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In this example:

circuit

I want to calculate the voltage that exists between point a and b.

Of course this is not about getting some homework done, I really want to understand this.

So, this is my reasoning:

The tension $U_{ab}$ is the tension from a to b.

It has the same value as the tension from b to a, just with an opposite sign.

So, the value will be a positive or a negative volt (there is one volt difference, we just don't know the sign).

Now, I suppose this is a common scenario where electrons are the ones carrying the charge.

Electrons carry a negative charge and they move from negative charged zones to more positive charged zones to feel more relaxed there / to reach an equilibrium with their environment.

So, electrons move from the - to the +.

If I go from a to be I am doing the opposite thing, going from + to -.

So, the math go as follows:

$$U_{ab} = - (-2 V) - 3V = -1 V$$

The solutions sheet in this example say it's +1 V, but not why.

May you please help me to understand it?

Visualizing it:

In the next image from Wikipedia we can visualize the situation.

image

Suppose it's a real battery where electrons are going out at the negative side and being attracted in the positive side.

The arrow that represents voltage is very clearly drawn as a pushing force from - to +.

Sign conventions:

I only found passive and active sign conventions. Both talk about what is considered positive for current. Current in or out. But it does not talk about tension. Actually, it represents tension in both cases going from $-$ to $+$.

Passive: current is being consumed.

enter image description here

Active: current is being created.

enter image description here

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  • $\begingroup$ Would you say $U_{eb}$ is -3 V or +3 V? $\endgroup$
    – The Photon
    Nov 30 '19 at 23:06
  • $\begingroup$ @ThePhoton I would say it is $-3 V$ because it goes from + to -, and electrons want to carry the charge from - to + $\endgroup$ Nov 30 '19 at 23:41
  • $\begingroup$ And how do you describe the effect of the voltage source connected between those nodes? $\endgroup$
    – The Photon
    Dec 1 '19 at 0:20
  • $\begingroup$ What does “tension from $a$ to $b$” mean? The voltage “between” $a$ and $b$ can be defined as the potential at $a$ minus the potential at $b$ or vice versa. $\endgroup$
    – G. Smith
    Dec 1 '19 at 0:57
  • $\begingroup$ Take your pick and be consistent, or use the definition in your book for $V_{ab}$. $\endgroup$
    – G. Smith
    Dec 1 '19 at 1:07
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I want to calculate the voltage that exists between point a and b.

I'm just going to try to answer this question without digging in to any of the side issues you raised.

It will help to remember a schematic is a highly abstracted view of an electric circuit. You can think of it as a way of visually representing a set of equations.

For example, a resistor designated R1 with value $R$ connected with its (arbitrarily chosen) positive node at a and negative node at b is a visual representation of the equation

$$I({\rm R1}) = \frac{V_a-V_b}{R}.$$

(For your problem this is actually irrelevant, since they haven't given you the value $R$, they've just told you one terminal is at -2 V relative to the other terminal)

Similarly, an ideal voltage source with value 3 V connected with its positive terminal at node e and negative terminal at node b is shorthand for

$$V_e - V_b = 3\ {\rm V}$$

The advantage of using abstract models like these schematic diagrams is that it saves you having to consider numerous physical details like whether the charge carriers are positively or negatively charged, what electric fields are present around the devices, etc. You should take advantage of this to focus your attention on the information presented in the schematic diagram and how it can be used to solve the problem, rather than complicate the problem by bringing in details not needed to find the solution.

May you please help me to understand it?

So in this specific problem you want to find the voltage between a and b. You have a diagram that shows you that

$$V_e - V_b = 3 V$$

and

$$V_a - V_e = -2 V$$

From simple arithmetic you know

$$ V_a - V_b = (V_a - V_e) + (V_e - V_b)$$

so

$$ V_a - V_b = -2\ {\rm V} + 3\ {\rm V} = +1\ {\rm V}$$

No information about the type of charge carriers in the system, or the passive sign convention, or even the actual behavior of resistors is needed to solve the problem from the given information.

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  • $\begingroup$ Thanks a lot sir Photon. This is a great answer . I have one more question regarding this topic. So $V_e - V_b = 3V$ is valid for / independent from Conventional Current / Electron Flow because it’s true for both cases. But is it passive or active sign convention or does it also not matter? $\endgroup$ Dec 1 '19 at 21:56
  • $\begingroup$ @AlvaroFranz, active/passive sign convention is about how we define the direction of current and voltage for each branch. It doesn't affect this problem because you don't have to consider any currents at all to solve it. $\endgroup$
    – The Photon
    Dec 1 '19 at 22:00
  • $\begingroup$ Much clearer now. Your help is very appreciated. Good day. $\endgroup$ Dec 1 '19 at 22:02
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Now, I suppose this is a common scenario where electrons are the ones carrying the charge. Electrons carry a negative charge and they move from negative charged zones to more positive charged zones to feel more relaxed there / to reach an equilibrium with their environment.

This is a recipe for lots of sign errors, including probably the one you're asking about here.

The rules for predicting how electric circuits work were mostly discovered before we knew about atomic structure, before the discovery of the electron, and in ignorance of the sign of the dominant charge carriers. Apart from a few esoteric exceptions, circuit analysis is exactly the same when you consider positive charges moving from positive to negative voltages as when you consider charges with the other sign moving the other way. Except that with two negatives you have extra opportunities to make a sign error.

There ain't nuthin' to be gained by "doing the opposite thing," as you write, especially if you change the sign but not also the direction of the charge flow.

A positive charge that starts at $b$ and goes to $e$ would gain 3V from the battery, then lose 2V across the resistor between $e$ and $a$. So the potential at $a$ is 1V more positive than the potential at $b$. Don't be distracted by the fact that we guessed the wrong sign for majority charge carriers in metals in the distant past.

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  • $\begingroup$ Thanks a lot for answering. I can understand the fact that it took some time for electricity to evolve as a science, and that maybe brought in this sign confusion. But I do not want to just cover the situation with: "Okay it's going to be the same whatever direction it goes". I would like to understand the reality. I have been taught that electrons carry the charge from the battery end that spits out the electrons (marked with -) to the battery end that attracts electrons (marked with +). If + means more potential... why would electrons move from less potential (-) to more potential (+)? $\endgroup$ Dec 1 '19 at 19:32
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    $\begingroup$ @Alvaro, because the electron has negative charge, so the force on it due to an electric field is in the opposite direction to the field vector. $\endgroup$
    – The Photon
    Dec 1 '19 at 20:06
  • $\begingroup$ Appreciate this answer too but The Photon expressed it in a way that is clearer to me. Thank you. $\endgroup$ Dec 1 '19 at 22:00

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