Quiz Answer: The multimeter will show about 10 ohm


quiz

As many of you guess correctly the multimeter will show about 10 ohm as it measures the DC resistance not the impedance and in this case it measures just the resistance of 10 ohm plus the resistance of the PCB tracks which is not much in this case.

 

7 Comments (+add yours?)

  1. Lachlan
    Nov 26, 2015 @ 10:16:30

    The Ohmmeter is correct, the impedance for a for static dc test signal is 10 ohms.
    for other wave’s form’s which are not DC of course, it will be some other value.
    related to the layout of the tracks ground plain’s.. etc, and the waveform and circuit load’s none liner.. etc.

    Reply

  2. tinkererguy
    Dec 01, 2015 @ 01:45:18

    I have read a long time ago a good thought experiment that helped me understand the characteristic impedance concept:

    let’s take a situation like the quiz above, but the PCB above goes all the way from here TO THE MOON. what will the multimeter show?

    the interesting answer is, it will show *65 Ohms* for approximately 3 seconds, then it will show 10 Ohms. The reason is that it takes that time for the wavefront in the trace to propagate all the way to the moon and back. before we receive the reflection from the termination (10-Ohm resistor), we only sense the impedance of the cable.

    Reply

    • OLIMEX Ltd
      Dec 01, 2015 @ 08:26:13

      not very accurate though, the characteristic impedance is the capacitive resistance at high frequency, which can’t be measured with DC voltmeter, so the voltmeter will always show 10 ohm no matter how long the lines are🙂

      Reply

    • LinuxUsers
      Dec 13, 2015 @ 07:34:59

      It wouldn’t show 65 ohms for 3 seconds. During these seconds, field spreads over the medium at speed of light. You would not register DC current before wave front arrives to another end of wire.

      Imagine you have water surface and stone. You throw stone into water. Ripple wave spreads. If you measure water level at distance, you only get idea about ripple when wave front reaches this place some time later. The very same idea works for electromagnetic waves. Electromagnetic waves also take some time to spread. Level change or flow change takes time to propagate. Wave would reflect from obstacles as well. Only notion of obstacle differs, and impedance is all about describing “obstacles”. In water you can see what happens when reflected wave meets original wave. Similar thing happens to electromagnetic waves as well. That’s why we do not make 90 degree turns and care about impedance. Else wave front would reflect back. Then it would mix with next pulse we’re trying to transmit. Which isn’t good thing most of time. Though some special cases exist, e.g. dipole antennaes are better when they actually match wave size🙂

      To make it more funny: if process is not static, voltage and current values do not have to change synchronously, it goes beyond of Ohm law for DC. In circuits dominated by inductance, current changes are delayed against voltage changes, and in capacitance-dominated circuits, current changes are ahead of voltage changes. That’s where things can get somewhat complicated. DC is really most simple part. But whole notion of DC is simplification and subject to some constraints. Traveling wave front isn’t DC in first place. It moves and spreads.

      Reply

  3. tinkererguy
    Dec 13, 2015 @ 18:49:24

    I would like to defend the result I described in the thought experiment with the cable to the moon.

    first, the easier part is to refute OLIMEX LTD’s comment, that the multimeter will always show 10 Ohm. During the first 3 seconds, the multimeter *can’t* show 10 Ohm, because it will violate a basic physical law: that information cannot travel faster than the speed of light. Your result implies that the information about which resistor is connected on the moon have been transmitted somehow to earth in zero time (since you say the multimeter shows this value immediately), and that is impossible.

