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Topic: Resistor Cube (Read 2106 times) |
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william wu
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Resistor Cube
« on: Jan 22nd, 2003, 12:34pm » |
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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.
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BNC
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Re: Resistor Cube
« Reply #1 on: Jan 22nd, 2003, 12:52pm » |
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Here is my answer (hidden):
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 Ohms law, if 1Amp -> 5/6V, then the equivalent resistance is 5/6 Ohm.
QED
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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