# Conductivity of water, and risk of shock? (fact checking)

Through simple researching I have discovered the unit of siemens (being a reciprocal of ohms).

Now I wish to apply this to a (silly) scenario where the conductivity of tap water is 800µS/cm (from this)

If I were to throw an extension line (assuming no circuit breaker, 120 volt) in to a theoretical pool, I have calculated something like this:

1/(800 µS) = 1250 Ω


Now using the formula I = E / R, I find out that:

I = 120 V / 1250 Ω = 96mA


Essentially, at 1 cm, 96 mA of current could flow through it and theoretically kill you?

And at 100 cm:

I = 120 V / (1250 Ω/cm 100 cm) = 0.96 mA


Theoretically allowing you to survive?

Now from my basic knowledge I assume the current would not even try to go through you unless you had a path to ground (a pipe, metal surface, ladder...) so would that not matter? If I were connected to the floor if that were a ground I would not die at 1 metre according to my calculations?

I promise I keep it theory!

• S - See addition to my answer – Russell McMahon Oct 3 '11 at 0:34
• you are more conductive than tap water, so the current would actually preferentially travel through you. – JustJeff Oct 3 '11 at 1:03

• You have a good chance of dying in a real world situation.

• You have a moderately good chance of dying in a theoretical situation that properly reflects what you'd see in reality in your example.

You have misunderstood the units.
If the units were ohms/cm you would expect the resistance to rise as cm increased.
As siemens are the inverse of ohm then it makes sense that as cm increase the siemens will increase so the ohms will decrease.

Used in the manner that you are using it the formula would indicates that shock hazard will increase with cm, not decrease. This is because of a basic misunderstanding as mentioned above. This is because the unit relates to a fundamental property of the material per init volume. The "per cm" is because you have length on the top line of the equation and area on the bottom line and they cancel.

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.

So far so good - your formula still suggests that resistance rises per unit length of material. And, this is strictly true for the situation that your numerical description relates to in practice. ie a linear cross section sample of water whose length is varied. That's like taking a hose of water BUT not a pool.

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

After someone read the above text, questions were asked about why resistance does not increase with distance. To understand this, carefully read again starting at "so far so good". ie IF you have a hose then more length = more resistance. BUT if you have a pool of constant depth (see below), the greater the pool area, the more water there is in "parallel" to conduct the current, so resistance remains APPROXIMATELY constant with increasing size.

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 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.

• Thank you a lot, what I just do not get is however to calculate "at what point" the current will not be harmful. My resistance per cm going up made sense to me, as 10cm away would have a higher resistance and thus lower the amperage to safe levels (such as applying a 1.5v battery down a 100km line, where does it stop in accordance to its ohms/cm?) Maybe I am thinking it wrong, resistance doesn't change, however I do not get how the current goes down and how to calculate this. – Hobbyist Oct 2 '11 at 17:57
• See addition to answer. – Russell McMahon Oct 3 '11 at 0:14