MultiMeter Help!

Thats all great and stuff, but the water is not the voltage source. The equipment is. The only way to bring the potential to 0 is to eliminate the source or bring the ground to the source voltage. Even if you bring the water to ground it's potential voltage is still less than the source voltage.

Did I mention I am loving this....;D

 

Well, that depends on whether the voltage is induced or a direct short. Now we're moving on from Georg Ohm to Daniel Faraday, nice!!

If the water's voltage is induced voltage, it is being created by inductance of an electromagnetic field coming from the equipment. If the water is connected to ground, however, the potential for voltage is still zero.


If the equipment is faulty (should be replaced), this is an electrical short (not induced voltage). If the water and ground are connected using a conductor with little to no resistance (ground probe), and there is an electrical short between water (ground) and "live" voltage, then one of a few things can happen:

1. GFCI will trip detecting a ground fault (why you should install a GFCI)

2. (no GFCI) a breaker may trip when the current draw in the short circuit equals MORE than the breaker (this does not mean it will trip the breaker every time)

3. One conductor will overheat until it fails open, or a fire or other serious heat damage may occur. This conductor could be one of several, including the equipment causing the short, the ground conductor (ground probe) or the electrical wiring to the equipment.

In the above scenario, voltage originates from a power plant and is conducted through electrical wiring to your equipment. When you short this to ground, it will create heat energy and because the conductors are not perfect, they will begin to break down. Still in this case we use ohm's law.

I = V/R

Where I, the current cannot be solved because the voltage (V) is shorted to ground at zero resistance (R). You can't solve an equation by dividing by zero, which is a sort of mathematical impossibility.

As long as the current is directly shorted to ground perfectly, there will be no measurable voltage difference, but this will create heat and begin to break down conductors until the circuit fails.

Now, take the above scenario and do NOT connect a ground probe. With no probe connected, the water voltage will go high because it is directly shorted to the equipment. You will probably not know it, though the fish might because the electrolyte mix in their tissue creates a voltage gradient! Again, use ohm's law:

I = V/R

Where the current (I) is applied to the fish's body (R) at a given voltage (V) shorted into the water through the failed equipment!! (they don't like this, BTW)

If you were to touch the water, you would be between the voltage and ground. Using the following equation again:

I = V/R

Where the current (I) passing through your body's resistace (R) to ground at a given voltage (V) in the water!!! Electrocution city! Just hope that your body's resistance (R) is high enough and the voltage (V) is low enough so that little current (I) passes.

me too! time for a beer... cheers!

 

This is indeed a true statement. Consider first the aforementioned receptacle that we were testing for 120V. It had 120V, did it not? Was there current flowing through the receptacle? Of course not. Viola, voltage without current!

Also consider a capacitor. A capacitor based on its physical construction does not allow current to pass through it, however a capacitor can have a voltage across its terminals. If you believe that you cannot have voltage without current than I suspect you would be more than happy to touch both of the terminals of this 1F capacitor that I just disconnected from my car battery where it had been connected for several hours...

To use your water analogy, consider a pipe that is under pressure at one end (voltage source) and open to the atmosphere at the other (ground) with a steady stream of water flowing (current). Now close the valve. No more water flowing (current) but the pipe is still under pressure (voltage). Again we have voltage without current.

I can go on if you wish...

 

You make a reasonable argument. However, the voltage would not be set in motion until current comes into play, thus V=IR. So, I'm still wondering how you can explain away ohm's law.

 

Ah folks.. I'd be very careful sticking probes or wires into your tank. Wire & most probes are made out of COPPER that is somewhat coated!!! Any nicks on the probe or sticking the whole probe to where the copper wire gets wet will be very BAD!! Good way to poison your inverts.

I'd recommend using something made of stainless or even stainless steel cable of some sort.

 

ok, so licking 9v battery is bad. Got it!! LOL..... Interesting thread. Love the debates...

 

I beg to differ.. Here is just one example.. Most of these experiments contain Magnesium which we try to maintain in our tanks. With the right mixture anything is possible.

How to Make Electricity From Saltwater & Air | eHow.com

 

Well this turned out to be an interesting thread!
Thank you all for your help and knowledge, I will see what device was causing my 2v and also get a ground probe in the tank.

I will bow out and let the debate continue.

Thanks Again!
K+

 

Please apply Ohm's Law to this circuit :


OK, LET ME GET A DIAGRAM FOR YOU...

 

Solve for Vo and I in this circuit: