# Electrical circuits and calculating resistance [closed]

Can anyone give me some ideas about how to simplify this circuit....

## closed as off-topic by Alfred Centauri, Aaron Stevens, ZeroTheHero, John Rennie, user191954 Oct 25 '18 at 8:10

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Okay there's one tip: bend the wires in order to make a rectangular circuit. For some reason, it's usually perceived as easier like that.

You can also add wire in between of two resistances.

If you do that, you'll see that the battery is splitted in 3 branches: each one with resistasnces of 3, 6, and again 3 $$\Omega$$.

The first two ones are rejoined in a single branch to a $$4\Omega$$.

So it is bassically a parallel circuit. One branch is simple ($$3\Omega$$), and the other one is a sum of "parallel 3//6" plus $$4\Omega$$ in series.

It really looks simpler now.

You can do this because we consider wires as ideal conductors, so adding more lenght means nothing, they add no extra resistance. Real wires do.