Re: very low voltage -- dangerous?

Floyd L. Davidson wrote:
Tony Hwang <dragon40@xxxxxxx> wrote:

Floyd L. Davidson wrote:


John McGaw <nobody@xxxxxxxx> wrote:


Put a load on the circuit, turn off the breaker, then measure -- it will
almost certainly show you 0 volts as it should.

That is not necessarily true. A "load", could for
example be provided by switching on a few lights some
appliance plugged into that particular circuit.
It might, or might not, affect the voltage reading,
depending on the wiring between the "load" and the point
where a reading is taken.


If you see voltage under
these circumstances _then_ you have a problem of some sort. As others
have pointed out, you can not blindly rely upon a high-impedance meter
in home-wiring situations since they can measure capacitive leakage
across open contacts and between adjacent conductors.

It isn't "capacitive leakage", either across open
contacts or between adjacent conductors. It's
induction.
And note that regardless of the load, the amount of
induction is exactly the same. That is, the power
induced into a given circuit by another is not going to
change when the load is changed. It's just a matter of
E = I * R, and reducing the load means the current goes
up and the voltage goes down. Since the meter is
reading voltage, it show a lower reading.


Hey,
It is E = I * Z since it is AC circuit. Your formula is for DC circuit.


Don't worry, that's the exact *same* thing.

Hi,
Not really. If they are same, why R and Z? My understanding is Z has two component, pure resistance plus reactance by inductance and capacitance
combined. No? Are there many non-reactive AC load?
.

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