GFCI current balance question

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danickstr

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I was wondering if an engineer could tell me how quickly the current balances in a GFCI circuit since it seems to me that the load fluctuating from a drill motor for example would be tripping the thing, although it doesn't. Also, why does the GFCI read current as equal, when that to me would imply that the energy is not being used in the device but rather just floating downstream to the POCO return.

In my mind, it seems that a 120v SP 15 amp device is using 15 amps. So how can 15 amps show on the neutral if it was consumed in the device?
 
OK sure, that is a good point, I did use the wrong term, but if the current and voltage is x watts going out to the load, then how can it be the same coming back? Since the voltage and current measure the same on both leads.
 
The balance of current between the energized (ungrounded) conductor and the grounded conductor occurs at the speed of light (speed of electrons). The GFCI doesn't have a chance to see the 'in' different than the 'out', unless the out is going someplace it shouldn't.
 
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If you measure voltage from ungrounded ('in') to grounded ('out') you see 120V, if you measure grounded ('out') to grounded ('out') there is 0 volts.

Remember it is not really 'in' and 'out', because the voltage and current are constantly going 'forward' and 'backward' 60 times per second. The 'in' and 'out' is just an analogy.
 
I'm not sure if you understand the function of a GFCI. It doesn't return to current balance.

A GFCI device compares the current on the hot conductor with the current on the neutral. When it sees a difference between these values it will open the circuit, stopping current flow.

Why these values might get unballanced would be if you get a shock from the plug. Instead of all of the current returning on the neutral wire, some of it is passing thru you to ground. Having a GFCI device then will minimize the shock by opening the circuit.

15 amps should show on the neutral if the hot also has 15 amps. If there were zero amp flow then your drill wouldn't work.
 
danickstr said:
OK sure, that is a good point, I did use the wrong term, but if the current and voltage is x watts going out to the load, then how can it be the same coming back? Since the voltage and current measure the same on both leads.
Click to expand...

Current flows in a circular path, out to the load then back to the source. current does not stop anywhere. In a normally working circuit the path is "out" on the black wire and then "back" on the white.
 
I did not know that current was the same on both sides, but now it makes sense that BW circuits would get brighter on the other side when a load was present.

But at the speed of light/60 times per second (3,000 miles?), it seems that the electrons would have to travel a long way and back again. Amazing that they know which way to go through the house neutral system.
 
danickstr said:
But at the speed of light/60 times per second (3,000 miles?), it seems that the electrons would have to travel a long way and back again. Amazing that they know which way to go through the house neutral system.
Click to expand...

Technically, current flows near the speed of light. The electrons move a lot slower. An example I found:

"For a copper wire of radius 1 mm carrying a steady current of 10 amps, the drift velocity is only about 0.024 cm/sec."

I think the best way to visuallize it would be that toy with 5 balls hanging in a row. You drop the ball on the end (ball 1) and when it hits (ball 2) the energy is transfered thru the next few balls even though they don't appear to have moved. The ball on the other end (ball 5) moves because it can't transfer the energy anywhere.

Since this is AC, the electron flow keeps changing direction back and forth.
 
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