Hypothetical question

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domnic

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Can you run a wire say ( 12-2wg) long enough to trip a 20 amp or any gfci with no load on it ?
 

Little Bill

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Can you run a wire say ( 12-2wg) long enough to trip a 20 amp or any gfci with no load on it ?

Are you asking if you could run the conductors so long that the resistance would be enough to trip a 20A breaker? In effect turning the resistance of the wires into a load?
 

don_resqcapt19

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No insulation is perfect, there is some leakage current with all conductors. If you have enough length the leakage current from the hot to the EGC will be enough to trip the ~5 mA GFCI.
 

junkhound

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If one calculates out characteristic impedance, it probably falls between 100 and 200 ohms for 12-2 w/g for black to white and about 70% that value for black to ground.

Thus an infinitely long wire will draw about 1.5 to 2A on the breaker, and about 60% or so of that on just the ground wire. Say about 3 pf per foot IIRC, so 5mA = 120/24K; 24K = 1/377*C, so C to trip GFCI is about 0.1uF, 0.1u/3p = 5 or 6 miles of 12-2 wire to trip GFCI., will not trip a normal breaker. Probably under 2 miles for wires in emt.

If you really want to know, stick a capacitance meter on 1000 ft of 12-2 and measure the capacitance <G>
 

PetrosA

Senior Member
Not sure about breakers, but I seem to remember that beyond 75' of a 2-wire cable (NM or UF) GFCIs can trip from leakage or capacitive coupling. I think there's something in the instructions for various GFCIs on the matter. I've only seen it happen once where a customer had a 12-2 UF feeding a receptacle on an island in the middle of a lake and the GFCI feeding it from a shed on land would trip constantly.
 

gar

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Romex from 1965 12-2 wg is about 19 pfd per foot from hot to the EGC. This would be the uncompensated path for leakage current seen by a GFCI.

With 120 V applied a current of 5 ma results from an impedance of 120/0.005 = 24,000 ohms. Xc = 1 / (2*Pi*f*C). For 24,000 ohms at 60 Hz C = 1 / (6.28*60*24*10^3) = 10^-6 / (377 * 24 * 10^-3) = 10^-6 / 9.05 = 0.11 mfd. At 110,000 pfd to equal a reactance of 24,000 ohms at 60 Hz and 19 pfd per foot the cable length has to be at least 5800 ft long.

It may be possible to trip a 20 A breaker with the above cable at a length somewhat less than 1/4 wavelength at 60 Hz with an unterminated line (open circuit). My guess about 75% of 1/4 wavelength because of the slower propogation in a cable than in free space. 1/4 wavelength at 60 Hz is 186,000 / (60 * 4) = 775 miles.

An open circuit transmission line at 1/4 wavelength reflects back to its input a short circuit. Losses in the line alter this. This is because the reflected signal is 180 degrees out of phase with the input signal.

mivey may be able tell us what a real world power distribution line with an open circuit at about 700 miles would do at the source. Was this part of the problem in the great eastern blackout?

To see the effect at the input of an open circuit transmission line with a low impedance source see my P1 photograph at http://beta-a2.com/cat-5e_photo.html . The waveform that starts up immediately at the left is the input to the transmisson line. The waveform starting at about 200 nS is the wave at the end of 150 ft of CAT-5E cable.

The free space velocity of light and radio waves is about 982 ft per microsecond. To travel 150 feet would take 0.153 microseconds. Thus, the propogation velocity of the electric wave in the CAT-5E cable is about 77% of the free space rate.

When you look at the input end waveform you see a big rise at about 400 nS. This is the result of the reflected energy from the open end of the transmission line.

Wires in a conduit filled with water will probably somewhat increase the capacitance per foot.

.
 

steve66

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Illinois
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Engineer
I assume you are talking about wires on the load of the GFCI protection device.

Yes, if the wires are long enough, stray capacitance can cause the GFCI to trip.

I know Square D GFCI breakers list a wire length limit. I think it is 200' (or maybe 150').
 

fmtjfw

Senior Member
160121-1003 EST

Romex from 1965 12-2 wg is about 19 pfd per foot from hot to the EGC. This would be the uncompensated path for leakage current seen by a GFCI.

With 120 V applied a current of 5 ma results from an impedance of 120/0.005 = 24,000 ohms. Xc = 1 / (2*Pi*f*C). For 24,000 ohms at 60 Hz C = 1 / (6.28*60*24*10^3) = 10^-6 / (377 * 24 * 10^-3) = 10^-6 / 9.05 = 0.11 mfd. At 110,000 pfd to equal a reactance of 24,000 ohms at 60 Hz and 19 pfd per foot the cable length has to be at least 5800 ft long.

