How far will a Fluke Ohm meter read?

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NE (9.06 miles @5.9 Degrees from Winged Horses)
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EC - retired
Won't the capacitance of the UF cable change once it is buried in earth?

Before Megger I used a Greenlee/ Beha tester that would audibly chirp as it charged a capacitor. We used this feature to charge buried cables. A good cable would charge and hold that charge. Some for quite a while. Faulted ones would either beep continoulsy or charge and not hold it for any length of time. Tested & repaired a lot of underground to wells & pivots with this method.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
100918-1009 EST

ptonsparky:

Yes the capacitance will be somewhat influenced by the earth. If you have one break, and measure from both ends, and ratio, then most of the effect probably washes out.

A capacitance measurement might at least give you an estimate. If you have a meter that can measure capacitance, then it is a cheap way to get an estimate of distance.

If the capacitance meter can measure dissipation, then you can get additional information on the cable condition, but not near as good as a megger.

.
 

Mule

Senior Member
Location
Oklahoma
Consider getting one. Things like this are so much easier with it.

Imagine walking back out there with a megger, disconnecting the load and hooking it up, and watching the results - when you hook it up, try to charge it to 500v and see that it can only load it to 3v, you have conclusive evidence that the cable is toast. As you're repairing the cable, you can test at the break and determine conclusively that in this direction you're fine, but in the other direction there's more trouble. No guesswork.

I walked into a house that was tripping an arc fault. I meggered the wires, and found a reading of 32.1MΩ. Found a problem, fixed it, and only improved it to 48.4 MΩ. I knew to keep looking, and found another problem. After that fix it grew to 1484MΩ, so I knew that the second fix was a substantial contributor to the problem.

It tells you so much (accurately) about how things are behaving, it's priceless, IMO.

George, you are a man that knows his circuit.....Knowing the integrity of the circuit and its insulation values is very valuable. In day to day continuity checks its really not needed, but one should have a megger around for sure.
In my case, I was just curious if a ohm meter measuring 300ft out and 300ft back would have enough VD to skew the reading. I dont like spending the customers money unless Im rock solid in my diagnosis. But I really should have known better as the cable is obviously open.
Thanks everyone for the valuable input.....!!

Mule
 

iwire

Moderator
Staff member
Location
Massachusetts

I don't know how to explain my thoughts any clearer.

You seem to want to try to verify an open conductor based on an assumption about the conductors insulation. I would stick with conductor testing.

A 'Megameter' is an insulation resistance tester, IMO it is not the right 'go to' tool to verify an open conductor.

Truthfully, given the OPs description I probably would have stopped testing once I did voltage checks at the pump end. :)
 
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iwire

Moderator
Staff member
Location
Massachusetts
I was just curious if a ohm meter measuring 300ft out and 300ft back would have enough VD to skew the reading.

Luckily voltage drop is as dependent on current as it is on length.

Your DMM uses so little current that voltage drop will not be a problem for most applications. Certainly not for a 600 foot long 10 AWG circuit. If the circuit was closed you could read it.


I dont like spending the customers money unless Im rock solid in my diagnosis.

I hear you, it is not a good thing to say they need a few thousand dollar repair only to find out that was not the problem.
 

SAC

Senior Member
Location
Massachusetts
100918-0848 EST

However, besides TDR

.

It looks like hand-held TDRs that can locate the distance to a power cable fault are available for around $350. Does anyone use TDRs for problems such as this? Or is this too expensive given how often it would actually save labor or materials (e.g., in the OPs case it seems like the cable needs replacing regardless of where it is open).
 

Speedskater

Senior Member
Location
Cleveland, Ohio
Occupation
retired broadcast, audio and industrial R&D engineering
Years ago, I had no problem measuring resistance on a pair of 22AWG signal wires about 2 or 3 miles long.
 

iwire

Moderator
Staff member
Location
Massachusetts
It looks like hand-held TDRs that can locate the distance to a power cable fault are available for around $350. Does anyone use TDRs for problems such as this? Or is this too expensive given how often it would actually save labor or materials (e.g., in the OPs case it seems like the cable needs replacing regardless of where it is open).

I know member 'e57' uses and is a fan of TDRs.
 

kwired

Electron manager
Location
NE Nebraska
George, you are a man that knows his circuit.....Knowing the integrity of the circuit and its insulation values is very valuable. In day to day continuity checks its really not needed, but one should have a megger around for sure.
In my case, I was just curious if a ohm meter measuring 300ft out and 300ft back would have enough VD to skew the reading. I dont like spending the customers money unless Im rock solid in my diagnosis. But I really should have known better as the cable is obviously open.
Thanks everyone for the valuable input.....!!

Mule


Voltage drop is a result of the resistance of the conductor, in your case the resistance was very high due to open circuit so most or all of the voltage did not make it back to the testing meter. You were just second guessing something you probably already knew.
 

massfd

Member
If you are starting at the panel do not forget that there is a pressure switch somewere in the circuit that could be open. Sometimes they are at the pump, other times the switch is located in the basement with the pressure tank.

