Whoa! I'll drink to that! 🍺 🍺 🍺
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Hi Todd, Thanks for your comments and your interest. Steve’s approaches, such as what you’ve described, sound like good ones to me. Regarding your questions: Wider spacing of the two conductors will increase inductance and decrease capacitance. The reason that inductance decreases when the conductors are twisted together, or at least brought closer together, is that since the currents in the two conductors are moving in opposite directions (one toward the load and one toward the source, at any given time, with the directions alternating between each half cycle of the signal in the case of AC), the magnetic fields created by those currents are in opposite directions and tend to cancel each other if the conductors are close together. To the extent that those fields do not cancel, the impedance presented to the signal will increase as the rate of change of the signal (i.e., its frequency) increases. For further explanation the Wikipedia writeup on Lenz’s Law may be helpful. Regarding capacitance, a capacitor consists basically of two conductive plates separated by a non-conductive dielectric, with each of the two terminals of the capacitor connected to one of the plates. As explained in the Wikipedia writeup on capacitance the closer those plates are to each other the greater the amount of capacitance, everything else being equal. Similarly, cable capacitance presents itself as a shunt (aka parallel) phenomenon between the two conductors, and therefore increases as the conductors become closer together. While inductance and resistance present themselves as series phenomena. In general, inductance is most likely to be significant in the case of speaker cables, to an increasing degree if the impedance of the speaker is low at high frequencies. (See my first post in this thread). Resistance may also be a significant factor in a speaker cable, of course, especially if the speaker has low impedance at many or most frequencies. Capacitance will usually be unimportant in the case of a speaker cable, unless it is very high, in which case it can cause problems for the amplifier. Especially if the amplifier is solid state and therefore most likely has low output impedance, and uses significant amounts of feedback, in which case even destructive oscillations can occur if the capacitance is extremely high. In general, resistance and inductance will be unimportant in the case of a line-level interconnect, aside from the possibility that the resistance and inductance of the ground/signal return conductor can make a difference with respect to ground loop-related low frequency hum or high frequency buzz or noise, if the components involved are susceptible to ground loop issues. Capacitance can be important in the case of a line-level interconnect, especially if the output impedance of the component providing the signal is high. The interaction of that output impedance and cable capacitance will form a low pass filter, whose rolloff will usually begin in the ultrasonic or RF region, and therefore be inconsequential, but if those parameters are too high the filter can start rolling off and/or introducing phase shifts at frequencies that are low enough to have audible consequences. For a given cable type, all of those parameters are of course proportional to length. All of this, btw, pertains to analog cables. Completely different considerations and effects come into play in the case of digital cables. And yes, crossing cables at right angles, or at least minimizing how much of their lengths are parallel and closely spaced, is good practice and will minimize or eliminate any effects the corresponding signals might otherwise have on each other. Best regards, -- Al |
@almarg this has become a fascinating thread. I’ve a quick question: I’ve built a few cables in consultation with Steve (williewonka) and his ideas. Basically, the neutral in his cables is doubled and wrapped around the signal or hot conductor. My question is this: to lessen inductance, typically cables are tightly twisted, since the induced magnetic field is strongest further out. If the two conductors are spaced widely enough apart, will this also lessen the effects of induction? What about orientation? I'm under the impression that cables crossing at right angles won't induce currents in each other. Thanks. I’m trying to make sense of all this as well. I do like the performance of Steve’s design. I’d like to understand the ’why’ a bit more. |
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Hi Jim, As often occurs when this kind of subject comes up, ambiguity and/or imprecise use of terminology muddles the issue. If you replace his use of the word "charge" with the words "charge carrier," I think what he says then becomes pretty much correct. As explained by Kijanki with the balls in a tube analogy, and as alluded to in my long post in this thread dated 8-23-2017 at 7:08 p.m. EDT (although what I said in that post was stated in terms signal energy rather than charge), charge propagates at near light speed, while charge carriers (electrons, in the case of a metallic conductor) move very slowly. And current, defined in terms of amperes, is proportional to the average number of charge carriers traversing a given cross-section of the conductor per unit time. Best regards, -- Al |
Al (almarg), http://amasci.com/elect/elefaq1.html#aelist Is this guy wrong? regards, Jim |
Allow me to summarize. It doesn’t really matter what the velocity of electrons is since they are only the charge carriers. It doesn’t really matter what direction electrons are traveling since they’re only the charge carriers. And compared to the velocity of the EM - even if one considered electrons were moving at Fermi velocity - the relative velocity of electrons is negligible. Recall that the velocity of light (photons) is constant even if it’s measured from a rapidly moving rocket ship. 🚀 |
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Thanks, Kijanki. Yes, that explains it pretty well. Although I can see how that statement could be misinterpreted. The reference to "in one direction" should say something like "in one direction for a given direction of the electric field," the direction of the electric field of course alternating every half-cycle in the case of AC. Also, the reference to "average velocity" is a bit misleading, because it could be interpreted as meaning that the much faster Fermi velocity of 1570 kilometers/second or so is numerically averaged in, even though (as I mentioned earlier) it cancels out of the average (the "net flow") since it is in random directions. Best regards, -- Al |
