1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The difficulty of understanding voltage

  1. Mar 4, 2013 #1
    Hi all, first post, I hope this is the right place.

    As I understand, voltage is a measure of the energy given to each coloumb of charge. Therefore, a cell with a higher emf will give more joules of energy to each coloumb of charge.
    However, voltage is also the driving force behind getting a current to flow. When a voltage is applied to a conductor a current can start flowing through it as it pushes the current round the circuit? (This last part seems a little cloudy and I do not fully grasp how it does so.)

    However, when the voltage is increased the current also increases. This does not seen to make sense to me because in my mind applying a higher voltage would only give each coloumb of charge more energy meaning the current flowing is the same but has more energy? - I fail to see the relationship between voltage and how it affects the current.

    I have read through books and countless websites yet I still cannot grasp this fundemental part of electricity! Any help would be much appreciated.

    Thanks in advance.
     
  2. jcsd
  3. Mar 4, 2013 #2
    Studying at the university is the best way to learn. These forums will only confuse you if you are in the beginner stage. Many highly skilled people are on these forums, willing to give you good info. But there are also low-skill people anxious to play professor. The result is that the low-skill posters get into heated debates with high-skill posters, and eventually the thread either drives people away, usually the good people, or gets locked by a moderator.

    I recommend limiting your discussions to a few people in the know. Someone who has developed circuits, electronics, as well as energy conversion (motors, generators, etc.) should be reliable. Their info is based on education and practice. Many self taught people have knowledge, but it is fragmented, in random order, and has big gaps. I recommend directing questions to an EE or physics professor, or a Ph.D. EE/phy working in R&D. An MS degree is pretty reliable if a Ph.D. is not available.

    You want to get the basics right. The first few courses are the most critical. A forum will utterly confuse beginners. Don't ask me why, but many people w/o a degree, or with very limited credentials, get a rush lecturing people about science. It's what they live for. If a person w/ more education/practice tells them " I beg to differ, this point you made needs correction ---", it becomes a battle. People with less education generally do not yield to those with more. Instead they draw a line in the sand and stand their ground.

    The OP is then left wondering who to believe. They may wish to believe the MS/Ph.D., but there is always the possibility that the lesser educated person could be right, albeit rarely. Here is my answer to your question.

    Voltage is work per charge. A battery rated at 12V imparts 12 joules of energy to every coulomb of charge it transports. The energy is provided by chemical means.

    Voltage does NOT "drive current". The chemical reaction does so, and the voltage is merely a mathematical ratio of work done per charge transported. But elaboration is too involved. I recommend you use reference texts peer-reviewed and written by experts, used in uni courses. Best regards.

    Claude
     
  4. Mar 4, 2013 #3
    Thank you for your answer Claude. I realise what you are saying about using the forums to further my knowledge and will hopefully have the chance to pursue physics further in the future, and by browsing a few threads I have certainly witnessed what you are talking about! However, I feel that I can not understand the physics I am dealing with at the moment without having a slightly better understanding of what voltage is. I (think I) understand what you are saying regarding voltage. However, what I fail to see, is why increasing the work per charge (by raising the voltage) will increase the current. Using the example you gave, If the battery in the circuit was changed to one that was 15v it would impart 15 joules of energy to every coloumb of charge it transports. This would result in a higher amount of energy being transported around the circuit assuming the ammount of charge stayed the same. However,why would raising this work done per charge increase the current, or the flow of charge? Surely it would have no bearing on the charge?

    Sorry if this is obvious but I can not seem to get my head around it.
     
  5. Mar 4, 2013 #4
    Thank you cwilkins. I think that explanation has eased my mind, for now. It always mentioned in the text books P.D was the energy per charge and also the driving force, but it never elaborated and left me stumped with two seemingly conflicting ideas.
     
  6. Mar 4, 2013 #5

    cjl

    User Avatar

    That depends on the situation. For a constant resistance circuit, increased voltage absolutely increases current.
     
