I'm at work, having a brain cramp and can't remember the answer to
If a GFCI breaker trips, can you tell if it tripped due to a ground
fault or due to an over-current situation just by looking at it?
Is there any way to tell which situation caused it to trip?
I'm talking about a GFCI *breaker*, not a receptacle.
Unlike a GFCI receptacle, a GFCI breaker will trip due to both
overcurrent and fault conditions.
I'm trying to determine if you can tell which "part" of the GFCI
That is not a "GFCI" circuit breaker. The OP question was regarding a GFCI
breaker. A combination AFCI/GFCI is not used for class A ground fault
There is no GFCI breaker, new or old, that has an indicator to determine
the cause of the trip
Correct. More specifically, they trip when the current in the Hot wire is x
mA more/less than the Neutral wire. I forget the actual mA ranges. In
reality, the earth ground has nothing to do with it; GFCI's can also be used
on 2-wire, ungrounded systems and work fine.
I think the "ground fault" in the name is because in order to have an
imbalance in the Hot/Neutral, the missing current has somehow gone into the
ground (fault condition) at some point other than the device itself. e.g.
inside conduit, junction boxes, frayed insulation, bad fixtures, miswires,
etc. etc. etc..
No, he is not correct, at least not in this situation.
As per the subject line of this tread, the question was related to
GFCI *breakers*, not GFCI receptacles.
A GFCI *breaker* will trip for both over-current and fault situations.
In a way, yes. Watt is originating from James Watt who invented steam
engine. It relates to Horse Power. In Ohm's law symbol of Watt is P
which means power. P=E x I, P=I^2 x R, P= E^2/R Amount of P is depending
on current and/or voltage. So if either one is high over the limit any
breaker will trip. Watt used for real work is consumed by resistive
load. Lost false Watt is used by inductive, capacitive load.
(Remember impedance Z?)
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