Under the influence of an applied voltage/electric field, I believe that ALL electrons do NOT travel axially, in a direction corresponding to the polarity of the applied voltage, just SOME of them do. I believe that some of them continue to move in a random manner, at the Fermi velocity. Al, Wikipedia definition, I posted before, explains it: The drift velocity is the average velocity that a particle, such as an electron, attains in a material due to an electric field. It can also be referred to as axial drift velocity. In general, an electron will propagate randomly in a conductor at the Fermi velocity. An applied electric field will give this random motion a small net flow velocity in one direction. |
Geoffkait: "Therefore in this wire the electrons are flowing at the rate of 23 µm/s. At 60 Hz alternating current, this means that within half a cycle the electrons drift less than 0.2 μm. In other words, electrons flowing across the contact point in a switch will never actually leave the switch." to which kijanki replied, "You’re talking AGAIN about electrons. Electric current moves with the speed of electric charge (electric field) and not the speed of electrons or drift velocity. When you flip a switch electric charge moves thru conductor at the speed of light (remember stacked balls?) magnetic wave follows at the same speed. If electric current moves at the drift velocity then in very long cable electrons at the end would not even move since it would take hours or days for charge to get there." >>>>>I never said electric current moves with the velocity of electrons. I already said I think electrons are the charge carriers. You are confusing me with someone else perhaps. then kijanki wrote, "All electrons along the wire move together instantly like stacked balls. Electric and magnetic fields have to move with the same speed (electro-magnetic field) because one doesn’t exist without the other. Again, imagine pipe filed with ping-pong balls. When you push them at one side of the pipe they will start coming out at the other instantly. When there is no change (DC) electrons move at drift velocity. DC current is proportional to drift velocity while drift velocity is proportional to magnitude of electric field. Any sudden change at the one end of the wire will travel thru the wire at the speed of light and it will arrive almost instantly and not a few days later. It will travel as wave of electric charge inside of the wire (stacked balls) and wave of magnetic field outside of the wire at the same light speed (or close to it)." >>>>>EM waves travel at light speed but magnetic fields are stationary. Magnetic fields are induced by the current traveling through the wire. It’s the right hand rule. I never said electrons carry the current. And I’ve always said the EM waves are comprised of photons. All EM waves are comprised of photons. That’s why they travel at lightspeed. |
Geoffkait 9-3-2017I believe I did answer that, Geoff: Almarg 9-3-2017So yes, the electrons moving at drift velocity do move back and forth with alternating current, never moving very far from their original location, assuming no DC is present. The Wikipedia writeup you quoted even alluded to that: "... electrons flowing across the contact point in a switch will never actually leave the switch." Also, Geoffkait 9-3-2017Again, I believe that is incorrect. Under the influence of an applied voltage/electric field, I believe that ALL electrons do NOT travel axially, in a direction corresponding to the polarity of the applied voltage, just SOME of them do. I believe that some of them continue to move in a random manner, at the Fermi velocity. If that were not true, how much voltage would have to be applied for ALL of the electrons to suddenly cease moving in a random manner at Fermi velocity, and obediently start moving in an axial manner at drift velocity? 1000 volts? 120 volts? 1 volt? 1 millivolt? 1 microvolt? 1 nanovolt? 1 picovolt? How much current is necessary to be able to say that "current is present"? 10 amperes? 1 ampere? 0.0000000000001 amperes? I hope you see my point. In any event, barring further questions from others I’m done with this discussion. Regards, -- Al |
No one has answered my question, why is there a "net velocity" for electrons? Assuming electrons move back and forth with alternating current, which I’m actually not convinced they do. Also, the Fermi velocity is the *directionally random* quantum mechanical velocity of electrons when no current is present. To refer to a Fermi velocity when current is present makes no sense since electrons then travel axially, I.e., not randomly. |
Kijanki, I of course agree with your post. I suspect, though, that Geoff intended his "Exhibit A" post to be a rebuttal (mainly in its second paragraph) of the second paragraph of my post which immediately preceded it. To recapitulate the relevant paragraphs: Almarg 9-3-2017 Geoffkait 9-3-2017 (quoting Wikipedia)My response to that, assuming it was intended as a rebuttal of my statement: The quoted Wikipedia paragraph, referring to Fermi "velocity in the absence of electric current," says nothing about whether or not random movement of electrons at Fermi velocity occurs when a current is present. And I believe that such movement does in fact occur when a current is present, which is why drift velocity corresponds to **net** electron movement, past any given point. Regards, -- Al |
Therefore in this wire the electrons are flowing at the rate of 23 µm/s. At 60 Hz alternating current, this means that within half a cycle the electrons drift less than 0.2 μm. In other words, electrons flowing across the contact point in a switch will never actually leave the switch.You’re talking AGAIN about electrons. Electric current moves with the speed of electric charge (electric field) and not the speed of electrons or drift velocity. When you flip a switch electric charge moves thru conductor at the speed of light (remember stacked balls?) magnetic wave follows at the same speed. If electric current moves at the drift velocity then in very long cable electrons at the end would not even move since it would take hours or days for charge to get there. All electrons along the wire move together instantly like stacked balls. Electric and magnetic fields have to move with the same speed (electro-magnetic field) because one doesn’t exist without the other. Again, imagine pipe filed with ping-pong balls. When you push them at one side of the pipe they will start coming out at the other instantly. When there is no change (DC) electrons move at drift velocity. DC current is proportional to drift velocity while drift velocity is proportional to magnitude of electric field. Any sudden change at the one end of the wire will travel thru the wire at the speed of light and it will arrive almost instantly and not a few days later. It will travel as wave of electric charge inside of the wire (stacked balls) and wave of magnetic field outside of the wire at the same light speed (or close to it). |
