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This doesn't immediately look so good, but if you think about it, why shouldn't it be? If the extension cord drops in, why should the electricity choose to flow through any of them, instead of just the ~1 inch of water to the other pole?

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I wouldn't get into that pool, but this isn't as bad as it might first appear:

  1. If the rubber is intact, then there is no path to ground. Electricity can only flow between the two conductors. There would be a strong field between them, but that would diminish rapidly with distance. The remaining currents thru the water (and you) a foot away from two conductors spaced normal outlet distance apart should be pretty negligeable.

  2. The real issue is leakage from the hot side thru the water to some ground connection. In this pool, it appears that would only be due to a pinhole leak in the pool floor (presumably a bigger opening would be found and fixed to prevent water loss). That would be some distance from where the power cord touched the water, which should be a significant resistance. For the current to be substantial, the distance would have to be close, making it less likely your body is near any significant current path. For longer distances, the current would be less due to the resistance, and more spread out anyway.

  3. Since the most serious danger is from hot to ground conduction, a ground fault interruptor should shut off the power feed if there was such condution. Of course anyone dumb enough to float a outlet strip in a pool, and then get into the pool, probably didn't think of plugging the thing into a GFI circuit.

  4. What's the real problem here? We've got 7G people on this planet and rising. If some of them want to do the rest of us a favor in reducing this problem, who are we to object? I'm all for Darwin Awards as long as I don't get one.

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    \$\begingroup\$ +1 for point 4! ;) \$\endgroup\$ – Count Zero Feb 6 '12 at 21:37
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    \$\begingroup\$ However, unless you plan on jumping in or out of the pool, or unplugging and throwing the extension cord out before getting out, I'd bet your two legs would make a nice path from the water to ground. \$\endgroup\$ – pfyon Feb 6 '12 at 21:55
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    \$\begingroup\$ @pfyon: Yes, and what about when the outlet strip falls in? I can imagine one of the geniuses reaching over to pull it up. Or adjusting the volume knob could be interesting depending on what kind of leakage. In short, electricity, pools, beers, and morons don't mix well, See my point #4. \$\endgroup\$ – Olin Lathrop Feb 6 '12 at 22:20
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    \$\begingroup\$ Or, from my point of view, they should be mixed more often ;) \$\endgroup\$ – Wouter van Ooijen Feb 6 '12 at 22:41
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    \$\begingroup\$ @pyfon - Your legs are an inferior conduction path when compared to chlorinated pool water. You'd have to be holding the strip, or occupying most of the volume (imagine the pool was 18" diameter) for a meaningful current to flow. Your legs are a high resistance in parallel with a lot of low resistance water. \$\endgroup\$ – Kevin Vermeer Feb 6 '12 at 22:43
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See diagram below:

  • Death a real risk but not a certainty.

    - Potential difference across the water in contact with you is what matters.
    Its potential becomes your potential !!!

  • More conductive water makes things worse.
    (Chlorine, salt chlorinator, ...)

  • Having the water conductive enough tot be BELOW your impedance is worse not better. If the breaker or fuse does not rip then you are shunted across a resistor with voltage on it and your body resistance is higher than its resistance so you don't shunt down the voltage much. If you were in a high imepdance solution you would form a low Z shunt across part of the circuit and most of the voltage drop would occur in the water not in contact with you. ____________

Have a look at the Wikipedia article on conductivity

This translates into the more useful (here) formula

  • Resistance = Resistivity x length / Area

Conductivity = 1 / resistivity

  • So Resistance = length / Area / conductivity

where length is the length of a sample of material and area is the cross section of the sample. This is where your 1/cm figure came from.

What you end up with is a resistor of about constant resistance between the live wire entry point and grounding points, with a voltage drop across the resistance. It's well beyond discussing here, but if you draw "curvilinear squares along the discharge path you can get some idea of the potential that will exist across a length of water in contact with YOU. If this potential is more than about 50V you are liable to be in trouble in a total contact situation.

See the diagram of a "pool" below with curvilinear squares drawn.
There is a 110 VAC contact point and a grounded point - both on a wall in this case.

Voltage contours in this case are as shown. What matters here is "scale".
In the scale 1 case at top the body is small in length compared to the pahse to ground distance and experiences about 1 to 15 Volt potential drop depending on orientation. Anyone floating stretched out along the arrow show would experience an unpleasant sensation but have little chance of dying.

If we rescale the pool so a body is the length of the bottom arrow "there will be trouble" - for a short while anyway :-). An "arrow length" person floating longwise along the arrow length would quite possible suffer ventricular fibrillation before they could stand up AND would probably be above the can't-let-go-limit so could not stand up. Nor cry out possible, nor breath :-(. Worst case (after immediate fibrillation) they may suddenly be paralysed and sink soundlessly and drown. I think that's worth another :-(.

enter image description here

In a pool of increasing size the area available increases with increasing size. The end result is that resistance tends to be somewhat constant as size increases - distance is longer but area increases. (This is where the concepts of amperes per square unit (mentioned in above page) and ohms per square) come from.

