wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> Resistor Cube
        
(Message started by: william wu on Jan 22^nd, 2003, 12:34pm)

Title: Resistor Cube
Post by william wu on Jan 22^nd, 2003, 12:34pm

Imagine a cube where each edge is a 1 ohm resistor. Find the resistance between opposite corners of the cube.

Many ways to solve this problem, but some ways are smarter than others.

Title: Re: Resistor Cube
Post by BNC on Jan 22^nd, 2003, 12:52pm

Here is my answer (hidden):
[hide]
In order to solve this, let us picture the cube as placed on the x-y-z axis, and assume each resistor is one “unit” long. Lets denote the resistors as follows:
R1 is between points (0,0,0) (0,0,1)
R2: (0,0,0) – (0,1,0)
R3: (0,0,0) – (1,0,0)
R4: (0,0,1) – (0,1,1)
R5: (0,0,0) – (1,0,1)
R6: (0,1,0) – 0,1,1)
R7: (0,1,0) – (1,1,0)
R8: (1,0,0) – 1,1,0)
R9: (1,0,0) – (1,0,1)
R10: (1,1,0) – (1,1,1)
R11: (1,0,1) – (1,1,1)
R12: (0,1,1) – (1,1,1)

We shall now connect a 1Amp current source between points (0,0,0) and (1,1,1).
Due to symmetry, 1/3 Amp will go through resistors R1, R2 and R3. Again, due to symmetry, 1/6 Amp will go through R4-R9. The currents merge again to form 1/3 Amp in R9-12.
Now, choose any route, and sum the voltage drops. For example: V1+V4+V12 = 1/3*1+1/6*1+1/3*1 = 5/6V.

According to Ohm’s law, if 1Amp -> 5/6V, then the equivalent resistance is 5/6 Ohm.

[/hide]
QED