You realy need to know the cable path before you condem it.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
100919-0928 EST

Mule:

By definition an ohmmeter is an instrument that measures resistance. In its simplest form it will be a DC instrument that supplies excitation to the resistance being measured. The ability to measure the resistance of a cable has nothing to do with the length of the wire in a long cable other than maybe the time you have to wait for the circuit to reach steady state.

To measure resistance with what is called an ohmmeter a DC current is applied to the resistance being measured and the voltage across the resistance is measured. From the current and voltage the resistance is determined.

In a simple instrument like a Simpson 260 there is essentially a battery of known voltage, a fixed series resistance, and a series microamp meter. When the meter leads are shorted together the resistance or source voltage is adjusted a minor amount for full scale deflection of the meter. This correspond to zero external resistance. With infinite external resistance the meter current is zero. In between these two limits the current is nonlinear vs resistance and thus the nonlinear resistance scale on the meter. Analysis of the series resistors provides an equation of Runknown = Vinternal/Ireading - Rinternal. For my newer, but still old, 270 half scale is 12 on the resistance range. The internal voltage for the 10,000 times range is 6 V. Meter full scale is 50 microamps. So at 25 microamps the total series resistance across the 6 V is Rtotal = 6/(25*10^6) = 240,000 ohms. Thus, Rinternal is 120,000 ohms. At 1/5 of meter full scale current what should the ohms scale read? See if your calculation agrees with the Simpson scale marking.

A Fluke 27 works in some different fashion. At a 10 megohm load the voltage is about 0.761 V, at 100K about 0.130 V, at 10K 0.114 V, at 1K 0.114 V, at 100 ohms 0.0478 V, and at 22.1 ohms about 0.0108 V. I suspect this is approximately a constant current source below a terminal voltage of about 0.114 V and some other characteristic above this inflection point. In turn it implies a ratioing measurement of voltage to current to calculate resistance.

You can measure any length cable you want with an ohmmeter. Knowing the resistance per foot of the wire you can estimate the cable length. However, for wire like #12 copper and a 100 ft circuit, 200 ft of wire, the resistance is about 1.6/5 = 0.3 ohms and hard to read on either a Simpson 260 or Fluke 27 with adequate accuracy to make a good estimate of length.

However, with 10 amps thru 10 ft of #12 I can estimate its length to better than a fraction of a foot with two Fluke 27s. Probably estimate to 1 inch.

.
 

Pitt123

Senior Member
100919-0928 EST

Mule:

By definition an ohmmeter is an instrument that measures resistance. In its simplest form it will be a DC instrument that supplies excitation to the resistance being measured. The ability to measure the resistance of a cable has nothing to do with the length of the wire in a long cable other than maybe the time you have to wait for the circuit to reach steady state.

To measure resistance with what is called an ohmmeter a DC current is applied to the resistance being measured and the voltage across the resistance is measured. From the current and voltage the resistance is determined.

In a simple instrument like a Simpson 260 there is essentially a battery of known voltage, a fixed series resistance, and a series microamp meter. When the meter leads are shorted together the resistance or source voltage is adjusted a minor amount for full scale deflection of the meter. This correspond to zero external resistance. With infinite external resistance the meter current is zero. In between these two limits the current is nonlinear vs resistance and thus the nonlinear resistance scale on the meter. Analysis of the series resistors provides an equation of Runknown = Vinternal/Ireading - Rinternal. For my newer, but still old, 270 half scale is 12 on the resistance range. The internal voltage for the 10,000 times range is 6 V. Meter full scale is 50 microamps. So at 25 microamps the total series resistance across the 6 V is Rtotal = 6/(25*10^6) = 240,000 ohms. Thus, Rinternal is 120,000 ohms. At 1/5 of meter full scale current what should the ohms scale read? See if your calculation agrees with the Simpson scale marking.

A Fluke 27 works in some different fashion. At a 10 megohm load the voltage is about 0.761 V, at 100K about 0.130 V, at 10K 0.114 V, at 1K 0.114 V, at 100 ohms 0.0478 V, and at 22.1 ohms about 0.0108 V. I suspect this is approximately a constant current source below a terminal voltage of about 0.114 V and some other characteristic above this inflection point. In turn it implies a ratioing measurement of voltage to current to calculate resistance.

You can measure any length cable you want with an ohmmeter. Knowing the resistance per foot of the wire you can estimate the cable length. However, for wire like #12 copper and a 100 ft circuit, 200 ft of wire, the resistance is about 1.6/5 = 0.3 ohms and hard to read on either a Simpson 260 or Fluke 27 with adequate accuracy to make a good estimate of length.

However, with 10 amps thru 10 ft of #12 I can estimate its length to better than a fraction of a foot with two Fluke 27s. Probably estimate to 1 inch.

.

