Resistance of wire changes between tests?

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Testing the resistance of an Earth wire using different machines revealed discrepancies, particularly between the MEM ELEC-3M and Bio-Tek 601 Pro. A colleague suggested that applying a higher current (10A) could temporarily reduce resistance by improving contact at the plug, which was confirmed through testing. However, concerns were raised about the accuracy of the measurements and the potential influence of contact resistance. The discussion emphasized the importance of proper measurement techniques, including four-wire resistance measurements, to minimize errors. Further testing is planned to determine the permanence of any observed changes in resistance.
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I've got a power cord on which I am testing the resistance of the Earth wire. I've used 2 different test machines, a MEM ELEC-3M and a Bio-Tek 601 Pro. The 601 can use 2 different test currents, 10A and 1A. The 3M uses only 1 test current, which is slightly less than 1A.

The minimum standard for a pass is a resistance lower than 0.2 ohm between the 2 ends of the Earth wire. The wire is 3 metres long. I normally use the 3M for testing. Sometimes, I get a wire that has more than 0.2 ohm and I usually declare it a fail. But recently, a colleague showed me that using the 601, I could send 10A through the failed wire and hence "fix" it somehow. He explained that a large current would somehow free the electrons and make the wire more conductive, reducing the resistance. I was not sure, but I've tested his theory and its true. When I checked the failed wire using a 10A current, the recorded resistance is lower. Then, when I bring the wire back to the 3M and test using a 1A current, it seems the resistance is now well under 0.2 ohm.

In addition, if I use the 1A current on the 601, I get a lower resistance reading versus using the 10A. I believe the machines, which are basically digital multimeters, simply divide the voltage across the ends of the wire by the current through the wire to get the measured resistance. In that case, why the difference in the measurement on the same test machine? I'd also appreciate if someone could tell me if my colleague's theory of more current fixing the wire is correct.
 
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Copper wire does not need to have its electrons "freed", if that was possible. It has ample free electrons anyway.

Using a higher current in the wire might have a heating effect, but this would increase the resistance of the wire. It would dissipate 20 watts over a 3 M length, so it could get slightly warm if not hot.

It might be possible to get a temporary improvement in the contact resistance of the connection of the wire to the plug by this heating. Maybe the plug contacts expand and grip the wire more firmly.
You could check this by re-testing a wire after 20 minutes or so, when it has cooled down to room temperature.
If it returns to its previous resistance, then the effect is only temporary.

It seems unlikely that 10 amps would be able to weld the copper wire to the plug contacts, but I guess that might happen.
 
vk6kro said:
Using a higher current in the wire might have a heating effect, but this would increase the resistance of the wire. It would dissipate 20 watts over a 3 M length, so it could get slightly warm if not hot.

The actual power dissipated would only be around 2W at best. The green Earth wire in the power cord is only about 2mm thick approximately. Cross sectional area would then be 3.14 x 10^-6 square metres. Taking the resistivity of standard copper, the resistance comes to 0.0161 ohms. Power = R.(I^2) = 0.0161 x (10^2) = 1.61W
 
Power = I^2 * R

I= 10 A
R = 0.2 ohms

Power = 10 * 10 * 0.2 = 20 watts
 
Also, 2mm thick wire is roughly 12 gauge (about 0.16 ohms per 100 ft), which is rated for about 10 amps, so shouldn't heat up very much. Measuring small resistances takes some care. Unless you are doing a 4-wire measurement, contact resistances may be the biggest source of error.
 
vk6kro said:
Power = I^2 * R

I= 10 A
R = 0.2 ohms

Power = 10 * 10 * 0.2 = 20 watts

This is correct if we assume that the measured resistance is correct. At the moment, due to the discrepancies between the 601 and the 3M, I am not even sure if the test equipment is correct. I thought calculating using the thickness, length and resistivity would be more foolproof in this case.
 
I've decided to try some further tests with the faulty and a brand new wire. Procedure as follows:

For each wire,

Step 1. Measure resistance on 3M
Step 2. Measure resistance on 601 at 1A
Step 3. Measure resistance on 601 at 10A
Step 4. Measure resistance on 601 at 1A
Step 5. Measure resistance on 3M

Record all 5 values. Repeat entire test at 15 min intervals.

The goal is to check and see whether the effect of the "fix" is purely temporary as vk6kro mentioned earlier. My colleague says its permanent. He says applying 10A make the wire "nice and soft" hence fixing it. I am not sure what he means by that because the melting point of copper is over 1000 degrees Celsius.

If someone knows of this experiment already done before, kindly post the link so that I may save some effort. Plus, my testing conditions are probably not perfectly ideal.
 
Why not measure the resistance at 10 amps, then don't turn it off, but keep recording it every 5 minutes to see how much it changes?

Try this with just a piece of the same wire and then with any plug that is normally on it.

I think you have an instrument problem or a procedural problem, because the resistance of copper does not have a memory of how it has been treated in the past.

I hope it doesn't, anyway.
 
Sounds like you need to take four wire resistance measurements. Most likely you're seeing more resistance in youo connections than is in the actual wire.
 
  • #10
If these are newly fabricated cords, could there be some chemical residue from the process in the connections which might burn off with the higher currents?
 
  • #11
It is also probably time to bring in some reliable multimeters to check the calibration of these instruments.

Put an ammeter in series with the test wire and see what actual currents are being produced.

Put a voltmeter across the wire to check the internal meter.

Calculate the resistance yourself to see if it agrees with the test instrument.
 

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