## Electrical 101 question.. – page 3 – electrician talk – professional electrical contractors forum u save gas station grants pass

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So this is what I’m having trouble with. Voltage is present within the ungrounded conductor, which helps push the current through the conductor. When it gets to the appliance, voltage drops to 0? Then the current is somehow able to flow through the neutral conductor without the presence of the voltage to help it flow through it? That is what I cannot even fathom

Voltage is similar. You’re still not seeing voltage as BETWEEN, voltage is only meaningful BETWEEN two points. Note that meters have TWO leads for voltage. You don’t have voltage at a point, you have voltage between two points. You’ll notice there are no one-wire electrical services, or transformers that only have one primary terminal, etc. etc. etc.

Consider the circuit with several bulbs in series on a battery. Check voltage with both leads of your meter on any one piece of wire you’ll see ZIP. That is because any two points on a wire are called "equipotential" – they would all be at the same voltage. Likewise any point on the neutral wires, neutral splices, any point on any EGC or any EGC splice, the main bonding jumper, the GEC, the ground rod, the ufer terminal – all equipotential. When things are bonded they are equipotential.

Again, I suggest you’re reading too much into the language, you’re thinking too much about how the words are used in regular English, that isn’t going to work out for you. Also, the analogies are not perfect for voltage, it isn’t quite just like water pressure or etc. etc.

Also, my journeyman taught me just last week about resistors and transformers and how meters and things like CT’s will basically convert amps to a readable amount, that there are coils inside that act as a resistance? Thus will lower the amount of amperage based on how many coils there are. Anyways

But I’m still confused about why you wouldn’t read 0 from neutral to your hot when everything is made up. If your voltage acts as a force and your amperage is flowing equally throughout the complete circuit, why would it not be the same as your hot? Why would you not read 120 from neutral to ground? This is what I’m super confused about now

You must study AC theory. In AC theory , voltage and current are always changing, between 2 points (Hot and Neutral ) but they always average the phase voltage and current , and that’s is 120 volts. So no matter you measure voltage between hot and neutral, even switch the prong of the

the same , Neutral are wire straight to the service panels and bonded (connected) by a main bonding jumper with the ground, so there is no voltage difference. It is like measuring voltage of a staright conductor end to end and you get no voltage difference

This has to be one of the most entertaining posts I’ve read in a long time. OP all your questions were answered in the posts. I understand your question, If you have voltage on the line side neutral to ground with appliance neutral open then the N is being shared somewhere else with a live load. At the same time if the appliance is turned off you will not get voltage between H & N but will with G if you turn on appliance you will get a voltage to N but it will not be the same as to G, if I understand you right that is what’s confusing you. Way to much to explain but try to google the purpose of neutral in electrical system and get a little more in depth on theory. That neutral can in some cases kill you just a the hot wire can if you are not careful.

No that part I understand, an open neutral. On the load side of the neutral if the appliance is on, you will read 120 volts from ground to neutral, you will also read 0 volts from hot to neutral because there is no potential difference. If you ‘un-plug’ the appliance, you will read 0 volts from ground to neutral and will read 120 volts from neutral to hot.

What I don’t understand is when the appliance is plugged in, and everything is made up properly, why you read 120 volts from hot to neutral. So I understand in this scenario, you will still read 0 volts from neutral to ground (b/c they are bonded), but there must be voltage on the neutral wire for current to flow through it, and also there must be voltage on the hot for the current to be flowing through it, so when you put your meter up to your hot and your neutral, why do you read 120? Shouldn’t it read 0… That is what is blowing my mind lol

Edit: One last thing, this is the thing that I’ve been confused about all along. So when it hits your neutral, it is essentially at 0 volts. But the current still flows through the neutral conductor, so don’t you need voltage to help force current through the neutral wire, so if that’s the case I don’t see why it would be at 0 volts, wouldn’t it have to be higher voltage so current can flow through the neutral?

Very good question. There is a voltage drop across the neutral conductor, that is to say, the voltage isn’t truly at 0v at the load side of the neutral wire, since that would imply there is absolutely zero resistance from that point, down the neutral, all the way to absolute ground. If that were the case, there would not be any current flow as you suspect. The voltage drop from that point to absolute ground is very low, but the resistance is likewise very low, therefore the current can remain consistent with the current going through the complete circuit (current remains the same for the entire circuit). Another way to think about it is that since the voltage drop in the complete circuit is proportional to resistance, and there is a tiny amount of resistance down the neutral wire, there must therefore a tiny amount of voltaic potential remaining on the load side of the neutral wire and this is what drives the current down that line.

Let’s say you have 120v source, 100 ohm load, and 10 ohms in the source and neutral wires (numbers chosen for simplicity sake). Total resistance will be 120 ohms (100+10+10), total current will be 1 amp(120v/120ohms). Voltage drop across the load will be 100v, and 10v across each cable.

You can still apply Ohm’s law at the neutral cable, 10v=10 ohms * 1 A. In reality, the resistance and voltage drop will be much less, but the idea is that yes, the voltage drop on the neutral side is very low, but so is the resistance, and therefore the current remains high and consistent with the rest of the circuit.