Could there be a point for a very long cable distance where the resistance of the cable was too large and the voltage supply in the meter would not be large enough to push the current though this resistance?

Can a meter be used for charging any sized capacitor as well for testing? Including testing 5kV caps?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
100919-1708 EST

Pitt123:

So long as Ohm's law applies to the circuit, then it does not matter how small the voltage is, or how large the resistance is. But at the extremely low or high ends there will be a failure of Ohm's law or its application.

Sure any size capacitor could be charged theoretically, but at extremes there will be problems.

.
 

cadpoint

Senior Member
Location
Durham, NC
100919-1708 EST

Pitt123:

So long as Ohm's law applies to the circuit, then it does not matter how small the voltage is, or how large the resistance is. But at the extremely low or high ends there will be a failure of Ohm's law or its application.

Respectfully; there is no fault in a stated law, the instruments used might not be able to detect it to the finite appliable values that one is for testing for or againest what's in the instrument thats trying to determine some value of it; but there is no fault of a stated Law. Ohms law is a simple A=B*C equation or a A=B/C equation depending on what one is proofing for. It's the limits of the testing equipment thats the problem.

Sure any size capacitor could be charged theoretically, but at extremes there will be problems.
.

Since you adding a capacitor; I would say that the equation for determining Ohms would be needed to to be changed to refect a capacitor.

I also wanted to state that I found your previous statement to Mule very precise and exacting, I could only wish I knew as much as I do now to have the same conversation with my Father, about the aspects that he did show me with my limited insight at the time! :D
 
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Mule

Senior Member
Location
Oklahoma
If you are starting at the panel do not forget that there is a pressure switch somewere in the circuit that could be open. Sometimes they are at the pump, other times the switch is located in the basement with the pressure tank.

You realy need to know the cable path before you condem it.

You wont believe this, the previous owner had the pump, bladder tank, pressure switch and associated valves buried !! So the plumber had to dig all that up, and set a new system on a pad. They are going to build a little house over it that is insulated and heated.....
 

Mule

Senior Member
Location
Oklahoma
100919-0928 EST

Mule:

By definition an ohmmeter is an instrument that measures resistance. In its simplest form it will be a DC instrument that supplies excitation to the resistance being measured. The ability to measure the resistance of a cable has nothing to do with the length of the wire in a long cable other than maybe the time you have to wait for the circuit to reach steady state.

To measure resistance with what is called an ohmmeter a DC current is applied to the resistance being measured and the voltage across the resistance is measured. From the current and voltage the resistance is determined.

In a simple instrument like a Simpson 260 there is essentially a battery of known voltage, a fixed series resistance, and a series microamp meter. When the meter leads are shorted together the resistance or source voltage is adjusted a minor amount for full scale deflection of the meter. This correspond to zero external resistance. With infinite external resistance the meter current is zero. In between these two limits the current is nonlinear vs resistance and thus the nonlinear resistance scale on the meter. Analysis of the series resistors provides an equation of Runknown = Vinternal/Ireading - Rinternal. For my newer, but still old, 270 half scale is 12 on the resistance range. The internal voltage for the 10,000 times range is 6 V. Meter full scale is 50 microamps. So at 25 microamps the total series resistance across the 6 V is Rtotal = 6/(25*10^6) = 240,000 ohms. Thus, Rinternal is 120,000 ohms. At 1/5 of meter full scale current what should the ohms scale read? See if your calculation agrees with the Simpson scale marking.

A Fluke 27 works in some different fashion. At a 10 megohm load the voltage is about 0.761 V, at 100K about 0.130 V, at 10K 0.114 V, at 1K 0.114 V, at 100 ohms 0.0478 V, and at 22.1 ohms about 0.0108 V. I suspect this is approximately a constant current source below a terminal voltage of about 0.114 V and some other characteristic above this inflection point. In turn it implies a ratioing measurement of voltage to current to calculate resistance.

You can measure any length cable you want with an ohmmeter. Knowing the resistance per foot of the wire you can estimate the cable length. However, for wire like #12 copper and a 100 ft circuit, 200 ft of wire, the resistance is about 1.6/5 = 0.3 ohms and hard to read on either a Simpson 260 or Fluke 27 with adequate accuracy to make a good estimate of length.

However, with 10 amps thru 10 ft of #12 I can estimate its length to better than a fraction of a foot with two Fluke 27s. Probably estimate to 1 inch.

.

Thanks Gar that's might nice of you....I still have my Simpson around here somewhere,mmmm...I said somewhere !! then I remember the old Tripplet's also.....haha I seen a Simpson smoked one time.... :<O
 

dbuckley

Senior Member
Could there be a point for a very long cable distance where the resistance of the cable was too large and the voltage supply in the meter would not be large enough to push the current though this resistance?
If you had a bit of #10 to the moon and back most any DVM would satisfactorally measure that resistance, and be able to tell the difference when Neil un-shorted the ends.

(I make it about 3M ohms)
 
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