Series RC circuits in parallel

Rats! A Champion got there before me. My answer was going to be this (I'll post it anyway because it's a different approach):

If the two sides are identical, the voltages at the nodes between the resistors and capacitors are identical too. That means you could join those points with a piece of wire and nothing would change - there can't be a current flow in the wire because the end potentials are the same. Your new circuit is then simply two resistors in parallel and two capacitors in parallel. So the top resistor is r/2 and the bottom capacitor is 2 x c.

Whilst it sounds very artificial in this question, it's essentially why having a lot of identical small ceramic capacitors spread around a high-speed logic board is such good value for decoupling. The Cs add, the ESRs divide down and the inductances divide down too. [At high speed it's a bit more complicated because you also have to start thinking about things like the rate at which changes propagate across the board - not all the capacitors will be able to participate equally.] If you're interested in this kind of thing, read Johnson and Graham's High-Speed Signal Propagation - it's very good. They're academics who I can recognise as being engineers, if you know what I mean, which is much rarer than you might suppose.

If you want to try out the 'equal potential' method, have a go at this.

If the resistors in the cube are all 6 ohms, what is the resistance between A and B? (You should be able to do it in your head.)

I've just noticed that all of yesterday's Champions have turned into today's Top Members. That's a pity - I won't now be able to use the jokes about horses

https://en.wikipedia.org/wiki/Champion_the_Wonder_Horse

and breakfasts

https://en.wikipedia.org/wiki/Breakfast_of_Champions

That's positively evil.

Thank you, perhaps I'll make a Minion yet.

It's a very old puzzle - I think I first saw it back in the 1970s.

I'm serious about being able to do it in your head, but you have to think about it in the right way. Don't try and solve it by equations, consider the voltages at the nodes if you put a potential across A-B and how they relate to each other given the symmetries involved, then it's just a little simple arithmetic.

Thank you, perhaps I'll make a Minion yet.

It's a very old puzzle - I think I first saw it back in the 1970s.

I'm serious about being able to do it in your head, but you have to think about it in the right way. Don't try and solve it by equations, consider the voltages at the nodes if you put a potential across A-B and how they relate to each other given the symmetries involved, then it's just a little simple arithmetic.