Exhibit A from the wiki page on Drift Velocity: Therefore in this wire the electrons are flowing at the rate of 23 µm/s. At 60 Hz alternating current, this means that within half a cycle the electrons drift less than 0.2 μm. In other words, electrons flowing across the contact point in a switch will never actually leave the switch. By comparison, the Fermi flow velocity of these electrons (which, at room temperature, can be thought of as their approximate velocity in the absence of electric current) is around 1570 km/s.[2] |
Geoff, re your pop quiz, as you appear to realize there is no **overall** net movement of the electrons, assuming that the DC component of the applied voltage is zero. However, within each half-cycle of the applied voltage there is net electron movement and net velocity in one direction or the other, the direction corresponding to the +/- polarity of the applied voltage at that instant. I had said that in one of my early posts in this thread. Also, I believe that your statement that "Fermi velocity (random) applies only to materials when no current is applied" is incorrect, and that there is always random movement of some electrons, at Fermi velocity and in random directions. That is why the word "net" comes into play. Since the movements at Fermi velocity are in random directions, that velocity does not factor into (or average into) the drift velocity. Regards, -- Al |
Kijanki wrote, Drift velocity is average electron velocity since it is "net" axial velocity in one direction while electrons move in different directions. https://en.wikipedia.org/wiki/Drift_velocity Fermi velocity (random) applies only to materials when no current is applied. As I stated previously the very low Drift Velocity indicates that electrons do not (rpt not) travel rapidly at any time in the conductor. If they did the net velocity or average velocity whatever would be much higher than the centimeter per hour velocity observed. Pop quiz, if electrons are changing direction with alternating current, why is there a net velocity in one direction along the axis? Shouldn’t there be zero net velocity? And why is the net electron velocity in one direction, not the other direction? Why do electrons favor one direction over the other, assuming vector of Drift Velocity is always in the same direction? |
Geoffkait 9-2-2-2017Yes, I had read the paper you are referring to. There is nothing in it that is inconsistent with what I have said. If you'll notice, it deals with a hypothetical situation in which the wire ***is*** the load. In other words, a single piece of "long" wire is connected directly across the terminals of a voltage source. (The numerous references in the paper to the wire being "long" presumably imply that its resistance is high enough to limit the resulting, um, current, to an amount that can be provided by the voltage source, and that would not cause the wire to melt). In that situation the Poynting Vector would point inward to the conductor, at all points along its length, as shown in Figure 1 of the paper. The energy carrying photons would therefore enter the conductor, causing the conductor would heat up. Note the references to energy flowing **into** "the cylinder," resulting in "Joule heating." "The cylinder" referring to the geometry of the wire. As the paper says: The picture that emerges from these considerations is that the electromagnetic field around a current carrying wire is such that the energy dissipated in the wire is brought into it by the corresponding Poynting vector through each point of its surface. That is all perfectly consistent with what I have said on the subject previously, assuming a more real world scenario involving low resistance wires conducting energy to a resistive load. In that situation the electromagnetic wave, and the photons comprising it, travel outside the conductors, aside from (as I said in my previous post) "the very small fraction of the photons corresponding to the very small amount of energy that is absorbed by the resistance of the conductor and converted to heat." Regards, -- Al |
Al, Herman is right - electric current (electricity) is a flow of electric charge. Current does not flow, current is - charge flows. The same is true in the river - water flows and current is. Unfortunately improper usage of "current flows" (instead of electricty flows) is so common, that I found myself using it. Improper became common. |
Drift velocity is average electron velocity since it is "net" axial velocity in one direction while electrons move in different directions. https://en.wikipedia.org/wiki/Drift_velocity |
almarg wrote, "now regarding, Geoffkait 9-2-2017 One thing I will sign up to is that if anything is traveling down the conductor it’s photons, not electrons. Free free to concur with comment, concur without comment or non concur. As I’ve stated on previous occasions, I agree fully that the energy of an electrical signal (or power) being conducted via wires is conducted at near light speed in the form of an electromagnetic wave that is comprised of photons. We’ll have to agree to disagree, however, as to whether those photons propagate within or outside of the conductor, aside from the very small fraction of the photons corresponding to the very small amount of energy that is absorbed by the resistance of the conductor and converted to heat. >>>>Uh, I’ve already stated that it’s a tie. As indicated by the mathematical paper from the Journal of Physics on the dodgy subject of whether the energy of the signal is located outside or inside the conductor the energy is actually partly outside and partly inside. And the mathematics for that conclusion is provided in the first couple of paragraphs. Don’t tell me you didn’t read it. GASP |
