# Why must the voltage drop over resistors in parallel be the same?

I am wondering why it is the case the when resistors are connected in parallel the voltage drop over them must be the same?

The reason I have seen arguing for this is that the electrons before the resistors appears in the same place, and the electrons after the resistors appears at the same place. And that means that the electrons that goes through resistor $$\R_1\$$ and those going through resistor $$\R_2\$$ are at the state before and after.

But why does this argument hold? Voltage is energy per charge, why couldn't it be the case that even though the electrons are at the same place before and after going through the resistors, the work done on the electrons going through $$\R_1\$$ is higher than the work done to the electrons going through resistor $$\R_2\$$?

• I have the impression that you think, the current through the resistor would cause (produce) the voltage across the resistors.Right? This would be totally wrong. No current without driving voltage! A current within a resistor needs an E-field - otherwise the electrons cannot move ! When we have two resistors which are - at the same time - connected to the same voltage source, we say they are in parallel. That is the correct sequence. – LvW Mar 4 at 14:28
• @LvW Thank you very much. Can we say that the voltage drop must be the same over both resistors because they have the same electric field over them? – user394334 Mar 4 at 14:49
• Have you learned Kirchhoff's Voltage Law? (It's essentially the simplified version of Math Keeps Me Busy's answer). – The Photon Mar 4 at 15:46
• Because Voltage is a state function, not a path function. – Mitu Raj Mar 4 at 15:49
• Suppose you and your friend start at a certain place and climb a mountain. But you hike up the north face of the mountain and your friend climbs the south face. When you both get to the top of the mountain have you both climbed the same height difference? – The Photon Mar 4 at 18:34

The reason I have seen argumenting for this is that the electrons before the resistors appears in the same place, and the electrons after the resistors appears at the same place. And that means that the electrons that goes through resistor $$\R_1\$$ and those going through resistor $$\R_2\$$ are at the state before and after.

Except when one is being precise, that is a fairly good explanation.

But why does this argument hold? Voltage is energy per charge, why couldn't it be the case that even though the electrons are at the same place before and after going through the resistors, the work done on the electrons going through $$\R_1\$$ is higher than the electrons going through resistor $$\R_2\$$?

[It sometimes is the case that the work done is different depending upon path. In particular, when there is a time-varying magnetic field involved. We will get to that. But first:]

$$\nabla \times E = -\frac{1}{c}\frac{\partial B}{\partial t}$$

If there is no time-varying magnetic field, this implies that

$$\nabla \times E = 0$$

which means that E is a conservative field. Being a conservative field implies that E is the gradient of a potential function, (which we call V)

$$E = \nabla V$$

V, being a potential function implies that the work done to move an electron from one point to another does not depend upon the path taken. That is why the voltage across two resistors in parallel are said to be equal. And, this is true whenever there is no time-varying magnetic field present.

[However, if there is a time varying magnetic field affecting the circuit, one needs to model this as if there were a transformer, or voltage source, present in one or both of the paths through the resistors. The voltages induced by these transformers will affect the voltage "dropped" by each of the resistors.]

• To whomever downvoted this answer, could you provide a rationale for your downvote? Thanks. – Math Keeps Me Busy Mar 4 at 14:33
• I think that if this is possible, the software should only allow downvoting if there is a reasoned comment. That is, it should make the downvoter make an effort and not just click on the button. – Circuit fantasist Mar 4 at 16:37
• @Circuitfantasist This has been debated in various forms before across the SE network. The consensus is against your proposal. – Matt Mar 4 at 17:26
• Thank you very much! But is it possible to actually prove that the field is conservative? That is, if we assume we have no time-varying magnetic force is it possible to derive the expression $\nabla \times E = 0$? – user394334 Mar 4 at 19:12
• If the magnetic field is not varying, then $\frac{\partial B}{\partial t}=0$. Those are just two ways of saying the same thing. Then the question is, is Faraday's law true? That is, does $\nabla \times E = -\frac{1}{c}\frac{\partial B}{\partial t}$? For that, I can only offer that humanity's rather extensive experience with electro-magnetics seems to confirm its accuracy. – Math Keeps Me Busy Mar 4 at 19:23

I think you're over-thinking this.

Here's how the voltage across two resistors in parallel is measured:

simulate this circuit – Schematic created using CircuitLab

Since R1 and R2 are in parallel they are each connected between the same nets, net A and net B. That means both R1 and R2 experience the same voltage, that is the voltage between those nets. The voltmeter also experiences that voltage and displays it as well.

Your misunderstanding could result from misunderstanding the relations between voltage, current and resistance.

I prefer to compare voltage to "pressure", in the case of parallel resistors, the voltage (pressure) on both is the same.

That voltage (pressure) will make charge / electrons flow through a resistor, compare it to how a liquid or a gas flows through a pipe. In electronics we call that flow of electrons (or charge) current and that current is directly a measure of how many electrons flow through a conductor or a resistor.

If the same voltage is applied to two resistors because they're in parallel (see above) and one resistor has a high value while the other resistor has a low value, then the low value resistor makes it "easy" for the electrons to flow through it so more electrons flow (per second) so the current will be higher (for the low value resistor). For the high value resistor it takes more effort so less current flows.

• I do kind of understand what you mean, and your explanation does resemble what I have seen before. But what I don't understand is even though the voltage is measured that way, why must the work/charge be the same for charges going through $R_1$ and $R_2$? – user394334 Mar 4 at 14:39
• why must the work/charge be the same for charges going through R1 and R2? Maybe you mean: "Why do the same currents flow through R1 and R2?" That is ONLY the case when the values of R1 and R2 are the same. If R1 and R2 do not have the same value then the currents will be different. The most current will flow through the resistor with the lowest value. Realize that current = charge flowing per second. – Bimpelrekkie Mar 4 at 14:44
• @Bimpelrekkie, no the OP is not asking about current but about "work per charge", which is how the physicist sees voltage. – Math Keeps Me Busy Mar 4 at 14:47
• @Bimpelrekkie No I mean work/charge(work per charge, joules per coloumb), the voltage, not the current. What I am wondering why this quantity must be the same for charges going through $R_1$ and $R_2$. – user394334 Mar 4 at 14:47
• @Bimpelrekkie "over-thinking" seems to be understatement here. Maybe you need to remove "rest of circuit" and net names from your schematics and just add a second voltmeter? – Maple Mar 4 at 14:49

Here's a more elementary explanation:

In an ideal wire with 0 resistance, the electric potential is the same everywhere in the wire. In other words, the voltage between two parts of that wire is 0. This is a direct consequence of Ampere's Law (or alternatively, Kirchoff's Voltage Law).

This is true of both ends of the resistors. Thus, since the two ends of the resistors are the same two wires, both ends have the same difference in potential, aka. the same voltage.