Hi Richard and John:
> >[R_Hughes >] I know it as prospective short-circuit current, but maybe I
go
> >boldly whereas you boldly go?
>
> Do we know what the current values are in various countries, for
> equipment connected to wall-sockets? Assuming that the mains lead has
> the resistance that can be deduced from the maximum permissible
> resistance of its PEC.
The short-circuit current value is limited by
the sum of the resistance of the phase conductor
and the resistance of the neutral/PE conductor
(assuming all three conductors are the same
size).
The resistance is a function of the physical
distance (length of wire) from the distribution
transformer and the contact resistances of the
various connections and overcurrent devices in
the current path.
Since the distance from the distribution
transformer to the load can vary from relatively
small to quite long, the deduction of the
maximum permissable resistance of the PE and
other conductors seems indeterminable.
However, according to my USA EE colleagues, the
design of distribution systems is based on no
more than 3% voltage drop (wrt nominal voltage)
at maximum rated load, regardless of distance
>from the distribution transformer to the load.
If we can assume this to be true throughout the
world, then we can deduce both the mains lead
resistance and the PE conductor resistance.
See my paper, "Equipotentiality and Grounding,
Derivation of Grounding Resistance for Equipment,"
http://www.ewh.ieee.org/soc/emcs/pstc/tech-spk.htm
Click on "equipotentiality and grounding."
Page 20 gives the source resistance for mains
systems as a function of system voltage drop and
maximum rated current.
For 3% voltage drop at maximum rated load, the
source resistance is about 0.2 ohm. So the
maximum rms current would be system nominal
voltage divided by 0.2 ohm, or 600 amps for 120
V and 1150 amps for 230 V.
At these currents, the overcurrent device will
operate in about 0.01 second maximum, which is
less than 1 cycle at either 50 Hz or 60 Hz. So,
the current will never rise to the calculated
currents.
So, the maximum current is set by the overcurrent
device characteristics, not by the resistance of
the supply system.
If we assume a time of 1 cycle (0.02 s), then
a typical overcurrent device will trip at a
maximum of 20 times the rated current. So the
maximum current for a 15-amp overcurrent device
for a duration of 1 cycle will be about 300 amps.
If we add in the resistance of a 2-meter power
cord (0.05 ohm per wire) and the maximum
resistance of the equipment (0.10 ohm), then the
maximum fault current is less than 300 amps.
Indeed, it drops to about 150-200 amps.
Best regards,
Rich
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