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Resistance in Ohm's Law

  1. Mar 30, 2012 #1
    I have some problem understanding the concept of the voltage changing across a resistor.

    Given V=IR (Ohm's Law)

    Ok so according to Ohm's Law, if I increase the resistance of a circuit, the potential difference of across the resistor will change.

    I know that potential difference is simply the difference between the potential of 2 position(where it crosses the resistor), so if the potential difference increases as resistance increase, there is only two ways it can be done.

    Given potential difference V = V1-V2

    either V1 increases or V2 decreases will cause the potential difference to increase.So which potential will change, V1 or V2?(and how if relevant)

    Another thing is, how does the increase in resistance increase the potential difference physically, or does it even happen in reality?

    I know that resistors work by going against the current flow, but I can't seem to get how the voltage will change.
  2. jcsd
  3. Mar 30, 2012 #2

    Doc Al

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    A better way to think of it is: In order to keep the current constant across a resistor, if I increase the resistance I must also increase the voltage across it. (If you use the same voltage source and just increase the resistance, the current will drop.)
  4. Mar 30, 2012 #3
    Potential Difference is the only thing physically meaningful. Absolute potentials have no meaning. So it is the potential difference that physically changes, not V1 or V2. If this is to hard to visualize, just define V1 = 0 as the ground and then measure everything relative to it. In that case, it would be V2 that is changing.

    I like to think of the potential landscape as analogous to a mountain landscape. Higher potentials would be the mountain peaks and lower potentials the valleys. Charge feels a force and moves where there is a potential difference. This is like a ball on the side of the mountain rolling downhill. The resistor is like mud on the side of the mountain slowing it down. Current is proportional to the speed. So in order to keep the speed of our rolling ball constant (like the current of our electric circuit) when switching from slightly mountain to very muddy mountain (like higher resistance), we have to use a steeper incline (higher voltage).
  5. Mar 30, 2012 #4


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    Hi Kurokari! The simplest way I can think of is to illustrate it with two simple setups (a power supply and one OR two resistors). Let's say our components are one 1V power supply and two 1 ohm resistors.

    Setup 1; (power supply (+)=1 V, (-)=0 V and one 1Ω resistor):

    (+)-V1-|resistor 1|-V2-(-)

    V1 = 1 V
    V2 = 0 V, and
    I=V/R=(V1-V2)/R1=(1-0)/1=1/1= 1 A

    Setup 2; (power supply and two 1Ω resistors):

    (+)-V1-|resistor 1|-V2-|resistor 2|-V3-(-)

    I=V/Rtotal=(V1-V3)/(R1+R2)=(1-0)/2=1/2= 0.5 A, and
    V1 = 1 V
    V2 = R2*I = 1*0.5 = 0.5 V
    V3 = 0 V

    (I hope you understand my crude drawings:))

    From this you see that neither the potentials V1 nor V2 change with only one resistor in the setup. But if you add another resistor, the potential V2 between the resistors will change; the voltage across the first resistor is V1-V2=1-0.5=0.5 V, and the voltage across the second resistor is V2-V3=0.5-0=0.5 V.

    The two voltages will be different from each other if one resistor is different from the other. To calculate them, you can do the following:

    1) Calculate total resistance for the resistors in series; Rtotal=R1+R2
    2) Calculate the current I = (V1-V3)/Rtotal (the same current I goes through both resistors)
    3) Voltage across first resistor; (V1-V2) = R1*I
    4) Voltage across second resistor; (V2-V3) = R2*I

    Calculation of 3) or 4) will give you the potential V2. I hope this helps; ask if anything is unclear. (if anyone spots any typo, please say so, I'm a little tired at the moment :smile:)
    Last edited: Mar 30, 2012
  6. Mar 31, 2012 #5
    Actually I'm quite familiar with the calculations.

    My problem is that when I increase the resistance of a variable resistance, the current should drop(since resistor work by resisting current flow), so how did it end up that in an ideal series circuit that the current is the same all along?

    That would mean that the potential difference increased in order to compensate for the increase in resistance. I mean, does the charge gets "congested" before the resistor because the resistor is resisting the current flow, and finally an increase in the absolute potential before the resistor? ( it's just some ramblings, but I hope you get what I want to ask :D)

    and if anyone can explain the above situation(physically, cause im quite familiar with the formulas already), mind shedding some light on how the voltage varies when I add another resistor?

    sorry if i sound im going in circles :D
  7. Mar 31, 2012 #6

    Doc Al

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    What do you mean by "the current is the same"? If you hook a battery to a variable resistor, when you increase the resistance the current will drop.

