I think you should both leave this argument here before it gets too far off track. You are both highly experienced and talented electronics engineers (as am I), but appear to be having a little trouble communicating your exact thoughts on the subject.

*Originally posted by alphnumeric *

... I=E/R you can't divide 0 into E. R can never be 0 anyway, unless you have it cooled to absolute zero. That equation will work with any value of R other than zero.

You just proved that the equation also works for R=0, as anything divided by zero tends towards infinity, as you say here ...

*Originally posted by alphnumeric *

... If the short is the only load in the circuit you get maximum current flow ...

Just to be fair, here are some anomalies from StormBringer's calculations (apologies for pointing this out) ...

*Originally posted by StormBringer *

... For example, lets take a look at three conditions of the same circuit; first, we look at it in operating condition(no faults)

we have a 12v source that is rated at 15A max

E=12v

R=0ohms

I=0A

In a *real* power supply this is correct (except E would drop to zero also), because the short-circuit protection will have kicked in, or the fuse would have blown. Real power supplies are also voltage regulated, and current limited. If we were talking purely theoretical, and we did maintain 12v across 0 ohms, then the current would be infinite.

*Originally posted by StormBringer *

... Now we look at it with an "open" condition.

E=12v

R=Infinity(since the circuit is broken)

I=MAX

It would appear that you've made a simple typo, and gotten your 'open-circuit' and 'short-circuit' currents mixed up (see previous example). That appears to have led to this discussion moving a little off track. Now that its cleared up, maybe we could all try doing something a little more productive.

Edit: One final comment ...

*Originally posted by StormBringer *

... If you measure across a short, you'll read NO drop in voltage and NO current(since there is no change from one side to the other(since it is essentially a wire and the potential is the same on both sides) ...

This is correct, but then you wouldn't have 12v across the zero ohm load, which makes the example misleading. Also, current can flow through zero resistance, as that is what superconductivity is about. However, there would be no power dissapated. Any real power source has its own internal impedence (hence its finite max current), so zero ohms will never be a reality in any real power circuit.

*Originally posted by StormBringer *

... If you have an open(broken circuit) and measure across the break(which will complete the circuit) you'll measure source voltage(source being whatever is measured before going into the load) and the current flow will be MAX(since your meter is now the load) ...

With a perfect (infinite impedence) voltmeter, and no load, the current will be zero (through the meter).

With a perfect (zero impedence) current meter, and no load, then the current will max out (through the meter) and PSU fuse will blow.