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timberfinn

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I am taking this HVAC class and have been studying up on the basics of electricity, magnetic induction, AC/DC, Ohms Law etc. I have a basic question concerning the flow of current in the neutral wire of a single phase 120v circuit. I ran a few questions by my instructor (Mr. Samuels). He was a tech on a nuclear sub for 7 years. He checked out the internet for answers to my questions since he admitted to me that he was stumped. What he showed me from the internet was actually lame ... did not address my questions. One of the questions was that I noticed that Ohms Law seems to contradict Kirchoffs Law in that there is the same amount of amps in all parts of a series circuit (Kirchoffs Law) but typically a voltmeter shows zero voltage in the neutral wire of the circuit going to ground. Ohms law states that Amps = volts divided by resistance. Obviously, if there are zero volts, per the meter, then amps should be zero also. But that is not the case. There are always amps per Kirchoff's Law. But common sense tells us that you cannot have current flow (amps) without pressure (volts). Mr. Samuels is stumped. He asked the master electrician for the whole compound, a Mr. Sword. Apparently Mr. Sword was stumped too. Several inquires by Samuels to different websites/blog sites have failed to answer my question as well. The question being "How can you have significant amps (per the amp meter) in the neutral wire of a single phase 120v circuit and yet have zero volts/pressure (per the volt meter) pushing that current. Mr. Samuels has said that he could not disagree with me but could give me an answer. Also if there is zero voltage at the end of the load, what powers another device if you add one in series?

Any help would be appreciated.
 
No breaking of ohms law or kirchoffs
Its all the calculations and the example.

The examples in theory application are based upon prefect conductors ie no resistance. So there will be no volt drop across them. V=IR =Ix0= 0

Now in real life that length of cable you measure will have a resistance even on the neutral cable and will have measurable voltage drop across it you just need a better multimeter to measure the volt drop. That voltage drop will be tiny compared to your desired load.

For instance a meter length of 1mm2 copper will have a resistance of 0.021365ohms so say a 32 amp current flow will only drop 0.68v

This why we always wire with the same size neutral as earth.
 
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In the case of two the loads they spilt the voltage so presuming the two loads are the same you would get 60v across each load rather than 120v across a single load
 
Electricity is like water, it can only flow when it has potential. So water at the top of a slope has potential to flow down. It won't flow sideways because sideways is at the same potential as it is already.

Similarly electric can only flow from a high to a lower potential. If you take your voltmeter and a live conductor and place your probes at two points along that conductor it won't show a voltage because the two points are at the same potential. Ie. you are not giving the electric anywhere to flow to.

It's the same with the neutral. If you put one probe on your neutral and the other on another neutral or the earth, then you won't get a reading because there is no where for it to flow. You are asking it to go sideways and it can't.
 
seems to be some confusion between volts and amps. ohm's law states V=IR as we all know. V means potential difference, so I=V/R where V is not just volts, it's the difference between the voltage on the hot wire and the voltage on the neutral wire. e.g. 120V on hot wire and 0V on the neutral gives a potential difference of 120V. therefore current will flow , it's magnitude being inversely proportional to the resistance ( or impedance in an A.C. circuit ).
 
Single phase supply -if really on its own ?
In isolation the pd in both Phase and Neutral ,
should be the same ( actually measuring them may be another story -choosing some reference point and -adding a load).
--- Now a big but ,
Most supplies have at least 1 more "Phase" involved .
(in the uk we have 3 that overall sum to zero)
The distribution network -saves money in copper having Opposing phases that when
balanced will lower copper losses in the neutral wire. (due to opposing currents from different loads / phases within the network.
...Laundry appliances for USA are typically 240V and use 2X "Hot" wires (!) for their power. (not considering Earthing today).
The neutral wire coming to your house ,
may well be closer to having no volt drop in it , but the cause is not 1 single phase user ,but the sum of several users -balancing .
 
