Find the equivalent resistance between the two points a and b.... all the resistances in the picture has equal value of "r".

Take care of the nodes,meshes and branches....please go through the picture in details for better view of the problem.

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## Comments

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TopNewestApply a voltage of $+v$ to point $A$ and $-v$ to point $B$. The voltage along the horizontal dividing line in the middle is $0 V$. Now look at the four resistors that connect the outer circle to the inner circle. The outer ends of these four resistors are all at a voltage of $0 V$. Hence, those four resistors, the four resistors of the inner circle, and the four resistors of the inner cross are all shorted out of the circuit.

Removing those twelve resistors yields a circuit with 3 sections in series, each of which has a resistance of $(R+R) \parallel (R+R) = R$. Therefore, the total resistance between points $A$ and $B$ is $3R$.

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Correct ! Those resistors will get eliminated , as the voltage will remain the same on the wires with no resistance between them . Voting you up !

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aaa

This is the circuit transformation where each resistance are 'r'. If you can't view this image, go to bigger image

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now I have to convert the star-delta conversion for any one triangle....isn't it?

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There are two shortings in this circuit which will not allow any current to flow to the inner branches. Thus, simplifying it we have three resistances in series and thus the answer is 3R.

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the R

eq of the 2 triangles above and below is R. If you restructure the diagram you will sea that it a wheatstone bridge.So you can eliminate the circuits inside the main circle and evaluate the Req for the outer circle=R. So , the final result will be R+R+R=3RLog in to reply

Ahhh!!A classical case of 1)WHEATSTONE BRIDGE 2)STAR- DELTA TRANSFORM....

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The 2 triangles have obviously r as eqln. resistance In case of circle setup inside the circle is shorted. Effectively outer perimeter of circle accounts for resistance and that also comes out to be r. So ++'R'effective=r+r+r=3r++

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