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Can you have a resistance less than one? Is this allowed? I've only ever seen circuits with one ohm resistance at least.

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5 Answers 5

Is this allowed?

It is often the case that resistors in the milli-Ohm and micro-Ohm range are required especially for current sensing applications. One need look no further than Digi-Key for examples.

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Certainly you can. Ordinary lengths of wire have resistances of the order of milli-Ohms. There is nothing special or discrete about one Ohm. It's just a historical unit, like a meter.

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Resistance is defined by $$R = \rho \frac{L}{A}$$

If you follow this equation, then the resistance can have any value. However, if you go down to the microscopic scale, there is a minimum unit of resistance, called a quantum of resistance. $$ R= \frac{h}{2e^2} = \frac{6.635 \times 10^{-34} Js}{2 \times 1.6 \times 10^{-19}C }= 12906.4037217 Ω$$

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Clearly whatever fundamental meaning $h/2e^2$ has it's not a minimum unit of resistance (nor a maximum)... –  Michael Brown Oct 24 '13 at 1:07
Well, it actually is. It is the resistance that one electron flowing through a quantum wire gets. –  mcodesmart Oct 24 '13 at 1:12
How can you buy kilohm and megohm resistors then? –  Michael Brown Oct 24 '13 at 1:14
I get what you're saying by the way, though it's probably easier to think in terms of minimum conductance rather than maximum resistance. At least for me it is. –  Michael Brown Oct 24 '13 at 1:19
Yes. You are correct. It is always referred to as the minimum conductance, but I thought the inverse would be valid, but leads to confusion. –  mcodesmart Oct 24 '13 at 1:25

Most certainly you can have fractional resistances, not only are various resistances that you can purchase in market lesser than 1 ohm; whenever you connect resistances in parallel, fractional resistances are obtained, sometimes those fraction turn out to be whole number but you can even get fractions which are non terminating repeating, you can not even buy such resistances and yet you can obtain them by a particular kind of connection.

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I'm answering here because I have not enough reputation to comment. What Programmer wrote here as a limit of resistance ($R=\frac{h}{2e^2}\sim12907\Omega$) is in fact maximum not minimum resistance. And then only in case of ballistic transport.

We have ballistic transport when mean free path of electron is much larger than size of conductor, so only electron collisions we get are collisions (or should we say reflections from) with walls of conductor. In semiconductors to achieve this you would normally have to go to low temperature, sometimes even lower than liquid helium, and have ultra high purity of semiconductor.

It's quite a lot easier to observe this in metals. That's what I did for my masters few years ago. To observe quantized conductivity you have to obtain metal nanowires. And this is a lot easier than it sounds. I achieved it by placing golden plate on piezostack, and a golden needle on a micrometer screw above it. I was driving piezostack with very unsophisticated driver so that plate would constantly come in contact with needle and retract. When needle was barely touching plate nanowires would form. Then very simple tools were used to measure conductance - I had reference resistor in series with this junction connected to precise voltage source, and an oscilloscope measuring voltage on this reference resistor.

Now this is a bit of cheating, because we do not see pure effect here. Conductance is quantized due to atomic configuration of wires. But energy of open mode is so high that configurations with conductances in multiple of above minimum are a lot more stable. And by mode I mean transverse momentum mode. Figuratively but not very precisely you could think of it as electrons not bouncing on walls, bouncing once, twice and so on. And as wire gets thicker and thicker more modes are possible. If it's too thick you get a lot of scattering and situation gets back to normal $$R=\sigma\frac{L}{A}$$

Sorry for lengthy (and unwieldy) comment, but I couldn't resist - it was my master's after all.

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