Aren't we confusing things a bit here ? Back to basics ... it will probably help to draw a circuit as we go, starting with an idealised AC generator (wiggly line inside a circle) connected by two wires to a resistive load -see the diagram in the section on Kirchoff's Voltage Law in this wikipedia page Kirchhoff's circuit laws - Wikipedia - https://en.m.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws (though that's using DC, it doesn't matter).
Yes, Kirchoff says that the current in the neutral will be the same as the current in the line - assuming no leakage paths. No problem there.
Ohms law says that V=IR, and this is the bit the OP is getting confused about.
Draw out the whole circuit - generator, 2 wires, and load. Ignore earth for now. Measure voltage between any 2 points in that circuit and Ohms law will apply. Measure along either of the cables (points a-b or c-d) and the voltage will be low because the resistance (R1 or R3) is small, measure across the load (points b-c) and the voltage will be high (almost the same as the supply voltage) because the resistance (R2) is large. Also, Kirchoff's 2nd law tells us that the voltsges across R1, R2, and R3 will add up to the generator voltage.
Hopefully everything is clear so far ?

Now what happens when we introduce "earth" - whether that be the mass of soil beneath our feet or the "tin can" of a submarine. In the absence of any connection to it, nothing happens !
Now connect the neutral of the supply (point d) to earth, and again nothing happens. There's no current flowing through earth so no voltage drops. Measure between any point in the original circuit and earth and you will get the same voltage as measuring between that point and the point where the supply neutral is earthed.
So we get back to the original query - why is there no voltage between neutral and earth at the load ? It should be clear now that there will be a small voltage - the same small voltage that you'd measure between neutral at the load and neutral at the generator.

Naturally real life is a bit different because of the massive mesh of different earth connections and parasitic currents flowing through them - especially in a PME environment. But for the purposes of explaining why the OP is seeing the results he does, we can ignore that.

All clear now ?
 
So we get back to the original query - why is there no voltage between neutral and earth at the load ? It should be clear now that there will be a small voltage -
....
All clear now ?
In some UK situiations your so called Earth reference point .
Is actually created by TNC-S .
The earth point is created at the customers premises by hanging it on the Neutral.
(I have no idea if a similar thing is done in the US )

( Broken Neutral worries - pose a problem for earthed equipment used outside -Several alternate earthing options exist)
 
I'm very surprised your Master Electrician could not explain this. Perhaps he misunderstood the question?
 
From what I've found, and I'm probably wrong as I've only dealt with a dozen or so electricians in the US directly.

However, the training seems to be to learn the "code", rather than the theory behind the "code".

They'll quote regulation numbers like rainman, but don't really understand why aside from the obvious ones.

UL don't help, the amount of times I've argued directly with them, on points where reasons behind their regulations break simple laws of physics numerous times.
 
Ohms law states that Amps = volts divided by resistance. Obviously, if there are zero volts, per the meter, then amps should be zero also.

Usually when established laws seem contradictory and you are using them correctly, you discover that you are not applying them to the same question. Ask yourself: Zero volts between which two points? What is the current flowing between those two points?

There might be zero volts between neutral and ground at the panel, but you are not measuring the current flow between neutral and ground. You are measuring the current flowing along the neutral cable, and there will be a corresponding (small) voltage between one end of it and the other, in agreement with V=IR.
 
QUOTE - if there is zero voltage at the end of the load, what powers another device if you add one in series?

Current powers the load NOT voltage, your measuring voltage, put an amp meter in the neutral line, then you will see some action.

little grass hopper needs much learning !
 
In some UK situiations your so called Earth reference point .
Is actually created by TNC-S .
The earth point is created at the customers premises by hanging it on the Neutral.
(I have no idea if a similar thing is done in the US )

( Broken Neutral worries - pose a problem for earthed equipment used outside -Several alternate earthing options exist)
In some UK situiations your so called Earth reference point .
Is actually created by TNC-S .
The earth point is created at the customers premises by hanging it on the Neutral.
(I have no idea if a similar thing is done in the US )

( Broken Neutral worries - pose a problem for earthed equipments this used outside -Several alternate earthing options exist)
Static zip in the US the power company runs 3 wires to use for domestic it has bare wire with it. We take a #6 bare wire attach to the power company wire, we bond that wire to the panel, we drive 2 10 foot ground rods at least 6 foot apart and bond both the rods. If it has a gas line bond it and galvanized water pipe
 
QUOTE - if there is zero voltage at the end of the load, what powers another device if you add one in series?

Current powers the load NOT voltage, your measuring voltage, put an amp meter in the neutral line, then you will see some action.

little grass hopper needs much learning !
Kirrchoff's law for series circuits states that the sum of the volt drops across the series loads will be the same as the generated voltage.
The measured Voltage across the generator will equal the sum of the voltages across each of the loads/resistances. Add an extra resistance in series and sum of the new voltages will equal the generated voltage.
The current flow through the whole circuit will be the same through each load.
 

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