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Constant Current in a series circuit

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    I have learnt that current is constant throughout a series circuit, with or without resistors. Even if the no. of resistors in the circuit increases, the current is the same through each resistor. Resistors resist current flow, even though the current flowing through any point in the circuit remains the same as charge cannot"disappear", will the speed of the current remain the same through each resistor? An ammeter measures current, but does it measure the speed of current or the flow of current? When a circuit is complete, do the charges start flowing out from the battery cell or are there already charges in the wires in the circuit? I understand the current drawn and the current flowing through a series circuit is basically the same idea, but how does the current "see"the no. of resistors in the circuit and adjust its speed to fit the resistance in the whole circuit? Assuming voltage is constant.
     
  2. jcsd
  3. Sep 26, 2011 #2
    Hey there, welcome to the forum!
    Let me just ascertain that I follow you, first:
    You mentioned that:
    But you must recall that:
    [itex]
    \normalsize
    I = \frac{V}{\sum{R}}
    [/itex]
    Meaning that unless V grows along with the number of resistors, that total current will not remain that same.
    The velocity of the charges in a conductor, to answer your query, depends on a characteristic factor, known as the conductivity. The drift velocity of the charges(which makes up the bulk of their speed), upon the application of a potential, is reliant upon the current travelling through(excepting fringe effects, and huge Vs),The amount of charge passing, per second, is what the Ammmeter measures.
    For a full description on how it works, try:
    http://en.wikipedia.org/wiki/Ammeter" [Broken]
    Hope that helps somewhat,
    Daniel
     
    Last edited by a moderator: May 5, 2017
  4. Sep 26, 2011 #3

    PeterO

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    Homework Helper

    The battery in the circuit establishes an electric field through the wires and resistors, so charges all the way around the circuit begin to move at the same time.
    The ammeter measures the flow rate - typically in coulombs per second. Given that it takes over 6 x 1018 electrons to make a whole coulomb of charge, the number of electrons flowing is probably best described as unimaginable. Even if the current is only several micro amperes, that is still over 6 x 1012 electrons per second - still an unimaginably large number, though we are happy to say 6 trillion.
     
  5. Sep 27, 2011 #4
     
  6. Sep 27, 2011 #5
    Thanks peter for the answer! Hmm by this do you mean that there are already charges in the wires, basically everywhere in the circuit?
     
  7. Sep 27, 2011 #6
    Hi,
    Perhaps a more atomic scaling of the elucidation could help you make sense of it.
    The Current, "decides" as you say, how large it should be, based on the number of resistors connected, due to the need to conserve energy. The current takes the form that is most restrictive in wasting its power on producing heat in a resistor. Each resistor R_i, carrying a current I, will produce a potential difference V_i. And as a consequence, you're right, the charges travel at an hoarsely slow speed, this is because they sweep, "drift" towards the various ends of the potential source, so, an individual electron, if we were able to monitor its exact location at any given time, could take several hours or even days, to cross some wires. But since a single electron does not produce the current passing through the circuit, but rather the entire agglomeration, so some charges which were initially closer to the source, would reach it faster(as the distance is shorter) and so forth.
    What must be considered however, is that by virtue of the classical theories on conductivity, the metal sees the electric field attached to it "instantly", and uniformly across it(due to the continuity principle), and so every particle experiences roughly the same force on it.
    Hope that helps,
    Daniel
     
  8. Sep 27, 2011 #7
    hmm may i ask what you mean by"metal"? do you mean the resistor? Ok so the current decides its form, but then my question still remains, how does it know how many resistors there are in the circuit?

    hey can i ask u another question? when i conducted an experiment in the sch lab, i connected the end of the circuit to each terminal of the battery. But i realised i connected the wrong end to the wrong terminal. so i switched the ends to their correct terminals. but the bulbs for both experiments had different brightness, and even had the same order of brightness for both experiments.bulb x was the brightest, bulb y next and followed by bulb z.
    although bulb x remained as bright as the previous bulb x, bulb y remained as bright as the previous and so on. can you explain what technical fault there was? I don't think there was a problem with the bulb caused we tested them before the experiment and they were of the same brightness.
    Thanks a lot!
     
  9. Sep 27, 2011 #8
    Hi...
    The current "does" so, mainly because the Sum of all Vs in the circuit has to equal the source, i.e Kirchhoff's Voltage law. By increasing the number of resistors, you get more V_is, and therefore the total Potential difference scatters across more hubs as it goes along.
    As for the brightness, that's an interesting point, and if you can, please elaborate on it some more.
    You should note that the blubs' intensity depends on the power, or P = R*I^2;
    Since the current is one for the entire series circuit, assuming the resistances are the same, they should all light up with equal brightness.
    Please tell us exactly what had happened,
    Daniel
     
  10. Sep 28, 2011 #9
    Daniel. i'm so sorry but i still don't get how more potential difference allows the current to "decide" its speed, (this is what you mean right?)oh and by the way i don't understand kirchoff's voltage law either.>.<
     
  11. Sep 28, 2011 #10
    Firstly, my group tested if bulb x, bulb y and bulb z were working properly by connecting then with wire to the battery. They all lit up with about the same brightness. However when we connected them in a series circuit, with bulb x first, followed by a resistor of one ohm, then bulb y, then a resistor of 2 ohm, then bulb z. Bulb x lit up the brightest and bulb y and z were dimmer but of the same brightness.
    we then switched the places of the resistors but that had no effect on the brightness of the bulbs.
    After that, we realised we connected the wires to the wrong terminals and the bulbs were not in the correct order. So we changed the connection and connected the wire (which was first connected to the postive terminal)to the negative terminal and vice versa however, the brightness of the bulbs were not affected.
    Can you explain why please? we came up with things like current flow from postivive terminal, electron flow from negative terminal, but do these actually affect the brightness?


    oh yeah, some people say that current flows from positive terminal so in a series circuit, whichever bulb gets the current first will use up the current and then the current will travel to other bulbs and whatever current passes through other bulbs does not affect the previous bulb because the latter is located behind it. this is a wrong conception, right?
     
