No current flow on a balanced neutral?

[ronaldrc]

Weress

I don't get it about the AC question, the neutral conductor would be magnetized regardless it would attract the iron filings to it if current was flowing.

Rattus and Glene

That is the theory of why the neutral works the way it does is because there are currents flowing and opposing each other at the same time.

 

[rattus]

Nope, not so:

Nope, not so:

Weress

I don't get it about the AC question, the neutral conductor would be magnetized regardless it would attract the iron filings to it if current was flowing.

Rattus and Glene

That is the theory of why the neutral works the way it does is because there are currents flowing and opposing each other at the same time.

Doesn't work that way Ronald. You can perform this experiment with batteries. Make a coil in the neutral and tweak the load currents until you read zero current in the neutral.

Don't use any iron because it will have some residual magnetism which will give a false result..

 

[ronaldrc]

Rattus

I no there is no current flow on a balanced neutral.

But I can not think of any way to prove it other than the heat it would generate in the neutral.

What do you mean it doesn't work that way?

 

[Besoeker]

Rattus

I no there is no current flow on a balanced neutral.

But I can not think of any way to prove it other than the heat it would generate in the neutral.

Why? Isn't just as simple as sticking an ammeter in series with it?
Or a CT round it?

 

[ronaldrc]

Besoeker

I think if we go by theory if we have the same exact load on lines #1 and #2 even if current was flowing in the neutral the magnetic fields created by the neutral would be equal and opposite and would cancel each other out.

I need to make it clear that I do not believe there is any current on the neutral conductor of a perfectly balanced circuit to the neutral.

But my question is can it be proved without a dought?

 

[Besoeker]

Besoeker

I think if we go by theory if we have the same exact load on lines #1 and #2 even if current was flowing in the neutral the magnetic fields created by the neutral would be equal and opposite and would cancel each other out.

I need to make it clear that I do not believe there is any current on the neutral conductor of a perfectly balanced circuit to the neutral.

But my question is can it be proved without a dought?

Stick an ammeter in series with it and the reading is zero Amps, then it would be reasonable to conclude that there is, in fact, zero Amps.

 

[glene77is]

Rattus,
Could this be a problem:
Harmonics could have been coming from one phase, and going out the neutral.
Ordinary Amp meters are not very sensitive to to higher frequency harmonics.
Use of an Oscilloscope would show harmonic voltage on the neutral.
Glen

 

[glene77is]

Ronald,
Your drawing is very clear.
A Balanced circuit will have no Neutral Current.
Given a pure 60 Hz signal.

However, just to complicate things, and make things more interesting,
let me add this comment:

IF given that harmonics are present, from one phase only,
THEN harmonics would pass through the Neutral conductor.
To observe these harmonics you may need an oscilliscope.

Most electricians do not get a lot of practice observing harmonics,
and may believe that simply connecting their Fluke or Wiggy
would give some indication,
although I find them not sensitive to high frequncy signals.

Having dealt with R.F. signals, and VFD hash, and Dimmer hash signals,
I expect the observation to require special equipment.

To the point,
I was tracing circuits in a warehouse, and the computer power supplies were making such a hash signal that my little yellow signal Receiver was tripped on by one circuit,
and also by the other circuit to which my little yellow Transmitter was attached.
So, I switched to a different 'wide band' tracer tool.
I could hear the hash, and read its relative amplitude.
The difference between the two circuits signals was readily distinguishable.
Clearly, the solution is to use appropriate tools for observing signals.

...

 

[Besoeker]

Rattus,
Could this be a problem:
Harmonics could have been coming from one phase, and going out the neutral.
Ordinary Amp meters are not very sensitive to to higher frequency harmonics.
Use of an Oscilloscope would show harmonic voltage on the neutral.
Glen

If it's a single-phase circuit, as given in the OP, the most predomininat harmonic is likely to be the third (or 180Hz in the US). A Fluke multi-meter accuracy is specified at up to 400 or 500Hz depending on model.

 

[glene77is]

Besoeker,

You are right. Fluke has a very useful meter.

I expect your Fluke meter could be enough for an answer in my scenario,
and during a Hands-On Experiment would answer the OP's question.

I bet the OP's question has been thought provoking for this group.
What I see is two similar perspectives, using two similar sets of words.

(1) The Kirchoff (equation) method answers itself.

(2) The Experimental method is still not tried,
has several variants,
but should provide its own type of practical answer.

...

 

[nosratrouhani]

No Current in Neutral wire

No Current in Neutral wire

There will be no current flowing in the neutral of a balanced 120/240V circuit. the circuit becomes a 240V circuit with the voltage equaly divided across each load (120V). you can remove the neutral and the circuit will keep working, this would be your practical proof.

