# ohm-meter (Ω.m) for resistivity units and other unit questions

#### Mr.Potential

Joined Dec 4, 2019
7
Hi all,

I'm having a real headache understanding something very simply. SI units for resistivity.

Now the thing is I can understand the resistivity equation and understand it's proportional to length and area, however the bit I just can't grasp is the ohm-meter unit as it's not ohm per meter but seemingly more ohm per area. I have a fluid whose resistivity is stated as 1.5Mohm.m, however that fluid is tracking across a gap of 1mm.

According to online calculators, 1 ohm.m = 1000 ohm.mm. 1.5Mohm.m = 1.5e+9 ohm.mm.

How is that division working? I just can't get it straight in my head other than a smaller area means higher resistance.

The other question I have is to do with unit affixes. Take these images as examples:  What does the -1 above each of these mean physically and mathematically? Is it meant to show the reciprocal of the actual number? These aren't powers or exponents, so what do they represent?

Cheers for looking and thanks in advance.

#### Papabravo

Joined Feb 24, 2006
13,759
The -1 indicates the the unit is part of the denominator. For example kinetic energy has units of:

$kg \cdot m^2\cdot sec^{-2}\;=\;kg\cdot (\frac{m^2}{sec^2})$

#### Delta prime

Joined Nov 15, 2019
337
• The -1 means "per" unit
• area, measured in m2, or metres squared
• volume, measured in m3, or metres cubed
• pressure, measured in N m-2, or Newtons per metre squared
• velocity, measured in m s-1, or metres per second
• acceleration, measured in m s-2, or metres per second per second
• It does not make sense to replace the exponent of −1−1 with e.g. −2−2 as that would result in a different unit (e.g: kg⋅m2⋅s−2kg⋅m2⋅s−2 is joule, the unit of energy, while kg⋅m2⋅s−3kg⋅m2⋅s−3 is watt, the unit of power).
• it does not make sense to remove the negative sign from the exponent as that would result in a different unit (e.g. 10 Hz=10 s−110 Hz=10 s−1 corresponds to a frequency — ten times per second — while 10 s10 s obviously corresponds to a duration).
• one has to choose between either the slash or the negative exponent as both would cancel each other out.

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#### sparky 1

Joined Nov 3, 2018
186
The variety of resistors each have their own properties. For a specific model you can find that manufacturers values.

A laser trimmed resistor and a volume of liquid use specific methods because the materials are different.

From a known calibrated liquid a divider can be made.

#### Mr.Potential

Joined Dec 4, 2019
7
The -1 indicates the the unit is part of the denominator. For example kinetic energy has units of:

$kg \cdot m^2\cdot sec^{-2}\;=\;kg\cdot (\frac{m^2}{sec^2})$
Hi Papabravo.

Is your statement correct as Delta Prime stated -1 means "per" unit which would seemingly make more sense in this context?

Take the Kelvin example, that's K-1. Kelvin is usually always used as a delta value in conjunction with Celsius and is per unit, so it's Δ°C*K as the step changes between Celsius and Kelvin are linear relatively.

#### Mr.Potential

Joined Dec 4, 2019
7
• The -1 means "per" unit
• area, measured in m2, or metres squared
• volume, measured in m3, or metres cubed
• pressure, measured in N m-2, or Newtons per metre squared
• velocity, measured in m s-1, or metres per second
• acceleration, measured in m s-2, or metres per second per second
• It does not make sense to replace the exponent of −1−1 with e.g. −2−2 as that would result in a different unit (e.g: kg⋅m2⋅s−2kg⋅m2⋅s−2 is joule, the unit of energy, while kg⋅m2⋅s−3kg⋅m2⋅s−3 is watt, the unit of power).
• it does not make sense to remove the negative sign from the exponent as that would result in a different unit (e.g. 10 Hz=10 s−110 Hz=10 s−1 corresponds to a frequency — ten times per second — while 10 s10 s obviously corresponds to a duration).
• one has to choose between either the slash or the negative exponent as both would cancel each other out.
Hi Delta Prime,

Your 'per' unit explanation does make sense to me and the reply I sent to Papabravo has helped clear this up in my head.