  7. Mar 4, 2013 #6
    Thank you Inquisitive one. That explanation was definatly very helpfull. However, it seems a little counter intuitive that an increase in the voltage would not increase current. As current is just the amount of charge that flows past a point in an amount of time, doubling the voltage would surely have the effect of doubling the speed of the flow which would cause twice the previous current to flow in the same amount of time.
     
  8. Mar 4, 2013 #7
  9. Mar 4, 2013 #8
    Will do! Thanks.
     
  10. Mar 4, 2013 #9

    berkeman

    User Avatar

    Staff: Mentor

    Please disregard the post by InquisitiveOne. It contained errors, and has been deleted by the Mentors.
     
  11. Mar 4, 2013 #10

    davenn

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    As Berkman say disregard Inquisitive one's post



    No ... the speed Doesnt change. there is a larger charge going past a given point in a given time ... not the same amount travelling faster :)

    Dave
     
  12. Mar 4, 2013 #11
    davenn,

    Yes, the speed (drift velocity) does change. The only way a larger charge can go past a point in a given time is for the drift velocity to increase. I hate to use a water anology, but if you increase the water flow rate, the velocity of the water in the pipe will have to increase. If you cut the current to zero, then the drift velocity will be zero also.

    Ratch
     
    Last edited: Mar 5, 2013
  13. Mar 5, 2013 #12
    I would agree with that. An increase in drift speed and energy is consistent with observed behavior in conductors. If the carriers have increased velocity, then collisions w/ lattice structure result in more photon emission (I2R heating) since more energy transfer is taking place. Also a larger voltage is accompanied by a larger E field resulting in more charges drifting per Ohm's Law J = σE. Ohm's Law is mAcroscopic, whereas lattice collisions, drift velocity, energy transfer, etc., is mIcroscopic. Either view gives valid results.

    For a beginner I'm not sure if the mIcro approach is recommended, but it doesn't hurt to discuss it. Hopefully the OP is benefiting from said treatise.

    Claude
     
  14. Mar 5, 2013 #13
    claude,

    No need to get involved with lattices, energy, fields, or Ohm's law. One only needs to realize that a conductor, like say copper, has a constant finite density of free mobile charge carrier electrons. If a larger number of charge carriers per unit time (current) exists in a conductor than before, the drift velocity of the charge carriers just has to increase. If the density is constant, which it is, then a higher velocity is the only way more charge carriers can move per unit of time.

    Ratch
     
  15. Mar 5, 2013 #14
    I would agree but with the caveat the we fully understand the meaning of "density". In Ohm's law, J=σE, since σ remains constant while E has increased due to V increasing, then J must increase. But J is current "density", in amp/meter2. So in a given cross sectional area of Cu, there is increased charge flow in coulomb/second*meter2.

    There are 2 ways this can happen, either the "density" (number of charges in the unit area) of charges over the area has increased w/o an increase in speed, or the same "density" (number of charges in the unit area) as before w/ increased speed. Either or a combo of both can result in an increase in "J".

    I like to think of the Lorentz force law, Fe = q*E. If E increases due to V increasing, then the force acting on each carrier increases resulting in higher carrier speed. Collisions occur and more energy is radiated as long IR (heating per I2R). I just thought I needed to clear up the "density" concept. Density can mean charges per area w/o regard to motion. Or it can include motion. Either way can result in changes in J.

    Claude
     
  16. Mar 5, 2013 #15
    Claude,

    I believe I defined what density I meant when I said "finite density of free mobile charge carrier electrons". If one multiplies the density of Cu by Avogadro's number, and divides by the atomic weight of Cu, the number of free electrons per unit volume is 8.4E28 electrons/m^3. That density of electrons per unit volume is constant no matter what the current is. It is not the same as the current density (J).

    Basically that is correct, with the exception that "increased charge flow" is in coulombs/sec. Let's see where that leads us.

    If you are talking about current density (J), then it is the number of charges per (unit area * time) or current/unit area. Since the conductor does not change its area, the only way the current can increase is for the electrons to move faster.

