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Suppose we have the following circuit:

parallel circuit

If voltage is work done per unit charge, why voltage is same across each resistor if the charge has to do more work in resistor R2 than in resistor R1?

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    $\begingroup$ Why do you think it is more work for the electron to "cross" the $R_2$ than the $R_1$? $\endgroup$
    – hyportnex
    Commented Mar 13, 2021 at 23:12
  • $\begingroup$ Because R2 is 200 Ω,100 Ω more than R1. $\endgroup$
    – Neptuno
    Commented Mar 13, 2021 at 23:18
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    $\begingroup$ OK, then my question is why do you think that passing through a resistor having larger resistance takes more work than passing through a smaller resistance? Hint: when you move a charge in empty space what resistance the charge has moved through? What is the resistance of space? $\endgroup$
    – hyportnex
    Commented Mar 13, 2021 at 23:21
  • $\begingroup$ I don't know. I guess you are referring force of attraction?? I really don't know. $\endgroup$
    – Neptuno
    Commented Mar 13, 2021 at 23:34
  • $\begingroup$ I wonder if your question is a little different than phrased here. I think what you want to know is: "Consider two resistors $R_1 = 100\Omega$ and $R_2 = 200\Omega$. Resistance is essentially how difficult it is for charges to flow. How is it possible, then, that more work is needed to push an electron through $R_2$ when they're connected in series, but the same work is needed to move an electron through $R_2$ as $R_1$ when they're connected in parallel? Based on my understanding of resistance, it seems like the work needed should always be proportional to the resistance." Is that your Q? $\endgroup$
    – jdphys
    Commented Mar 14, 2021 at 15:18

3 Answers 3

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The work is the same. Both resistors would require the same work per coulomb to move a test charge through the resistor. One way to get an intuition for why this is true is from a physical picture. This won't be true always, but let's imagine the reason why the resistor $R_2$ has double the resistance is because it's twice as long. What happens is that the electric field generated in $R_2$ is half as strong as the electric field in $R_1$. Consequently, the electric force on a test charge in $R_2$ is also halved. But the work is the same, because charges move twice as far in $R_2$, and $W = Fd$.

Ultimately, the reason the voltage drop is the same across both resistors is that total energy is conserved. Any individual charge moving through $R_1$ starts at the positive terminal of the battery and ends at the negative terminal (assuming conventional current). The exact same thing is true of any single charge moving through $R_2$. If the two charges start at the exact same place (the positive terminal), then they must begin their journey with the exact same amount of electric potential energy. If they end at the same place (the negative terminal), then they must end their journey with the exact same electric potential energy. Hence, both charges must lose the same amount of electric PE along their path. But if the wires are ideal, then the only place they can lose their electric PE is inside the resistor that they traveled through.

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  • $\begingroup$ But then, what about resistors in serie? Suppose two resistors in serie: R1 =100 Ω and R2 = 200 Ω. According to you, let's imagine the reason why R2 has double the resistance is because it's twice as long. Then the work is the same because R2 is twice as long and the electric field generated is half as strong as the electric field in R1. But that is not true. Voltage in resistors in serie is different. $\endgroup$
    – Neptuno
    Commented Mar 14, 2021 at 7:15
  • $\begingroup$ My statements in the first paragraph are applicable for the specific example you asked about in your original question. The electric field strength is only proportional to length in scenarios where the work to move a charge through the resistors is constant (as is the case with parallel resistors). If the resistors are in series, then the work would not be the same and $R_2$ would not have half the electric field strength. $\endgroup$
    – jdphys
    Commented Mar 14, 2021 at 14:45
  • $\begingroup$ I think we've identified the issue here. You're taking the idea that "the work done is proportional to the resistance" and applying it too broadly. The statement $W \propto R$is not true in general--it's only true for series resistors. Your misconception above about parallel resistors has arisen because you're applying the idea that $W \propto R$ to parallel resistors, but we can't do that. $\endgroup$
    – jdphys
    Commented Mar 14, 2021 at 14:55
  • $\begingroup$ To think about it generally, look at what each resistor's ends are connected to. In parallel, the resistors' "front sides" are electrically in contact, and so they're at the same electric potential V. It's analogous to say the front sides are at the same "height" (if you want to make an analogy to gravitational PE). In parallel, their "back sides" are also electrically connected and thus at the same potential V. Hence, the change in electric PE (and thus the work) are the same through both resistors. But in series, this isn't the case. $\endgroup$
    – jdphys
    Commented Mar 14, 2021 at 14:58
  • $\begingroup$ In series, the front of $R_1$ is connected to the positive terminal and thus at the same potential $V$ as the positive terminal. The back of $R_1$ is connected to the front of $R_2$, and so those two locations are at the same potential $V$. The back of $R_2$ is connected to the negative terminal, and so those two locations are at the same potential $V$. These contacts are important--in static scenarios, those electrical contacts determine how charges get distributed, which contributes to the strength of the electric field. $\endgroup$
    – jdphys
    Commented Mar 14, 2021 at 15:04
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The entire question invites us to think in analogies. And analogies can be wonderfully illuminating but can also be bad pitfalls leading our minds astray. Still, let's go with it and do a little free association.

I think of voltage as potential energy. The electron-volt eV is a unit of energy.

The actual work done moving charges through is manifest in the current. This is why Ohm's law says that $I = U / R$. Intuitively you seem to think that the work done pushing the electron through the higher resistor is higher, but is that really so? We intuitively think of friction when pushing an object along a path and so the analogy might offer itself. But that idea of "it takes more work to push electrons through a higher resistor" is too simple a thought.