  12. Sep 28, 2011 #11
    Well, physically speaking, there shouldn't be any difference at all in their illumination; You better verify the equipment used and the entire setting as a whole.
    The only influence on the current would arrive from the different resistors you used, as you connected them in series. Since they have different values, different power in each, that could sway the whole thing either way.
    For further reading, it might avail you to see this(In case I've harangued you sufficiently, and grown wearisome :)):
    http://wiki.answers.com/Q/What_happens_to_light_intensity_of_lamp_in_series_circuit_when_more_lamps_are_added_in_circuit" [Broken]
    Hope that's more edifying,
    Daniel
     
    Last edited by a moderator: May 5, 2017
  13. Sep 28, 2011 #12
    If the wattage of the lamps varied, the brightest lamps would be those with the lower wattage rating (yes, this is not a mistake!).

    why is this not a mistake? are you saying power decides the brightness of the lamps too?

    is it cos the voltage in higher when there are more watts?
     
  14. Sep 28, 2011 #13
    i'm sorry, i don't really understand what you're saying. can u explain this more simply? thanks a lot!
     
  15. Sep 28, 2011 #14
    .
    Yes, that's exactly what happens.
    Sure, as you yourself had mentioned, you used,
    , meaning that the bulbs were interspersed by resistances of different values, that can affect their luminosity.
    The power of any bulb, will be measured by: R*I^2 or V^2/R. The wattage directly depends on R, meaning, the smaller R is, the lower the wattage(Less heat produced for the same current). Since the power/intensity of each bulb, is its power, V^2/R, the smaller R is, the greater its brightness; But only should the resistances of the bulbs be different. There should be no effect if they're the same!
    Daniel
     
  16. Sep 28, 2011 #15
    ok, so you mean the smaller the resistance, the lower the wattage, thus the brighter the bulb?

    i think i get you now! so maybe the bulbs in the series circuit of my experiment were affected by the different resistors! cos my teach just said" maybe your setup is wrong". -.-/like that helps. Thanks so much!^^
     
  17. Sep 28, 2011 #16
    oh gosh i just realised i have a huge problem. current flows from positive terminal of battery, electrons flow from negative terminal of battery, so exactly what causes the bulbs to light up? the charges(current right)? then what are the electrons for? are they actually the same thing? my teach said current flows from the battery, but ain't there already charges in the wires? or is that just a term?


    ARGHHHHHH I'MMA TOTAL IDIOT.
     
  18. Sep 28, 2011 #17
    The current is defined as the local rate of change of the charges per unit time, or [itex] \Large I = \frac{dq}{dt} [/itex]
    By convention, The positive current was thought to originate from Positive charges, but of course, as the molecular theory of conductors, and indeed matter in general developped, it was quickly established that only electrons carry the current, in particular, those present in the so-called "conduction/valence" bands of the material.
    These, start to "flow", or drift, towards the terminus, actively, when an external electric field is applied, such as with a potential difference...
    The changing of the terminii, in such constructs as the light bulbs does nothing to change their luminosity, except for the sign of current due to the above convention(that is, current flows from plus to minues).
    But, in Semiconductors, this is a tad more complicated. Diodes are one such characteristic beings, http://en.wikipedia.org/wiki/Diode" [Broken], that only allow a passage of current one way.
    That's Great, I do hope I really did explain this thoroughly.
    Never berate yourself.
     
    Last edited by a moderator: May 5, 2017
  19. Sep 28, 2011 #18

    NascentOxygen

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    Staff: Mentor

    In electronics, the word "constant" has a meaning different to how you are using it. A better word for what you mean is "identical", meaning that the current in each resistor is identical with that each other resistor in that series string in your arrangement.
     
  20. Sep 29, 2011 #19
    um i don't really get this line. what do you mean by the sign of current? why does it suddenly become negative? if i understand you correctly, the electrons are the one who carry the positive current right, and they flow from the negative terminal right?



    oh by the way, my teacher mentioned that because the moment u close a circuit the current decides the path with the keast resistance which is the path they will take, as i mentioned in one of my earlier posts, but i have a new problem. if the current decides what path to take, then why are there charges in the circuit in the first place? won't the curren(charges) flow out from the battery only when the circuit is closed? or does it mean that the charges will simply go the reverse direction if they are in the branch with a resistor to the branch with no resistor?


    thanks so much to both of you for your help!
     
  21. Sep 29, 2011 #20

    NascentOxygen

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    Staff: Mentor

    Your teacher is wrong. Current will flow in all paths; it's just that more current will flow in the path with least resistance.

    Because we've deliberately built the circuit using elements/materials that have lots of available charges. The circuit comprises conductors in one form or another. All conductors have charges which are readily moved by the application of a voltage. Metals have oodles of electrons that can move and produce a current. Copper and gold are especially good for this. But in a pinch you could improvise a conductor out of a wet piece of string. :smile:

    You could construct a circuit using insulating materials, but then it wouldn't do anything very spectacular, except NOT conduct electricity. That may or may not be handy. Often we really do want to stop electricity flowing from some point to another. It's useful, but not spectacular. :smile:
     
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