 

[rattus]

Rattus,
Could this be a problem:
Harmonics could have been coming from one phase, and going out the neutral.
Ordinary Amp meters are not very sensitive to to higher frequency harmonics.
Use of an Oscilloscope would show harmonic voltage on the neutral.
Glen

Maybe, but consider this possibility. Consider a malfunctioning Triac where only one lobe is chopped. This would create a DC component of current which would not register on many ammeters. Something like a half-wave rectifier. You think?

 

[ronaldrc]

Rattus

That would be the same thing as pulsating DC are you sure a regular amp clamp wouldn't
read it? I know it wouldn't read it correctly but seems like it should read something, I have never tried it I don't know.

 

[gar]

081124-1934 EST

A pulsating DC signal will have an AC component to it. A true RMS will read the RMS value of the AC component. Take this pulsating DC signal thru a capacitor and the DC component is removed and the output is the AC component. To get a good representation of the AC component the cutoff frequency must be low enough relative to the fundamental of the AC component.

Many AC voltmeters have an input capacitor that strips the DC component. The Simpson 267-270 does not unless you connect thru the Ouput terminal.

.

 

[Besoeker]

Rattus

That would be the same thing as pulsating DC are you sure a regular amp clamp wouldn't
read it? I know it wouldn't read it correctly but seems like it should read something, I have never tried it I don't know.

A clamp-on ammeter would read, but maybe not give a very meaningful reading if the iron saturates.
I have sometimes done this on a three-phase SCR bridge.
It's a quick and simple method to show up a missing firing pulse which results in DC on one of the input phases.
The clue is that the jaws stick together and it is hard, or harder than normal, to open which is what happens if you put it in a conductor carrying a significant DC current.

 

[glene77is]

Gar,

If there is no Voltage across the Neutral,
and if the grounded end of the Neutral is at ground potential,
Then is the ungrounded end of the Neutral at ground potential?

So, then,
is the junction of the L1 and L2 (at the top of the Neutral)
also at ground potential?

In analog electronics, dealing with Operational Amplifiers,
this is called a 'summing junction',
where all currents balance to zero
(as fast as the active impedance of the IC chip can respond).

Any comments?
Or maybe, am I off the subject?

...

 

[gar]

081125-0756 EST

glene77is:

If there is no Voltage across the Neutral,
and if the grounded end of the Neutral is at ground potential,
Then is the ungrounded end of the Neutral at ground potential?

Yes.

So, then,
is the junction of the L1 and L2 (at the top of the Neutral)
also at ground potential?

I do not understand the question because I do not know where the drawing that you are referring to is located. However, if a point P1 is "ground" and this is your measurement reference point, then any other point that has zero potential difference from P1 will lso be at "ground potential".


In an operational amplifier circuit with negative feedback and very high gain under steady state conditions the difference voltage between the two input terminals of the amplifier is very small and therefore assumed to be 0. Also the input current to the amplifier is assumed to be very small. Thus, all input currents to the - input are equal to the feedback current and this becomes a summing circuit of the input currents. If the input currents are from resistors from voltage inputs, then the circuit sums the input voltages.

If the feedback element is a pure capacitor instead of a resistance, then the circuit is a mathematical integrator of the inputs until saturation occurs.

.

 

[glene77is]

The RMS of a Half-Wave rectified signal would be deceptive, if you expect a full wave signal, indeed.

Speaking of which, up on my shelves is a Voltmeter which I designed in the 80's.
Depending on the switch settings, the Vin passes through a pair of Active diodes, switchable for positive and/or negative. Then I switch between Peaking or Logrithmic or RMS amplifiers. I had thought it would be a useful project for measuring the vertical trace voltage in television circuits. (When I was young, I thought a lot of useless things!)

Anyway, your comment makes a good point
concerning the 'Unknown' characteristics of signals.

...

 

[gar]

081126-0832 EST

The RMS value of a half wave rectified sine wave is 0.707 (sq-root of 1/2) times the value of the full wave sine wave rectified or unrectified. Without calculus you can figure this out from P = V^2/R or I^2*R. Some might expect it to be 1/2, but the waveform is more peaked and thus greater power dissipation.

On the other hand if you use a Simpson 260 to make the measurement then the voltage should drop in half because the Simpson is basically an average reading meter.

.

 

[LarryFine]

On the other hand if you use a Simpson 260 to make the measurement then the voltage should drop in half because the Simpson is basically an average reading meter.

The same can be said of any standard moving-coil meter, which behaves like a motor.