Ground Fault Isolators (GFI's)

fourthclassC wrote:
Simple solution. Replace all GFI receptacles with old fashioned ones. (good quality ones)

Simple, but not especially wise. GFCIs are required by code in many areas (basically where outlets are in areas that could be damp) because they detect and shut off current that could very well be electrocuting someone if it were not shut off.

You coiuld similarly avoid the annoyance of fuses that blew in the old days by replacing them with pennies. Some people have paid dearly for doing that.

Simple solution. Replace all GFI receptacles with old fashioned ones. (good quality ones)

Just like surge protectors, GFI devices deteriorate over time from the constant transient spikes present on almost all lines - nearby motors kicking on and off, power grid switching, etc.

One difference between the two types of devices is that GFI's will eventually trip at a lower threshold or not even reset. Surge protectors on the other hand will simply no longer offer any protection and should be replaced every few years.

Never buy a cheap GFI or a cheap surge protector. You'll never know whether they're working or not.

On a GFI protected circuit, if one were to install a 12k Ohm resistor from 120 VAC Hot to Ground (there by provide a 10 milliamp leak) the GFI should trip. because there would be 10 mA less current in the neutral then the hot. IMO

On edit: If we also added a 12k Ohm resistor between neutral and ground there would be no current flow through that resistor because the voltage across this resistor would be effectively zero volts.

There is a problem with the hypothesis. The GFCI compares the current flowing thru the hot connection to the current flowing thru the neutral connection. A difference of 5-6 milliamps will cause a trip of the GFCI. That current flow should be exactly the same in both. If there were 5 amps flowing thru the hot wire (and therefore 5 amps flowing thru neutral) and it induced 10 milliamperes current thru the ground wire, that does not mean there would be any change in the flow thru hot and/or neutral. Those current flows would still remain identical unless there were a leakage somewhere. And an induced flow in a ground wire is not leakage and would not cause s trip. The GFCI does not do a compare to the ground wire.

larry cad wrote:

Harvard wrote:
Keep in mind, it is a total Xc of ALL hot to ground wires in the RV, not just the shore power cord.

My first rough calculation using a value of 100 pf/meter gives me a length of about 2250 feet of close coupled wire to produce .005 amps of leakage current.

That would be at 60 Hz .... and not taking in consideration the presence of any RFI filters across the power input of things like the converter.

For 600 Hz it would be 225 feet and 6000 Hz would be 22 feet.

It then becomes a question of whether or not it is remotely possible to see high frequency noise that is present long enough to trip a GFI. Typically a spike of noise will set up an ringing series of oscillations, the frequency and duration of which is dependent on the inductance and capacitance of the network involved.

Harvard wrote:
Keep in mind, it is a total Xc of ALL hot to ground wires in the RV, not just the shore power cord.

My first rough calculation using a value of 100 pf/meter gives me a length of about 2250 feet of close coupled wire to produce .005 amps of leakage current.

Keep in mind, it is a total Xc of ALL hot to ground wires in the RV, not just the shore power cord.

Harvard wrote:
A connection to an RV presents many more feet of 3 conductor wire as compared to a normal appliance. The more feet of wire the more capacitance between the conductors. This capacitance will conduct AC current between Hot and Ground, sometimes enough to trip GFIs which is about, I believe, 6 milliamps.

That is an interesting hypothesis and it sounds believable. The GFCI compares current entering into the "hot" terminal to the current leaving from the "neutral" terminal. They must be equal within .005 amps or the device will trip. In order to use ohm's law to determine if .005 amps will flow between close conductors, you would first need to determine the capacitance of the power cord and then calculate the Xc and then use ohm's law to calculate the leakage current to determine if this is even possible. Anyone know the typical capacitance per foot of standard RV 30 and 50 amp shore power cord???