# Definition of Ohm in SI basic units in words

One way Wikipedia defines Ohm is (this is also teached in school): $$1\Omega =1{\dfrac {{\mbox{V}}}{{\mbox{A}}}}$$ They describe this definition in words, too:

The ohm is defined as a resistance between two points of a conductor when a constant potential difference of 1.0 volt, applied to these points, produces in the conductor a current of 1.0 ampere, the conductor not being the seat of any electromotive force.

The definition of Ohm in SI basic units is: $$1\Omega = 1{\dfrac {{\mbox{kg}}\cdot {\mbox{m}}^{2}}{{\mbox{s}}^{3}\cdot {\mbox{A}}^{2}}}$$ It's really hard for me to get that this definition is correct. It's clear that mathematical calculations confirm this definition. But how do you describe the definition of the SI in words like that paragraph on wikipedia?

Edit: How would you describe it? Although it is not common to do it that way, I think describing it that way, could be very interesting.

I think the short answer is, you don't. The reason we call the unit of force a Newton and not a kg m/s$^2$ is because it is convenient and it expresses the relation you want to convey when used elsewhere (e.g., $F=-kx$ for a spring).

Similarly, it is convenient to "hide" the MKS base units into a single term, the potential $V$ in this case, so that the formula is easier to remember and that the relation is conveyed, in this case the relation between potential difference, current, and resistance.

• You are totally right. This is the common way to refer to other formulars by using their symbol. But assuming you want to descibe it anyway. So I correct my question to: How would you describe it? I think describe it that way, could be very interesting. Commented Feb 15, 2014 at 19:00
• @andrew: How would I describe an ohm? It's the unit that describes how much the flow of charge is hindered by the material. Commented Feb 15, 2014 at 19:08
• Yeah. This is an good answere. But what do you mean with "flow of charge"? Do you mean the Amperage $I$ ? Commented Feb 15, 2014 at 19:19
• @andrew: If you feel more comfortable with it, you could replace "flow of charge" with "bulk motion of electrons," as it is the electrons that are moving in an electric circuit. Commented Feb 15, 2014 at 19:23
• So physically the motion of electrons you mean is $I=\frac{Q}{t}$, isn't it? Commented Feb 15, 2014 at 19:26

I'm not sure there's much of a point to what you're asking. The intuitive way to understand an Ohm is to use $\Omega = V/A$. If you want to use SI units, you can, and the math indeed tells you that your other definition is correct, but you're not gonna get much out of it. Indeed, the most you could do is to separate it like this:

\begin{align}\Omega &= \frac{\text{kg} \cdot \text{m}^2 }{ \text{s}^3 \cdot \text{A}^2 }\\ &= \frac{ \text{kg}\cdot\text{m}^2}{\text{s}^2}\cdot \frac1{\text{A}\cdot\text{s}}\cdot \frac1{\text{A}} \\ &= \frac{\text{J}/\text{C}}{\text{A}}\\ &= \frac{\text{V}}{\text{A}} \end{align}

This is just a proof of the equivalence between the two definitions, but don't expect to get any nice word description of the SI definition.

Not sure whether this is correct, but if you have to do it, I think you can say that it is:

the work done by the conductor per unit charge per unit current through the conductor, or in terms of SI units, $\mathrm{\frac JC\cdot \frac1A}$

which is the same as:

the work done by the conductor per unit current per unit time per unit current, $\mathrm{\frac J{A\cdot s\cdot A}}$

We know that the work done is equal to the dot product of the force and the displacement, so it is:

the electric force multiplied by the displacement of the charge carrier per unit time per unit charge squared, $\mathrm{\frac{N\cdot m}{s\cdot A^2}}$

and we know force has SI units $\mathrm{kg\;m\;s^{-2}}$

So I guess you can say that the ohm is the resistance when one newton of electric force causes a charge carrier to displace one meter in one second with a current of one ampere.