Hi Jim, I always have great respect for Herman’s opinions and insights, and as usual I don’t disagree with any of his comments that you quoted. And as you’ve no doubt gathered over the years, like him I happen to be someone who generally tries to be as precise as possible with words and terminology, which he is certainly trying to be in the quoted passages. However, it seems to me that there are circumstances which can make some latitude in the use of terminology appropriate. Including this one, given the extremely widespread use (or arguably misuse) of the term "current." As well as the fact that for nearly all practical purposes, other than perhaps providing fodder for Internet debates about wire directionality (which Herman has not expressed a strong opinion about either way), the more widely used concept of "current" works fine. So as I said earlier: Almarg 9-1-2017Now, regarding: Geoffkait 9-2-2017As I’ve stated on previous occasions, I agree fully that the energy of an electrical signal (or power) being conducted via wires is conducted at near light speed in the form of an electromagnetic wave that is comprised of photons. We’ll have to agree to disagree, however, as to whether those photons propagate within or outside of the conductor, aside from the very small fraction of the photons corresponding to the very small amount of energy that is absorbed by the resistance of the conductor and converted to heat. Regards, -- Al |
almarg 7,451 posts 09-02-2017 1:14pm I don’t know what the gauge of the leads is, but given that the total length of the two leads is about 8 feet I suspect the lead resistance is a significant contributor to the 0.1 ohms, together with round-off due to the limited resolution.I got the same reading from my Fluke 87. Regarding fuse resistance, you might find the information on page 2 of this Littelfuse datasheet to be of interest. For the 4 amp 250 volt slow blow 6.3 x 32 mm glass fuse which is among the many listed, the "cold" resistance (meaning the resistance with negligible current being conducted) is indicated as 0.0311 ohms. So for a design which puts say half the rated max current through it the voltage drop would be a bit more than 0.06 volts.Al, I will check out the link you provided. So for a design which puts say half the rated max current through it the voltage drop would be a bit more than 0.06 volts.There’s that current again..... Just a guess the more current, the more heat produced by the energy the load is consuming, the more the resistance of the fuse the greater the VD. Correct? Or is it the more energy the load is consuming the greater the current. Which came first the chicken or the egg? At any rate my understanding, it is, the energy the load is consuming, that if it, increases above the rating of the fuse, (given by the manufacture in amperes), the fuse link will melt breaking the circuit. IT IS the energy that melts the link, not the current. Sound about right? I hope. I am still confused on the discussion of current in a closed circuit. Here is part of a response herman posted in response to a post of mine. herman Quote: As stated above current does not move. Current means something is moving. If we switch to charge instead of current then those don’t move to the load either. The charges in an AC circuit merely sit there and vibrate. Later on down the page herman posted a responded again to a post of mine. If you say the AC fuse blew because there was too much current flowing through it everybody nods in agreement even though that isn’t true. If you say the wire in the fuse melted because it got too hot after absorbing energy from the electromagnetic wave people look at you like you are insane and want to argue that vibrating electrons constitute current flow.https://forum.audiogon.com/discussions/directional-cables?page=3 Jim |
Al, sorry, but I prefer to not (rpt not) sign up to your explanation, either. For the same reason, actually, that I gave for not (rpt) agreeing with the drift velocity being defined as the average velocity. I.e., it doesn’t make sense. Please don’t put words in my mouth. If you guys want to agree to that explanation feel free to knock yourselves out. One thing I will sign up to is that if anything is traveling down the conductor it's photons, not electrons. Free free to concur with comment, concur without comment or non concur. |
Gentlemen, I believe that at this point we are all on the same page regarding what occurs when an electrical signal propagates. To the extent that there is disagreement I believe it just revolves around terminology, and its interpretation. Regarding "how could Drift Velocity be the average electron velocity?" I think that if the word "average" is changed to the word "net" we could all agree. At least I hope so. The word "net" in this context implying that random electron movements at Fermi velocity would cancel out of the drift velocity calculation, with electron movement caused by the applied voltage remaining in the calculation. Also, regarding the mention in the article that Jim quoted to the effect that drift velocity is a function of wire thickness, that is correct, and related specific calculations can be seen in the Wikipedia article on drift velocity I linked to earlier. Also, Kijanki, thanks for the excellent and very informative perspective you provided a few posts back on the Poynting Vector, E and H fields, etc. Jim, re your Fluke 87 multimeter, I have an 87V I purchased a couple of years ago, which I assume is a similar but more recent model. Great meter, although certainly not cheap (I think I paid around $375 for it). When the tips of the leads on mine are held together it reads either 0.1 ohms or 0.2 ohms, depending on exactly how the tips are held against each other. I don’t know what the gauge of the leads is, but given that the total length of the two leads is about 8 feet I suspect the lead resistance is a significant contributor to the 0.1 ohms, together with round-off due to the limited resolution. I previously had a small Triplett model 310 analog multimeter, which was ridiculously inaccurate (e.g. it indicated my AC as being around 95 volts; the Fluke indicates about 118 or 119 depending on time of day, etc). Which was surprising because I had read that many electricians use that particular Triplett model. Guess I just had a bad example of it. Regarding fuse resistance, you might find the information on page 2 of this Littelfuse datasheet to be of interest. For the 4 amp 250 volt slow blow 6.3 x 32 mm glass fuse which is among the many listed, the "cold" resistance (meaning the resistance with negligible current being conducted) is indicated as 0.0311 ohms. So for a design which puts say half the rated max current through it the voltage drop would be a bit more than 0.06 volts. Best regards, -- Al |
