# What's the reason for this symmetry argument?

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.

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)

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.

• 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.
Commented Mar 6, 2019 at 16:08
• @Dvij Mankad Can you tell me is there's a way to directly transfer the question to Electrical Engineering Stack Exchange? Commented Mar 6, 2019 at 16:10
• @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? Commented Mar 6, 2019 at 16:25
• 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.
Commented Mar 6, 2019 at 16:27
• @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. Commented Mar 6, 2019 at 17:14