First, to think "work" we must consider current. The current through the higher resistor is less given equal voltage, and therefore, power dissipated is higher on the lower resistor. $P = U^{2} / R$ and because work is related to power as $W = P \cdot \Delta t$, in actuality the work through the lower resistor is higher during the same time! Since $P = U \cdot I$ and therefore $W = U \cdot I \cdot \Delta t$ with $I = Q \cdot \Delta t$ you have $W = U \cdot Q \cdot \Delta t$ with $I$ being lower, not higher, at the higher resistor. When we pay our electric bills, we pay for kilowatt-hours, i.e., the amount of $P \cdot \Delta t$ which is "electric work" and that electric work is obviously higher through the lower resistor!

If we take that mechanical work where that intuition comes from as $W = \int F_s \cdot ds$, the force is at all times determined by the electric field times the charge $F = E \cdot q$ of the electron. So that force is constant, and the work is constant if the distance is constant. But what does "distance" man here? What is it with that distance? Isn't the electric field dependent on distance?

Whereas the electric field in the parallel plate experiment is dependent on the distance of the plates $E = U / d$, in a conductor this is different! The length of the conductor does not change the electric field pushing the electrons through the conductor. Imagine you add just more wire from the plate to the battery in the parallel plate experiment, it doesn't change the electric field between the plates.

And when I say "velocity of the electrons" I think that's only an intuitive analogy, in reality a conductor will have a certain number of movable electrons per cross section area, and so, if you have twice the cross section, the same electrical field will push (or pull) twice the electrons through the cross section (if you remember, the resistance of the same conductor material is inversely proportional to the cross section), hence twice the current. It's not even that the electrons necessarily move faster, it's that there are more that are moving.

I would challenge the idea that your best analogy of one resistor twice greater than the other is a wire twice its length, although that is also true. The cross section area consideration I just gave is the right one, since what makes a material conductive is the availability of movable electrons, and the actual speed of each electron doesn't really matter when what matters is how many go through a cross section during a given time interval. Why then does the length of the wire increase the resistance? Because if you have any given cross section area, that has one infinitessimal resistance, then each of those infinitessimal resistances will add up during the length of the wire.

Of course in the end, the best we have to understand what's going on is our laws and formulas about $\vec E, Q, U, I, \Delta t, W, P, \vec F_s, s$ and all that good stuff, and this and only this tell us whether or not something does more or les "work" in one or the other part of the circuit.

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    $\begingroup$ Consider that the second resistor has twice the resistance because it's twice as long. The force wouldn't be equal in the two resistors, or else the work wouldn't be equal in the two resistors. $\endgroup$
    – jdphys
    Commented Mar 14, 2021 at 2:33
  • $\begingroup$ Why do you downvote my answer if we both are saying similar things? I started out with voltage as energy and you say "energy is conserved" -- same thing! You have "twice as long", I have the complementary intuition "half as thick". The distance of the electric field E = V / d applies to the parallel plates, but it does not apply in a conductor, because you are not interested in an electron moving from one end to another. All that matters is how many charges are passing through a cross section per unit of time. $\endgroup$ Commented Mar 14, 2021 at 15:05
  • $\begingroup$ This statement is not correct: "The work done pushing the electron through the higher resistor is higher." The exact same work is needed to push an electron through two parallel resistors. I thought these statements were phrased too broadly: "So that force is constant. The distance is constant." The force and distance won't always be the same for the two resistors in parallel. Your answer doesn't mention "half as thick," and I think it would be valuable to edit the answer to clarify that F and d are not always the same in parallel resistors--only for the specific example of half thickness. $\endgroup$
    – jdphys
    Commented Mar 14, 2021 at 15:15
  • $\begingroup$ I was echoing the intuition of the question in my first paragraph and then I questioned exactly that in my second paragraph. You will notice that this was a debate in the initial comments under the question. Developing a thought while discussing it. $\endgroup$ Commented Mar 14, 2021 at 15:42
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    $\begingroup$ I do include an example, but I state that it's an example. I very specifically said "This won't always be true, but let's imagine..." I'm looking forward to reading the edited version. $\endgroup$
    – jdphys
    Commented Mar 14, 2021 at 21:07
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I will explain this in terms of the water analogy. Consider a river bed that follows the same setup as the circuit. The water level is the voltage. Higher voltage = higher water level = more energy per unit of water/charge. The resistors are dams that restrict the water flow. The battery is like a pump that works to keep the water level at a certain height.

You can imagine that any part of the river bed that is connected directly will very quickly assume the same height as the rest of the water. If any part is lower water will rush in to equalize the water height. You can translate this back to the circuit any wire that touches the supply voltage will quickly take on the same voltage. Any part of an (ideal) wire will have the same voltage because if it doesn't, charge will redistribute itself until it is. This changes when you encounter a resistor: resistors restrict the flow of charge and allow one end to have a higher voltage then the other. Similar to how the dam allows the water on one side to be higher.

So using this reasoning it makes more sense that the voltage difference over $R_1$ is the same as the difference over $R_2$. Let's go back to the water analogy one more time: we have two dams connected to the same lake. One dam has a large opening near the top (low resistance) and the other one a small opening (large resistance). When a water particle moves through the opening it loses potential energy because it falls down. It doesn't matter through which dam it moves: it loses the same amount of potential energy i.e. it falls the same height. But more water will move through the larger opening so low resistance = more current.

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