Convert resistivity units to conductivity units
Q. Hi!
How do I convert resistivity measurements (in water samples) to conductivity units. For example, my readings were 18.4 Mohmcm to _____ microS/cm?
Thanks,
Merlyn Lebron Toa Alta, Puerto Rico
A. Hi, Merlyn. One is simply the inverse of the other. If the resistivity is 18.4 million ohmcm, then the conductivity is 1/(18.4 million ohmcm) or 1/(18.4 million) S/cm or 1/18.4 microS/cm. Regards,

Electricity and Electronics 
Q. You explained how conductivity is the inverse of resistivity (surface), but is there any relation between conductivity and volume resistivity?
Brendan Phairstudent  New Hyde Park, New York
A. I don't think the previous answer is restricted to surface resistivity vs. surface conductivity, Brendan. These terms are intended to be the inverse of each other regardless of the units or dimensions. But I feel your pain about obfuscatory terms like "volume resistivity" which seem to defeat the application of our everyday intuition :)
I think the easiest way to keep it from going hopelessly abstract on us is to start with Ohm's Law, i.e., that current equals voltage divided by resistance. We know intuitively that if one wire is twice as long as another, it's resistance will be twice as great, which is to say that its conductance is half as great. If the diameter of one wire is twice the diameter of another, it's area is four times as great, so it's like having four wires in parallel, and the conductance is quadrupled, which is to say the resistance is quartered. If two wires are the same length and diameter, but one is made of a metal that carries electricity only half as well as the other, its resistivity is twice that of the other material, which is to say that its conductivity is half as great.
So, if we express resistivity in Ohminches, and length is expressed in inches, and area is expressed in square inches, then resistance is expressed in Ohms. Checking the logic of this by checking the units: to get the resistance in Ohms, we multiply the resistivity of the conductor in Ohminches by the length of the conductor in inches and divide by its area in inches^{2}, and all the units cancel out except the Ohms.
Ted Mooney, P.E. RET finishing.com Brick, New Jersey 
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Q. I am asking a question of water quality. I do not understand the relationship of resistivity to conductivity.
500 mmhos is how many megohms?
Thanks,
manufacturing  California
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A. Again, conductivity is simply the inverse of resistivity, Paula. That is: Conductivity = 1 / Resistivity. Or Resistivity = 1/ Conductivity. In fact the term 'mho' is not the name of any scientist, but was invented just to be the mirror image or 'inverse' of Ohm. These days people tend to use Siemen rather than mho though.
But you can't get sloppy with 'the units' or you'll stay hopelessly confused forever... mhos is not a valid unit for conductivity and megohms is not a valid unit for resistivity! Please start at the top of the page and study slowly. Also be careful with expressions like 'mmhos' because a leading m is usually an abbreviation for milli or 1/1000 when working in the metric system, whereas I think you are treating it as an abbreviation for micro or 1/1,000,000. Since we don't like our answers to be wrong by 3 orders of magnitude, I don't think you should take the chance of using that abbreviation :)
Good luck.
Ted Mooney, P.E. RET finishing.com Brick, New Jersey 
August 27, 2009
A. Ted Mooney,
All of your answers are correct. The reason that resistivity or conductivity are volumetric is that the number represents a value given a medium of area unit and length unit, therefore volume.
Soil in a box 1 meter wide, 1 meter deep, and 1 meter long will measure OHM*meter. The value can then be converted to OHM*cm, or OHM*mm or OHM*inch. Resistivity of a sample of one cubic cm, or cubic mm, or cubic inch.
 Pittsburgh, Pennsylvania
October 13, 2009
Q. "How do I convert resistivity measurements (in water samples) to conductivity units. For example, my readings were 18.4 Mohmcm to _____ microS/cm?"
Please clarify the answer to the above question, is the answer 0.054 microS/cm ?
 Elstree, Hertfordshire, UK
October 13, 2009
A. Hi, Chris. Conductivity is the opposite (or, more properly, the reciprocal) of Resistivity.
Years ago a Siemen/cm was called a "mho/cm"; something nice about that terminology was that it implied that a mho/cm is the reciprocal of an ohmcm (which it is).
