Another infinite circuit

@Pranav Arora – I too am not convinced as the calculations are very bad and generally these questions are not very calculative but have some nice trick.

I think we will just have to wait for someone to post a proper solution.

@Pranav Arora – Listen, Pranav I have been trying to understand a similar question given in that Hungarian website (the same has been shared by Mr. Jon Haussmann above).

I can't understand exactly how the solution works and hence I thought I would discuss it with you.

In that question we have to find the equivalent resistance between points $A$ and $C$ , and between points $A$ and $B$ for the following circuit (which is quite similar to ours):

Here each resistance is of value $1\Omega$ .

After a lot of back calculation and translation this is what I have been able to make of it:

According to the solution the equivalent circuit can be drawn as follows:

I don't know which point $O$ is but I guess it is the point symmetrically located in the last branch of the circuit (refer the first image that I have put up).

Here $R_{AO}=R_{CO}=R$ and $R_{BO}=r$ .

Thus $R_{AC}=2R$ and $R_{AB}=R+r$

The solution says that as the circuit is infinitely long removing one unit will not change the equivalent resistance.Hence $R_{AC}$ and $R_{AB}$ have been computed as follows:

For $R_{AC}$ they have drawn the equivalent circuit as follows:

Note that middle branch can be taken out due to symmetry.

Thus $\displaystyle\frac{1}{R_{AC}}=\displaystyle\frac{1}{2R}=\displaystyle\frac{1}{2}+\displaystyle\frac{1}{2R+2}$

For $R_{AB}$ they have drawn the equivalent circuit as follows:

Thus $\displaystyle\frac{1}{R_{AB}} =\displaystyle\frac{1}{R+r}=1+{1\over\displaystyle(1+R)+{\strut1\over\displaystyle{\strut1\over\displaystyle1+r}+{\strut1\over\displaystyle2+R}}}$

What do you think about it? Please share your views.