# Homework Help: Cube of Resistor Problem - Nice one

1. Aug 25, 2013

### luiseduardo

1. The problem statement, all variables and given/known data
Asymmetrical Cube - Determine the equivalent resistance between points A and C’ of the circuit below.

2. Relevant equations

U = R.i

3. The attempt at a solution

This problem was my professor that created. I know how to solve it, but I would like to know if other people have a better idea how to solve it. I tried to use Kirchhoff's Laws and I found 6 equations and was a bit complicate to find the final answer. I think it's a good challenge.

$$\frac{1}{4} \left(R_1+R_2+R_3+\frac{1}{R_1^{-1}+R_2^{-1}+R_3^{-1}} \right)$$

Last edited by a moderator: Aug 29, 2013
2. Aug 26, 2013

### Staff: Mentor

3 equations are sufficient, as the cube still has a symmetry. If you add the potential of opposite points in the cube, you always get the same value.

I like the result.

3. Aug 28, 2013

### rude man

The easy way is to recognize that, by symmetry, the voltage at the three corners connecting A by one resistor each is the same. Similarly, the voltage at the three corners connecting C' by one resistor each is the same.

Nodes of equal voltage can be shorted together without affecting the circuit. This degenerates the cube into three parallel sections connected in series:
Requiv = R1||R2||R3 + R1||R1||R2||R2||R3||R3 + R1||R2||R3.
No KCL or KVL equations needed!

Last edited: Aug 28, 2013
4. Aug 28, 2013

### Staff: Mentor

That symmetry does not exist with different resistors.

5. Aug 28, 2013

### rude man

Yes, I hadn't noticed the resistors could be of different values. Such an old problem, never seen it with differing values before.

6. Aug 28, 2013

### Staff: Mentor

7. Aug 28, 2013

### rude man

OK, but I resent a physicist being worth only 2/3 of a mathematician ... I suppose only 1 point per EE, or less even? :-)

8. Sep 15, 2013

### luiseduardo

My solution:

http://www.luiseduardo.com.br/electricity/electrodynamics/cubeofresistors.htm [Broken]

Last edited by a moderator: May 6, 2017