Induced voltage causing GFCI breaker to trip

[gar]

140920-1212 EDT

Some experiments:

Using a 250 ft coil of #14 romex with EGC. This was not uncoiled.

Capacitance @ 1 kHz:
0.0050 ufd D=0.22 --- Black to white with EGC floating
0.0066 ufd D=0.28 --- EGC to white with black floating
0.0072 ufd D=0.26 --- EGC to black with white floating

Voltage --- 10 Megohm input impedance:
0.019 V --- EGC to white residual with 47 ohm meter shunt
26 V --- EGC to white with 124 V from EGC to black, not easily predicted.
60 V --- EGC to white with 124 V from white to black, expected.
0.030 V --- EGC to white with 47 ohm meter shunt 124 V white to black

0.030 V --- EGC to white with 47 ohm meter shunt 124 V white to black and a 10 A load at the far end.

Note the secondary loop is about 1/2 the area of the primary loop.

0.03 V across 47 ohms is less than 1 mA. This is in the mud so we can not tell much, but it is not a large current.

I do not want to uncoil the loop and try to do a more controlled experiment.

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[gar]

140921-2128 EDT

An additional experiment:

I cut a 25 ft coil of 14-2 w/g romex in half. The EGCs are left floating. The two pieces are taped together.

One piece that I shall call the primary has a switchable 10 A resistive load at the far end. A short AC cord is connected to the primary input end to apply 120 V nominal to the primary input end.

The second romex piece I shall call the secondary. An AC voltmeter is connected to the secondary input end.

First, the secondary residual open circuit voltage with no input voltage and nothing grounded is 0.49 V.

Second, with 120 applied to the primary input with no load and the secondary open at both ends the voltage reading is 20.3 V. Now via the AC plug the pieces of romex have their white wires connected to earth.

Third, the 10 A load is applied. no change in the secondary voltage.

Fourth, without the 10 A load and the secondary shorted at the far end the secondary voltage is 0, with or without the primary connected to 120 V.

Fifth, with the 10 A load, the secondary far end shorted, and the secondary input open circuit the secondary input end reads 0.0029 V.

Sixth, same as fifth except meter input is shunted with 47 ohms. Same result. Secondary current is no greater than 2.9/50 mA.

At 500 ft I might expect 40 times this voltage or 0.116 V. The current thru a 47 ohm resistor would be 0.0025 A or 2.5 mA. I believe this is very much a voltage source until the 47 ohm shunt resistor is made much smaller.

Twisting the secondary pair will greatly reduce the magnetically induced voltage. Twisting both cables at different pitches will also reduced the coupled voltage.

in this experiment the two cables were tightly coupled, but the individual wires are close together. A wider spacing would produce more flux coupling.

I have not proofread for mistakes. You can run your own experiemnts.

.

 

[Sierrasparky]

140921-2128 EDT

An additional experiment:

I cut a 25 ft coil of 14-2 w/g romex in half. The EGCs are left floating. The two pieces are taped together.

One piece that I shall call the primary has a switchable 10 A resistive load at the far end. A short AC cord is connected to the primary input end to apply 120 V nominal to the primary input end.

The second romex piece I shall call the secondary. An AC voltmeter is connected to the secondary input end.

First, the secondary residual open circuit voltage with no input voltage and nothing grounded is 0.49 V.

Second, with 120 applied to the primary input with no load and the secondary open at both ends the voltage reading is 20.3 V. Now via the AC plug the pieces of romex have their white wires connected to earth.

Third, the 10 A load is applied. no change in the secondary voltage.

Fourth, without the 10 A load and the secondary shorted at the far end the secondary voltage is 0, with or without the primary connected to 120 V.

Fifth, with the 10 A load, the secondary far end shorted, and the secondary input open circuit the secondary input end reads 0.0029 V.

Sixth, same as fifth except meter input is shunted with 47 ohms. Same result. Secondary current is no greater than 2.9/50 mA.

At 500 ft I might expect 40 times this voltage or 0.116 V. The current thru a 47 ohm resistor would be 0.0025 A or 2.5 mA. I believe this is very much a voltage source until the 47 ohm shunt resistor is made much smaller.

Twisting the secondary pair will greatly reduce the magnetically induced voltage. Twisting both cables at different pitches will also reduced the coupled voltage.

in this experiment the two cables were tightly coupled, but the individual wires are close together. A wider spacing would produce more flux coupling.

I have not proofread for mistakes. You can run your own experiemnts.

.

Interesting , something to try myself.

 

[gar]

140922-1156 EDT

A correction to my post numbered 22.

The statement
"Sixth, same as fifth except meter input is shunted with 47 ohms. Same result. Secondary current is no greater than 2.9/50 mA."
is incorrect.

It should read something like:

"Sixth, same as fifth except meter input is shunted with 47 ohms. Same voltage result.

Secondary current will increase as the shunt resistance decreases."

In the experiment of yesterday with 47 ohms the measured current was 2.9/47 = 0.062 mA. A measurement today with 2.7 ohms for the shunt produces a current of 2.8/2.7 = 1.03 mA. Using a 10 A 50 mV shunt, .05/10 = 0.005 ohms, the voltage is 0.2 mV, or a current of 0.04 A, 40 mA. The shunt resistance is now below the source impedance of the secondary.

The equivalent impedance of the GFCI current transformer may have a significant effect on what happens in the actual circuit in combination with the line impedance.

I do not believe the original post indicated any load current on the circuit that was first energized of the two circuits. No load current, then no magnetically induced voltage.

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[gar]

140922-1329 EDT

Another useful calculation to perform is to assume the open circuit source voltage of my secondary is about 0.0029 V. Then for a wire resistance in the secondary of about 2.5*25/1000 = 62.5/1000 = 0.0625 the short circuit current is 0.0029/0.0625 = 0.046 A = 46 mA. The above is with primarry input of 120 V nominal to a load that I measured today of 13 A. Previously I assumed the heater was 10 A, but the heater I am using now is different than what I have previously used.

If we increased the pairing of these cables from 12.5 ft to 400 ft, and assume the primary current is constant, then what do you expect the short circuit current to do?

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