I would go on and say that it is the resistance when a charge carrier of one kilogram accelerates at one meter per second squared, and this acceleration causes the charge carrier to displace one meter in one second, producing a current of one ampere. But I'm not very certain about the "charge carrier of one kilogram" part.

• +1 These are also very interesting definitons. Commented Feb 16, 2014 at 10:44
• @andrew actually this is just an attempt to expand the definitions until they fit the SI base units. Be aware that these definitions are all equivalent. Commented Feb 16, 2014 at 12:41
• Yeah. But they are kind of different because their context is different.. Anyway, I am with you. Commented Feb 16, 2014 at 13:53

I would describe it as (example) 120 joules per coulomb (120 volts) divided by 60 coulombs per second (60 amps) equals 2 (ohms) of resistance "which means you have 1/2 or 2 times less the amperes then voltage". so maybe an ohm can be n of VpA (# of volts[SI] per amp[SI] or in this case, # of N Kg per charge for every charge per second). But that's still essentially giving the formulae.

I know I'm answering 3 years late but here are my thoughts.

Resistance in a cable means how much the electrons of each cable crash each other.If they dont crush the are free to move with top speed and waste no energy in anything else other than their movement. If they crash each other, half their energy becomes heat and they lose half of their speed.

Now let's assume that you have three cables from three different materials, and since the material is different, the resistance is also different.

Lets also assume that you have 1 million electrons which you arrange in a line, one behind the other.

The first one has little resistance so when you put a difference of 1V between the start and the end of the cable you get 1A of current which is a high number, which means that electrons move very quickly from start to end. At any point in the cable you have 1 c/s passing through which is (randomly picking a number) around 500 electrons, one behind the other passing by every second. The cable gave you 1A=1c/s for a voltage of 1V so it has 1Ω of resistance.

The second cable is a worse conductor so when you put 1V you only get 0.5 A passing by which 250 electrons one behind the other passing by any point of the cable at any second

for 1V you got 0.5A so you have 2Ω because the electrons moved slowlier.

i hope you get the point. Resistance is how many amperes you get if you put 1V of difference at the sides of a cable.

I think that the problem is not a mathematical understanding of ohms, but that the units are not obviously meaningful. However, resistance is not an intuitive concept in any system. If you think of potential difference for gravity, it is J/kg. You could get that by pumping water up to some height. It would have units of $$m^2/s^2$$ or J/kg. When the water fell down through a pipe or tube, it would have a flow rate or current and you could measure the kg/s that left the tube at the bottom. As the tube got wider the flow rate would increase, but the flow rate would also depend on the initial height or potential. If you look at the units for the tube's resistance, it would be $$m^2/(s \ kg)$$ or Js/kg^2. The units aren't descriptive there either. Instead we simply think of it in terms of the ratio of potential/flow rate. In other words resistance really is best thought of as volts/amp, the ratio of work or heat loss per unit charge flow. To continue the water analogy, if you only raise water 1 m in the air, you need a much wider pipe to get 1 kg/s out the end than if the water were raised 10 m in the air. So 1 V needs a tiny 1 ohm resistor to allow 1 amp. If you raise the voltage to 10 V you get 10 times the current passing through the same resistor. That makes sense and that's really what resistance is describing.

Resistance $$R = \frac{mL}{nAe^2t}$$

where m = mass of electron L = Length of conductor ( metre) A = Area of conductor ( sq.metre) e = charge (coulomb) t = average time for electron to travel the distance L ( seconds) n = number of electrons per volume of conductor ( per cu.m)

Resistance $$R = \frac{Kgm^2}{C^2s}$$

Since q = I t ; C = Ampere Sec

Therefore $$R = \frac{Kgm^2}{A^2s^3}$$

OR $$R = \frac{N-m}{A^2s}$$

$$R = \frac{N-m}{A . C}$$ Since work done per unit charge is the Volt ( Force x distance /charge = Volt)

$$R = \frac{V}{A}$$

• Are you mixing names of units and letters used for units? Volt/Amp = R ? A is area or ampere....? Commented May 12, 2017 at 13:27