How could Drift Velocity be the average electron velocity? That would mean some electrons travel *even slower* than one cm per hour. That’s not a typo. One cm per hour. And it would also mean that no (rpt no) electrons travel very fast. Otherwise, the average velocity would be much higher. If it doesn’t make sense it’s not true. In fact, electrons don't have to move at all for the whole thing to work. |
The "electron drift" IS the electron velocity. No, it is not. Electron drift is AVERAGE speed of all electrons. Individual electrons move fast at about 1% of speed of light even without electric field (Fermi Velocity). No, actually that analogy intimates that current travels instantaneously, which cannot (rpt cannot) be true. Line of balls is simplification of lattice of electrons that is disturbed at one end. As I mention before connection between electrons is not a physical one. Electrons poses electrical charge and repel each other. It takes time for disturbance of the lattice to travel thru wire. Thus electrons are charge carriers, they are not the charge per se. Electrons poses the charge - it is called Elementary Charge. There are two forces between electrons - gravitational and Coulomb (electrostatic) force. Gravitational is attractive but is very, very small in comparison to Coulomb repulsive force, that exist exactly because electrons have charge. |
kijanki wrote, "Back to our analogy with balls stacked in the tube - last ball will start moving the same moment as first ball (they push each other). That’s electric charge moving (electric current)." No, actually that analogy intimates that current travels instantaneously, which cannot (rpt cannot) be true. I mean unless you’re invoking action at a distance. Current moves at *near lightspeed* which means current must be photons, no? The electrons are only charge carriers, they're not (rpt not) the charge per se. |
The "electron drift" IS the electron velocity. Drift velocity is on the order of cm per hour. In other words they, the electrons, are virtually stationary. The "drift velocity" is not (rpt not) the net velocity due to back and forth motion of the electrons. Whatever moves at lightspeed or near lightspeed are photons. You know, the only particles that can - and must - travel at lightspeed or near lightspeed. If you want to say current is an EM wave and therefore comprises photons I have no problem with that. Thus electrons are charge carriers, they are not the charge per se. |
Jea48, It is all very confusing. AFAIK electric current (as motion of electric charge) in wire moves very fast - close to speed of light. Individual electrons also travel fast at about 1% of the light speed (2000km/s) but they move in different directions. What moves really slow is average speed of all electrons (drift velocity). Back to our analogy with balls stacked in the tube - last ball will start moving the same moment as first ball (they push each other). That's electric charge moving (electric current) at the speed of light. |
Correction: In my post dated 09-01-2017 8:33pm The only fuse I have on hand is a 4 amp slow blow fuse. I have an older model Fluke 87 True RMS multimeter and I checked for resistance across the fuse link end caps.That should read, meter reads 000.1 ohm. . |
@kijanki, Thank you for your informative response. From what I understand the movement of the current in the conductor is quite slow.... Correct? Quote from link below. Electric current is not a flow of energy; it's a flow of charge. Charge and energy are two very different things. To separate them in your mind, see this list of differences. The term "Electric Current" means the same thing as "charge flow." Electric current is a very slow flow of charges, while energy flows fast. Also, during AC alternating current the charges move slightly back and forth while the energy moves rapidly forward. The energy does not flow back to the battery again. At the same time, the electric current is different; it is a very slow circular flow, and the electric charges flow through the light bulb filament and all of them flow back out again. They return to the battery.http://amasci.com/miscon/eleca.html#cflow After reading your post I went back and checked again what I had read. Then I clicked on the blue high lighted "slow" word. And this came up. The quick answer Then, The complicated answer Quote: "In other words, the speed of the charges is proportional to the value of electric current; small current means slow charge-flow, large current means high speed. Zero current means the charges have stopped in place. Note however that an electric current does not have just one speed within any circuit. Charges speed up whenever they flow into a thinner wire. The high current in a large flash-lantern's lightbulb will be much faster than the same current in the other conductors in the lantern. Even though an electric current is a very slow flow of charges, we can't know the actual speed of flow unless first we know the thickness of the wires, as well as the *value* (the amperes) of the current in the wires. " And then he says, The speed of electric current Since nothing visibly moves when the charge-sea flows, we cannot measure the speed of its flow by eye. Instead we do it by making some assumptions and doing a calculation. Let's say we have an electric current in normal lamp cord connected to bright light bulb. The electric current works out to be a flow of approximatly 3 inches per hour. Very slow!http://amasci.com/miscon/speed.html Wow! I'll have to reread it again tomorrow. But I think he is saying the same thing you said. At least some parts of what he is saying. But not others? Thanks again for your response, Jim |
