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enter image description here

I understand that current flows from high potential to low potential.

As you can see in the picture, the top of the circuit has a potential 2V higher than the bottom.

However, since the current is shown as -2A, it actually flows from the low potential side (bottom, -) to the high potential side (top, +).

Is this possible case?

(※The Square is simply a 'Device'. In this case, the Device supplies energy (-4W).)

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  • $\begingroup$ What is that square symbol? Is that supposed to be a voltage source, resistor or some unknown component? $\endgroup$
    – Puk
    Commented Mar 21, 2022 at 9:00
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    $\begingroup$ The Square is simply a 'Device'. In this case, the Device supplies energy (-4W). $\endgroup$
    – ringo
    Commented Mar 21, 2022 at 9:04

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Current can flow from low potential to high potential. In fact, in any circuit where potential isn't the same everywhere and there is current flowing, current must flow from low potential to high potential in some part of the circuit. Note that when you hook up a resistor to a battery, current is flowing from high potential to low potential in the resistor, but from low potential to high potential in the battery.

At steady state (when potentials and currents are not changing in time), this can happen in devices that are sources of some type of electromotive force (emf), such as a battery, solar cell, thermocouple (through the Seebeck effect) or a piece of circuit moving in a magnetic field (motional emf).

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  • $\begingroup$ Thank you for answer. Then, since this case is a part of the whole circuit (due to the grayed out part in the figure), is my question a possible case? Conversely, if there are no grayed out parts, is this something that can never happen? Thank you. $\endgroup$
    – ringo
    Commented Mar 21, 2022 at 9:38
  • $\begingroup$ This case is certainly possible. For example, this will happen if the gray part is a 1 Ω resistor. I'm not sure what you mean by "no grayed out parts". If you mean there is nothing connected to your device (i.e. the device is open-circuited), then this couldn't happen because there would be no current flow. $\endgroup$
    – Puk
    Commented Mar 21, 2022 at 9:46
  • $\begingroup$ I will correct it. If the gray part is a battery connected with the + pole facing down, is this possible or not? $\endgroup$
    – ringo
    Commented Mar 21, 2022 at 10:02
  • $\begingroup$ Not if it's an ideal battery: Kirchhoff's voltage law would be violated (the sum of voltage drops along the loop must be zero). It is possible if e.g. it's a 2 V battery with a 4 Ω resistor in series (which could be the internal resistance of the battery). $\endgroup$
    – Puk
    Commented Mar 21, 2022 at 10:08
  • $\begingroup$ Okay, I'll try to understand. Thank you for answer. $\endgroup$
    – ringo
    Commented Mar 21, 2022 at 10:10
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By convention, the electric current positive direction is the direction in which positive charge moves. It is unclear in your schematic what the "rectangle on the right" represents: a voltage source, a resistor, or something else?

I would read the current in your schematic as follows:

  • current magnitude is $2 \text{ A}$
  • positive charge moves from right to left; hence, potential on the right is higher than potential on the left

If you are still confused by the negative sign, just flip the arrow of the current direction (rotate by 180$^\circ$) and put positive sign next to the current.

Can current also flow from low potential to high potential?

In the absence of external forces, positive charge always moves from higher to lower potential, much like objects fall from higher to lower altitude. However, charges can go from lower to higher potential if there was something providing the force necessary to overcome the potential difference; e.g. a voltage source (battery). This is the same as you climbing to a higher altitude by using an elevator.

It is also possible for positive charge to move from lower potential to higher potential by means of power converters. These use semiconductors (diodes and transistors), inductors and capacitors to transfer energy (power) from input to output. Physics still apply to power converters, but it is the way how they work that enables positive charge to flow from lower to higher potential.

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  • $\begingroup$ Thanks for your Answer, Gulin. The square is the device that supplies 4W in this circuit. (Power is -4W) Also, the following facts never change: Fisrt, The top of the circuit is 2V higher than the bottom. Second, The square cannot change the direction of the current. So, in the gray part on the left side of the circuit (the omitted part of the circuit), there could be a device that reverses the direction of the current flow you're talking about? $\endgroup$
    – ringo
    Commented Mar 21, 2022 at 9:19
  • $\begingroup$ @ringo Why do you think there should be something that reverses the current direction? The arrow indicates where it is "assumed that positive charges go", and the negative sign indicates that the true direction is opposite to the assumed direction. If you draw arrow from right to left and put $+2 \text{ A}$ for the current, is it clear in that case? $\endgroup$
    – Marko Gulin
    Commented Mar 21, 2022 at 9:21
  • $\begingroup$ Uh... Since '-2A' is the opposite of the assumed direction, the current will actually flow from the bottom of the circuit up. However, the upper part of the circuit is obviously 2V higher than the lower part. (I think this fact is absolute.) So, I was asking if current can flow from a low potential (bottom of the circuit) to a high potential (top of the circuit). (I'm not familiar with English yet, so please understand that the context may be strange.) $\endgroup$
    – ringo
    Commented Mar 21, 2022 at 9:28
  • $\begingroup$ @ringo Voltage source (the rectangle in your schematic) provides force to charges which enables them go in the "opposite direction". The equivalent would be you climbing to a higher altitude using an elevator. $\endgroup$
    – Marko Gulin
    Commented Mar 21, 2022 at 9:32
  • $\begingroup$ In fact, when I took this class, the key was only calculating the power of the device. (It didn't matter if the direction of the current was possible or impossible.) So it wasn't even mentioned whether the device (square) could change the direction of the current. Are you saying that the device can change the direction of the current because it supplies energy to the circuit? $\endgroup$
    – ringo
    Commented Mar 21, 2022 at 9:49

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