# Voltage between two points on an empty wire

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1. Feb 9, 2017

### TemporaryMan1233

I'm a computer science student. I'm currently learning electronics so that I can program embedded devices effectively. I'm reading from the free book at allaboutcircuits.com.

Why is the voltage between two points on a wire in a closed circuit equal to zero? Note that between these two points there is no resistance; Think about it as a wire with no other electric devices attached in between these two points.

Here is an illustration of what I mean: (taken from allaboutcircuits.com)

My question is, why is the voltage between 1 and 2 (or, 3 and 4) zero, but that of 2 and 3 not? I can't understand how these two have different voltages.

Here is what I understand:
• The current in the whole circuit is affected by the resistance of the lamp, because the potential energy of the battery is trying to "push" electrons with a certain amount of force per unit charge, but the resistance of the lamp will make this push "slower". Think about electrons in the circuit as marbles arrayed along the wire. At any point, if there is a resistance, the whole "push" of electrons (i.e. marbles) in the whole circuit will be "slowed". That results in a lower current (i.e. the whole array of marbles which spans the wire will be slowed in movement).
• Voltage is the measure of electric potential energy per unit charge (Columb). It is a result of imbalance of electrons (one matter has an excess of electrons from another matter which now has a deficiency of electrons-that's why it measured between two points). This imbalance creates that electric potential energy, which is "actualized" in the form of a force pushing electrons back to their original matter to balance the excess/deficiency.

2. Feb 9, 2017

### cnh1995

Considering the wires to be ideal (zero resistance), the electrons need no energy to maintain their motion through the wire. But the lamp has a resistance and hence, some energy is required to maintain the current through the bulb. So all the voltage appears across the bulb.

This explanation may not be accurate, but I think it will give you a basic idea about voltage drop.

3. Feb 9, 2017

### TemporaryMan1233

I still don't understand how this fits with the definition of voltage.
I think I know what voltage is, but I can't understand what makes voltage zero and what doesn't.

Voltage across a battery is nonzero. Voltage across anything that has resistance is nonzero. Voltage across anything that has no resistance is zero. Why is that? Why is Ohm's law the way it is?

It seems that most people are satisfied by memorizing Ohm's law, which results in them proving physical behavior in a mathematical sense. (Wire has negligible resistance, V = RI, R ~ 0, thus V ~ 0)

Last edited: Feb 9, 2017
4. Feb 9, 2017

### cnh1995

Voltage V= energy/unit charge=W/q.
To move a charge from point a to point b, the electrostatic potential energy needed by the charge can be written as W=q*Vab.
As you can see, this energy is proportional to the potential difference between a and b.

In your circuit, this energy is zero for ideal wires, hence no voltage appears across the wires. All the voltage appears across the bulb. Because of the resistance of the bulb, electrons in the bulb filament require more electrostatic potential energy to maintain the same current as through the ideal conductors. Hence all the voltage appears across the bulb.

5. Feb 9, 2017

### TemporaryMan1233

1. Why is the potential energy zero for an ideal wire (in a closed circuit, which has running electrons)?
2. AFAIK, voltage gives a constant push of electrons to the circuit. If there is a lamp in the circuit, the push is the same but the current will be less, because the resistance in the lamp slowed the whole push of the electrons. Therefore, I do not understand your statement: Because of the resistance of the bulb, electrons in the bulb filament require more electrostatic potential energy to maintain the same current as through the ideal conductors.
I'm here to understand, correct me when wrong.

6. Feb 9, 2017

### cnh1995

Because there is no resistance. Imagine a ball rolling on an infinitely long frictionless surface. It will continue to roll with the same velocity forever. But if the surface had friction, it would stop after a while. So if you still want the ball to move with the same speed, you need to apply a force which is equal and opposite to the friction. You need force to "overcome" the friction. Similarly, you need voltage (electric field precisely) to overcome electrical resistance. You can think of voltage as an electrical analogy of mechanical force.
I don't know what you mean by 'push' here.

Battery establishes a constant voltage across the circuit. How that voltage gets divided among the components depends on their electrical resistances.

7. Feb 9, 2017

### Rive

Because one definition of 'ideal wire' is the zero voltage drop regardless of the current?
Kinda' first rule of the Tautology club.....

The 'same' here means: the same current which flows through the loop, and not the 'same as it would be with ideal wire instead'.

Intended to be constant current I think?

8. Feb 9, 2017

### cnh1995

No. Battery is a voltage source. Current depends upon the resistance of the network.

9. Feb 9, 2017

### Rive

That way it's too easy to misunderstand. Voltage is usually not constant 'across the circuit'. It constant only on the poles of the battery.

10. Feb 9, 2017

### TemporaryMan1233

As someone who has a weak base in physics, I'll go the mathematical way of understanding electronics. I can't handle physics anymore.

11. Feb 9, 2017

### Rive

Take it easy.