    Can you describe the exact arrangement you are thinking of?
  8. Mar 31, 2012 #7
    Oh my, I'm so sorry for making such an ugly mistake, I was rambling through what I want to ask and I got confused. Yes current does decrease when the resistance is increased through a variable resistor.

    Probably the last question, how does the potential difference gets divided evenly in a circuit with 2 resistors and a battery.

    Say, I have a 10V battery, 1 6 ohm resistor and 1 4 ohm resistor, am I right to say that the voltage across the 6ohm resistor is 6V and 4V for the latter resistor?

    If so, how is it done?
  9. Mar 31, 2012 #8


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    If the two resistors , the 6 Ohm and the 4 Ohm are in series between the positive and negative terminals of the battery, it's simple Ohms law to define the current through each resistor and then the voltage drop across each resistor.

    You say you know the formula... So do the math and show us your working. :)

  10. Mar 31, 2012 #9
    Hmm, let me have a try at this.

    Using V = (R1/(R1+R2))V0 , where R1 = 6 and V0 = 10
    I get V = 6V

    same goes for the second resistor where R2 = 4 and V0 = 10, I get V = 4V

    It's just that I find it puzzling as to how the potential difference is so evenly divided. (or perhaps I'm using a wrong approach in understanding it?)
  11. Apr 1, 2012 #10

    Doc Al

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    You might find it less puzzling if you look at it in terms of current. Once you understand what the current must be (given the total resistance) you can more easily understand what the voltage drops must be across each resistor.
  12. Apr 1, 2012 #11


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    Hi again, Kurokari! I'd say the same as Doc Al. When you have the current (which will be constant and the same everywhere in the examples discussed), the voltage drops can be calculated. I reread your posts and I'd like to give an example of how you can view the voltage drop in more detail;

    Consider a setup with one power supply and one resistor. Furthermore, you probably know that a wire has a resistance which is proportional to the length of the wire (see e.g. here; R = ρ*l/A; R = resistance, ρ = electrical resistivity, l = length, A = cross-section area)

    Now, imagine that the resistor is actually a very long wire, that is


    is equivalent to


    In this example Vp represents a measuring point that can be moved along the wire. First, let's define (-) as ground/earth, i.e. V2 = 0 V. We know that the current I will be I = V/R = (V1-V2)/R = V1/R. We also know that I will be constant (it's a DC setup), and that I will be the same at any point in the setup (it's a series setup).

    Furthermore, we know that voltage is the difference between two potentials, and since V2 = 0 V, this means that the potential Vp at any point on the wire will be equal to the voltage between Vp and V2, that is (Vp-V2) = (Vp-0) = Vp.

    Now, let's measure Vp at different points on the wire:
    If you measure Vp close to V1, it will be Vp = V1.
    If you measure Vp close to V2, it will be Vp = 0 V.
    If you measure Vp at any point in between (from left to right), it will be linearly decreasing from V1 to 0 V (see graph example in attached picture below, y=Vp, x is proportional to distance; please note that this is only an illustration of linear decrease). E.g. if you measure Vp at half the length of the wire, Vp = (V1-V2)/2 = (V1-0)/2 = V1/2.

    Furthermore, a variable resistor is similar to my example above; with a variable resistor Vp is connected to ground (V2) in my example;


    You can vary the resistance by adjusting the 'wire length' between V1 and Vp (longer length = more resistance = smaller current). Please note that in this example current will go from (+) to (-); there's no current going to the right of Vp.

    It's really not more complicated than this. Please note that this example is meant to illustrate how you can view the voltage drop "inside" a resistor. Adding more resistors in series is equivalent to a longer total 'wire length' in my example, and the potentials between the resistors will be equivalent to certain fixed points at different positions on the total 'wire';

    (+)-V1-|-------resistor 1-------Vp1-----resistor2-----Vp2---resistor 3---|-V2-(-)

    (sorry for another long reply, but I'd rather be thorough than confusing, I hope I was successful :smile:)

    Attached Files:

    Last edited: Apr 1, 2012
  13. Apr 1, 2012 #12
    Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance,[1] one arrives at the usual mathematical equation that describes this relationship:[2]
    [itex] I = \frac{V}{R}[/itex]

    From Wikipedia http://en.wikipedia.org/wiki/Ohm's_law

    Changing a value of resistor in a circuit will change the TOTAL resistance of the circuit thus result in different current. Since voltage across a resistor is proportional to the current, new voltage appear across the resistor.
    Last edited: Apr 1, 2012
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