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I will try to explain where my question is coming from.

Knowing that voltage is the electric potential energy per unit of charge at a given point. And knowing that electrons are pushed from - to go to +

To me that means that the electrons won't have potential energy when arrived at + And they have the most when starting from - They should have more potential energy where they are leaving from and less where they are arriving.

That would mean that electrons have more potential energy at - than at + So in an AA battery (1.5V) I would expect that the voltage at the anode (-) is 1.5V and the voltage at the cathode (+) is 0V

But I am learning the opposite from my course and I do not understand why it is so. enter image description here

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    \$\begingroup\$ This is a great question which still has not received a convincing answer unfortunately. The answers and the comments have all further confused the issue. I think the answer to this question will first have to define its terms very specifically, and a preface on what electricity will have to be added. First we need to settle what electricity from a battery is even doing as a ground for the answer. \$\endgroup\$
    – user7420124
    Jan 22 at 1:37

4 Answers 4

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Conventional current is "upside down" with regard to electrons, due to some arbitrary choices of labelling made centuries ago. That's why I normally suggest ignoring them and focusing on conventional current and voltage analysis.

Electrons are negatively charged. Each electron has a charge of minus 1.602176634×10−19 coulombs.

You could label the "negative terminal" -1.5V and the "positive terminal" 0V. Then your negatively charged electrons flow away from the negative side of the field (because similar charges repel) towards the positive side of the field.

(Electrons themselves do not really "carry" energy, the energy is embodied in the field, and the ability of the field to do work on electrons is the energy.)

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  • \$\begingroup\$ Yep, I know that physically electrons go from - to + due to being negatively charged. My question is more related to the schematic representation of Voltage in a circuit. While I understand why we use an arrow going from - to + (like the direction of the electrons) I am not getting why in this schema Va = 1.5V and Vb = 0V \$\endgroup\$
    – MiniYuuzhan
    Jul 29, 2020 at 13:08
  • \$\begingroup\$ You need to see what the implication of electrons being negatively charged is: that they move in the opposite direction to the sign of the field of voltage. \$\endgroup\$
    – pjc50
    Jul 29, 2020 at 13:13
  • \$\begingroup\$ Just as a gravitational field with objects of positive mass causes objects to roll downhill, towards the field, electrons with negative charge "roll uphill" in an electric field. \$\endgroup\$
    – pjc50
    Jul 29, 2020 at 13:14
  • \$\begingroup\$ Noted, so it seems to be the takeaway from this post for me, since I am working with positive voltage (+1.5V) The point with the higher positive potential must be the + side of the battery to follow the direction of positive charges. In your example of putting -1.5V at the negative terminal (A) and 0V at the positive end (B) Still validate Vb > Va \$\endgroup\$
    – MiniYuuzhan
    Jul 29, 2020 at 13:32
  • \$\begingroup\$ ? in your picture A is the positive terminal and B is the negative terminal. \$\endgroup\$
    – pjc50
    Jul 29, 2020 at 13:58
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0v point of a circuit is completely a matter of choice.

In the above drawing, yo can as well put 0v at point A and then you'll have -1.5v at point B.

It is only the difference that matters.

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  • \$\begingroup\$ So it is because we want to use positive voltage through the circuit that we use Vb = 0V and Va = 1.5V ? What if we keep these A and B and say Vb - Va = 1.5V ? Is it still correct ? Then Voltage at B becomes 1.5V \$\endgroup\$
    – MiniYuuzhan
    Jul 29, 2020 at 13:04
  • \$\begingroup\$ No, you've changed sign. \$\endgroup\$
    – pjc50
    Jul 29, 2020 at 13:05
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The battery positive terminal is called the cathode: -

enter image description here

Picture from Battery university

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  • \$\begingroup\$ It depends on the context. In a lot of cases (e.g. electro-chemistry), the anode/cathode depends not on the potential, but on the current direction and changes on charge/discharge. \$\endgroup\$
    – fraxinus
    Jul 29, 2020 at 13:18
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As I learned, the arrow of the voltage is commonly drawn from + to - like this:

 -   + |
| |    | 1.5V
| |    |
|_|  - ˅

As the technical direction of current is defined as from + to - it totally makes sense to label the Cathode the higher potential because the energy of a positive charge is higher there.

Edit: Image of a complete schematic:

Image of a complete schematic

Note that all of U_0, U_R, I are positive and of course U_0 = U_R.

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  • \$\begingroup\$ No, the arrow is normally drawn in the direction of increasing potential - the opposite of your drawing. Current is draw in the direction of conventional flow, from positive to negative but through the load, not the power source. \$\endgroup\$
    – Transistor
    Jul 29, 2020 at 13:50
  • \$\begingroup\$ I study electric engineering and information technology at KIT (Germany) and we always use the convention in my answer. \$\endgroup\$
    – Gedobbles
    Jul 29, 2020 at 13:57
  • \$\begingroup\$ Can you post an example of a circuit showing a power source and a load and the voltage arrows? Danke! \$\endgroup\$
    – Transistor
    Jul 29, 2020 at 14:05
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    \$\begingroup\$ @Transistor From what I've seen of this notation, Gedobbles's usage is correct. I don't like using arrows for voltage for exactly this reason -- there's no obvious interpretation for the direction of the arrow. \$\endgroup\$
    – Adam Haun
    Jul 29, 2020 at 14:50
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    \$\begingroup\$ @Gedobbles I label them with + and - signs, like the ones on the battery on the left side of your schematic. Here's an example: chegg.com/homework-help/questions-and-answers/… \$\endgroup\$
    – Adam Haun
    Jul 29, 2020 at 15:29

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