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If copper is more conductive than water (at any reasonable PH), submerging copper electronic circuits in should have no effect, as the electricity should continue to follow the path of least resistance (the highly conductive copper PCB paths, for example) rather than shorting into the mildly conductive water. However, dumping water on electronics clearly shorts them out, even though copper is the superior conductor.

Why does this occur when the path of least resistance should be the copper rather than the water?

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Fallacy correction:

"Electricity" NEVER 'follows the path of least resistance'.

Electricity is NOT like a single large Zorb bounding down a hillside.

Electricity would be better modelled in this context as
a large reservoir of water being poured down a rough hillside.
Most of the water will follow major gullies and channels but some will find smaller side channels.

The question is not
"which path will the electricity follow?"
but
"how much current will flow along this given path?"

Low resistance paths will conduct higher currents.

Higher resistance paths will conduct less current.
Only infinite resistance paths will conduct zero current.

There are NO infinite resistance paths.

Plus:

When voltage is applied to water above a certain critical voltage, the water will decompose into Hydrogen and Oxygen - - known as "electrolysis". Any components in the water are liable to also decompose to produce eg Chlorine gas from chlorine products in the water. Even wholly pure deionised water can be decomposed in this manner.

Once you get even a small amount of current flow you get ions formed which promotes more current flow, lower resistance, more breakdown ... zap.

I have dropped a "pager" into sea water and a portable phone into concentrated chlorine solution* and "saved" both with no long term damage by immediately taking out the batteries and washing them in copious quantities of fresh water and then leaving them to dry completely (a very important step). And I have seen equipment destroyed by exposure to water with the batteries then left in.

  • Note to self: Remove portable phone from top pocket before stirring large bucket of Sodium Hypochlorite solution.

    When splashing along sea edge and generally having fun your pager should be in a salt-spray protected bag or, better still, left at home.

    Note to the young: Pager's were things we used to have before they invented cell phones. Effectively a receive only SMS system with no voice or any other features. Doctor's still seem to use them.

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    \$\begingroup\$ Beautifully answered. This was easy to understand. Thank you. \$\endgroup\$ – Asker Jan 24 '13 at 3:19
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    \$\begingroup\$ +1 For explaining this mysterious "Pager" reference. Thank you. \$\endgroup\$ – DrFriedParts Jan 24 '13 at 5:43
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    \$\begingroup\$ Pagers: volunteer fire halls are still HUGE users...some ambulance services...support staff in hospitals, food servers and also given to people waiting for tables at restaurants so they can wonder away until the table is ready. It's a tech on it's last breathe but going down fighting. \$\endgroup\$ – Chef Flambe Jan 30 '13 at 2:31
  • \$\begingroup\$ Great answer, Russell! Question: how come the low 4.2VDC of portable devices manage to ionize water and pass such a huge current (miliamps) in order to burn out electronics? Could you please provide a sample simple calculation? For example, I believe one needs 10kV to ionize 1 cm of air gap! \$\endgroup\$ – Vorac Apr 4 '13 at 7:09
  • \$\begingroup\$ @Vorac Starting with "It happens" + deionised water doesn't conduct the most common reason is liable to be even small trace contaminants initially plus impurity of the contaminating water. I have seen and heard of various things destroyed by liquids plus battery action and, as mentioned above, I have two amazing examples of how prompt action can stop major damage when massively conductive contamination is present (seawater and pool chlorine). I had a friend who had a camera in a "waterproof" plastic container destroyed by row-boat bilge water in the time it took to row out to a moored yacht. \$\endgroup\$ – Russell McMahon Apr 4 '13 at 10:28
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The problem is that although the traces have a very low resistance, the traces are not a fully meshed network. The PCB is a collection of networks. There isn't a low resistance path between any two nodes, but only between specific nodes which are understood to be on the same network, as dictated by the circuit design which was transferred to the PCB artwork.

Between some other nodes that are not on the same network, the resistance or impedance may be, in fact, very high. Close to infinite, in some cases. Moreover, there may be a voltage between these points. A prime example of this are power rail networks.

Ion-rich water will create conductive paths between nodes which are not supposed to be thus connected, and the voltage difference will cause a current to flow, creating short circuits or near short circuits.

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  • \$\begingroup\$ Another really good answer. These are way more enlightened than the "yahoo answers" threads google returns in response to this query. Hopefully this thread out-pageranks them someday. \$\endgroup\$ – Asker Jan 24 '13 at 3:21
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A circuit board has traces on it. These traces are meant to be insulated from one another. Proper circuit operation requires it.

If you dump water on the board, the traces are no longer insulated.

If the "circuit" was just a solid plane of copper, the water would have no effect.

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  • \$\begingroup\$ To be physically precise, this part of your statement needs tighter qualification: "If the "circuit" was just a solid plane of copper, the water would have no effect." -- (1) Water is polar; It effects the impedance experienced by currents in the copper (so you must specify low frequency), (2) it creates an electrostatic field (so you must specify low voltage), (3) the copper reacts with water to form copper(II)hydroxide due to the self-dissociation of water given enough time (so you must specify DI water and short contact time). \$\endgroup\$ – DrFriedParts Jan 24 '13 at 7:17
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It is following the path of least resistance. In this case, that path is the copper and the ions in the water that allow some more electrons. You need to look at water as a very high resistance. It isn't infinitely high, so some current will flow there, usually with bad results. Just as in the case with two resistors in parallel, the copper is the low resistance, and most current flows through that path, but some will flow in the other high resistor (ions in the water). Note that deionized water cannot conduct due to the lack of ions.

Another thing usually ignored is that ICs are not completely hermetically sealed, so water and other elements can go in and wreak havoc inside, especially with time when you take into account oxidation and ions that can short the IC.

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  • \$\begingroup\$ Appreciate your insights. \$\endgroup\$ – Asker Jan 24 '13 at 3:23
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    \$\begingroup\$ This statement is not correct: "Note that deionized water cannot conduct due to the lack of ions." DI water has ions, just no net ions and (hopefully, no impurities). All the time a small percentage of the water molecules self-disassociate and recombine. Water is also polar so a current will create a field. And relatively small voltages (as little as >1.55V) can split the water molecules apart creating large amounts of ions. When you apply a electromotive force (voltage) to a metal wire, the wire itself doesn't change much. When you do the same to a solution all bets are off. \$\endgroup\$ – DrFriedParts Jan 24 '13 at 7:28

protected by W5VO Jan 24 '13 at 5:14

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