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Voltage Drops across Resistive Elements

  1. Jan 18, 2009 #1
    I have been having a very difficult time grasping the concept of voltage drops across resistive elements.

    In a series configuration....How is it that the greater the value of the resistor, the greater the voltage drop across that resistor? Why does voltage distribute itself that way, in fragments?
  2. jcsd
  3. Jan 18, 2009 #2


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    It's not exactly "fragmented". There is a potential drop in the connecting wires in between resistors as well, so it's actually continuous but because the resistance of the connecting wires have comparably minute resistances, we can approximate their potential drops to be 0.

    And I thought you asked a very similar question some time back:
  4. Jan 18, 2009 #3
    Yes I did, and I put some thought into it for a few months and tried to give it a shot but still I can't wrap my head around it.
  5. Jan 18, 2009 #4
    Allow me to rephrase....The way I like to think about resistors in series is imagining one big long resistor.... with segments of more dense atoms in sections representing the heavier consentrations of resistance or resistors.

    Like you said there is a "continuous" drop across the entire length of the circuit, I understand that, but the heavier consentrations of resistance or resistors have a greater voltage drop associated with them. Why does voltage drop distribute itself more or less over some segments of the entire resistance of the circuit.
  6. Jan 19, 2009 #5
    I am not sure what i am saying is correct. please forgive and correct me if say any thing wrong. Actually the voltage drop is due to energy waste (in the form of heat for the resistor) in a circuit. this will differ for each resistor which is connected with the circuit. it means each resistor will waste the energy in different amount of energy(It is called load) and also when calculating for the entire circuit after every resistance the voltage will be different due to the lose of voltage in that resistor.
  7. Jan 28, 2009 #6
    v = ir
  8. Jan 28, 2009 #7
    V=IR is the relationship..... yes.
    but that doesn't answer my question.
  9. Jan 29, 2009 #8
    One of the things that I learned while in college for electrical engr is that sometime it is best not to try to understand something and just go along with it. Eventually with experience and thoughts, you will get an "ah ha" moment and you will understand it. Unless you'll get a test question on the concept of voltage drop across a resistor, maybe it is best to let it marinate in your head for a while. It's good that you want to understand it though.
    Last edited: Jan 29, 2009
  10. Jan 29, 2009 #9
    You are absolutely right that it helps but I am "looking" for the understaning of it.
  11. Jan 29, 2009 #10


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    Remember that there are different "levels" of understanding. As long as you stick to circuit theory the "reason" is simple: it is just the constant of proportionallty in Ohm's law that related current and voltage.

    However, if you REALLY want to understand what is going on, meaning why the potential energy (which is what voltage is) drops as you go from on terminal of a real resistor (as opposed to an idealized) to the other you need a very good understanding of solid state physics/many body physics; this is post-grad level physics and not something you can expected to have a deeper understanding of unless you happen to work in the field of condensed matter theory.
  12. Jan 29, 2009 #11
    i agree with david90. things like this used to trouble me, like why the lower the resistance, the higher the load and more power you consume. it just didn't sit right with my gut instinct.

    but one thing you might try is looking at it from the current's point of view, i=v/r. for a given current, you have to "push" harder through higher resistances, and that push is voltage.

    and consider that in most of our circuits, the power comes from a voltage source. the resistances aren't so much determining the voltages as they are determining the currents.
  13. Jan 29, 2009 #12
    Well I don't know solid state physics but at least its a sensible answer.
    Could you elaborate a little on the concept of this in relation to my question. Thanks.

    Yes I understand all of this but here is where I'm at......

    Consider a short circuit condition with a load.. the voltage drop across the internal resistance of the source is the same but has a greater voltage drop associated with it.

    My only problem with the statement...."Voltage is pushing harder"....is that voltage is NOT pushing harder. The source voltage is the same, and because of this...its something that I am having trouble understanding.

    Is there a basic unit relationship with voltage drop to resistance? (other than Ohms law)

    I know, obviously the whole applied voltage drops proportionally across the entire resistance of a circuit.

