When you have one resistor in a series circuit, the amount of voltage taken up by that resistor is 100%. Adding another resistor (same Ohms) makes that voltage drop 50%. Why, since only current should be changing, and the voltage (joule per coulomb) shouldn’t depend on current (coulombs per second)? It’s like the voltage knows that there’s another resistor (before it hits the first one) and decides to split its energy for each. But obviously that isn’t the case because voltage isn’t a sentient being, so what causes it to split its energy rather than imparting all of the energy on the first resistor as it normally would if there wasn’t another? At first I thought the answer was that the current decreases as a result of the resistors and that’s what causes the voltage to decrease for each, but that doesn’t make sense since the amount of electrons shouldn’t control the energy they carry.

  • \$\begingroup\$ V = I*R the pressure (voltage) of the circuit is related to the flow (current) and the way to control flow is through resistance. Electrons don't flow unimpeded through a material, there are imperfections (on the atomic level) because of these imperfections the electrons bump into each other and atoms, they heat up the material and slow the flow of other electrons. \$\endgroup\$ – Voltage Spike Jul 15 '16 at 19:47
  • \$\begingroup\$ Think of it like having a pipe with two restrictions and a pump. The pump is like a voltage source, the restrictions are like resistors. If you started measuring the pressure from the intake of the pump you would notice that you would get a positive reading from after the pump and between the restrictions, but not after the restrictions, because the pressure is the same in after the restrictions (and before the pump). If you measured the flow, it would be the same no matter where you measured it. A pump is like a voltage source, a resistor is like a restriction, and flow is like current. \$\endgroup\$ – Voltage Spike Jul 15 '16 at 19:53
  • \$\begingroup\$ Which is the "first" resistor? Do you use conventional (positive) current, or Electron (negative) current? \$\endgroup\$ – Peter Bennett Jul 15 '16 at 21:50
  • \$\begingroup\$ @PeterBennett 1. say the constant source voltage is 8V, and the first resistor is 4 Ohms (Thus the amps are 2A), I don't get how adding another 4 Ohms resistor causes the voltage to be 4V for each. 2. Negative current \$\endgroup\$ – Edward Seley Jul 15 '16 at 22:43
  • \$\begingroup\$ @MADARAMINOSHI - I've rolled-back to the previous version (i.e. I've "undone" your edit) because, despite what you said in the edit reason, the diagram you added did not match the words in the question (nor in the comment immediately above this one, from the OP). Therefore, it was more likely to confuse readers, than to help them. \$\endgroup\$ – SamGibson Jan 16 '19 at 20:29

When you have one resistor with 10V on one end and 0V on the other end, if you could measure the voltage half way up the resistor it would be 5V.

Now, cut that resistor in half and electrically join the ends you have just cut. It's still 5V so what's the problem?

  • \$\begingroup\$ But if you were to remove that second half, the voltage drop would be 10V for just that first half of the resistor, which doesn't make sense to me. \$\endgroup\$ – Edward Seley Jul 15 '16 at 23:46
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    \$\begingroup\$ If you removed the 2nd resistor there would be NO current flowing (because it opens the circuit) and the voltage drop would be zero i.e. 10V on each side of the resistor due to zero current flow. The problem you have is basically the inverse of my problem... I understand almost intuitively so I find it difficult to see your problem!! Keep struggling until you understand then, keep struggling on the new bigger problem until you understand. Then reach a level when you don't recognize a higher problem. Good luck. \$\endgroup\$ – Andy aka Jul 16 '16 at 0:30

In the case that you add another resistor the voltage is not doubled because you are implying the voltage is fixed (so it provides whatever current maintains that voltage). If instead the current was fixed (relative to the current produced by the single resistor), the voltage WOULD double.

Think of water going downstream in a pipe, at some angle. If you make this angle closer to flat, the water will flow slower. Does the water 'know' this? I am unsure, this is more of a philosophical question.


Another (perhaps useful) explantation is via the electrical field / potential.

A charge flow in a resistive material can only be established if you have a potential gradient which equals an electrical field. (induction let aside). Imagine having the full voltage drop over your first resistor and 0V over your second. This would lead to no electrical field in the second one thus preventing the flow of charges in the second resistor, which effectively prevents any current. This is contradicting the voltage drop over the first resistor.

In fact by assuming a voltage drop of 10V over the first and 0V over the second resistor the field equations get unsolvable.

If you put it the other way round, we can make an experiment: Take some homogenous resistor material with some dimensions: A=cross section and l=length. The resistor material is connected by two metal plates with area A on both sides and voltage is applied. The resulting electrical field is (nearly) as homogenous as in vacuum with the difference that in between the plates there is a (bad) conductor with a specific conductivity. A current density will form on every point of the conductor and only defined by the field and the conductivity. If you integrate the current density over A you get your current.