    The number of charges in the unit area makes no sense. How would you compute that? There are an infinite number of cross section unit areas along a conductor. The amount of current in the unit area does make sense. That is the number of charges that pass through a unit area per unit of time.

    No, it can only happen the one way I described.

    You don't need to know that to discern that more current means a higher drift speed.

    I am afraid you did not clear that up. Current density without motion is always zero. Charge per area is a surface charge, and not applicable to this discussion.

    Ratch
     
  17. Mar 5, 2013 #16
    But when I said "charge per unit area" I inferred (albeit I wasn't clear enough) that the charges were in motion. In other words let's say J1 is 2.00 amp/meter2. Then an increase takes place where J2 is 3 amp/meter2. Say that J1 at 2 A/m2 was for a cross section of 1e-6 m2. If every charge carrier was accounted for throughout the cross section, such as the case you described with a conductor like Cu, then I would agree. But for a semiconductor where all available charges do not necessarily provide conduction, an increase in current density can be realized w/o an increase in speed if a larger number of carriers conduct at the same speed. Did the OP stipulate conductor like Cu, or any material?

    Anyway, if we restrict the discussion to conductors, such as Cu, then I would agree that increasing the input voltage source value with resistance held constant would result in an increase in current, which includes an increase in drift velocity. That would explain increased current. Since the cross section is constant but the carriers incur faster speed, then the number of carriers flowing per second has increased.

    I was only pointing out that for materials such as semiconductors, which may not be germane to this discussion, an increase in J could happen w/o an increase in drift speed. But for Cu conductors, I would agree that speed increases. Of course, with semiconductors, the resistance would be different if differing J values occur at the same drift speed. So getting back to the OP, I must stick with the conditions of the OP, that would not be germane. If R is constant and V increases, so does I increase. With constant R, then drift speed had to increase.

    Claude
     
  18. Mar 5, 2013 #17

    davenn

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    OK
    thats interesting .... everything I have learnt on this forum didnt comment on a change in drift velocity


    I definately need some clarification on that then :)

    Dave
     
  19. Mar 5, 2013 #18
    davenn,

    I did not realize it either until I started to think about it.

    Just ask.

    Ratch
     
  20. Mar 6, 2013 #19

    davenn

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I was under ... and obviously the mistaken idea ... going by your comments, that electron drift was relatively constant, rather there was just more electrons going past a given point ... not because they were moving faster, but because the "power" source was supplying more of them

    so does this mean that all the comments by various people that the electron drift velocity is a few mm / sec
    isnt correct ?

    I'm always willing to learn :)

    D
     
    Last edited: Mar 6, 2013
  21. Mar 6, 2013 #20

    There is some misunderstanding here. The power source DOESN'T supply the circuit with electrons.The electrons are already in the wire . In a first crude approach you can depict it as an ideal Fermi gas of electrons moving almost free of electric forces.
    When you increase the voltage each individual electron will increase its drift velocity by very little ( the order of magnitude is ~mm/s) . But the voltage affects many electrons. So you have many electrons , each of them affected infinitessimaly by the voltage.
    Now, the current is :

    I = (charge per unit volume)*(drift velocity)*(wire cross sectional area)
    I = (electron charge)*(nr of electrons per unit volume)*(drift velocity)*(wire cross sectional area)

    the drift velocity per electron increases with voltage so, (electrons)*(drift velocity) increases therefore current increases with voltage
     
  22. Mar 6, 2013 #21
    davenn,

    e.chaniotakis has it correct. Electron drift velocity is very, very slow. Increasing it several times is still very slow. A voltage source takes in as many electron as it expells, so the net electron gain/loss of the source is zero. There is an uncountable sea of electrons in a good conductor like Cu, so the drift velocity does not have to be very much to supply large currents. And the finite density limit of free electrons in a conductor (8.4E28 electrons/meter for Cu) insures that the drift velocity has to speed up in order to support larger currents.