I am however still struggling with the Ω.m query. Can you help me any further with that?

#### crutschow

Joined Mar 14, 2008
24,997
I am however still struggling with the Ω.m query.
An ohmmeter just measures resistance in ohms.
The "per" part is determined by what you are measuring.
E.G. if you measure 1 ohm through 10 feet of wire, then the wire resistance is 0.1ohm/ft.

#### Papabravo

Joined Feb 24, 2006
13,759
Hi Papabravo.

Is your statement correct as Delta Prime stated -1 means "per" unit which would seemingly make more sense in this context?

Take the Kelvin example, that's K-1. Kelvin is usually always used as a delta value in conjunction with Celsius and is per unit, so it's Δ°C*K as the step changes between Celsius and Kelvin are linear relatively.
Yes

#### Mr.Potential

Joined Dec 4, 2019
7
Then surely my understanding is still clear as mud?

#### Mr.Potential

Joined Dec 4, 2019
7
An ohmmeter just measures resistance in ohms.
The "per" part is determined by what you are measuring.
E.G. if you measure 1 ohm through 10 feet of wire, then the wire resistance is 0.1ohm/ft.
Hi crutshow,

That's the thing though, it's not per in this case otherwise it would be written as Ω/m and I would totaly agree with you. That's the mistake I originally made as the fluid stated a value of 1.5MΩ.m so I assumed it was 1.5MΩ*0.001mm=1.5k per mm,

1Ω.m (not 1Ω/m) = 1000Ω.mm and 1.5MΩ.m = 1.5e+9Ω.mm

The values go up, not down as is the case if it was Ω/m. This is where I'm confused.

I still think it's because resistivity is based on length and area and I think the area is why it increases. This is why smaller CSA conductors have much higher resistance per meter as the resistivity due to the area is so high.

#### bogosort

Joined Sep 24, 2011
459
The values go up, not down as is the case if it was Ω/m. This is where I'm confused.
This is by design. When using a system of units (such as SI), whatever we do, the physical dimensions must agree. We can't, for example, reasonably compare a mass to a length. Which is bigger, 10 kg or 2 m?

Let's say that we've defined resistance as the ratio of voltage to current. Experimentally, we find that different materials exhibit different resistance, and we conclude that resistance is a property of materials. We give this property a name, resistivity, and a symbol $$\rho$$.

Experimentally, we note that whatever material we test, the material's resistivity is proportional to its cross-sectional area $$A$$, and inversely proportional to its length $$\mathcal{l}$$: $\rho \propto \frac{A}{\mathcal{l}}$ Testing a bunch of materials, we infer a general relationship between a material's resistivity, its geometry, and the measured resistance $$R$$ when applying a voltage across it: $R = \rho \frac{\mathcal{l}}{A}$ All that's left to do is give $$\rho$$ appropriate dimensions so that both sides of the equation are dimensionally equal.

By convention, we say that current $$I$$ is a base dimension. Other base dimensions include length $$L$$, mass $$M$$, and time $$T$$. The dimensions of voltage are derived from the base dimensions; there are many ways to do this, but we'll just say that $[V] = M L^2 I^{-1} T^{-3}$ In other words, voltage has dimensions of mass-area per cubed-time current. Therefore, resistance has dimensions of $[R] = \frac{[V]}{I} = \frac{M L^2 I^{-1} T^{-3}}{I} = \frac{M L^2}{I^2 T^3}$

Now all we have to do is make the dimensions work on both sides of the resistance equation: \begin{align} \frac{M L^2}{I^2 T^3} &= [\rho] \frac{[\mathcal{l}]}{[A]} \\ &= [\rho] \frac{L}{L^2} \\ &= [\rho] L^{-1} \end{align} What dimensions must $$\rho$$ have in order to make the right side look like the left? Well, dimensions of resistance-length do exactly that, as the length cancels the inverse-length on the right: \begin{align} [\rho] L^{-1} &= \left( \frac{M L^2}{I^2 T^3} L \right) L^{-1} \\ &= \frac{M L^2}{I^2 T^3} \\ &= [R] \end{align} As for what physical intuition we might or might not have for dimensions of resistance-length, well, consider how intuitive "mass-area per cube-time square-current" is, yet we don't blink an eye when we call it "resistance".

