Finding Equivalent Resistance Using Symmetry and Kirchhoff's Laws

[sktg]

Homework Statement

Twelve resistors form the edges of a cube as shown
in the figure. If the potential difference between
opposite corners A and A' is 18 V, find the equivalent
resistance between A and A'.(Figure is in 'Attempt at a Solution' section)

Relevant Equations

There are no equations in for this question.

By the use of symmetry, I have found that equal currents will flow through opposite edges like A-C and C'A', but still, the use of kirchhoffs current and voltage laws lead to four equations in four variables which I am not able to solve, neither I am sure that the four equations which I have created are independent since I have used symmetry in dividing currents.I have also tried using star delta conversion but its going too long. I think there should be some shorter method to solve this, can anyone please give a hint?(Not complete solution)

 

[etotheipi]

I haven't worked it through, so full disclaimer this might not work, but have you tried labelling the currents AB, AC, AD with $i_1, i_2, i_3$ respectively, and also B'A', C'A', D'A' with $i_1, i_2, i_3$ respectively (due to the symmetry you mention), and then labelling the potential of each node in terms of $V$ and $i_1, i_2, i_3$ (where $V = 18V$)? You should be able to then work out some more currents; that might give you sufficient Kirchhoff current law equations? I'll probably try it tomorrow since I'm a bit tired now...

Let's see if someone has a nicer method

 

[sktg]

I haven't worked it through, so full disclaimer this might not work, but have you tried labelling the currents AB, AC, AD with $i_1, i_2, i_3$ respectively, and also B'A', C'A', D'A' with $i_1, i_2, i_3$ respectively (due to the symmetry you mention), and then labelling the potential of each node in terms of $V$ and $i_1, i_2, i_3$ (where $V = 18V$)? You should be able to then work out some more currents; that might give you sufficient Kirchhoff current law equations? I'll probably try it tomorrow since I'm a bit tired now...

Let's see if someone has a nicer method

I have tried but calculations are very long

 

[etotheipi]

I have tried but calculations are very long

I think if you follow through the calculations I mentioned you should be able to get at least 3 simultaneous equations from applying Kirchhoff's current law at at least 3 junctions without too much difficulty. It's tedious, perhaps, but such is life sometimes .