What is the equivalent resistance between $A$ and $B$ in the given circuit? [closed]

Hi, I've been doing some current electricity problems by using Kirchhoff's laws. I've tried applying KVL(Kirchhoff Voltage law) to this circuit, but to no avail. There happens to be too many variables to work on with my approach. I would appreciate if I could get some help regarding solving this problem and similar ones.

closed as off-topic by Emilio Pisanty, ZeroTheHero, John Rennie, Jon Custer, M. EnnsJan 8 at 16:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, ZeroTheHero, John Rennie, Jon Custer, M. Enns
If this question can be reworded to fit the rules in the help center, please edit the question.

• This is a symmetry question, often the central resistor carries no current due to symmetry, i.e. both side are at the same potential. – PhysicsDave Jan 6 at 20:43
• But here the central R does have current. The way to solve is to see 2 paths. Path 1 is A to D to F to B, note that D to F equates to 0.5R so this paths 2.5R. Path 2 is A to C to E to B, in this path we get 3R (note central R has zero I). Combine the 2 paths in parallel. – PhysicsDave Jan 6 at 20:52

The way to solve is to see 2 paths. Path 1 is $$A$$ to $$D$$ to $$F$$ to $$B$$, note that $$D$$ to $$F$$ equates to $$0.5R$$ so this path is $$2.5R$$. Path 2 is $$A$$ to $$C$$ to $$E$$ to $$B$$, in this path we get $$3R$$ (note central $$R$$ has zero $$I$$). Combine the 2 paths in parallel.
• Well, it gives the right answer, but how does it work? Why does D to F equate to 0.5 $R?$ Why do you say that the "central $R$ has zero $I"?$ None of the resistors has zero current. Would you please spell out your method more fully? – Philip Wood Jan 6 at 23:13
Here are two hints. (1) Apply Kirchhoff I at the junctions on the diagram itself, by labelling the currents $$x,$$ $$y,$$ $$(x-y)$$ (or whatever is right), and so on. In other words, don't waste time writing formal Kirchhoff I equations, and don't call the currents $$"I_1",$$ $$"I_2"$$ etc.