    What would hapen if you had an infinite resistance with a voltage source of lets say 100v potential.....You would have zero current flow (no kinetic energy) and since there is no kinetic energy being dissipated ...you would have a zero voltage drop associated with the infinite resistive load, right?
  14. Jan 29, 2009 #13
    How are you going to have a voltage drop with no current? You say there is a short circuit, that means there is no voltage drop anywhere because there is nothing anywhere.

    A perfect crystal lattice, with low enough thermal motion and no deviations from periodic structure, would have no resistivity, but a real metal has crystallographic defects, impurities, multiple isotopes, and thermal motion of the atoms. Electrons scatter from all of these, resulting in resistance to their flow.
  15. Jan 29, 2009 #14
    not sure what you're saying. if your source is modeled as an ideal voltage source with an internal series resistance, then shorting the outputs causes the entire voltage drop to occur across the internal resistance. an internal resistance means the source is not ideal (none are), and so the voltage output at the terminals is highest under no load, and decreases according to Vo = V_ideal -I*R_internal. shorting the output means Vo=0= V_ideal -I*R_internal, or: V_ideal = I*R_internal, and solving for short circuit current, Isc=V_ideal/R_internal.

    i'm not quite sure what you're thinking, but there would be no "drop" in the voltage from the source. that is, Vo=V_ideal from above because there is no internal drop in the source series resistance R_internal. however, there is a voltage of 100V across the infinite resistance. and even though there is no current flow, the electrical field varies continuously
    across the resistor from 100V potential at one end, to 0V potential at the reference node.

    it's a bit like a gravity field, where a higher voltage represents a higher elevation.
  16. Jan 29, 2009 #15
    I can see you guys are confused with my wording. Let me draw something up. Ill get back with you.
  17. Jan 30, 2009 #16
    Think of it like this:

    Imagine a pool of water at sea level. Adjacent to this body of water is a terraced mountain, like the ones used for farming in Chile. The top of this mountain is at, say, 100m. A pump is used to bring water from the pool at sea level to the top of the terraced mountain. This pump is analogous to a voltage source establishing a difference in potential across elements in an electric circuit. The water at the top of the terraced mountain then flows down the series of slopes and back into the original pool of water. Each of the individual slopes on the terraced mountain is analogous to individual resistive elements in an electric circuit in the sense that they cause a drop in potential from one terraced level to the next. The steeper the slope, the higher the resistivity, and the larger the drop in voltage potential. The whole system described is best thought of as representing a 1-loop, ideal, closed DC circuit with a single voltage source.

    I hope that helps in visualizing and making sense of the voltage drops. If not, let me know and I will try to clarify.
  18. Jan 30, 2009 #17
    In this case, the voltage across the infinite resistive load is still 100v, but there is no voltage drop attributed to this infinite resistance. If there was an associated drop in voltage across the resistive element, then a current inversely proportional to the resistivity would be drawn by the load, making the load conductive. An infinitely resistive load is NOT conductive at all.

    The circuit you described is equivalent to a single voltage source in an open circuit condition. There is no current flow, but there is still a difference in potential between the positive and negative terminals of the voltage source and any non-conductive load the leads happen to be in contact with.
  19. Jan 31, 2009 #18
    Thank You!

    This is "exactly" what I thought. Can you believe a group of Engineers had told me otherwise.

    Now...is it safe to assume that because Ex=(Rx/Rt)Et......The fraction of Voltage drop (Ex) across a single resistor (Rx) is actually represented as if that same resistor (Rx) under load was dividing the total Voltage (Et) units consumed across itself with some sort of "proportion"?
  20. Jan 31, 2009 #19
    Yes, the proportion being the ratio of the resistive element to the total resistance, (Rx/Rt). This is a circuit analysis technique called voltage division or voltage divider.
  21. Jan 31, 2009 #20
    there's still a "voltage drop" across the infinite resistance measured in volts per meter. if the infinite resistive element in your example is 4 meters long, then it has an E field across it of 25 V/m. at 1 m from the reference node, the voltage is 25V, at 3 m it is 75V.
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