# Using symmetry, to solve more complex circuits.

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17 views [AO and OB segment are same as any other segment]. Each segment have same resistance r.

So I know that potential at C and E is same due to symmetry, B and D is same due to symmetry, and so on. Current in CB is same as ED and current in AO is same as OB and so on. But now what? Upper and lower part are not isolated because current in CO can go to OB and also go to OD .

I have solved this problem by brute force and answer is $\frac45$r.

related to an answer for: How to use symmetry?
asked Jul 24
edited Jul 24

## 1 Answer

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Best answer

By symmetry the current in branch CO equals the current in branch OB. Therefore the two branches CO and OB can be detached from O, while remaining connected to each other.

Branch CB then consists of two parallel branches, one of $r$ and one of $2r$. This combination is equivalent to $\frac23r$. The resistance across ACBB' (where B' is the output node) is then $r+\frac23r+r=\frac83r$.

The same trick can be applied to symmetrical branch ACDB' to reduce it also to $\frac83r$.

The network between input and output nodes is made up of 3 branches in parallel (ACBB', AOB', AEDB') with resistances $\frac83r, 2r, \frac83r$ respectively. The two resistances of $\frac83r$ in parallel are equivalent to $\frac43r$. This combination is in parallel with $2r$ so the total resistance is $$\frac{\frac43r \times 2r}{\frac43r+2r}=\frac45r$$ answered Jul 24 by (28,746 points)
selected Jul 30 by Swarnim