (UTC)
- With the neon lamp biased at a point midway between A and B by a current through a high value resistor as I described [3], apply a small voltage through a capacitor. That should demonstrate that the current and voltage swings are in the opposite direction consistent with the negative I/V slope. Also you may see the lamp brightness modulated with decreasing brightness with increasing voltage swings and vice versa. However a constant voltage supply is not good for testing the lamp because of the danger of the current leaping into the possibly destructive segment B->C. It's better to make a constant current supply be the controlling variable, which enables straightforward plotting of the whole V = f(i) characteristic. Note that this is not an efficient way to create a fluorescent lamp Dimmer. This is better achieved by a solid-state circuit that delivers a variable pulse-width voltage. DroneB ( talk) 20:25, 14 October 2018 (UTC)
- The question has been answered by now, but I will still leave a link to electrical breakdown. (Though not exactly the same thing, it helps understand how resistance can decrease by the application of strong voltage.) Tigraan Click here to contact me 11:46, 16 October 2018 (UTC)