To find equivalent resistance across A and B here: enter image description here

I thought a lot on this, but I don't get how to even approach it. Nodal analysis is too long for this. It there a symmetry trick I am missing out?

A hint will just do, that'll provide me direction to work on it.

This is a follow-up question for my attempt on Why does the superposition principle for calculating equivalent resistance work for asymmetrical objects?

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    $\begingroup$ Besides parallel and series rules there is a transformation between a Y junction of three resistors and an equivalent triangle of there resistors, call the Y-delta transform (or equivalence). Applying this may recast the network into one where series and parallel rules can be applied. $\endgroup$ – ggcg Jun 23 '18 at 12:45
  • $\begingroup$ You missed a direct 1R connection between A & B. $\endgroup$ – Gyromagnetic Jun 23 '18 at 13:22
  • $\begingroup$ Oh thankyou @ggcg. I didn't know that method yet, any other methods are also welcome :) $\endgroup$ – Ice Inkberry Jun 23 '18 at 14:03
  • $\begingroup$ @Gyromagnetic I deliberately missed it because that one was in parallel with this one, and I wanted to know if I can solve this one separately. $\endgroup$ – Ice Inkberry Jun 23 '18 at 14:03
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    $\begingroup$ It is not usually taught in basic physics books. I learned in a junior level course. It's on wikipedia, look it up and give it a try. $\endgroup$ – ggcg Jun 23 '18 at 15:52

I can't figure out how to draw a circuit with reasonable effort, so you'll just have to imagine the circuit being completed with a voltage source from $B$ to $A$ with voltage $v$, positive end left. Then if the resistance of each resistor is $R$, we can solve the circuit via mesh analysis the way an EE would.

Put a mesh current running clockwise in each mesh of the circuit with $i_1$ in the loop we just completed at 6:00, $i_2$ at 8:00, around to $i_6$ at 4:00 and $i_7$ in the center mesh. Then applying Kirchoff's voltage law around each mesh we get the system of equations $$\begin{bmatrix}3&-1&0&0&0&-1&-1\\ -1&4&-1&0&0&0&-1\\ 0&-1&4&-1&0&0&-1\\ 0&0&-1&4&-1&0&-1\\ 0&0&0&-1&4&-1&-1\\ -1&0&0&0&-1&4&-1\\ -1&-1&-1&-1&-1&-1&6\end{bmatrix} \begin{bmatrix}i_1\\i_2\\i_3\\i_4\\i_5\\i_6\\i_7\end{bmatrix} =\begin{bmatrix}v/R\\0\\0\\0\\0\\0\\0\end{bmatrix}$$ With solution $$\begin{bmatrix}i_1\\i_2\\i_3\\i_4\\i_5\\i_6\\i_7\end{bmatrix}=\begin{bmatrix}67\\29\\19\\17\\19\\29\\30\end{bmatrix}\frac{v}{113R}$$ The resistance of the network is thus $$R_{eq}=\frac v{i_1}=\frac{113}{67}R$$ If you insisted on performing node analysis, then instead we could label the outer node at 9:00 as $v_1$, around to $v_4$ at 3:00. The node at 7:00 has voltage $v$ and that at 5:00 is ground. The inner nodes are numbered from $v_5$ at 7:00 around to $v_{10}$ at 5:00. Then applying Kirchoff's current law at each node we arrive at $$\begin{bmatrix}3&-1&0&0&0&-1&0&0&0&0\\ -1&3&-1&0&0&0&-1&0&0&0\\ 0&-1&3&-1&0&0&0&-1&0&0\\ 0&0&-1&3&0&0&0&0&-1&0\\ 0&0&0&0&3&-1&0&0&0&-1\\ -1&0&0&0&-1&3&-1&0&0&0\\ 0&-1&0&0&0&-1&3&-1&0&0\\ 0&0&-1&0&0&0&-1&3&-1&0\\ 0&0&0&-1&0&0&0&-1&3&-1\\ 0&0&0&0&-1&0&0&0&-1&3\end{bmatrix} \begin{bmatrix}v_1\\v_2\\v_3\\v_4\\v_5\\v_6\\v_7\\v_8\\v_9\\v_{10}\end{bmatrix} =\begin{bmatrix}v\\0\\0\\0\\v\\0\\0\\0\\0\\0\end{bmatrix}$$ With solution $$\begin{bmatrix}v_1\\v_2\\v_3\\v_4\\v_5\\v_6\\v_7\\v_8\\v_9\\v_{10}\end{bmatrix} =\begin{bmatrix}84\\65\\48\\29\\75\\74\\63\\50\\39\\38\end{bmatrix}\frac{v}{113}$$ The current through the voltage source is thus $$i=\frac{v_4}R+\frac{v_{10}}R=\frac{(29+38)v}{113R}=\frac{67v}{113R}$$ as before.

  • $\begingroup$ Would the downvoter care to explain what he thought was lacking in this answer? $\endgroup$ – user5713492 Jul 1 '18 at 14:54

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