In a message dated 7/5/2006 7:44:38 AM Eastern Daylight Time,
g3rzp@g3rzp.wanadoo.co.uk writes:
This is an interesting one. Is it assumed that you can't get a fault such
that the full fault current flows down the ground (i.e. the protection) wire?
Such fault current being greater than the normal load current? Intuitively,
I'd sort of expect the protective ground conductor to be of at least the same
gauge, if not bigger, than the current carrying conductors.
Also, do we assume that in the case of two lots of 110 volt appliances, the
currents will be approximately in anti phase, and so cancel in the neutral?
This would fall down with bad power factors of opposite sign. (OK - opposite
sine or rather, cosine, if you want a bad pun)
73
Peter G3RZP
Peter, the way things work if all wiring is done to code is that the bond
wire could carry a fault current as great as the load current or more, but
since the bond wire would not carry normal current it would never heat up. A
fault current would be and is considered to be when the chassis becomes
energized by one of the regular current carrying conductors in which case it
would
be a dead short which will throw the breaker in a fraction of a second. The
wire can easily handle the surge for the time it takes to throw the breaker in
a dead short. SInce the neutral is suppossed to be wired so that it is the
only wire to carry current back to the source it will be most capable. In
the case of a 240 v single phase the neutral only needs to carry the
unbalanced
load. If both phases A&B are carrying X amps of current, there is no
current flow on the neutral. Lou
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