    Ratch
     
  23. Mar 6, 2013 #22
    We must be sure to understand which quantities are held constant and which vary. If a conductor and/or fixed load resistance is connected across a CVS (constant voltage source), then the source, assuming it has high compliance (low internal series resistance & adequate energy conversion ability), will maintain a fixed voltage. If the resistance is fixed as well, which it appears to be in this example, then we can compute current per Ohm's law.

    If the source voltage is increased to a new fixed value, but the resistance is the same as before, than the current must increase per Ohm's law. So we arrived at the question "does the increase in current incur a higher drift speed, or the same drift speed with increased charge carrier count in motion?"

    I believe we've come to a consensus here that the drift speed would have to increase in order to increase the current, which must happen since voltage was increased with resistance held constant. To understand why at a mIcro scale involves solid state physics. When I have free time (hardly ever) I can refer to my Kittel text on said subject. But the model of a Cu conductor and a sea of electrons appears correct.

    The no. of available carriers (electrons) in Cu is a fixed physical constant. These electrons will drift as soon as an external source provides an E field. All available electrons are in conduction, since Cu is a metal. An increase in source voltage incurs an increase in E field with constant resistance. The increase in current can only come about via an increase in drift speed. The percentage of free electrons participating in conduction is already at its maximum value, a fixed quantity.

    Thus the drift speed must increase to realize an increase in J the current density. I believe the explanations above are correct. I initially pondered the thought that a material with current density J1 can undergo an increase in current to a new value J2, by 2 means. But in this example the R value does not change, so the only way to increase J is by increasing drift speed. We seem to have consensus.

    Claude
     
  24. Mar 6, 2013 #23
    Thanks for all the replys, they certainly made for some interesting, at least the parts I could get my head around did. I have another question regarding voltage that confuses me, I have tried to use a practical example to explain it, hopefully you can follow my logic and spot any places I have slipped up.

    There is a component of a circuit, it has a P.D of 12V. This means that every coloumb of charge that is moved through the component does 12J of work. The current through this component is 2 Amps = 2 coloumbs per second. The resistance of this component could be worked out with R=V/I. The resistance with the above valies would be 12/2 = 6 ohms.
    The P.D is changed from 12v to 10v and the resistance is kept the same. This leads to an decrease in current. Why exactly does this happen? If the P.D across this component falls does it mean (excuse my rather crude terminology) that the "driving force" of the current has decreased, yet the charge still has to face the same amount of resistance, meaning the current that gets through decreases? This is how I see it at the moment, but I do not really understand why the P.D across the component ( the work that is done when charge moves through it), affects the flow of current. If it was the EMF I would understand it, but does the P.D across a component also equate to the "driving force" across the component?
     
  25. Mar 6, 2013 #24
    Instead of "driving force", the terminology in use is "Lorentz force". If a charge carrier is in the region where there are E & B fields present, and it is moving at a speed "u", then the 2 forces incurred by the charge carrier are "Fe & Fm" (electric & magnetic.

    Fe = q*E, and Fm = q*(uXB). F = Fe + Fm.

    Of course charged particles have their own E field, so that when collisions occur, and a layer of charges builds up in a region, then incoming charges feel this force due to accumulated charges. Likewise, a circuit possesses self inductance so that when charges move, this is current, a magnetic field exists which tends to oppose a change in current.

    All of these characteristics enter in when circuits are studied. In your example, decreasing the source from 12V to 10V results in a decrease in E field, and a decrease in Lorentz force as well as decreases in current. J = σE. so since σ is constant for the material, a decrease in E results in a decrease in J the current density.

    How this decrease occurs was already discussed. The drift speed decreases.

    Claude
     
  26. Mar 6, 2013 #25
    Thank you for your reply. However, I think I worded my question incorrectly. I am wondering about lowering the PD of a component, for example using a 10V bulb as opposed to a 12v bulb. If the resistance of the two bulbs are the same the current through the bulb must decrease, why does the P.D across a component affect the current through it. I feel I now have an understanding of the rlationship between the the current and voltage of the source but still lack information on why reducing the PD of a component would change the current across a component?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...