If you integrate the E-field along the way from one plate to the other you will get the voltage. And moving from one plate along l to the other you will measure and calculate a linearily increasing voltage. The voltage is thus proportional to the position of a probe inside the resistor block.

Now slice the resistor block halfway of l (area is again A), and insert an infinitesimal thin layer of a perfect conductor. As the field was normal to this cross section before you inserted the conductor, nothing will change by doing this. So you can perfectly assume that the voltage potential at this inserted conductor plate is the same as in the untouched resistor at the same position. If you measure the voltage relatively to one of the outer plates or integrate the field from one of those there will be half of the full voltage.

Now you can increase the thickness of the intermediate conductor. If it is an ideal conductor, the electrical field inside it is zero. So it makes no difference how thick you let it grow (i.e. we make a wire here). The field keeps being split into the two parts and keep the same value (in V/m) as in the undivided resistor. But in those half resistors there's only l/2 left and from before we know, the voltage is half the voltage of your source.

You may say: "But I added another resistor of the same value instead of splitting mine into half" The answer is: Repeat the building of the resistor with a material with half the conductivity. The electrical field will stay the same and the current densitiy will be half the old value. But there can't be any difference in voltages, because they do not depend on anything but l and the voltage source in our experiment.

  • \$\begingroup\$ I get that the voltage drop lessens when you add a resistor, but I don't understand why. What directly causes the voltage drop to lessen its first resistor so it has enough voltage for the second? \$\endgroup\$ – Edward Seley Jul 15 '16 at 23:43
  • \$\begingroup\$ @EdwardSeley I tried To address it from a different point oft view in my edit. \$\endgroup\$ – Ariser Jul 16 '16 at 0:40

You are right, the voltage source doesn't know how many resistors there are in the circuit or which arrangement they have (series, parallel, a combination of both). What the source really sees is the 'equivalent' resistance.

You have been presenting an example of two resistors in series, of the same value. To the source, it looks as if there were only one resistor, that is, \$R_1\$+\$R_2\$. If there were another arrangement, the source would still 'see' just a single resistance, which is the equivalent of whatever circuit you have.

In a case where \$R_1=R_2=4\Omega\$ and they are in series, the source sees this as a single \$8\Omega\$ resistor. What determines how much voltage is being drop across each resistor is the current.

I infer that you want to know why the voltage drop changes when you 'suddenly' add in another resistor to a circuit. The thing is that the current changes in a series circuit, and therefore the voltage drop as well (\$V_R=I*R\$).

Consider the following circuit:


simulate this circuit – Schematic created using CircuitLab

Let's assume the potentiometer is at \$0\Omega\$ at the beginning. All the source sees at that time is the value of \$R=100\Omega\$. As you increase the resistance of the potentiometer, you are reducing the current and in consequence the voltage across \$R\$ has to drop (less current, less voltage across individual resistors in this circuit). The rest of the voltage is dropped across the potentiometer.

  • \$\begingroup\$ Thanks for your explanation, one thing still has me hung up (I hope you don't mind), how come less current leads to less voltage drop? I get the math behind this using Ohm's Law, but what is the intuition that causes a decrease in current to lead to a lesser voltage drop? Current is coulombs per second and voltage is joules per coulomb, so shouldn't that mean that voltage is independent of the amount of coulombs (and thus size in current)? \$\endgroup\$ – Edward Seley Jul 16 '16 at 0:51
  • \$\begingroup\$ @EdwardSeley You're welcome! Well, you just said "voltage is joules per coulomb," so voltage isn't independent of coulomb, it depends on it, it's part of its units. Let's put it this way: If there is less coulomb per seconds (current), then there should be less joules (voltage is joules per coulomb). You see what I am getting at? \$\endgroup\$ – Big6 Jul 16 '16 at 1:02
  • \$\begingroup\$ @EdwardSeley Also, Ohms is Joules-Seconds per coulomb squared (j-sec/C^2). So when you multiply I*R (which will give you the voltage drop) , you end up with units of joules per coulombs (voltage). So if current goes down, so does the voltage. \$\endgroup\$ – Big6 Jul 16 '16 at 1:09

When you add another resistor, current will be halved (You can think like the same voltage now have to "overcome" double resistance, so that is the reason it will flow "slower"). Like you said, current is coulombs per second, so you will have half of it per second. Voltage on other side is joule per coulomb, and you have half coulombs, so you will have half voltage.

Or by using Ohms low: $$ Voltage = Current * Resistance $$ Because your current is halved through the resistor, so is the voltage.


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