So, if you have
18,400,000 ohmcm, then the reciprocal is
1 / (18,400,000 ohmcm) or
1 Siemen / (18,400,000 cm)
.00000005.4 Siemen/cm or .054 microS/cm.
Regards,
Ted Mooney, P.E. RET finishing.com Brick, New Jersey 
November 25, 2009
A. If Conductivity = 0.054 µS
= 0.054 x 106 S
= 0.54 x 105 S
= 0.54 x 105 mho
Now Resistivity = 1 / Conductivity
= 1/ 0.54 x 105 mho
= 1.84 x 105 mho
= 18.4 Mmho
Sr. Manager  Mumbai, INDIA
June 15, 2010
Q. How can I convert soil thermal resistivity from (mK/W) to (ohm.m)
Meto HabElectrical Designer  Qatar
February 12, 2011
Q. The relation between a volume resistivity (VR) and surface resistivity (SR) can be defined as VR = SR * thickness if the material is fully conductive.
Can the relation hold good for conductive coatings on a substrate?
 Mumbai. Maharashtra, India
February 12, 2011
A. Hi Amol,
Please explain your situation that prompts you to ask. While resistivity is a meaningful physical quality, "SR" and "VR" are more in the nature of mathematical constructions than meaningful physical qualities, and the relationship between the two is a simple mathematical manipulation rather devoid of original physical insights, similar to converting length from meters to inches. Picture:
Take a short slice of a piece of wire; it's a cylinder. You can measure the electrical resistance of this cylinder in Ohms, or you calculate the resistance by dividing the voltage across it by the current that flows through it. If you look at the cylinder endon, you see a circle, and that circle has some specific area or surface area or cross sectional area. If you multiply that surface area x the length of the slice of wire (or you could call it its thickness) you have the volume of the cylinder.
Resistivity, is a property of the wire and if you multiply the resistivity by the length and divide by the area you get the resistance.
If a coating on a substrate is more conductive than the substrate, it has lower resistivity than the substrate, and mathematical manipulations, like whether or not you include thickness in your units, doesn't change the fact.
Ted Mooney, P.E. RET finishing.com Brick, New Jersey 
June 10, 2011
Q. How do you convert microohms per cm to ohms per m? Thanks, Jag.
Jagan Jemployee  Chennai, TN, India
December 3, 2012
A. Hi Jag. This is not something you calculate so much as something you must take the time to implicitly comprehend. A microohm is a millionth of an ohm and a cm is 1/100 of a meter. So you take the value you were given and insert the conversion factors (1,000,000 microohms)/ohm and (100 cm)/m.
Because these conversion factors are "equalities", always equal to 1, it doesn't matter whether you invert them. So you put them into your equation either as I have written them or 'upside down' as ohm/(1,000,000 microohms) and m/(100 cm) depending on which way will let you cross out the microohms term and cm terms you started with and leave you with the ohms and m terms you want to end up with. You must learn to always "watch the units", and such conversions will come easily and always work right.
Example, with 5000 microohms per cm
Try 5000 microohms/cm x (1,000,000 microohms)/ohm x (100 cm)/m ...
We have cm in the denominator of the first term and the numerator of the third term, so they can be crossed out, leaving ...
5000 microohms x (1,000,000 microohms)/ohm x 100/m, but this leaves us with microohms in the numerators of the first and second terms. They can't be crossed out; but it is valid to invert the second term since it's an 'equality'...
5000 microohms x ohm/(1,000,000 microohms) x 100/m; now we can cross out microohms since its in the numerator (of the first term) and denominator (of the second term), leaving ...
5000 x ohm/1,000,000 x 100/m = 0.5 ohms/m.
