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Just wondering why exactly the current is flowing in the directions depicted in the image below? It is my (rudimentary) understanding that current flows from regions of high potential to regions of low potential. In this case I imagined that the current would flow from the anode of the \$18V\$ battery to the top node and then there would be 3 courses the current could take:

  1. The cathode of the \$18V\$ battery
  2. The anode of the \$12V\$ battery
  3. the cathode of the \$12V\$ battery

My thinking here would be that the current would just go to the cathode of the \$18V\$ battery since that would appear to be the highest potential difference. Although, how do we know what the potential of the cathode for the \$12V\$ battery is set at? If voltage of a battery is just describing a difference between the positive and negative terminals, then couldn't it be the case that the difference between the anode of \$18V\$ battery and the cathode of the \$12V\$ battery is greater than that of the difference between terminals of the \$18V\$ battery? Which would then drive the (positive) current out of the \$18V\$ anode to the top node then down through the \$6\Omega\$ resistor to the bottom node, then through the \$4\Omega\$ resistor into the cathode of the \$12V\$ battery. Furthermore why does the current split off and go down each available path at each node? If there's one path that has the greatest potential difference I would think the current would simply take that path to the exclusion of the others.

I have italicized more specific hangups I have when viewing this diagram and thinking about circuits, more generally, but I would appreciate someone's step-by-step thought process on how to understand the general path of current in any circuit, as well. Thanks!

Circuit Schematic

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  • \$\begingroup\$ if I2 is negative the current flows in the opposite direction. if I2 is zero (I think it's zero) no current flows, \$\endgroup\$
    – Jasen Слава Україні
    Commented Sep 12, 2023 at 4:02
  • \$\begingroup\$ Numerical, Are you already comfortable with how voltaic batteries work, internally? (That page covers the two half-reactions that compose the full redox reaction.) \$\endgroup\$
    – periblepsis
    Commented Sep 12, 2023 at 9:42
  • \$\begingroup\$ From this Battery University page: "The electrode of a battery that releases electrons during discharge is called anode; the electrode that absorbs the electrons is the cathode. The battery anode is always negative and the cathode positive. This appears to violate the convention as the anode is the terminal into which current flows. ... however taking power away from a battery on discharge turns the anode negative. Since the battery is an electric storage device providing energy, the battery anode is always negative." \$\endgroup\$
    – periblepsis
    Commented Sep 12, 2023 at 9:44
  • \$\begingroup\$ "If there's one path that has the greatest potential difference I would think the current would simply take that path to the exclusion of the others." Well, one problem is that your schematic will have no current through the 12 V supply. So all the 18 V source current does go through the 6 Ohm resistor. But if you modified things so that this wasn't uniquely trivial, lowering the 12 V source to -15 V instead then 3 A would go through the 3 Ohm resistor from left to right and 1 A would go through the 6 Ohm resistor, top to bottom. \$\endgroup\$
    – periblepsis
    Commented Sep 12, 2023 at 10:36
  • \$\begingroup\$ What would happen if all 4 amps arriving from the 1 Ohm chose to go through either just the 3 Ohm resistor or else just the 6 Ohm resistor and didn't split up? One of the two would have 0 V across it. Not so? \$\endgroup\$
    – periblepsis
    Commented Sep 12, 2023 at 10:37

1 Answer 1

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The currents do not necessarily flow in directions as depicted.

The positive flow direction is set for sake of notation and you do the calculations based on the equations and the selected direction of positive current.

If the calculations indicate that the current is negative, then it does not flow in the direction chosen as positive.

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  • \$\begingroup\$ I'm a little confused. I (believe I) understand that current (in circuits) is the flow of electrons, which flow from the cathode towards more positive charge, and that we generally perform calculations based on the "positive current". What I want to know is this depiction for the "positive current" correct? Furthermore, if that is so, then how can one reason, from first principles, the direction of the positive current? In order to perform nodal or mesh analysis. Thanks for the reply! \$\endgroup\$
    – Numerical Disintegration
    Commented Sep 11, 2023 at 20:43
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    \$\begingroup\$ The point, like I said, is to select an arbitrary direction of positive current for making the calculation. Then the result of calculation tells you which way compared to your arbitraty selected direction the current flows. Sort of like you assume that a motor starts running clockwise when you start it, but after starting the motor you see it going counter-clockwise so against the direction you chose as positive. \$\endgroup\$
    – Justme
    Commented Sep 11, 2023 at 21:00

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