Read current from two ground-connected supplies

• 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.

In addition to stevenh's answer, I'd like to suggest 2 more alternatives:

1. You can have a separate ADC in each of the floating channels. The ADCs would be linked to the PIC through serial bus (such as SPI). The bus goes to the floating channel through digital optocouplers, which are cheaper than analog optocouplers. Once you have an ADC inside the channel, you can make as many measurements as the ADC has got channels without needing additional optocouplers. This approach was already mentioned in kenny's comment.
2. Each channel could have a dedicated PIC. Each channel with its PIC (and its display, if there are displays) would all float together, and independently of the other channel. You wouldn't need to install a an analog optocoupler for each signal you want to measure. Furthermore, the PICs could talk to each-other through serial link (UART for example). The serial link would go through digital optocouplers.