So at say 1250 ohm/cm, this would be the face to face resistance across a 1 cm side cube, or a 10 cm per side cube, or a 1 meter or 10 metre per side cube.

In the case of a body falling into a pool all sorts of complications arise. If you were holding the live wire you'd briefly be in trouble,. Then beyond trouble :-(. If the wire fell in a pool you were in then you'd need to know where the ground connections were.

Don't try this at home.

In the real world water will not be pure. In a chlorinated pool conductivity will be affected. And more ... . See electrolytic conductivity


Why does resistance NOT increase with distance as water volume increases?

The "side to side" resistance of a square of water of constant depth with sides of 1mm or 10 mm or 100mm or 1 metre or even 1 kilometre is the same !!!! As the distance goes up by N there are N times as many paths in parallel.

BUT as the size of a cube of water goes up by N the resistance DROPS by a factor of N.
Consider a 1cm per side cube and a 10 cm per side cube.Imagine (to make life easier) that the face to face resistance of the 1 cm sided cube is 1000 ohms.

The 10 x 10 x 10 cube has 10 times the path length so a 1 x 1 x 10 cm path across the cube from face o face would be expected to have a resistance of 10 x 1000 = 10,000 ohms. BUT as the area of th face has risen from 1 x 1 = 1 cm^2 to 10 x 10 = 100 cm^2 cm^2 there are 100 such strips in parallel so the resistance will be 100 times lower than for one strip. So resistance will be.

  • 1000 ohms x 10 / 100 = 100 ohms.

The bigger the cube the lower the resistance.
Resistance decreases linearly with increasing size of sides.

In a pool of constant depth one side of the above "cube" is instead held constant so this reduces the expression from area/volume to area/area and the resistance is constant.

In a pool with you and a live wire and some earth points at unknown locations the situation is confused. You don't know where the ground contact is to the water or "client" and more. so the problem is insoluble as stated. You need a more precise statement to tie things down.

Another related issue:

If you stand in water up to your neck, if there is 230 VAC from pool surface to pool floor, will you be electrocuted? All that is required for high current to flow in you is a low enough connection resistance to your neck and to your feet. If you are taller, and the water is deeper, the situation will be the same. The path through the body of the water is irrelevant here IF the resistance to neck and feet is low.


Fun only: Three phase cable in pool, yee ha !!!

enter image description here

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  • \$\begingroup\$ In the first diagram, is there return path adjacent to the hot? The premise of the other answers is that the adjacent hot and return (neutral and ground) in the power switch keeps the electric field localized. \$\endgroup\$ – Michael Pruitt Feb 7 '12 at 2:58
  • \$\begingroup\$ @MichaelPruitt - The adjacent return answer is akin to the fallacy that current takes the easiest path to ground. As long as a breaker doesn't blow, voltage will drop between a phase lead and any grounded point elsewhere. You COULD argue for a close earth connection causing a low potential right around the phase lead BUT you can just as easily argue that pahse forms a high voltage area around the ground lead (ie neither is true) and a bit of even quite rough curvilinear square doodling will see you backing away from the pool with your hands in sight. \$\endgroup\$ – Russell McMahon Feb 7 '12 at 4:43
  • \$\begingroup\$ I actually disagree, as michael said here, as long as you are a distance from the hot and ground lines(unlike something like a bathtub and something falling in) you are going to be somewhat protected by the surge protector being a great distance away. If the surge protector has no fuse other then its power lines it could dissipate enough power for you to feel it, possibly, but that chance is minor as the conductive water is putting such an unbelievably low resistance in parallel to you with something that will be current limited by the resistance of the leads. \$\endgroup\$ – Kortuk Feb 8 '12 at 8:49
  • \$\begingroup\$ (2) @Kortuk - apparently you disbelieve and or disregard my curvilear squares examples, which are how "potentials" distribute themelves in fields. I'm uncertain where the idfea that there may or will be a surge protector came from, and the suggestion that water has a very low resistance is wrong. While you can achieve low resistance by application of enough ionic material (salt, chlorine etc) plain tap water is quite non conductive and I imagine that many pools would fall in the middle ground. Olin's suggestion that no ground is available is not one I'd like to risk mty life on. \$\endgroup\$ – Russell McMahon Feb 8 '12 at 13:11
  • \$\begingroup\$ (1) The anonymous downvoter strikes again. (3) In Australia they sell or sold kettles which have a bare coiled coil element on a ceramic former which is immerced in water (NO insulation) and two stiff conducting leads feed it mains 230 VAC. Not surprisingly, the jug design does not allow you to open the lid when the cord is inserted. The two support wires apply mains AC across water with a gap of perhaps 40 mm . These jugs would be illegal here and may no longer be used in Australia. \$\endgroup\$ – Russell McMahon Feb 8 '12 at 13:14
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Even if it is a good resistive insulator (which I doubt in this high water pressure condition) the capacitance which formed by the bottom surface and ground may be big enough to be dangerous.

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