Current how it works.

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kjroller

Senior Member
Location
Dawson Mn
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Master electrician
If I have a 120v heater and I cut the Neutral so current can't make its way back to source will the hot wire still have any current on it. From how I understand it the answer is no as it won't trip a GFCI breaker as there is no imbalance. Let me know what you think thank you.
 
Basically, what I am asking is if I intentionally open the neutral and put my amp clamp on the hot with a load attached will I see current on the hot.
 
Certain conditions you could have a slight reactive current due to the hot conductor capacitive coupling to a neutral or a ground. But that would not be enough to matter.
 
Would be safer if you disconnect the energized conductor on a 120 volt heater or other systems. Somebody could look at your heater thinking power is off and get shocked Have replaced a GFCI that did not trip. If you disconnect the energized wire in a 120 volt system there can never be any current flow. If you only disconnected the neutral conductor and you had water intrusion you could have leakage current to ground.
 
In an open circuit there is voltage, but no current. That is why voltage is called “potential”. It has the potential, but until the circuit is completed, there is no current.

Think of current as the volume of energy.

If you had a simple 120 L-N circuit protected by a GFCI breaker, and you snipped the neutral conductor, you would have 0 current flow, not an imbalance of current.
 
Seven-Delta-FortyOne said:
That is why voltage is called “potential”. It has the potential, but until the circuit is completed, there is no current.
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Not a bad way to think of it. But the science behind the term goes deeper. Please allow me to offer this, if only for its entertainment value:

Consider a positively charged particle sitting somewhere in space. Point "A" is some distance away. Point "B" is further away along the same line. Also along the same line, much farther away (we don't need to know exactly how far), is another positively charged particle. The two positive charges repel each other, so it will require some energy to move the second particle closer to the first particle.

The amount of energy needed to move the second particle from wherever it started to point "B" is called the "potential energy at point B." I will abbreviate it as "PE-B." I will use "PE-A" to designate the "potential energy at point 'A' ." Clearly, PE-A is larger than PE-B, because the second particle would have further to travel, in order to make its way to point "A."

If you subtract PE-A minus PE-B, you get the "potential difference" between the two points. That "potential difference" is the basic meaning of "voltage." That is why we must always speak of voltage as a value taken between two points in a circuit.

The amount of energy needed to move the second particle depends of the amount of charge sitting on that particle. If the charge is higher, the repelling force it feels will be greater. Thus, the parameter "volt" is defined as "joules" (an amount of energy) per "coulomb" (an amount of charge).
 
charlie b said:
Please allow me to offer this, if only for its entertainment value:
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I was entertained…

I have used a similar description to explain the phenomenon of static discharge when you walk on a carpet with plastic soled shoes and touch a doorknob (or your bratty little sister). I know it’s not exactly the same, but similar.
 
Deltaforce said:
But how a dipole antenna work then?
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Gaps in circuits have capacitance, which means at high enough frequencies, they will eventually enable the circuit on one side of the gap to affect the other side of the circuit. Capacitors are gaps and no electron directly crosses that gap. But what really happens, is the electric field builds as the charges accumulate in equal and opposite amounts on each side. The aggregate effect of this, is called displacement current (because it has units of amps just like current), and produces the same magnetic field effects as if it were a current.

At low frequencies, capacitance is close to an open circuit while inductance is approximated as a short. Air gaps that don't arc, and standard insulation rated for the voltage, are good enough to consider an open circuit and not expect one side of the circuit to affect the other, in any significant amount. Not without a physically constructed capacitor, to intentionally add capacitance.

But at high enough frequencies, the behavior is opposite. Capacitance approximates as a short, and inductance acts as a gap (called a choke in trade slang). We can't assume a gap is a perfect gap anymore, and we have to consider how the capacitance of that gap affects the circuit, and its interaction with the rest of the impedance.
 
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