Find the Equivalent Resistance in Complicated Electrical Circuits

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Grievery
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Hi! I have been trying to understand how a "complicated" circiut can be transformed to show which resistors are connected in series and which are parallel... Sometimes I manage to do it but it's often too hard for me because I can't find any sort of algorithm about it. Could anybody show me "step-by-step" how to transform this circuit for example...?
Circuit schematic with terminals A and B; resistors labelled 2 Ω, 2 Ω, 2 Ω, 2 Ω and 6 Ω connected in a network of horizontal, vertical and diagonal branches.

We need to find the equivalent resistance...
Thank you in advance!
 
It's always a good idea to label the components in a diagram. In this case your resistors should be numbered R1, R2 etc. This allows people to identify what is being referred to and, if the OP doesit, doesn't waste 2 posts.
Alternatively you can label the connection nodes, A, B, C etc (which you have already started, but not finished).

I will comment (ask) if your diagram is correct since R1 and R2 are (almost) shorted out.

Can you see which resistors I mean by R1 and R2?

Oh and welcome to PF
 
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First of all, thank you for the very quick response... I have labeled the resistors... Now, how can I find the equivalent one.
P.S. The answer is 3 ohms.
Circuit schematic with resistors R1 (2 Ω), R2 (2 Ω), R3 (2 Ω), R4 (2 Ω), R5 (6 Ω) and R6 (2 Ω) between terminals A and B.
 
I find it often helpful to redraw the network to bring out the relationships. Sometimes you have to do this several times to get a clear result.

The attachment shows a sequence of single step series or parallel simplifications leading to the desired result.

Although I agree with the 3 ohm answer I have left you do do some of the work in the intermediate steps.

Hope this helps
 
Attachments
  • Series of six hand-drawn circuit diagrams between terminals A and B, showing successive series and parallel reductions of resistors R1 to R5, ending with a single 3 ohm resistor.
    net2.jpg
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Well.. thank you once again! :) I hope now I can figure out HOW to do it myself...