The literal answer might be a bit disappointing. It's because we engineers are lazy, so when the resistance of the wires does not effects the results we tend to replace them with 'ideal wires' to spare some time with the calculations.
And one definition of the 'ideal wires' is that regardless the current they has no voltage difference between any two points of them.

Ps.: usually it's also easier to explain the basics this way.

Last edited: Feb 9, 2017
12. Feb 9, 2017

### cnh1995

I was imagining the network as a black box with two input terminals directly connected across the battery. So 'across the circuit' and 'across the battery' were the same thing in my head.
Maybe you should refer some good physics book containing pictorial explanations of the equations. It is a bit difficult to explain circuits here from the very basic level (at least for me). You can study circuits using a good book and ask some specific questions here if anything is unclear.

I believe science advisors here will be able to explain better.

13. Feb 9, 2017

### TemporaryMan1233

I'm reading the book on allaboutcircuits.com. What book would you recommend for a programmer?

14. Feb 9, 2017

### cnh1995

As a beginner, you can refer Halliday-Resnick or Serway-Jewett.

15. Feb 9, 2017

Temp - I really think your hang up here is the "ideal" vs "real" --- in ALL real cases there is a voltage drop if there is any current. The purpose of using an ideal model is this Wire Voltage drop tends to be much lower ( factor of 100 or greater) than the voltages applied or drop(s) by the other circuit elements, or even the variance of the values of the other elements. For example a typical resistor will have 2 to 10% variation, there are higher precision ones, 6V battery has an internal resistance, and the bulb - the 3W is probably more like the Upper limit for the product, and the variation is close to 2.6W to 3.0W

However for most circuits delivering power or considerable current, the V drop in the wire is considered - like jumper cables, or household wiring, and this affects the correct or best choice for the size of the wire used.

In your circuit case you could have 6 V battery and small 3 Watt lamp, in fact this would be a very common early experiment. In this case the bulb when on, will use 3W/6V = 0.5 Amps of current - note the bulbs resistance increases dramatically as it is turned on, so when on ( @ 6V) the bulb has a resistance of V/I = 6V / 0.5A = 12 Ohms.

Use something silly small for the wire, like 30 ga wire, you have 0.103 Ohms resistance ( real) per foot of wire. In the circuit above ~ 1% of the total resistance if added to the lamp.

Typical lab hook up wire, like you would use for the circuit above, like 20 Ga solid copper, the resistance drops to 0.099 Ohms per foot ( 0.1%) . Something unnecessarily large - 10 Gauge 0.0099 Ohms per foot - 0r ~ 0.01% - one part in 10,000.

16. Feb 9, 2017

### TemporaryMan1233

I already know that there is resistance in everything in real life.

I'm not trolling but, how can it be that there is no simple explanation to my question...?
All those electrical engineers have taken the definition of voltage and never really understood it. I call it understanding by routine, which is really bad.

17. Feb 9, 2017

### TemporaryMan1233

Is this thinking valid:
Across the wire, there is no imbalance of electrons, and therefore there is no electric potential energy, and thus no voltage.

But now the question is, why is there a voltage across the lamp? Is there an imbalance of electrons at each pole of the lamp? (I think not!)

18. Feb 9, 2017

### cnh1995

That's not true. In fact, it is not possible to pursue engineering without understanding such a fundamental thing.
If you are interested in the exact mechanism of electrical conduction, look up 'surface charge feedback mechanism'. There's plenty of material available on the internet. (Refer 'Matter and Interactions' or D.J. Griffiths if you want to study from books) It will tell you why and how the voltages develop across the circuit elements (charge imbalance gives rise to surface charges on the components).
For energy transfer process, look up 'Poynting vector'. These two theories will answer most of your questions.

Meanwhile, I will ask some qualified science advisors here for a better intuitive explanation. I have my limitations.

19. Feb 9, 2017

### cnh1995

Just because one can't explain something satisfactorily doesn't mean they don't understand it.
Everyone here is trying to help you with your problem. If you expect us to teach you what voltage is from scratch, it is surely going to confuse you because everyone has their own way of explaining things.

Here is a thread from past about the same question. See if it helps..

20. Feb 9, 2017

### TemporaryMan1233

I appreciate your help. I did not mean that you don't know what voltage is.

21. Feb 9, 2017

### jim hardy

I think your trouble stems from definitions. I dont think your definition of voltage is correct.

This is close but not quite right
That's it's units allright
but it's actually a simpler concept than that.

And Rive is right, when we lazy engineers get accustomed to using it we broad brush past the details.

Here's how to think about it

Voltage is Potential Difference. Two words, not one. Difference is the easy one.
So what's potential?
Potential is the work required to bring a unit positive charge from infinity to wherever you're measuring potential.
That takes some thought.
Imagine yourself at Alpha Centauri(close enough to infinity for demonstration purposes)
with a one Coulomb sized bucket full of charge,
a force gage, and a ruler.
I grew up with dynes and centimeters but Newtons and Meters are easier....