• Delta prime and nsaspook

#### Mr.Potential

Joined Dec 4, 2019
7
This is by design. When using a system of units (such as SI), whatever we do, the physical dimensions must agree. We can't, for example, reasonably compare a mass to a length. Which is bigger, 10 kg or 2 m?

Let's say that we've defined resistance as the ratio of voltage to current. Experimentally, we find that different materials exhibit different resistance, and we conclude that resistance is a property of materials. We give this property a name, resistivity, and a symbol $$\rho$$.

Experimentally, we note that whatever material we test, the material's resistivity is proportional to its cross-sectional area $$A$$, and inversely proportional to its length $$\mathcal{l}$$: $\rho \propto \frac{A}{\mathcal{l}}$ Testing a bunch of materials, we infer a general relationship between a material's resistivity, its geometry, and the measured resistance $$R$$ when applying a voltage across it: $R = \rho \frac{\mathcal{l}}{A}$ All that's left to do is give $$\rho$$ appropriate dimensions so that both sides of the equation are dimensionally equal.

By convention, we say that current $$I$$ is a base dimension. Other base dimensions include length $$L$$, mass $$M$$, and time $$T$$. The dimensions of voltage are derived from the base dimensions; there are many ways to do this, but we'll just say that $[V] = M L^2 I^{-1} T^{-3}$ In other words, voltage has dimensions of mass-area per cubed-time current. Therefore, resistance has dimensions of $[R] = \frac{[V]}{I} = \frac{M L^2 I^{-1} T^{-3}}{I} = \frac{M L^2}{I^2 T^3}$

Now all we have to do is make the dimensions work on both sides of the resistance equation: \begin{align} \frac{M L^2}{I^2 T^3} &= [\rho] \frac{[\mathcal{l}]}{[A]} \\ &= [\rho] \frac{L}{L^2} \\ &= [\rho] L^{-1} \end{align} What dimensions must $$\rho$$ have in order to make the right side look like the left? Well, dimensions of resistance-length do exactly that, as the length cancels the inverse-length on the right: \begin{align} [\rho] L^{-1} &= \left( \frac{M L^2}{I^2 T^3} L \right) L^{-1} \\ &= \frac{M L^2}{I^2 T^3} \\ &= [R] \end{align} As for what physical intuition we might or might not have for dimensions of resistance-length, well, consider how intuitive "mass-area per cube-time square-current" is, yet we don't blink an eye when we call it "resistance".

Hi bogosort,

Many thanks for the reply, much appreciated.

I wish I was cleverer as I still don't get it even though I understand the principle of resistivity.

So are we saying it's definitely not 'per' meter then in the sense that if it was 1 ohm per meter length then 1mm would equal 0.001 ohm?

Regards

Mitch

#### Delta prime

Joined Nov 15, 2019
337
Hi there remember me? Many resistors and conductors have a uniform cross section with a uniform flow of electric current. It is therefore possible to create the more specific, but more widely used electrical resistivity formula or equation
Where:
R is the electrical resistance of a uniform specimen of the material measured in ohms
l is the length of the piece of material measured in metres, m
A is the cross-sectional area of the specimen measured in square metres, m^2
It can be seen from the equations that the resistance can be varied by changing a variety of different parameters.
For example, keeping the material resistivity constant, the resistance of the sample can be increased by increasing the length, or decreasing the cross sectional area. It can also be seen from the resistivity equations that increasing the resistivity of the material will increase the resistance assuming the same dimensions. Similarly decreasing the resistivity will decrease the resistance.
copper is a good conductor as it provides a low level of resistivity, its cost is not too high, and it also provides other physical characteristics that are useful in many electrical and electronic applications. The resistivity of copper is around 1.7 x 10-8 ohm metre (or 17. nΩm), although figures will vary slightly according to the grade of the copper