Al, Thanks again for your response. Thanks, Jim. Regarding...Well I knew watts is a unit of power and I somewhat understood energy and joules. But I guess I didn’t understand the real differences between the two. I do have a better understanding now thanks to you Al.Am I correct in assuming watts is a measurement of electrical energy?Watts is a unit of power, as you of course realize. Power is a quantity that is defined at a specific instant of time, although its average value over some interval of time can of course be calculated. Energy is defined as the product (multiplication) of power and time, and can be expressed as some number of joules, as well as in various other units. . Quote:The electrical energy will be greater at 120V than at 24V for a circuit using the same 2 amp fuse for overcurrent protection.The power and the energy being conveyed to the load will of course be much greater in the 120 and 240 volt cases than in the 24 volt case. But as I’m sure you realize but others may not, the only voltage that the fuse "knows about" is the one that appears between its two terminals, which when it is not blown corresponds to the amount of current it is conducting times its resistance. In the case of audio equipment operating normally that voltage will typically be a small fraction of a volt. "But as I’m sure you realize but others may not, the only voltage that the fuse "knows about" is the one that appears between its two terminals, which when it is not blown corresponds to the amount of current it is conducting times its resistance." " times its resistance." I have not ever heard it explained that way before. I honestly have never measured a voltage across the end caps or blades of a good fuse. A blown fuse on the other hand yes, as you stated. I have measured a slight voltage drop across the fuse holder clips, mostly cartridge fuses. A VD across the fuse holder clips indicates poor contact pressure and or corrosion, poor surface area between the fuse caps and fuse holder clips. The only fuse I have on hand is a 4 amp slow blow fuse. I have an older model Fluke 87 True RMS multimeter and I checked for resistance across the fuse link end caps. With the meter set on ohms auto first touching the two probes together the meter reads 000.01 ohm. I got the same exact reading checking the fuse. LOL, I even reversed the fuse and got the same reading. (You know who that was for) I have read posts of guys that buy audio grade fuses that say they do indeed measure a resistance across the fuse link end caps. What you said above does make sense though. I have a good basic understanding how the electromagnetic wave thingy works, I just need learn the lingo better how to express it. Me thinks when talking about electrical power issues and electrical safety codes, like NEC, I will stick with the old school way I was taught and have a good understanding of. Besides that is what the majority of people understand. Especially electricians. As for ICs and speaker cables the old school theory just doesn’t fit the reality of how the audio signal travels from the source to the load. Thanks again Al for all your help, Jim . |
From what I understand the movement of the current in the conductor is quite slow.... Correct? Electric current is a flow of electric charge and not the flow of electrons. (In fluids electric charge is carried by ions and not the electrons). Number of electrons crossing given point defines amount of electric charge (current) passing. Motion of electric charge is usually explained as a row of stacked balls in the tube - when you push them slowly they will move slowly but when you hit the first one with a hammer the last one will respond instantly - that's the speed of electric current (charge). Of course there is plenty of space between electrons but "stacking" is not physical but electrical (electric charge). As for the energy transfer on the outside of the conductor by electromagnetic wave - without it current in the wire alone would not explain energy transfer, since the same amount of electric charge comes and leaves the load (same current leaves and comes back to power supply). Poynting vector is defined by Electric Field E and magnetic field H. Amount of energy transferred is proportional to magnitude of both fields ExH. Electric field is proportional to voltage while magnetic field is proportional to current. Multiply them and you'll get the power P=V*I (power over time is energy). Current flowing thru the fuse causes voltage drop since fuse has resistance. This voltage drop creates electric field across the fuse - without this voltage drop (fuse resistance equal zero) there would be no electric field and magnetic field (current) alone cannot deliver energy to fuse. |
Thanks, Jim. Regarding... Am I correct in assuming watts is a measurement of electrical energy?Watts is a unit of power, as you of course realize. Power is a quantity that is defined at a specific instant of time, although its average value over some interval of time can of course be calculated. Energy is defined as the product (multiplication) of power and time, and can be expressed as some number of joules, as well as in various other units. The electrical energy will be greater at 120V than at 24V for a circuit using the same 2 amp fuse for overcurrent protection.The power and the energy being conveyed to the load will of course be much greater in the 120 and 240 volt cases than in the 24 volt case. But as I’m sure you realize but others may not, the only voltage that the fuse "knows about" is the one that appears between its two terminals, which when it is not blown corresponds to the amount of current it is conducting times its resistance. In the case of audio equipment operating normally that voltage will typically be a small fraction of a volt. If and when the fuse were to blow, however, the full 120 volts would then appear across its terminals. Although no current would be conducted then since the resistance of the blown fuse would be essentially infinite. Relevant to all of this, it’s worth noting that in the detailed specifications that are provided by the major fuse manufacturers, such as Littelfuse and Eaton/Cooper Bussmann, the "melting point" (i.e., the point at which the fuse is nominally rated to blow) is specified as i^2 x t (e.g., amperes squared x seconds). As you of course realize, power into a resistive load = i^2 x R, and i^2 x t is therefore proportional to energy. Best regards, -- Al |