It may be possible to trip a 20 A breaker with the above cable at a length somewhat less than 1/4 wavelength at 60 Hz with an unterminated line (open circuit). My guess about 75% of 1/4 wavelength because of the slower propogation in a cable than in free space. 1/4 wavelength at 60 Hz is 186,000 / (60 * 4) = 775 miles.

An open circuit transmission line at 1/4 wavelength reflects back to its input a short circuit. Losses in the line alter this. This is because the reflected signal is 180 degrees out of phase with the input signal.

mivey may be able tell us what a real world power distribution line with an open circuit at about 700 miles would do at the source. Was this part of the problem in the great eastern blackout?

To see the effect at the input of an open circuit transmission line with a low impedance source see my P1 photograph at http://beta-a2.com/cat-5e_photo.html . The waveform that starts up immediately at the left is the input to the transmisson line. The waveform starting at about 200 nS is the wave at the end of 150 ft of CAT-5E cable.

The free space velocity of light and radio waves is about 982 ft per microsecond. To travel 150 feet would take 0.153 microseconds. Thus, the propogation velocity of the electric wave in the CAT-5E cable is about 77% of the free space rate.

When you look at the input end waveform you see a big rise at about 400 nS. This is the result of the reflected energy from the open end of the transmission line.

Wires in a conduit filled with water will probably somewhat increase the capacitance per foot.

.
Wavelength is one of the factors in using DC for long HV power lines. More voltage using DC for given insulators strings is another. DC terminal stations synching with local wave timing a third. This (DC) can prevent reclosing problems that occur when you reclose an AC HV power line that is linking a generation-rich island with a load-rich island. The generation island speeds up, thus altering phase position, while the load island slows down -- BANG!
 

gar

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Location
Ann Arbor, Michigan
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EE
160121-1412 EST

steve66:

I belive the Sq-D criteria is based upon an assumed set of wires in a water filled conduit.


fmtjfw:

Phase differences can cause real big problems as you indicated.


Another experiment:

A 94 ft length of Belden CAT-5E cable. This experiment uses an adjustable sine wave source with a 50 ohm output impedance. The cable was open at the far end, an infinite impedance load. A cable pair was directly connected to the sine wave source, and a Rigol scope was used for voltage and frequency measurements.

As one adjusts thru the resonant frequency there is a pronounced drop in the transmission line input voltage. With no load on the source the voltage peak was 4 divisions and at null, resonance, 1/2 division. The resonant frequency was 1.77 MHz, or a full cycle period of 0.565 microseconds.

Cancelation occurs at the transmission line input when the line is electrically 1/4 wavelength long. Our two way travel time for 2*94 = 188 ft is 0.283 microseconds. It takes light in free space 1*188/982 = 0.191 microseconds to travel 188 ft. Thus, the velocity of wave propogation in this Belden cable is about 48% slower than the free space speed of light.

The approximate impedance at the line input is 50*1/7 = 7 ohms at 1.77 MHz.

.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
160121-2434 EST

Phil Corso:

Try truely to get two wires that are so balanced that there is no leakage outside of that supply-return path --- resitive, capacitive, or inductive.

The real problem is the real world and that is why, if my memory is correct, that Sq-D or someone else, did tests with wires in a water filled conduit.

.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
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My reading of the first post was that two questions were ask relative to a 3 wire wire cable. Romex with 2 #12 current carrying conductors, and a centered EGC wire.

One question related to tripping a 20 A breaker.

The second question related to tripping a GFCI device.

.
 

mivey

Senior Member
160121-1003 EST

mivey may be able tell us what a real world power distribution line with an open circuit at about 700 miles would do at the source. Was this part of the problem in the great eastern blackout?
I'll try to look today at a long circuit.

IIRC, The blackout was due to oscillations in the system due to dropped load, dropped generation, dropped load, etc. The automated systems were cascading and reacting faster than we could manually stop them. Anti-cascade was mostly a manual effort and by time a person figures out what is going on it is all over or developed even further.

We are supposed to have better systems in place today. Knock-knock-knock.
 

Smart $

Esteemed Member
Location
Ohio
Gentlepeople,

Unless there is a third wire (EGC) nothing will happen! The GFI compares phase to neutral current. If different it interrupts the circuit!

Regards, phil Corso
Doesn't take a third wire specifically, but rather one or more alternate pathways. However, the EGC was included in the discussion from the start...
Can you run a wire say ( 12-2wg) long enough to trip a 20 amp or any gfci with no load on it ?
 
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