The "this is not a short" and "we regret to inform you this IS a short"

[saturated]

Sorry just thinking out loud 

For the longest time I was thinking that making contact with components of an energized circuit with the leads of a (handheld) multimeter was gonna be a short
 
Well...now I know that IT IS NOT
And the (I presume) explanation for that is there is no current introduced between the leads only potential
(I'm sure there are limitations to this concept)

So I'm wondering how people can slip up and short something out  while probing with their multimeter leads and it occurred to me that what happens is they contact two parts of the circuit WITH THE SAME LEAD

 

And....that IS a short.   

[g1]

Yes that is it, touching 2 component leads with the same probe.
But also, the meter is reconfigured as a short when you are in the current (Amps or mA) ranges.  That is why we must 'break' the circuit and insert the meter in series to measure current.
If you do a current measurement, then forget and try to measure voltage with the meter still set for current, you are essentially putting a piece of wire across the points your 2 probes connect to.  Hopefully the meter fuse will save the piece of gear you are working on.

[joecool85]

Yes that is it, touching 2 component leads with the same probe.
But also, the meter is reconfigured as a short when you are in the current (Amps or mA) ranges.  That is why we must 'break' the circuit and insert the meter in series to measure current.
If you do a current measurement, then forget and try to measure voltage with the meter still set for current, you are essentially putting a piece of wire across the points your 2 probes connect to.  Hopefully the meter fuse will save the piece of gear you are working on.

For complete clarity, the meter set in amp mode will actually not be a short, but a VERY low ohm reading, something on the order of 0.1 ohm if I recall.  So, effectively a short, but not quite.  The reason this matters is because the meter will measure the voltage across that resistor and calculate the amperage draw.