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How is a voltage drop across a resistor caused?

  1. Aug 30, 2010 #1
    Preface: Newbie here. Sorry if this has been asked before - I did have search but couldn't quite find what I was after. Also, I'm not putting this into the electrical engineering section because I don't want what will inevitably be an electrical engineer's answer =)
    Here goes:

    Voltage across a circuit (say a circuit from A to B) is the potential for work to be done through that circuit due to charge. If the charge at A equals the charge at B then no work will be done; it is when there is a difference between these charges that work can and will be done (from the higher toward the lower).
    Yay? Nay?

    The amount of current (amount of charge per second) going into a resistor equals the amount of current coming out from that resistor. If, for example, several resistors of very different resistances are places in series, the current flowing through them should be equal.
    Yay? Nay?

    What actually causes the voltage drop across a resistor if, as the previous statement points out, there can be no accumulation or reduction of charge at either end of the resistor? If the amount of charge per unit time flowing on both sides is the same, where does the potential difference come from?

    Don't get me wrong - I know there *is* a potential drop and it wouldn't make sense for there to not be a drop, otherwise you could get work for nothing. But saying "it doesn't make sense not to" isn't much of an explanation to me.

    Thanks a bunch!!
     
  2. jcsd
  3. Aug 30, 2010 #2

    CompuChip

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    You can picture a current (in most cases) as a stream of electrons, each of which carries some amount of energy.
    Then two things are important: voltage U (measured in Volts) and current I (measured in Amperes). In our picture, the voltage corresponds to the energy the electrons carry, and the current to the number of electrons which passes. Of course, the more electrons pass by, and the more energy they carry, the more energy is available to do useful work (hence the formula P = U I).

    Now a resistor is simply something which takes energy off the electrons. In the simplest case, the energy is simply converted to heat (and is thereby "lost" as useful energy), but in most cases you want to use this energy to make coffee or power your computer so you can browse PF.

    If you put resistors in series, as you say, no electric charge can accumulate. Therefore, the rate at which the electrons flow through must be equal everywhere: I is constant. However, each resistor takes some energy off the electrons, therefore there is a drop in U over each separate resistor (U = U1 + U2 + .... + Un, for n resistors).

    If you make a parrallel circuit, then there will be some junction with a wire to each resistor. You can imagine the electrons distributing over these wires (in much the same way as cars distribute on a road which broadens from one to four lanes), so a different current will flow through each resistor, such that I = I1 + I2 + ... + In (for n resistors). However, in this case, each resistor doesn't have to "share" electrons with other resistors, so they can all take the same energy: U is constant.
    [Actually, in retrospect I realised I could have explained that better... reading my own post made me wonder why the current doesn't just split up in equal parts. Of course, this has to do with the possible difference in resistance of each resistor. Since electrons are "lazy" (current will always choose the most efficient route), the lower the resistance of a resistor, the easier it is for an electron to flow through, and the more electrons will want to go through (and all flowing through the resistor with the lowest resistance is not efficient, just like all driving in a single row on a four-lane highway is not efficient). Therefore, electrons will split up with more (i.e. higher current) going through "weaker" resistors (i.e. lower resistance) and vice versa, in correspondence to Ohm's law R = U / I. Of course, the voltage drop over each resistor should be the same, because measuring the difference over a single resistor is equivalent to measuring it between the two junctions (before and after the parallel circuit of resistors) and that measurement should be unambiguous.]

    Note that this story is very simplified, but it is important that you remember the difference between voltage and current.
     
    Last edited: Aug 30, 2010
  4. Aug 30, 2010 #3
    Actually Voltage across a circuit between points A and B is the work to be done on a unit charge to move it from A to B.
    voltage (across A and B) = work done on unit charge (to move from A to B)
    As you can see, this definition has nothing to do with presence of charges at A or B. If such a voltage is applied across series of resistors, due to closed path, charges do move from higher potential to lower. But how much charges flow? What is the rate of flow of charges? It depends on the resistance offered by the resistors. In fact resistor is a device in which following relation holds
    [tex]
    work \,done \,per\, unit \,charge \, \alpha\, \frac{dq}{dt}
    [/tex]
    where [itex] \frac{dq}{dt} [/itex] is the rate of flow of charges.

    The resistor provides friction to charge flow through it and some work has to be done here to move the charges through it. In other words (as per above definition) some voltage is lost across each resistor.

    Hope I explained a little bit about voltage drop across resistor to you.
     
    Last edited: Aug 30, 2010
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