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.
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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 |
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. |
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 |
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