Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Voltage between two points on an empty wire

  1. Feb 9, 2017 #1
    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)

    00030.png

    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. jcsd
  3. Feb 9, 2017 #2

    cnh1995

    User Avatar
    Homework Helper

    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.
     
  4. Feb 9, 2017 #3
    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
  5. Feb 9, 2017 #4

    cnh1995

    User Avatar
    Homework Helper

    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.
     
  6. Feb 9, 2017 #5
    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.
     
  7. Feb 9, 2017 #6

    cnh1995

    User Avatar
    Homework Helper

    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.
     
  8. Feb 9, 2017 #7
    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?
     
  9. Feb 9, 2017 #8

    cnh1995

    User Avatar
    Homework Helper

    No. Battery is a voltage source. Current depends upon the resistance of the network.
     
  10. Feb 9, 2017 #9
    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.
     
  11. Feb 9, 2017 #10
    As someone who has a weak base in physics, I'll go the mathematical way of understanding electronics. I can't handle physics anymore.
     
  12. Feb 9, 2017 #11
    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
  13. Feb 9, 2017 #12

    cnh1995

    User Avatar
    Homework Helper

    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.
     
  14. Feb 9, 2017 #13
    I'm reading the book on allaboutcircuits.com. What book would you recommend for a programmer?
     
  15. Feb 9, 2017 #14

    cnh1995

    User Avatar
    Homework Helper

    As a beginner, you can refer Halliday-Resnick or Serway-Jewett.
     
  16. Feb 9, 2017 #15
    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.
     
  17. Feb 9, 2017 #16
    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.
     
  18. Feb 9, 2017 #17
    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!)
     
  19. Feb 9, 2017 #18

    cnh1995

    User Avatar
    Homework Helper

    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.

    @sophiecentaur could you please help?
     
  20. Feb 9, 2017 #19

    cnh1995

    User Avatar
    Homework Helper

    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..
    https://www.physicsforums.com/threads/what-is-voltage-really.879740/
     
  21. Feb 9, 2017 #20
    I appreciate your help. I did not mean that you don't know what voltage is.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted