Why voltage is same across parallel circuit if it is work done per unit charge? [duplicate]

Question

Suppose we have the following 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?

Answer 1

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.

Answer 2

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.

Answer 3

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.