# Find equivalent resistance between two terminals connecting a face diagonal.

317 views

In a well-known cube network, each edge contains a 24 ohm resistor and
additional 12 ohm resistors along each face diagonal. These resistor are
only connected electrically at the corners of the cube. For the sake
readability, in the figure only some of these branches Without
resistors are shown. Find equivalent resistance between two terminals
connecting a face diagonal.

My try:

On solving this I am not getting the answer .

The answer is given as 7 ohm.

Can you explain your strategy? It is not obvious what you are doing.
I connected a battery of 100 volt between a and b and then defined potential of all points . Then I made a simplified circuit and showed their resistances in parallel . Now on solving it I am not getting the required answer .
How did you assign potentials ? How did you simplify the circuit? The cube has 8 nodes but your diagram has only 5. Where are all the other connections? Node A should be connected to 6 other nodes, your diagram shows only 2 of these.
In the first figure I haven't shown all resistances for the sake of readability.
After connecting a battery of 100 V I defined 50 V to all those points which in middle of the circuit ( i.e A and B ) . Then I assume one the remaining point as x V therefore the other one should be 100-x V by symmetry . Now to simplify the circuit I further marked points and joined resistances in between them . ( For a clear diagram I showed only one resistance between any two points and wrote the resistances ( i.e 12, 12, 24, 24 all in parallel)
Sorry. Yes I understand now what you've done. Your final diagram is correct, I think.

By symmetry, whatever current flows from x to 50V through the 4 ohm resistor flows from 50V to 100-x through another 4 ohm resistor. So you can disconnect these two 4 ohm resistors from the 50V node. They are together in parallel with a 12 ohm resistor. Then you have 3 parallel branches from A to B.

The answer I get is 44/10 ohms, not 7 ohms.

The simplified network is then as shown on the right. Current flows symmetrically left to right, so that current AC=CB and EC=CF. The connection at CDGH can be removed, leaving 3 parallel branches from AB, ACB, AE(C)FB, with resistances of $12, 8$ and $48+12||8$ respectively.
The equivalent resistance across AB is $\frac{22}{5}=4.4\Omega$.