# Homework Help: A simple question about the flow of the electricity

1. Dec 23, 2013

### DorelXD

Hey guys! I'm trying to get a bettter undesrstanding of how the electric circuits work. But I do have a rather simple question. So, in the internal circuit we give energy to the charge. Let's say that I have an electricity source and two resitors, like this:

So when passing through circuit elements the charge gives up energy, so that the sum of the amount of energy it gives to each resistor is equal to the initial energy. But why? Why dosen't the charge give all the energy to the first resistor it encounters ?

And why is there no potential difference between the two red resistors? I don't get that :( . I know it's a simple concept but I can't understand why the charge loses energy only when it passes through resistors ?

Last edited: Dec 23, 2013
2. Dec 23, 2013

### HallsofIvy

Because it doesn't have to. Think of "resistance" as friction as the current passes through the resistor. There will be a certain amount of friction for the length of the resistor which causes the charge to lose energy (force times distance). The amount of energy lost depends upon the friction and distance but is not, in general, all of the energy of the current.

3. Dec 23, 2013

### DorelXD

Ok, I think I get it. But, from what I read, and using the model you gave me, when the charge comes out of the second red resistor it should lose al its energy, right ? So if the charge loses all the energy, then how can it go back to the source ?

4. Dec 23, 2013

### SteamKing

Staff Emeritus
Why should the charge lose all of its energy after leaving the second resistor? The amount of energy lost is still proportional to the amount of resistance.

5. Dec 23, 2013

### nsaspook

Which resistor is the first resistor? Is it possible the energy in the circuit flows from green to red down both sides of the circuit at the same time?

6. Dec 23, 2013

### DorelXD

So the movement of the electrons has nothing to do with the energy they store ?

7. Dec 23, 2013

### nsaspook

Nothing is not the right word to describe the relationship in general but in a simple circuit like yours you might want to think about the electrical source and what it provides to produce current flow.

8. Dec 23, 2013

### CWatters

The wire is assumed to have zero resistance.

If that's not correct and it has some resistance then the distribution of energy around the circuit will be different.

Perhaps it helps to understand that resistors are just little rods of some material that isn't an ideal conductor. So two resistors in series are equivalent to one longer rod or any number of shorter rods.

Why doesn't the charge give all the energy to the first few mm or Î¼m of one of the resistors.

9. Dec 23, 2013

### DorelXD

I rephrase my question. I drew a wire, and I also chose to represent two cross-section through it.

In other, word, how do I know that in the same interval of time the same amount of charge passes through the AA' and BB' ? It is quite intuitive, but I would like a justification. I can't find a simple argument, and it's driving me nuts. I can't gasp some simple concepts about electricity.

Last edited: Dec 23, 2013
10. Dec 23, 2013

### Staff: Mentor

If it were to happen that in a single continuous circuit fewer charges were passing one point than another, then the only explanations could be that charge was vanishing somehow, or it was accumulating or bunching up at some point intermediate. There is no mechanism shown here that could allow either scenario.

http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]

Last edited by a moderator: May 6, 2017
11. Dec 24, 2013

### rude man

Charge cannot bunch up anywhere in a circuit. Not even in a capacitor. This fact is based on one of Maxwell's equations, which says that the total voltage drops around any circuit is zero. If current bunching took place then that statement would no longer be true. Kirchhoff recognized this and formulated his KCL and KVL laws.

12. Dec 24, 2013

### DorelXD

But what if the current bunched in some place ? Why isn't that possible ?

13. Dec 24, 2013

### Staff: Mentor

Like charges repel each other. They try to push each other apart, and the closer they are the stronger they push apart so it follows they actively work against bunching up.

There is a difference between curent and charge, but I'm sure you realize this.
http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]

Last edited by a moderator: May 6, 2017
14. Dec 24, 2013

### DorelXD

Ohhhh, right. :) :) :) :) :) :) :)

Well, current is charge that flows.

Now, please, help me to understand the KVL law. I tried to make an analogy with the gravitational field.

I would like to post my questions gradually. Here's my first.

