Remember that although you may not have an 'ohmic' or conductive path to
ground, you are likely to be capacitively coupled to ground. When measuring
the voltage from yourself to the 230V line with a high impendence voltmeter
you may see a small voltage. This is because you have created an AC voltage
divider with several elements: the voltmeter (1 or 10M ohm typical
impedance), you (? ohms) and the capacitive coupling between you and ground
(? ohms at 60hz).

Now, assuming for the sake of concrete numbers:

Rmeter = 1Mohm
Ryou = 0ohm    (worst case - highest current)
Rcap = 19Mohm
V = 240V

Then your meter would show a voltage of :
     1M/(1M+19M)*240V = 12V

Oh, and the current would be
    240V / 20Mohm = 12 microamps (should be quite safe)

Now, you can test this theory be placing a 1M resistor in parallel with the
voltmeter. Now the voltage read on the meter should be:

    500K/(500K+19M)*240V = 6.15V

And this time the current is

Now, as far as your burning the LED lead: in order to get enough current to
do that, you must have accidentally connected with a real ground somewhere,
I would think. Please describe in more detail the physical arrangement of
your experiment.

Bob Ammerman
RAm Systems

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