    it may be easier to correlate with your (correct) knowledge that the characteristic impedance is relevant only to high frequencies and therefore cannot be measured with a multimeter, by considering what “high frequency” means: let’s see, is 100MHz “high frequency”? (usually yes), how about 10MHz? (less often but occasionaly), how about 1MHz? (generally not, but sometimes yes, it depends…) so what is going on here? The answer is: it’s relative. relative to what? to *the physical size of the circuit*! if the wavelength corresponding to the frequency in question is significantly larger than the length of the longest conducting path in the circuit, then we can forget about characteristic impedances and transmission lines and only think in terms of DC resistances and so on. It’s called the Lumped Element Model. Now, let’s think again about my thought experiment: I enlarged the circuit in question to be huge: approximately 1.5 light-seconds long (several hundred thousand kilometers), so *everything* that happens faster than a few seconds (including what you briefly see in the multimeter for a second or two) is considered a “high frequency”. it’s as if we turned the multimeter into an oscilloscope…

    So, if we are in agreement that we can’t see “10 Ohm” in the first 3 seconds, what *will* we see? to prove that it is the characteristic impedance, as I claimed in my original comment, is a bit more challenging, and since this has been a long post already I will write about it later separately.

    Reply

  4. LinuxUser
    Dec 14, 2015 @ 04:00:25

    > what *will* we see?
    If we disregard capacity, inductance, Earth’s magnetic field, etc (which is not to be taken as granted for circuit of this size) IMHO it would look like an open circuit while field just travels across the moon. While wave front just travels, no DC current started yet, so it just does not looks like closed circuit at this point. So your device sees infinite resistance. Once wave front reaches far end, it would make circuit complete, DC current starts and you’ll see current corresponding to 10Ohms resistance of path. What you get is: ~3 seconds of previous, old line state, then your new, changed line state.

    You thrown stone into water (changed voltage by starting measurement). Wave front haven’t arrived to your remote measurement point yet. What do you see? Just still water sufrace, nothing else. Previous line state. If it looked like an open circuit and there was no current, so be it so. It wouldn’t change until wave front arrives to far end.

    Interestingly, if you quickly throw, say, logical 1 and then logical 0, this perturbation travels over wire, and impulse length haves some physical meaning, i.e. signal between transitions haves some physical length (expressed as C * T) and this wavefronts pinch travels over wire at speed of light. Arrival time is delayed by wire length. This leads to quite many “fancy” things. Track would gladly behave like antennae, especiylly if your you have high frequency, and especially if track length is comparable to 1/4 of wavelength. There are also some fancy things like zero-DC line coding, e.g. 8b/10b thing, which overcomes inability of low-power transmitter to recharge long wire of transmission line anyhow fast. Idea here is not to override global DC line state, but rather throw short bursts which only change wire state “locally” as signal travels. That’s where reflections can get really nasty. Ringing can somewhat remind you how wave in water dissipates. By periodically disturbing medium you can get e.g. wave, which would follow specific pattern. Like standing wave, interference patterns, etc. You can encounter all these in electric field as well. After all, it somewhat resembles gravity field.

    DC is a most simple and basic thing one can get, but it only makes sense in describing static processes which are settled down to degree we can make assumption things do not change over time. Full circuits descriptions are more complicated. Full 3D+time field descriptions are quite mindblasting. And lengthy transmission lines are quite a special case, actually. More or less lengthy lines of anyhow comparable lengts are worth of separate study in quite many regards you usually to not expect from just piece of wire. E.g. such a long wire would be an excellent antennae, even for low frequency signals. E.g. mains 50/60Hz waves have much smaller wave lengths. Such a long wire is going to transmit virtually everything and would also pick up all electromagnetic fields around as well.

    Reply

  5. tinkererguy
    Jan 29, 2016 @ 02:27:54

    I forgot about this thread and just remembered, and although I don’t have time to fully answer and explain what you missed, I do want to point out I looked it up and it turns out it’s actually a fairly common thought experiment taught in textbooks on the subject, with the answer as I originally described…

    you may want to look in the following link for more info:
    http://www.allaboutcircuits.com/textbook/alternating-current/chpt-14/characteristic-impedance/

    and also here:
    https://books.google.co.il/books?id=_IiONSphoB4C&printsec=frontcover#v=onepage&q=moon&f=false
    (in this book, check out page 230, or search for “moon”…)

    Reply

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