Regards,
Ted Mooney, P.E. RET finishing.com Brick, New Jersey 
December 3, 2012
Q. Can I convert sheet resistance to conductance/conductivity without knowing the thickness of the thin film?
Azhar Pirzado straqbourg, Alsace, France
December 3, 2012
A. Hi Azhar. Although the question is a bit abstract for me to follow, I'd say you can determine the conductance because it is the reciprocal of the resistance, but not the conductivity because you can't make the units match without additional terms like thickness.
Regards,
Ted Mooney, P.E. RET finishing.com Brick, New Jersey 
Hi Ted Mooney
Thanks for your answer!
 sSraqbourg, Alsace, France
December 7, 2012
Q. Hi All
I need to know how one can extract the value of resistance from IV graph in Origin. I just plotted V and I on xaxis and yaxis respectively. I wonder there is some easy way to extract the resistance?
Thanks is advance,
 Straqbourg, Alsace, France
May 14, 2013
A. Hi Azhar. Your question went unanswered a long time; perhaps other people had difficulty understanding it as well. Now that I very slowly reread it, I see that you are asking how you can determine the resistance from a graph of current vs. voltage. Ohm's Law says V=IR, which can also be expressed as R=V/I. So you simply take any point on the line you graphed, and the resistance in Ohms is simply whatever the voltage (in Volts) is at that point divided by the current (in Amps) at that point.
Regards,
Ted Mooney, P.E. RET finishing.com Brick, New Jersey 
May 14, 2013
Q. My resistance reading is 0.0428 mohm/m how to calculate the conductivity.
AJAY S. LINGE Pune, Maharashtra, India
May 14, 2013
A. Hi Ajay. You'll make best continuing progress on this if you carefully think it through, and then be very careful of your words and their precise meanings. The suffixes "ance" and "ivity" are very different and must not be confused. So your question cannot be answered because it is improperly asked; although a length of wire might exhibit a resistance of 0.04328 ohms for each meter of its length, please carefully study the thread and figure out what is wrong with your question, and I think the answer will come to you. Thanks and good luck.
Regards,
Ted Mooney, P.E. RET finishing.com Brick, New Jersey 
August 28, 2013
Q. I'm trying to find an electrically equivalent material to a Silver filled Silicon, Ablebond 1903. Original test method indicated Bond Joint Resistance: 0.03 ohm/0.5 in2. All available product's properties are now reported as Volume Resistivity, (i.e.: 0.01 ohmcm). This bond forms an EMI seal, and I need to be able compare the properties. By definition Volume Resistivity (rho) = R*(A/L). If I know the bond area, can I convert the Bond Joint Resistance to Volume Resistivity?
George F. Tirone
 Sylmar, California, USA
^ Privately contact this inquirer ^
September 3, 2013 A. George,
September 2013 A. Hi George. Sorry, I don't really know. My knowledge harks back only to definitions from school many years ago, and I have no actual experience with the term "bond joint resistance". But the units don't seem to work the way you would wish because the "bond joint resistance" seems to be independent of the thickness of the adhesive and to be controlled only by the surface area. It thus probably assumes a thin adhesive layer of under .010", with the exact thickness of the adhesive not being a critical factor in the resistance across the joint as long as it's in the accepted range. I strongly suspect that the other conductive filled adhesives you would like to use as a replacement are similar in application and properties, and I tend to doubt that "bond joint resistance" data is unavailable for them  but it probably depends more on the material of the two surfaces to be joined than upon the volume resistivity of the thin layer of goop between them. Good luck. Regards,

September 7, 2013
Thanks for your feedback. Understanding that Volume resistivity is Resistance/(Cross Sectional Area*Bond Line Thickness) the computation I made is at best, ambiguous. I don't have any of the original material for conducting a Volume Resistivity test. However, I do have a sample of the new material and the original manufacturer's test procedure used for "Bond Joint Resistance". Therefore, I will determine a "Bond Joint Resistance" for the new material and be able to compare it's electrical property. This seems to be the only way that I can get an answer that I can depend on. Also, the NSN for the original new material are the same, indicating that the Federal Government recognizes them to be suitable substitutions. I appreciate your comments and will post the conclusion when available.
George F. Tirone, PE
 Sylmar, California, USA
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