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Okay so I've learnt about various methods to find equivalent resistance in circuit. But there's one particular argument which I don't understand and I would be grateful if someone will prove it using symmetry arguments.

enter image description here

In this figure, for sake of clarity resistors have been represented by straight lines (not by jagged lines).

Symmetry argument that I don't understand: The current entering the resistor AC is the same leaving in resistor FB (same is true for other symmetrical parts of circuit).

I will tell what I mean by saying 'symmetry argument.' For example in this circuit (again for sake of clarity I represent resistors by straight lines)

enter image description here

It immediately clicks inside my mind that paths AHFB, ACGB and AEFB are equivalent so physics and nature will not discriminate between any path, thus same current will flow through resistor AC, AH and AE making potential at C equal to H and E. This is a satisfactory symmetry argument.

As for the top circuit I can't find a satisfactory reason by myself that why current entering resistor AC equals current leaving resistor FB. Maybe it involves changing polarities of the cell?

NOTE: Please don't tell me about the answer (i.e. equivalent resistance) I'm sure I will find the answer myself once I will find the reason of the question mentioned above. I want to feel that 'Aha' moment once will find the correct answer myself. So please help me.

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  • \$\begingroup\$ Hi Shivansh! Your question looks interesting but it is about engineering and not about physics per se as you probably realize. Thus, I am voting to close this question. You might want to ask this question on the Electrical Engineering Stack Exchange. \$\endgroup\$ – Dvij Mankad Mar 6 at 16:08
  • \$\begingroup\$ @Dvij Mankad Can you tell me is there's a way to directly transfer the question to Electrical Engineering Stack Exchange? \$\endgroup\$ – Shivansh J Mar 6 at 16:10
  • \$\begingroup\$ @Dvij Mankad I've heard that only users above 3000 reputation can import question to other website. As I don't have enough reputation, can you please transfer this question to Electronic Stack Exchange? \$\endgroup\$ – Shivansh J Mar 6 at 16:25
  • \$\begingroup\$ Yes, you probably can't transfer it yourself but I don't see an option for myself to transfer it either. I will flag it for moderator attention so that they can have a look and transfer if it is possible/desirable. \$\endgroup\$ – Dvij Mankad Mar 6 at 16:27
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    \$\begingroup\$ @TechDriod I know about Kirchhoff's junction rules, but as I said in question I prefer symmetry argument (which I also explained by giving an example). I know that ANY circuit can be solved using Kirchhoff's rules but it will make calculations tedious. Exploiting symmetry makes complex circuits (of certain kind, of course) very very fast to solve. \$\endgroup\$ – Shivansh J Mar 6 at 17:14
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If you reversed the battery you would have the same circuit. You could see this by reversing the battery, and then redrawing the circuit horizontally flipped - you would get the exact same circuit, but with different labels.

So FB in the reversed circuit would be in the same position as AC in the original circuit.

Since your circuit only has resistors, it has symmetry regarding the direction of current flow. Reversing the battery doesn't change the amount of current flowing through each resistor - it only reverses the direction of the current flow.

You should be able to see that the current flowing out of AC in the reversed circuit is the same as the current flowing into AC in the original circuit, because only the current direction is changed. But I also just showed you that AC in the reversed circuit is the same as FB in the original circuit. Therefore, the amount of current flow through AC in the original circuit, and FB in the original circuit, is the same.

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  • \$\begingroup\$ Exactly the answer I was looking for!!! Thank you !! When I read "Reversing the battery doesn't change the amount of current flowing in resistor-it only reverses the direction of current flow" I understood the argument. Thanks you again for making it so simple! \$\endgroup\$ – Shivansh J Mar 7 at 8:20
  • \$\begingroup\$ I upvoted your answer, but unfortunately my upvoted doesn't show in public because of my low reputation. \$\endgroup\$ – Shivansh J Mar 7 at 8:22

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