Al, (almarg), Thank you for your responses to my questions. I would agree, it is an electric field not a magnetic field.... if only a voltage, (potential), is present, an electromagnet field will exist outside of the conductor/s without there being current... Correct?I’m not 100% certain, but I believe in that situation an electric field would be present, but not a magnetic field. . Since the amount of energy that is absorbed from the electromagnetic wave by the conductor in the fuse and converted into heat (causing it to blow if excessive) is proportional to both the energy that is being conveyed by that wave and to "the current," it is reasonable (and of course far more practical) to analyze the situation in terms of amperes and ohms, rather than in terms of joules (a unit of energy) and Poynting Vectors. Your second paragraph has to be the logical case. And not just for the "why" the fuse blows. The electrical energy will be greater at 120V than at 24V for a circuit using the same 2 amp fuse for overcurrent protection. 120V x 2A = 240 watts 24V x 2A = 48 watts 240V x 2A = 480 watts Am I correct in assuming watts is a measurement of electrical energy? Jim . |
Hi Jim, As always you ask good questions. Regarding the first one, though... Could you please explain in more detail the relationship of the electromagnetic wave, that travels in the space outside of the conductor, (At near the speed of light), and the "current" that travels very slowly slightly vibrating back and forth at 60Hz in the conductor.... I'm not sure what I can add to what I said in my long post above dated 8-23-2017 at 7:08 p.m. EDT. Regarding your other questions: From what I understand the movement of the current in the conductor is quite slow.... Correct?Correct, assuming "current" is defined as the movement of charge carriers (i.e., electrons in a metallic conductor). An example described in the Wikipedia writeup on Drift Velocity indicates that for a current of 1 ampere in a copper conductor of 2 mm diameter the velocity calculates to 23 um/second ("um" = millionths of a meter). As noted in the writeup, btw, random movement of electrons even in the absence of "current" occurs at a far greater velocity (the Fermi velocity) than the "drift velocity" of current, although the Fermi velocity is still vastly slower than the speed of electromagnetic wave propagation. Am I correct in saying you can’t have the electromagnetic wave without having current?Yes, in the case of electrical energy that is being conveyed via wires. Electromagnetic waves can of course propagate in free space, as in the cases of radio waves and light waves. The bigger the load, the more current in the conductor. The more current in the conductor the larger the electromagnet wave.... Correct?Yes, assuming "larger" is interpreted in the sense of having "more energy." IF the wire is too small to handle the amount of current in the wire is it the current that causes the wire to overheat or is it the energy of the electromagnetic wave? Please explain in detail.The Poynting Vector, which describes the direction in which energy is being propagated, would be perfectly parallel to the conductor if the conductor's resistance were zero. Since that resistance is non-zero, the Vector will tilt slightly toward the conductor, resulting in a small amount of energy being transferred into conductor, absorbed by its resistance, and converted to heat. In effect, the resistance of the conductor causes it to become part of the load. ... if only a voltage, (potential), is present, an electromagnet field will exist outside of the conductor/s without there being current... Correct?I'm not 100% certain, but I believe in that situation an electric field would be present, but not a magnetic field. I know it is the energy, from the electromagnetic wave, that makes a heating element heat up and gives off its’ heat into the surrounding air around it. It is not the "current" directly causing the resistance element to heat up.... Correct?As I've said, in the case of electrical signals (or power) being conducted via wires the electromagnetic wave and "the current" go hand-in-hand, and one would not exist without the other. So the question is essentially just an academic/philosophical one IMO, not unlike the classical question of whether the chicken or the egg came first. Your succeeding statements involving E, I, P, etc. are of course correct. herman said it is the energy of the electromagnetic wave passing on the outside of the fuse element link that causes it to melt and blow open when the fuse is overloaded.It is energy absorbed **from** the electromagnetic wave by the non-zero resistance of the conductor in the fuse, which as I said causes the Poynting vector to tilt slightly toward the conductor, that causes it to blow. Is not P the energy of the electromagnetic wave?They are proportional, but strictly speaking energy corresponds to power x time. Here is where I get hung up. As you know a 2 amp 250V fuse can be used for any voltage 250V or less. It could be used where the voltage is 24V. The ampere rating of the fuse is still 2 amps. So to me the current has to be some component that causes the fuse to blow when the current that passes through the fuse link and exceeds 2 amps in the time curve set by the fuse manufacture. NOTE I did not say current flow.In my earlier long post I defined "the current" as follows: What can be referred to as "the current," as opposed to "the signal," can be considered as corresponding to the number of electrons traversing a given cross-section of a conductor in a given amount of time. One ampere of current, for example, corresponds to one coulomb per second, where one coulomb corresponds to the amount of charge possessed by about 6.2 x 10^18 electrons.Since the amount of energy that is absorbed from the electromagnetic wave by the conductor in the fuse and converted into heat (causing it to blow if excessive) is proportional to both the energy that is being conveyed by that wave and to "the current," it is reasonable (and of course far more practical) to analyze the situation in terms of amperes and ohms, rather than in terms of joules (a unit of energy) and Poynting Vectors. And correspondingly, since in the case of electrical signals (or power) being conducted via wires the slow moving "current" and the very fast moving electromagnetic wave go hand-in-hand (as I've explained), IMO it would be meaningless to think of one but not the other as being the cause of the fuse blowing. Best, -- Al |
@almarg, AC transmission using wire conductors.
Al, Could you please explain in more detail the relationship of the electromagnetic wave, that travels in the space outside of the conductor, (At near the speed of light), and the "current" that travels very slowly slightly vibrating back and forth at 60Hz in the conductor. From what I understand the movement of the current in the conductor is quite slow.... Correct?
