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Fall of Potential Test - Voltage/Resistance Rises Dramatically

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  1. BrickSalad is offline Junior Member Pro Subscriber
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    Fall of Potential Test - Voltage/Resistance Rises Dramatically

    I can not seem to make sense of this common drawing:

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    In it, we have a current supplying electrode and a ground electrode, which essentially makes a circuit. Then we can use the potential probe to measure the voltage between it and the ground electrode, giving us the resistance as well (via Ohm's law). We can see that near the ground electrode there isn't going to be much resistance, and we're not really measuring the resistance that a fault current would feel since it will dissipate much further into the earth.

    All that I think I understand. The part that's tripping me up is the right side of this graph, where voltage/resistance rises dramatically as the potential probe gets close to the current probe. What exactly is causing this sudden rise?

    A related question I have is with regards to those concentric shells. On some places, they are labelled as "effective resistance areas". What exactly is the definition of an effective resistance area?

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    fall of potential test

    Good question. The sudden rise in resistance at the end of the curve is probably due to a decrease in voltage drop between the two current probes. As for the "effective resistance area," I found this excerpt from "Getting Down to Earth" by Megger:

    Resistance of Surrounding Earth: An electrode driven into earth of uniform resistivity radiates current in all directions. Think of the electrode as being surrounded by shells of earth, all of equal thickness.

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    The earth shell nearest the electrode naturally has the smallest surface area and so offers the greatest resistance. The next earth shell is somewhat larger in area and offers less resistance. Finally, a distance from the electrode will be reached where inclusion of additional earth shells does not add significantly to the resistance of the earth surrounding the electrode. It is this critical volume of soil that determines the effectiveness of the ground electrode and which therefore must be effectively measured in order to make this determination. Ground testing is distinct when compared to more familiar forms of electrical measurement, in that it is a volumetric measurement and cannot be treated as a “point” property.

    Generally, the resistance of the surrounding earth will be the largest of the three components making up the resistance of a ground connection. The several factors that can affect this value are discussed in Section II on Earth Resistivity. From Section II, you’ll see that earth resistivity depends on the soil material, the moisture content, and the temperature. It is far from a constant, predictable value ranging generally from 500 to 50,000 ohm-cm3.

    Getting Down to Earth is a good read, I found this on page 12/13, here is a PDF you can download if you want to read more: http://www.biddlemegger.com/biddle-u...ToEarth-MC.pdf

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    Wouldn't an increase in resistance be a result from an INCREASE in voltage drop? R=V/I

    Quote Originally Posted by SecondGen View Post
    The sudden rise in resistance at the end of the curve is probably due to a decrease in voltage drop between the two current probes.

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    Re: fall of potential test

    Sorry it took me a week to follow up on my initial post. Thanks for the explanation about the Effective Resistance Area, that really made sense!

    However, I am still confused about that rise in resistance at the end of the curve. You say that it might be due to the decrease in voltage between the two current probes. But the current probes don't move, it's just one potential probe that we move around, right? Why would the voltage change between the current probes? And like madMAX said, wouldn't that decrease the resistance, not increase it?

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    Hey guys, sorry for the confusion. Max is correct, resistance increases as voltage drop increases. Maybe this drawing will help better explain:

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    The potential difference between Rod 1 (X) and Rod 3 (P) is measured by a voltmeter, and the current flow between rods X and C is measured by an ammeter. Using Ohm’s Law, R = E/I, you can obtain the ground electrode resistance R.

    It's the same method used in a DLRO or winding resistance test set. The current probes never move because the current is fixed. Potential probes measure voltage along the current path. Remember, C1 and P1 are separate but it's easy to think of them as one because they are always tied together at one end for a 3-point test.

    Another way to help explain might be to think of it as if you were doing a voltage drop test across a circuit breaker contact. If you have a fixed current of say 100A, you can put a voltmeter across the contacts and measure the voltage drop. Divide the two and you get a resistance.

    Think of the breaker line/load terminals as C1 C2 and your fluke meter as P1 P2, its pretty much the same thing!

    Example 0.052V / 100A = 0.00052 ohms.

    What changes with ground testing is that you move your voltage probe along the current path to map out the changes in voltage within the earth.

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  11. BrickSalad is offline Junior Member Pro Subscriber
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    Ah, I think you misunderstood my question. I understand how the ground resistance test works with the current probes, potential probes, Ohm's law, and all that good stuff (great explanation though!) The only question in my mind was why the resistance would increase more rapidly at the end than it does at any other point in the graph. One would naturally expect that the resistance keeps increasing at a slower and slower rate as we get further away from the test electrode.

    After talking to an engineer in my department, I've basically got it figured out though. I'm going to post an answer to my own question later today, complete with drawings, because I've never sen it anywhere else and this explanation might help out future techs down the line. Get ready for some exciting electric theory!

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  13. BrickSalad is offline Junior Member Pro Subscriber
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    All right, here goes. The problem with diagrams that kinda make it look like a circuit are that they make you think of a typical circuit, like the current is going straight from the current rod to the electrode. In reality, the current is going all over the place. All of the measured current is indeed going from the current rod to the electrode, but there are a practically infinite number of paths for it to take. So, I made a drawing to help visualize this:



    So, look how the lines are all compressed together near the rods, but spread apart in the middle. In the middle, for the same amount of current, there is a much greater area to travel through for the current, and therefore less resistance. It's kind like pushing the same amount of water through a big pipe and a small pipe, the small pipe will "resist" more because it will have more pressure.

    But we're electrical guys, so we need an analogy that we can relate to a bit more, right? In terms of circuits, you can think of it like parallel resistors. In the middle, we have lots of resistors in parallel, and near the ends we only have a few. Something like this:



    Now, what happens if you put the Ground electrode at "G", the current electrode at "I", and measure the voltage at each of the dots? Let's plot it in terms of distance from G... here's what it looks like:



    As you can see, it's steeper near the edges, and shallower in the middle... just like the diagrams we see in the textbooks!

    Ready for some fun? After making that graph, I decided I wanted to make it a lot more accurate. The more paths we have, the closer it is to ground, right? Well, this one is mighty complicated, and I'll leave the more technically minded folk to make sense of my assumptions and math if they care to. The rest of you should just look at the pretty pictures and appreciate that it looks even closer to the drawings in the textbook:



    Wow, that took me way too long to make!
    Last edited by BrickSalad; December 2, 2015 at 01:36 PM. Reason: Making images right

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    Based on your incredible explanation, I would say I definitely misunderstood your question. Very informative thread, this one is going in the bookmarks.

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    This was very informative.... good question!

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