Let the point A be at the groudn level. So, here I go: I move the object from A to B and it gains potential energy. Since the gravitational force is conservative the path dosen't matter. By moving the object back to A again it loses al the potential energy.

Now in a circuit with only a battery: I move the charge from - to + and it gains electrical potential energy. So, if I move it back to - it should get there with no potential energy, right ? So, for me it seems that the potential energy the charge gets is converted to kinetic energy, so that the charge can move back to where it came from. I don't see how we could use some of the energy of the charge to do something else with it. That is, I can't see how putting a resistor in the circuit will use the energy in another way, so that we do something useful with it.

What am I doing wrong? Is it because the charge dosen't actually convert the potential energy to kinetic energy ? That is, the movement of the charge isn't based on the potential energy it gained when it exited the internal circuity, but on the collisions with other charges ? But how's that possible? The potential energy should change while the charge is moving beacause it changes the position relative to the battery :( :( :( :( :( . I'm confused...

15. Dec 24, 2013

### nsaspook

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elevol.html#c1

16. Dec 24, 2013

### DorelXD

17. Dec 24, 2013

### Staff: Mentor

That is not a good analogy. A higher voltage battery doesn't cause a given quantity of charge to flow in the external conductor at a greater speed; it causes more charges to flow and with an unchanged speed (if you want to put it that way).

The model you are thinking of would be thermionic emission in a vacuum tube, where electrons are liberated from the cathode and fly through a vacuum to the anode terminal of the high voltage battery. The higher the voltage, the faster the electrons are accelerated and the more KE each gains. This energy can be liberated in spectacular fashion when the electron crashes into the metal anode.

But the unimpeded flight of a charged particle through the emptiness of a vacuum is totally different to the "bucket brigade" transfer of electricity that goes on in the sea of free electrons in a metallic conductor. The electron that enters the piece of wire at one end of the battery is not the same electron that disappears back into the other end of the battery.

18. Dec 24, 2013

### rude man

Plus, it would violate conservation of energy.

19. Dec 24, 2013

### rude man

When a mass moves from the top to the bottom the potential energy at the top changes to kinetic energy at the bottom: mgh → 1/2 mv^2.

In the case of charge q the potential energy 'at the top' changes to heat energy 'at the bottom': qE → Q
E is the battery emf.

20. Dec 24, 2013

### DorelXD

Wow, I begin to understand :) . How exactly is the conservation of energy principle violated, could you please explain in more detail ? I'm dumb and I don't understand.

21. Dec 24, 2013

### rude man

As I said, if charge bunched up at a node then the total voltage drops around a circuit would not be zero, but that would violate the conservation of energy since the electrostatic field is conservative.

22. Dec 24, 2013

### haruspex

It wouldn't violate conservation of energy. In fact, bunching can occur, but the positive and negative bunches cancel out, as in a capacitor. Total charge in the circuit (including the power source) is obviously conserved, as it is a closed system. Total charge within the power source tends to be conserved because of the way the power is generated.
Even without a capacitor as such, circuits have a small amount of capacitance, and inductance. If a resistance suddenly increases some charge bunching will occur, but since the capacitance is usually low even a small bunching will generate a large voltage, tending to spread the charge out again.

DorelXD, from the OP, it seems to me that your initial misconception was that the energy was a property of the charge, and the charge somehow carried the energy around with it, spending it along the way. Rather, it is charge x voltage. The resistors being the same, there'll be the same voltage drop across each.
The wires will also have some voltage drop, since they will have a little resistance, but it's much smaller.

23. Dec 24, 2013

### rude man

The OP's circuit consists of a battery and two resistors. There can be no bunching at any node therein.

Introducing parasitics in a first EE or physics course leads only to confusion by the OP.

24. Dec 26, 2013

### DorelXD

Thank you all for your time! It begins to make sense, but I'm still struggling a little.

25. Dec 26, 2013

### DorelXD

So let's put it that way: let's say I travel to the top of the building. Say that I'm a lazy person , and for reaching the top I take the elevator. But when I go back, I take the stairs. When I first reached the top my potential energy raised. When I climbed down the stairs my potential energy went back to 0. Is that what's happening inside a circuit ? I still can't picture it clearly in my mind..