The electromagnetic wave is caused by the applied source voltage and the "current", "charge", in the conductor? (Amount of current in the closed circuit determined by the resistance of the connected load. I = E/R)..... Correct? Am I correct in saying you can’t have the electromagnetic wave without having current? Install an on/off switch in series in the circuit. Close the switch the current passes through the switch contacts through the load and back to source.... Correct? The bigger the load, the more current in the conductor. The more current in the conductor the larger the electromagnet wave.... Correct? And of course the conductor, wire, must have a current, ampere rating, to safely carry the current in the wire so the wire will not overheat. IF the wire is too small to handle the amount of current in the wire is it the current that causes the wire to overheat or is it the energy of the electromagnetic wave? Please explain in detail. . Not to confuse things, if only a voltage, (potential), is present, an electromagnet field will exist outside of the conductor/s without there being current... Correct? . I know it is the energy, from the electromagnetic wave, that makes a heating element heat up and gives off its’ heat into the surrounding air around it. It is not the "current" directly causing the resistance element to heat up.... Correct? I know the amount of energy consumed,(in watts), by the resistance element is determined by the source voltage and the resistance, in ohms, of the resistance element. E / R = I and we know the current..... Correct? The Fuse..... E x I = P E = voltage I = Current, amps P = power, energy, measured in, watts, VA A fuse rated at 2 amps with a maximum voltage rating of 250V. herman said it is the energy of the electromagnetic wave passing on the outside of the fuse element link that causes it to melt and blow open when the fuse is overloaded. OK Isn’t the size, (for lack of a better word), of the electromagnetic wave energy determined by the applied source voltage and the current in a closed circuit? E x I = P. Is not P the energy of the electromagnetic wave? So say the load is 150 watts and a 2 amp 250V fuse is used to protect the load. The FLA of the 150 watt load is, 150W/120V = 1.25 amps. Here is where I get hung up. As you know a 2 amp 250V fuse can be used for any voltage 250V or less. It could be used where the voltage is 24V. The ampere rating of the fuse is still 2 amps. So to me the current has to be some component that causes the fuse to blow when the current that passes through the fuse link and exceeds 2 amps in the time curve set by the fuse manufacture. NOTE I did not say current flow. WOW,... I know,..... I sure have a lot of questions on my mind. Blame herman. Very best regards, Jim |
Chalmersiv, the answer to your question is of course not predictable with any kind of certainty, in part because cable effects are dependent to a significant degree on the technical characteristics of what is being connected. And in the case of a speaker cable, among other things on how the impedance of the speaker varies as a function of frequency, as well as the nominal impedance of the speaker, and perhaps in some cases also on the amplifier's output impedance and on how much feedback its design incorporates. But as I mentioned earlier the inductance of a speaker cable can be drastically degraded (i.e., increased) if the + and - conductors are not in very close proximity. Although whatever significance that may have will depend on the impedance of the speaker at high frequencies, and also on the length of the cable. And I note that the description of the T-14 cites low inductance as one of its key features, while the description of the Q-10 does not. That despite the fact that the materials used for their conductors and dielectrics are apparently the same. So for that reason, together with what has been said earlier about the + and - conductors having essentially equal importance, I would recommend against using any of the cheaper wires you listed for the negative conductors. What I would suggest that you try, if you already haven't, is using BOTH the T-14 and the Q-10 in parallel. And in each case with the conductors that are enclosed within a given cable jacket being used for BOTH + and - (which if I understand your last post correctly may not have been what you were doing), rather than for just one polarity. Also, I'm not sure if I understand whether your Q-10 is configured with a sufficient number of connections for biwiring, but if not use it in parallel with the T-14 for the bass connections, and use the additional T-14s alone for the mid/hi connections. On another note: Jim & Kijanki, thanks again for the nice words. Kijanki, in response to your question, no, I have never been a teacher, and I have not ever had any particular desire to be one. But in my career working in a corporate environment I have always found it advantageous to be able to communicate in as clear and precise a manner as possible. Also, speaking of being knowledgeable, I'll mention that Jim's (Jea48's) knowledge of all things electrician-related continually amazes me. And I've certainly learned more than a few things from his posts over the years. As well as having had the pleasure in various threads here of the two of us successfully resolving more than a few problems people have had with their systems. Best regards, -- Al |
I have 4- 8 ft lengths of dh labs q 10....each length has 2 x 12 ga and 2 x 14 ga... 8- 8 ft lengths of dh labs t-14......each length has 2 x 14 ga 4- 8 ft length of solid copper 8 ga grounding wire from lowes sheathed in heat shrink tubing shrunk only at ends... many other copper speaker wire from radio shack flat braid to 8 ga stranded how do I connect to 4 post bi wire speaker terminals.... I have run 4 of the t-14 to each speaker and it sounded better than 1 run of q-10...1 length of t-14 to each terminal if I use 1 length of q-10 to each pos terminal what do I use for neg terminal or suggestions for connection needed thanks, chalmersiv |
One last thing. Over, I think it is on the Directionality of wire thread, someone posted a link to a largely mathematical article from, I believe it was the Journal of Physics, that described how energy travels down a conductor. The article was used by the posted to support the popular idea that energy travels completely OUTSIDE the wire, not inside the wire. Yet, that very article - in the first couple of paragraphs - states very clearly that energy is traveling both inside the wire AND outside the wire. The mathematics for both energies are subsequently described. I already acknowledged that (some energy travels outside the conductor) might be true a couple of weeks ago. Furthermore, based on that evidence, the mathematical evidence, I hereby declare this current argument a tie. The peer review tribunal can take a break. |