Now start walking toward earth, measuring the force in Newtons exerted on your bucket of charges and tabulating it at every meter along the way.
So as you move toward earth you're going to tabulate the Newton-meters and keep a running sum.
When you've reached the top of your lightbulb you will write there what is that point's potential. That'd be its absolute potential.
Now repeat but this time walk to the bottom of your light bulb and again write its absolute potential.
The difference between those absolute potentials is the voltage across your light bulb.
When i grasped the concept was the day I imagined myself counting dyne-centimeters all the way from Alpha Centauri to my workbench in Miami Central High School's electronics lab, ca 1962 .

Now since we can't get to Alpha Centauri let alone infinity it's completely impractical to do that measurement,
and that's why we never know what is the absolute potential of anyplace.
So we just have to settle for the difference in absolute potentials between two places we can reach.
That's easily measured with a two wire voltmeter provided its leads are long enough to reach our two points of interest.
That difference in absolute potentials is "VOLTAGE" . The voltmeter reads that.

Whatever is the absolute potential at one end of a battery, it's different at the other end by whatever is the voltage of your battery. We can only measure that difference.

That's voltage. Forget about clouds of electrons.

Now, an electric field will cause charges to migrate along the field if they can. Inside a copper wire they migrate easily so a miniscule field will cause quite a bit of current . That's why the voltage between ends of a wire is miniscule, charges move equalizing local charge densities along its length..

This oversimplified layman's explanation should help you make sense of the concept. I don't mean to come across anti-academic; au contraire.
Don't just memorize formulas, understand what's happening and they'll become intuitive.
Looking up definitions is always a good idea - laying the foundation if you will. Then use your imagination to link them to your everyday experience. That's called "Memory Pegs" .

Working inside circuits is different from electrostatics, we have simplifications like no field along a wire and V=IR neglecting magnetic induction.
Poynting Vectors and Magnetic Vector Potential come later, after you've got used to working inside circuits.

Apologies for being so basic in an academic forum, corrections to any of above are welcome.

old jim himself

Last edited by a moderator: May 8, 2017
22. Feb 9, 2017

### jim hardy

23. Feb 9, 2017

### sophiecentaur

The thread has strayed outside the terms of the OP - who is not a Physicist but appears to want to get a couple of fairly basic things sorted out.
The reason for this is that no energy is expended in getting charges from one end to the other of an ideal wire. The energy that's expended in a load can be described in terms of the Potential Difference between the two terminals. That's how Volts (PD) are defined. Getting into the harder business of detailed descriptions of mechanisms is probably more than the OP wanted.
Why would you expect it to be simple? The rest of Physics is pretty hard if you want secure explanations of more or less anything.
People who can't do the routines correctly are never going to have a good understanding. It's so easy to forget that all the things you (and I) consider that you 'understand' were actually put together in your brain from stuff you had already 'learned', sometimes by rote or from familiarity. A 'simple' treatment involves accepting a few basics and delivers good answers but it won't lead to 'understanding. Isn't that like the rest of our lives? It's all a matter of layers and you make a choice.
You should just limit your 'explanation' to the relationship between the variables. Expressions like "trying to push" and "make this push slower" do not actually constitute an explanation or a description of the processes. Anthropomorphising doesn't often help in Science. Learn the routines and use them and, funnily enough, it all starts to make good sense - to whatever level we're talking at the time.

24. Feb 13, 2017

### Inventive

Keeping things simple, the resistance of the wire between points 1,2 or 3,4 is zero (in the ideal case) volts because the resistance is very small very small. The resistance will depending on the IR drop of the wire between those pairs of points. The farther
apart they are, the greater the IR drop. Electron flow decreases reducing the current with a increases IR drop. If you were to put a voltmeter lead at point 1 and the other at point 2 it would show zero voltage if the points are very close because there is no significant resistance to generate a potential difference. If the points move apart with the wire still connected the voltmeter will begin reading a potential difference. Sometimes it helps to get a non theoretical point of view. Hope this helps

25. Feb 14, 2017

### XZ923

Let me take a swing at this one. Like the OP, I'm not a physicist; I'm a mechanic and electrical hobbyist.

Assuming an ideal wire, which is what your original post seemed to stipulate, there is no "between 1 and 2" from an electrical standpoint. 1, 2, and the positive terminal of the battery are all the same thing. Same goes for three, four, and the negative terminal.

Voltage (as Jim explained) is a difference in electrical potential between two points. There is a greater concentration of electrons at the negative terminal of the battery, so when there's a pathway to the positive, electrons flow along the conductor until the "pressures" equalize (and yes I know that's not a correct application of the word "pressure"). Now, if the wires are ideal, there is no difference in potential between the positive terminal, point 1, or point 2 (vice versa for negative/3/4) since they're connected directly to each other. When you connect the lamp, one side of it will be connected to the positive/1/2 point of potential and the other will be connected to the negative/3/4 point. The voltage of the battery across the lamp will then induce the current determined by V=IR

In a real-world application, there will be a very small amount of voltage drop depending on the gauge of the wire.