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• Mr.Potential

#### bogosort

Joined Sep 24, 2011
459
So are we saying it's definitely not 'per' meter then in the sense that if it was 1 ohm per meter length then 1mm would equal 0.001 ohm?
Correct. The SI unit of resistivity is the "ohm meter", $$\Omega \cdot \text{m}$$, which is a multiplication.

Sometimes you'll see cables specified in units of "ohms per meter" $$\Omega / \text{m}$$, a division, to help calculate the resistance of a cable over a long run.

• Mr.Potential

#### Mr.Potential

Joined Dec 4, 2019
7
Hi there remember me? Many resistors and conductors have a uniform cross section with a uniform flow of electric current. It is therefore possible to create the more specific, but more widely used electrical resistivity formula or equation
Where:
R is the electrical resistance of a uniform specimen of the material measured in ohms
l is the length of the piece of material measured in metres, m
A is the cross-sectional area of the specimen measured in square metres, m^2
It can be seen from the equations that the resistance can be varied by changing a variety of different parameters.
For example, keeping the material resistivity constant, the resistance of the sample can be increased by increasing the length, or decreasing the cross sectional area. It can also be seen from the resistivity equations that increasing the resistivity of the material will increase the resistance assuming the same dimensions. Similarly decreasing the resistivity will decrease the resistance.
copper is a good conductor as it provides a low level of resistivity, its cost is not too high, and it also provides other physical characteristics that are useful in many electrical and electronic applications. The resistivity of copper is around 1.7 x 10-8 ohm metre (or 17. nΩm), although figures will vary slightly according to the grade of the copper
NOW I GET IT!!!

The penny has finally dropped having read that. I drew it in my head and voila, it hit me.

Makes total sense now. I knew all along really . . . . Thanks all for your help, really is appreciated.

Next on the list, quantum mechanics • bogosort

#### MrAl

Joined Jun 17, 2014
7,604
Hi,

In short resistivity is a property of the material without referring to any specific dimension or direction. This gives us a means to classify materials as to their resistance without having to specify any specific dimensions.
For example, at 20 degrees C copper is about 1.7e-10 Ohm Meters while silver is about 1.6e-10. This tells us, without having to refer to any dimensional measurements, that silver is a slightly better conductor than copper at that temperature.

The variable often used for resistivity is the Greek 'rho', and it is in for example Ohm Meters.
When we finally get a material of specific dimensions, we can then calculate the total resistance when it is oriented such that the current flows along the length. The actual total resistance would then be:
R=rho*L/A

and you see now we specified the length L and Area A but before that we could understand something about the material itself (such as copper) and we could easily do this same with silver just by knowing its resistivity.

It is informative to work out the units which i am betting WBahn will appreciate.
Working the above formula with units alone we have:
Ohms=(Ohms*Meters)*Meters/(Meters*Meters)

and here (Ohms**Meters) is the resistivity rho.
expanding and removing the parens, we get:
Ohms=Ohms*Meters*Meters/(Meters*Meters)
and of course Meters*Meters/(Meters*Meters)=1 so we have:
Ohms=Ohms*1
or just:
Ohms=Ohms

So we calculated the resistance knowing just the resistivity and later including the dimensions of the piece of material we are going to use for something practical.

So i guess you can think of resistivity as "intrinsic resistance", a property within the material itself with no other specifications yet.

Note that with modern materials it may be possible to have a resistivity for each direction current might flow though the material, and each resistivity may be different. I like to call this kind of material an "anisoelectric" type of material but there are variations.

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